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Engineering Chemistry-II

Department of Chemistry, Annapoorana Engineering College, Salem 2

Unit-I

Electrochemistry

Galvanic (or) Voltaic (or) Electrochemical cell

It is a device that produces electrical energy at the expense of chemical energy

produced in a reaction.

A cell consists of two half cells or electrodes.

A half-cell or electrode contains a metal rod dipped in an electrolytic solution.

Electrolytic cell: It is a cell in which electrical energy brings about a chemical

reaction.

Electrochemical cell: It is a device that produces electrical energy at the expense of

chemical energy produced in a reaction.

Differences between electrolytic cell and electrochemical cell.

Electrolytic Cell Electrochemical cell

1. Electrical energy brings about a

chemical reaction.

2. Anode is positively charged.

3. Cathode is negatively changed.

4. Electrons move from anode to

cathode through external circuit.5. The extent of chemical reaction at the

electrode is governed by Faradays

laws of electrolysis.

6. The amount of electricity passed is

measured by a coulometer.

7. One electrolyte and two electrodes of

the same element are generally used

in these cells.

1. Electrical energy is produced at the

expense of chemical energy.

2. Anode is negatively charged.

3. Cathode is positively charged.

4. Electrons move from cathode to

anode through external circuit.5. The emf of the cell depends on

concentration of electrolyte and

nature of the metal electrode.

6. The emf of the cell is measured by a

potentiometer.

7. Two different electrolytes and two

different electrodes are often used.

Electrochemical (or) galvanic cell: Daniel cell is a typical galvanic or

electrochemical cell. The cell consists of a zinc rod dipped in ZnSO4 solution. The

two solutions are interconnected by a salt bridge. The electrodes are connected by a

wire through a voltmeter.

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Engineering Chemistry

When the cell produces

cathode.

Zn Zn2+ + 2e

Cu+2 + 2e Cu

Zn + Cu2+ Zn2+ + Cu

Daniel cell is represented as

Zn ZnSO4 CuSO4(IM) (IM)

The Standard emf of the cell is 1.

Representation of a galvanic ce

A galvanic cell has two

points are kept in view when a c

1) Anode is written on the l

2) The electrode and the ele

phase boundaries. The p

electrode is written first a

Pt H2(g) H+ (1(1atm)

Zn ZnSO4 (1M)

For cathode, the electroly

CuSO4 (1

3) Concentrations of electrolytes

II

current, oxidation occurs at anode and reducti

(at anode)

(at cathode)

(net cell reaction)

u

1 volt.

ll

lectrodes namely anode and cathode. The follo

ll is represented:

ft hand side and cathode on the right side.

ctrolyte are separated by a vertical line to denot

hysical state is indicated in bracket. For anode

nd then the electrolyte. (e.g)

)

te is written first and then the electrode (e.g.)

) Cu

and pressures of gases are mentioned.

n at

ing

the

, the







EMF OF A CELL

It is the potential difference that causes flow of electrons from the electrode of

higher potential to the electrode of lower potential.

EMF of a cell is related to electrode potentials as follows:

EMF of Cell = [Standard reduction potential of RHS electrode] –

[Standard reduction potential of LHS electrode]

E cell = E right – E left

Measurement of emf of a Cell :

Using Poggendorfs compensation principle, the emf of a cell can be measured

using a potentiometer.

AB is a uniform wire of high resistance. A battery W is connected to the ends

of the wire through a rheostat. The unknown cell X is connected in the circuit. Its

positive pole is connected to A and negative pole to a sliding contact D through a

galvanometer, G. The sliding contact is moved along the wire AB, till the

galvanometer shows null deflection. The distance AD is measured. The emf of

unknown cell is proportional to AD.

Ex AD ----- (1)

Now the unknown cell is replaced by the standard cell ‘S and the sliding

contact is moved along the wire, until galvanometer shows null deflection. The

distance AD' is measured.

Es AD' ----- (2)

From equations (1) & (2)

Ex AD=

Es AD'AD

Ex = x EsAD'

Knowing Es, the emf of unknown cell can be calculated.







W

Applications of emf measurement :

1. Potentiometric titrations are performed.

2. Solubility of sparingly soluble salt is determined.

3. Hydrolysis constants of salts are determined.

4. The valency of ions is determined.

5. The standard free energy change of a reaction is calculated using the equation

- G = nFE

n = Number of electrons involved in the reaction

E = Standard emf of the cell

F = 96500 coulombs.

6. The equilibrium constant K of a reaction is calculated from Eo.

log K =nE

0.0591

7. pH of a solution is determined.

Electrode Potential :

When a metal M is dipped in its salt solution, one of the following reactions

occurs depending on the metal :







1. Positive metal ions pass into the solution :

M Mn+ + ne- (oxidation)

(e.g.) When Zn rod is dipped in ZnSO4 solution, Zn goes into solution as Zn2+.The electrons attach to Zn rod, giving it a negative charge. The negative charge on the

rod attracts positive ions from solution. Thus a double layer of ions is formed close to

the rod.

2. Positive ions from the solution deposit over the metal.

Mn+ + ne- M (reduction)

When Cu rod is dipped in CuSO4 solution, Cu+2 ions from the solution

deposit on metal rod. They attract negative ions from solution. Thus a double layer of

ions is formed close to the metal rod. This is called Helmholtz double layer.

As a result a potential difference is set up between the metal and the solution.

The equilibrium value of the potential difference is known as electrode potential.

Factors affecting electrode potential or emf of Cell :

(i) nature of the metal







(ii) temperature

(iii) concentration of metal ions in the solution

Single electrode potential (or) electrode potential :

It is a measure of the tendency of the metal electrode to lose or gain electrons,

when it is in contact with its own salt solution. It is developed due to the formation of a double layer around the metal rod.

Standard electrode potential :

It is a measure of the tendency of the metal electrode to lose or gain electrons,

when it is in contact with its own salt solution of 1M strength at 25C.

Nernst equation for electrode potential :

Let us consider the reactionMn+ + ne- M ----- (1)

The free energy change of this equilibrium G is related to the equilibrium

constant K by the Vant Hoff isotherm.

[Product]ΔG = ΔG° + RTln -----(2)

[Reactant]

We know, G = - nFE

G = -nFE ------ (3)

Substituting equations (1) & (3) in equation (2)

n+

[M]-nFE = -nFE° RTln -----(4)

[M ]

Dividing by – nF and using the fact that activity of solid metal [M]=1, we have

= −[ ]

( )

At 25C, R = 8.314J/K/mol

F = 96500 coulombs. So equation (5) becomes,

n+

n+

0.0591 1E = E° log ( )

n [M ]

0.0591E = E° log [M ]

n

or







This is Nernst equation for a single electrode potential.

Applications of Nernst Equation :

i) To calculate electrode potential of unknown metal.

ii) To predict corrosion of metals.

iii) To set up electrochemical series.

Nernst equation for a reversible cell :

Let us consider the reaction in a reversible cell :

A + B C + D ----- (1)

The free energy change G of this equilibrium is related to the equilibrium

constant K by the Vant Hoff isotherm.

[Product]ΔG = ΔG° + RTln -----(2)[Reactant]

We know, G = - nFE

G = -nFE ------ (3)

Substituting equations (1) & (3) in equation (2)

[C][D]-nFE = -nFE° RTln -----(4)

[A][B]

Dividing by – nFRT [C][D]

E = E°- ln -----(5)nF [A][B]

At 25C, R = 8.314J/K/mol,

F = 96500 coulombs. So equation (5) becomes,

0.0591 [C][D]E = E° log

n [A][B]

This is Nernst equation for a reversible cell.

Measurement of single electrode potential :

It is impossible to determine the value of a single electrode potential. But we

can always measure the potential difference between two electrodes using a

potentiometer, by combining the two electrodes to form a cell. For this purpose we







use reference electrode. Standard hydrogen electrode is called primary reference

electrode. Calomel electrode is called secondary reference electrode.

Standard hydrogen electrode (SHE) (Primary Reference Electrode)

It has a platinum foil connected to platinum wire and sealed in a glass tube.The platinum foil is dipped in 1M HCl. Hydrogen gas 1 atm pressure is passed

through the side arm of glass tube as shown in the figure. The standard electrode

potential of SHE is taken as zero. The electrode is represented,

Pt | H2(g) (1 atm) | H+ (1M)

The electrode reaction is

2H+ + 2e- H2

Limitations (or) drawbacks of SHE:

1) H2 gas reduces many ions like Ag+ and affects compounds of Hg, Ag etc.

2) It is difficult to get pure H2.

3) The pressure of H2 is to be kept 1 atm all the time.

4) It is difficult to set up and transport.







5) The electrode potential changes with barometric pressure.

6) A large volume of test solution is required.

7) It cannot be used in solutions of redox systems.

8) The solution may poison platinum surface.

So we use a secondary reference electrode.

Calomel Electrode (Secondary reference electrode):

It consists of a glass tube containing pure mercury at the bottom. A paste of

mercurous chloride covers the mercury. A solution of potassium chloride is present

over the paste. The bottom of the tube is sealed with a platinum wire. There is a side

tube for electrical contact. The electrode is represented as,

Hg | Hg2Cl2(s) | KCl(aq)

The electrode reaction is,

Hg2Cl2 + 2e- 2Hg + 2Cl-

The electrode potential is,

- 2

-

RTE = E° ln [Cl ]

2FRT

E = E° ln [Cl ]F

At 25oC,

E = E – 0.0591 log (Cl-)

For saturated KCl, E = +0.242 volt.







Measurement of single electrode potential using a reference electrode (saturated

calomel electrode):

The given electrode, say zinc electrode, is coupled with saturated calomel

electrode as in the figure. Since the reduction potential of zinc electrode less than that

of calomel electrode, zinc acts as anode and calomel as cathode. The cell reaction will

be

Zn/ ZnSO4 (1 M) // KCl (satd )/ Hg2Cl2 /HgZn + Hg2Cl2 Zn2+ + 2Hg + 2Cl-

The emf of the cell is measured using a potentiometer. The value of Ecell =

1.002 volt.

Now, Ecell = Eright – Eleft

= Ecal - EZn

1.002 = 0.242 - EZn

EZn = 0.242-1.002

EZn = - 0.76 volt.

Advantages of Reference Electrode (Calomel Electrode):

1. Easy to set up.

2. Easily transportable

3. Long shelf life

4. Reproducibility of emf







5. Low temperature coefficient

6. Electrode can be used in a variety of solutions.7. Eo value is accurately known.

Ion sensitive electrode :Ion sensitive electrodes have the ability to respond only to a specific ion and

develop a potential ignoring other ions in the solution.

Applications of Ion-sensitive electrode :

i) To determine ions like H+, K+, Li+, etc.

ii) To determine hardness of water (Ca+2 and Mg+2 ions)

iii) To determine concentration of F-, NO3-, CN- etc.

iv) To determine concentration of a gas using gas-sensing electrodes.

v) To determine pH of a solution using H+ ion sensitive electrode.

The types (classification) of ion-sensitive electrode :

1) Glass membrane electrodes

2) Solid state electrode

3) Pungor or precipitate electrodes

4) Liquid – liquid electrode

Glass Electrode (or) Measurement of pH using glass electrode :

Glass electrode contains a thin-walled glass bulb. The glass has low melting

point and high electrical conductivity. 0.1M HCl is present in the bulb. A platinum

wire is inserted in the acid.

When the glass membrane separates two solutions differing in pH, exchange

of H+ ions takes place between the solutions. As a result a potential is developed

across the membrane. The potential EG is given by,

EG = EG + 0.0591 pH







Measurement of pH :

The glass electrode is dipped in the given solution. This system is connected to

saturated calomel electrode as in the figure. The emf of the resulting cell is measured

using a potentiometer.

From the emf, the pH of the solution is calculated as below:

Ecell = Eright – Eleft

Ecell = Ecal – Eglass

Ecell = 0.242 – (EG + 0.0591 pH)

Ecell = 0.242 - EG - 0.0591 pH

pH = G cell0.242 - E - E0.0591







Advantages of Glass Electrode :

i) It is easily constructed and used

ii) Results are accurate

iii) Electrode is not easily poisoned

iv) Equilibrium is quickly attainedv) It can be used in strong oxidizing solutions, coloured solutions and in

presence of metal ions

vi) Using special glass electrode, pH can be measured from 0 to 12.

vii) It is used in chemical, industrial, biological and agricultural laboratories.

Disadvantages or Limitations :

i) Glass has high resistance. So special electronic potentiometer must be

used.

ii) It cannot be used in highly alkaline solutions, in pure ethanol or in acetic

acid. If the solution pH is more than 12, glass membrane is affected by

cations.

Electrochemical Series :

Electrodes are arranged in the increasing order of their standard reduction

potential values. This order is called electrochemical series.

Use or Application or Significance of Electrochemical series or emf series:

1. Calculation of Standard emf of a Cell :

We can calculate the standard emf of a cell, by noting the standard reduction

potentials from the electrochemical series.

E E ECell RHS LHS

2. Predicting feasibility of a reaction :

The feasibility of a reaction can be predicted from the E value of the

corresponding cell reaction.

If ECell

is positive, the reaction occurs spontaneously. If ECell

is negative, the

reaction is not feasible.







3. Hydrogen displacement behavior:

We can find out which metals displace hydrogen gas from dilute acids.

Metals with negative electrode potential liberate hydrogen from dilute

sulphuric acid.

(e.g.) Zn with E = -0.76 V displaces H2 from dil. H2SO4.Zn + H2SO4 ZnSO4 + H2

Silver with a positive E value of 0.8, will not displace H2 from dil.H2SO4.

Ag + H2SO4 No reaction

4) Determination of Equilibrium constant of a reaction :

Standard electrode potentials are used to determine the equilibrium constants

as follows : We know,

- G = nFE

- G = 2.303 RT log K

Hence, 2.303 RT log K = nfE

log K =nFE°

2.303RT

Knowing E, n, F, R and T, K can be calculated.

5) Displacement of one element by another :

Metals with a lower value of reduction potential will displace metals with a

higher reduction potential from their solution.

(e.g.) Zn with E = -0.76V can displace copper (E=+0.34V) or silver

(E=+0.8V) from their solution.

Zn + CuSO4 Cu + ZnSO4

Potentiometric Titrations :

Principle :

We know the potential of an electrode depends on the concentration of the

solution in which it is dipped. As the concentration changes, the emf also changes.







Electrolyte and its conductivity :

Electrolyte like NaCl completely ionizes in solution.

NaCl Na+ + Cl-

The ions formed move in an electric field to oppositely charged electrodes and

conduct electricity.

Strong electrolyte :

A strong electrolyte is completely ionized in solution at all concentrations.

E.g. NaCl.

NaCl Na+ + Cl- (100%)

Weak electrolyte :

A weak electrolyte is partially ionized in solution. (e.g.) CH3COOH.

CH3COOH CH3COO- + H+ (Partial)

Conductometric Titrations :

The conductance of a solution depends on the number of ions, nature and

charge of the ion and its mobility. During a titration, there is a change in the number

and nature of ions in solution. Hence there is a change in conductance. This can be

used to detect end point of a titration. This type of titration is called conductometric

titrations. The temperature should be maintained constant throughout. The titrant

should be 10 times stronger than the solution to be titrated so that volume change is

very small.

Titration of strong acid versus strong base :

Let us consider the titration of HCl versus NaOH.

A known volume of HCl is taken in a beaker. The conductivity cell is dipped

in the acid. The NaOH is added from the burette in small volumes and the

conductance is measured each time. When the alkali is added, the fast moving H+ ions

is replaced by slow moving Na+ ions. So the conductance decreases until all the acid

is neutralized.

H+ + Cl- + (Na+ + OH-) Na+ + Cl- + H2O







After neutralization, further addition of NaOH increases the conductance

sharply due to the presence of fast moving OH- ions in the solution.

A graph is plotted between conductance and volume of NaOH added. The

point of intersection of two straight lines gives the end point.

Titration of weak acid versus strong base:

Let us consider the titration of acetic acid versus NaOH.

A known volume of acetic acid is taken in a beaker. The conductivity cell is

dipped in the acid. The NaOH is added from the burette in small volumes and the

conductance is measured each time. When the alkali is added, the conductance of the

solution increases due to the formation of completely ionized sodium acetate.

CH3COOH + (Na+ + OH-) CH3COO- + Na+ + H2O

After neutralization, further addition of NaOH increases conductance sharply

due to the presence of fast moving OH- ions in the solution.

A graph is plotted between conductance and volume of NaOH. The point of

intersection of two straight lines gives the end point.

Titration of a mixture of weak and strong acids versus strong base:

Let us consider the titration of a mixture of HCl and CH3COOH versus NaOH.







A known volume of HCl and CH3COOH is taken in a beaker. The

conductivity cell is dipped in the acid. The NaOH is added from the burette in small

volumes and the conductance is measured each time.

First the strong acid HCl is neutralized and the conductance decreases until all

the acid is neutralized. Then the neutralization of CH3COOH takes place. Theconductance slowly increases, until all CH3COOH is neutralized. Further addition of

alkali increases the conductance sharply due to the presence of fast moving OH- ions

in the solution.

A graph is plotted between conductance and volume of NaOH. The first end

point corresponds to neutralization of HCl. The second end point corresponds to

neutralization of CH3COOH.

Advantages (Merits) of Conductometric titrations :

1. Indicator is not necessary.

2. Dilute solutions can be titrated.

3. Accurate end point is obtained.

4. No special attention is necessary near the end point.

5. Coloured solutions can be titrated.

6. Weak acid can be titrated against weak base.

QUESTIONS :

PART – A

1. Write about conductivity of electrolyte.

2. What are strong and weak electrolytes?

3. Define electrochemical cell.

4. Differentiate between Galvanic and Electrolytic cells.

5. Distinguish between reversible and irreversible cells.







6. Justify whether the reaction is irreversible or not.

Zn + H2SO4 ZnSO4 + H2

7. What is an electrode potential? How is it developed?

8. Define standard electrode potential.

9. What are the factors that affect electrode potentials?10. What are the factors that affect the emf of a cell?

11. Define emf of a cell.

12. What is an electrochemical cell?

13. Why is it not possible to measure the potential of an isolated half cell or

electrode?

Answer : Let us connect the metal rod to one arm of voltmeter. When we connect

the other arm to the solution through a wire, the metal wire dipping in the solution

will produce another double layer.

14. Represent a galvanic cell according to IUPAC convention.

15. What is a salt bridge? What are its uses or functions?

Answer : Salt bridge is an inverted U tube containing a paste of KCl or NH4NO3

and agar agar.

Functions or Uses : (i) It gives internal connection between the two electrode

solutions; (ii) It eliminates liquid junction potential; (iii) It maintains electrical

neutrality between electrodes; (iv) It helps accurate measurement of emf of the

cell.

16. Can we use a KCl salt bridge for a cell containing silver electrode?

Answer : No. Because the silver ions react with Cl- ions.

Ag+ + Cl- AgCl

So, the concentration of Ag+ ion changes and emf also changes.

17. What are the factors that affect emf of a cell?

18. Can we use a nickel and zinc spatula to stir CuSO4 solution?

Answer: No, Ni and Zn have more negative potentials than Cu. They displace Cu

from its solution.

Ni + Cu2+ Ni2+ + Cu

Zn + Cu+2 Zn+2 + Cu

19. How will you predict the spontaneity of a (redox) reaction using emf value?

20. Define reference electrode. Give example.







Answer : An electrode whose potential is accurately known or whose potential is

arbitrarily fixed is reference electrode. The potential is accurate, stable,

reproducible. The electrode has long shelf life. (E.g.) Calomel electrode.

21. Why cannot we use a voltmeter to measure the emf of a cell?

Answer : The cell current is drawn into the voltmeter. It may lower the cell emf.We cannot get accurate value.

22. Zn reacts with H2SO4 to liberate H2. But Ag does not. Explain.

23. Explain electrochemical series.

24. What are uses or significances of electrochemical series?

25. Write down the cell reaction and expression for the emf of the cell : Zn | ZnSO4

|| CuSO4 | Cu.

26. What are the applications of Nernst equation?

27. What are the applications of emf measurement?

28. Give two examples of reference electrodes.

29. What are the advantages of conductometric titrations?

30. What are the advantages of potentiometric titrations?

31. Give some advantages of glass electrode.

32. What are the disadvantages of glass electrode?

33. What are the differences between metallic conductance and electrolytic

conductance?

Metallic Conductance Electrolytic Conductance

1. It involves flow of electrons in a

conductor.

It involves movement of ions in

solution.

2. It does not involve transfer of matter. Transfer of ions occurs.

3. Conductance decreases with increase

in temperature. This is because the

flow of electrons is disturbed.

Conductance increases with increase

in temperature. This is because the

mobility of ions increases.

34. What is ion-sensitive electrode?

35. Write the Nernst equation for Daniel Cell.

Answer : Zn | Zn2+ || Cu+2 | Cu

Zn + Cu2+ Zn2+ + Cu







+2

+2

0.0591 [Zn ]E = E - log

2 [Cu ]

36. Write the Nernst equation for any one electrode reaction.

Answer : Mn+ + ne- M

n+0.0591E = E + log [M ]

n

37. What is Helmholtz double layer?

PART – B

1. Describe the construction and working of any galvanic cell.

2. Derive Nernst equation for the emf of a cell.

3. Derive the Nernst equation for the emf of an electrode.

4. What are the conventions followed in the representation of a cell?

5. What are the differences between electrolytic cell and electrochemical cell?

6. How is the emf of a cell measured?

7. Describe standard hydrogen electrode.

8. Explain the construction of calomel electrode. How is it used to measure the emf

of an electrode?

9. What is ion sensitive electrode? What are its applications?10. Write a note on glass electrode and its use in pH measurement.

11. What is electrochemical series? What are its applications?

12. What is the principle of potentiometric titration? Explain any one potentiometric

titration.

13. What is the principle of conductometric titrations? Describe the conductometric

titration of a strong acid with a strong base.

14. What are the advantages of conductometric titrations? Explain the conductometric

titration of a mixture of strong and weak acids with a strong base.






It is due to the direct at

moisture. Alkali and Alkaline ea

temperatures, most metals excep

Mechanism

i) Oxidation occurs at t

M M2+

ii) Oxygen takes up the

½ O2 + 2e-

iii) O2- ion reacts with M

M2+ + O2-

The metal surface is c

corrosion occurs by diffusion of

of oxide film is perpendicular to

Different types of oxide

(i) Porous and Non-Porous Oxi

(a) If the volume of the oxiconsumed, the oxide layer is por

(e.g.) The volumes of ox

the volume of the metal consum

II

tack of oxygen on metal surface in the absenc

rth metals are corroded at low temperatures. At

Au, Pt and Ag are oxidized.

e surface of the metal to form M2+ ions.

+ 2e-

lectrons. O2 is reduced to O2-

O2-

+ to form metal oxide.

MO

nverted to a monolayer of metal oxide. Fu

M2+ ion through the metal oxide barrier. The gr

the metal surface.

films are formed.

de Film (or) Pilling-Bedworth Rule

e layer formed is less than the volume of theus.

ides of alkali and alkaline earth metals are less

d. So the oxide layer is porous and non-protectiv

e of

high

ther

wth

etal

than

.







(b) If the volume of the oxide layer formed is greater than the volume of the metal

consumed, the oxide layer is non-porous.

(e.g.) The volumes of oxides of heavy metals such as Pb, Sn are greater than

the volumes of the metal consumed. So the oxide layer is non-porous and protective.

(ii) Stable Oxide Layer

A stable oxide layer is firmly adsorbed on the metal surface. The layer is

impervious and prevents further corrosion. So the layer itself acts as a protective

coating. (E.g.) Oxides of Al, Cu etc.

(iii) Unstable oxide Layer

This is mainly produced on the surface of noble metals such Ag, Au etc. The

unstable oxide decomposes to stable metal and oxygen.

Metal Oxide Metal + Oxygen

(iv) Volatile Oxide

The oxide film volatilizes as soon as it is formed. It leaves fresh metal surface

for further continuous attack. (e.g.) Molybdenum oxide MoO3.

(2) Corrosion by Hydrogen

(a) Hydrogen embrittlement

Definition

It is formation of cracks and blisters on the metal by hydrogen gas when the

metal comes into contact with H2S.

Iron liberates atomic hydrogen by reacting with H2S.

Fe + H2S FeS + 2H

Hydrogen atoms diffuse into the voids of metal matrix. When the pressure of

the gas increases, cracks and blisters develop on the metal.

(b) Decarburisation

It is the process of decrease in the carbon content of steel. At high

temperature, molecular hydrogen decomposes to atomic hydrogen.

2

High TemperatureH 2H







When steel is exposed to this environment, carbon in the steel reacts with

atomic hydrogen.

C + 4H CH4

Hence the carbon content in steel decreases. Collection of methane gas in the

voids of steel develops high pressure and causes cracking.

(3) Liquid Metal Corrosion

It is due to the chemical action of flowing liquid metal at high temperature. It

involves :

(i) dissolution of a solid metal by the liquid metal.

(ii) Penetration of liquid metal into the solid metal.

II. Wet (or) Electrochemical Corrosion :

It occurs under the following conditions.

(i) When two dissimilar metals or alloys are in contact with each other in

presence of an aqueous solution or moisture.

(ii) When the metal is exposed to an electrolyte with varying amounts of

oxygen.

Mechanism of Wet Corrosion

(1) Metal dissolution occurs at the anode.

M Mn+ + ne-

(2) Reduction reaction occurs at the cathode in different environments.

(a) Acidic environment : Here hydrogen gas is evolved at the cathode.

2 H+ + 2e- H2

(b) Neutral environment : In neutral or slightly alkaline medium, hydroxide

ions are formed at the cathode.

½ O2 + 2e-

+ H2O 2OH-

(a) Hydrogen Evolution type corrosion (In Acidic Medium)






All metals above hydrog

in acidic solution with simultane

(e.g.) When iron come

hydrogen evolution occurs.

At anode : Iron is oxidize

Fe Fe+2 + 2

At cathode : H+ ion is red

2 H+ + 2e-

(b) Absorption of Oxygen (or

neutral or weakly alkaline med

The surface of iron is nsome cracks develop on the film

the metal part acts as cathode.

(e.g.) When iron is in

oxygen, OH- ions are formed.


II

n in the electrochemical series tend to get disso

us evolution of H2 gas.

s into contact with non-oxidising acid like

d to Fe2+

-

uced to H2.

2

Formation of hydroxide ion type corrosio

ium)

rmally coated with a thin film of iron oxide. B, anodic areas are created on the surface. The re

ontact with an electrolyte solution in presenc

d to Fe+2

lved

Cl,

(In

ut if st of

e of







Fe Fe2+ + 2e-

At cathode : O2 is reduced to OH-.

½ O2 + 2e- + H2O 2OH-

Overall Reaction

Fe+2 + 2OH- Fe(OH)2

If enough oxygen is present, Fe(OH)2 is oxidized to Fe(OH)3.

4Fe(OH)2 + O2 + H2O 4Fe(OH)3

Differences between chemical corrosion and electrochemical corrosion:

Chemical Corrosion Electrochemical Corrosion

1. It occurs in dry condition It occurs in presence of moisture or

electrolyte.

2. It occurs by the direct chemical attack

on the metal by the environment.

It occurs by the formation of a large

number of anodic and cathodic areas.

3. Even a homogenous metal surface is

corroded.

Only heterogeneous or bimetallic

surface is corroded.

4. Corrosion products gather at the place

of corrosion.

Corrosion occurs at the anode, while

the products form elsewhere.

5. It is a self controlled process It is a continuous process6. It takes place by adsorption

mechanism.

It follows electrochemical reaction.

(e.g.) Mild scale formation on iron

surface

(e.g.) Rusting of iron under moist

atmosphere

Types of electrochemical corrosion

There are two types:

(i) Galvanic corrosion(ii) Differential aeration or Concentration cell corrosion

(i) Galvanic corrosion

When two different metals are in contact with each other in presence of

aqueous solution or moisture, galvanic corrosion takes place.






The metal with more ne

less negative electrode potential

In the Zn-Fe couple as s

potential, dissolves in preference

Example :

Steel screw in a brass m

E0 = -0.44V. For Cu E0 = +0.34

Prevention

Galvanic corrosion is mi

metals.

(ii) Differential aeration (or) co

Let a metal be partially

metal above the solution is more

part inside the solution acts as an

At anode : Corrosion occ

M M

II

gative electrode potential acts as anode. Metal

cts as cathode.

own in the figure, zinc with more negative elect

to iron. Zn acts as anode and Fe as cathode.

arine hardware easily undergoes corrosion. Iron

. Iron corrodes in preference to Cu.

nimized by providing an insulation between the

ncentration cell corrosion

immersed in a conducting solution. The part o

aerated and acts like cathode. The less aerated

ode and corrodes.

rs (less aeration)+ + 2e-

with

rode

has

two

the

etal






At cathode : Production o

½ O2 + 2e-

Water line corrosion

Let us consider metal tanwater line is exposed to higher

below water level. The metal le

called water line corrosion.

Examples of differential aerati

i) Pitting or localize

ii) Crevice corrosion

iii) Pipeline corrosion

iv) Corrosion on wire

(i) Pitting Corrosion

It is the localized attack r

Example : Metal area cov

The area covered by t

uncovered area exposed to air or

The rate of corrosion is

smaller. Thus more material is re


Fe Fe2+ + 2

At cathode : O2 is reduce

½ O2 + H2O + 2e-

Overall reaction :

Fe2+ + OH- F

II

f OH- ions (more aeration)

+ H2O 2OH-

k partially filled up with water. The metal area a concentration of oxygen (cathode) than the

s exposed to O2 acts as anode and corrodes. Th

n corrosion

corrosion

fence

sulting in the formation of a hole due to corrosi

ered by a drop of water, sand, dirt etc.

e drop or dirt acts as anode and corrodes.

O2 acts as cathode.

ore if the cathodic area is larger and anodic ar

moved from the same area and a pit is formed.

d to Fe+2

-

to OH-.

2OH-

(OH)2

ove etal

is is

n.

The

a is






(ii) Crevice Corrosion

Let a crevice or crack b

with a liquid. The crevice acts li

The exposed area acts as cathode

(e.g.) rivets, joints.

(iii) Pipeline Corrosion

Buried pipelines or cableanother type (sand, more aerated

(iv) Corrosion on wire fence

In a wire fence, the wire

fence. So corrosion takes place a

Factors influencing corrosion

1. Nature of the metal

(i) Position in emf series

Metals above hydrogen i

they have negative reduction p

active metal with a higher negati

(ii) Areas of anode and cathod

II

etween two different metallic objects be in co

e anode due to less oxygen availability and corr

.

s passing from one type of soil (clay, less aerateget corroded due to differential aeration.

at the crossings are less aerated than the rest o

the wire crossings, which become anodic.

n the electrochemical series corrode easily bec

tential. When two metals are in contact, the

e potential corrodes.

tact

des.

) to

the

ause

ore







Corrosion will be severe if the anodic area is smaller and cathodic area is

larger. The larger cathodic area demands more electrons. So the anodic area corrodes

faster.

(iii) Purity

100% pure metal will not corrode. (e.g.) Pure Zn does not corrode. If the metal

has trace amount of impurity, it corrodes. (e.g.) Zinc metal with iron or copper

impurity forms an electrochemical cell. The base metal Zn acts as anode and corrodes.

(iv) Over Voltage

Corrosion rate is inversely proportional to the over voltage of the metal in a

corrosive surroundings. (e.g.) The hydrogen over voltage of Zn in 1M H2SO4 is 0.7V.

So the rate of corrosion is low. But when some Cu impurity is present, the over

voltage is reduced and corrosion rate increases.

(v) Nature of the Film

Nature of film formed on the metal surface determines extent of corrosion.

(e.g.) In the case of alkali and alkaline earth metals, the oxide film formed is porous.

The corrosion continues. In the case of heavy metals, the oxide film is non-porous.

The film acts as a protective layer.

(vi) Nature of corrosion product

If the corrosion product is soluble in the corroding medium, corrosion rate is

faster. Similarly if the corrosion product is volatile (e.g. MoO3), corrosion will be

more.

2. Nature of Environment

(i) Temperature

Increase of temperature increases corrosion rate because the rate of diffusion

of ions increases.

(ii) Humidity






Rate of corrosion is more

solvent for O2, CO2 etc, to prod

cell.

(iii) Corrosive gases

Acidic gases like CO2,

corrosion.

(iv) Presence of suspended par

Particles like NaCl, (NH

and increase rate of corrosion.

(v) Effect of pH

Generally in alkaline me

medium.

The effect of pH on the

diagram as indicated in the figur

The figure shows zones

which pH=7 and correspondin

corrosion zone. So iron rusts und

The rate of corrosion can

1) If the potential is changeimmune to corrosion.

2) If the potential applied is

3) If the pH is increased to

4) If the pH is reduced to le

II

, if humidity of environment is high. Moisture ac

uce electrolyte necessary for formation of corr

SO2, H2S etc, produce electrolytes and incr

icles

)2SO4 along with moisture are powerful electro

ium, the rate of corrosion is less compared to a

corrosion of iron in water is shown in the Pour

.

f corrosion, immunity and passivity. Z is the poi

electrode potential is E= -0.2V. This is in

er these conditions.

be altered by shifting the point Z to different regi

to -0.8V by applying external current, iron bec

positive, iron becomes passive.

ore than 7, corrosion rate decreases.

s than 7, rate of corrosion increases.

ts as

sion

ease

lytes

idic

baix

nt at

the

ons.

mes







CORROSION CONTROL

The rate of corrosion can be controlled by modifying the metal or

environment. Some control methods are

1) proper selection of metals

2) Use of pure metals3) Use of metal alloys

4) Cathodic protection

a. Sacrificial anode protection

b. Impressed current cathodic protection

5) Changing the environment

6) Use of inhibitors

a. Anodic inhibitors

b. Cathodic inhibitors

7) Applying protective coatings

1) Proper selection of metals

Noble metals are used in ornaments and in surgical instruments, because they

do not corrode. Contact of dissimilar metals far away from each other in

electrochemical series should be avoided.

2) By using pure metals

Pure metals have high corrosion resistance. Even a trace of impurity will lead

to corrosion, the base metal becoming anode.

3) Use of alloys

Use of metal alloys is a good method of protection against corrosion. (e.g.)

Stainless steel containing chromium forms a coherent oxide film which protects steel

against further attack.

4) Proper designing

i. Complicated designs with more angles, sharp edges and corners should be

avoided.






ii. Direct contact of dissimi

material between the two

iii. Smaller area for cathode

iv. Tanks and containers s

drained off completely.v. Crevices should be avoid

vi. Bendings should be smo

vii. Annealing minimizes cor

5) Cathodic Protection

The metal to be protectedways.

a) Sacrificial anodic protection

Here the metal to be pr

active metal (anodic metal) call

II

lar metals lead to galvanic corrosion. So insul

metals should be inserted.

nd larger area for anode must be provided.

ould be designed such that the liquid shoul

d or they should be filled using fillers.

th.

rosion.

is made to act like a cathode. This is achieved in

tected is made cathode by connecting it to a

d sacrificial anode. Only the more active metal

ting

be

two

ore

will






be corroded, protecting the par

method is called sacrificial anodi

Applications

i) Protection of buried p

ii) Protection of ships an

iii) Calcium metal is useiv) Magnesium sheets ar

formation.

b) Impressed current cathodic

Here an impressed curr

corrosion current. Thus the corro

The negative terminal of

positive terminal is connected toin a ‘back -fill (containing a mix

increase electrical contact.

Mg

Applications

II

nt metal. Since the anodic metal is sacrificed,

c protection. Mg, Zn are used as sacrificial anode

Metal to be protected

ipelines, cables

d boats

to minimize engine corrosion inserted into domestic water boilers to prevent

protection method

nt is applied in an opposite direction to annul

ding metal is converted to cathode from anode.

battery is connected to the metal to be protected.

an inert electrode like graphite. The anode is b ture of gypsum, coke breeze and sodium sulphat

Metal to be protected

the

s.

rust

the

The

ried ) to







i) Protection of tanks, transmission line towers, underground water pipes, oil

pipe line, ships etc.

Limitations

i) It is costlyii) It fails when current is switched off.

Corrosion inhibitors

A corrosion inhibitor is a substance that reduces corrosion, when added to the

corrosive environment. There are three types of inhibitors.

i) Anodic inhibitors - chromate, nitrate

ii) Cathodic inhibitors - amines

iii) Vapour phase inhibitors - benzonitrile.

i) Anodic inhibitors

(e.g.) chromate, nitrate, phosphates, tungstate.

The inhibitors form insoluble compound with the newly produced metal ions

and prevent corrosion. This compound is adsorbed on the metal surface to form a

passive film. Anodic inhibitors are used to repair

i) the crack of oxide film on metal surface

ii) pitting corrosion

iii) porous oxide film on metal surface

ii) Cathodic inhibitors

There are two types depending on the nature of cathodic reaction in an

electrochemical reaction.

a) In acidic solution

Example : amines, thiourea, mercaptans act as inhibitors.

Here evolution of H2 is the cathodic reaction.

2 H+ + 2e- H2







Paint is a mechanical dispersion of one or more fine pigments in a medium

(thinner + vehicle). When a paint is applied to metal surface, the thinner evaporates.

The vehicle undergoes slow oxidation to form a pigmented film.

Requirements or requisites of a good paint

A good paint should,

i) have good covering power

ii) spread easily on the surface

iii) not crack on drying

iv) adhere well to the surface

v) give a glossy film

vi) be corrosion and water resistant

vii) have stable colour

Constituents of Paint

Pigment

Vehicle

Thinner

Drier

Filler

Plasticizer

Antiskinning agent

1. Pigment

It is a solid that gives colour to the paint.

Functions:

To give colour and opacity to the film.

To provide strength to the film.

To protect film by reflecting U.V. rays.

To provide resistance to abrasion and weather.

Example:

White pigment - White lead, TiO2

Blue pigment - Prussion blue







Green pigment - Chromium oxide

Red pigment - Red lead, Fe3O4

2. Vehicle (or) Drying Oil

It is the film-forming liquid. It holds the ingredients of the paint. It is a non-volatile high molecular weight fatty acid of vegetable or animal.

Function

To hold the pigment on the surface.

To form a protective layer by oxidation and polymerization.

To impart water repellency, toughness and durability of film.

To improve adhesion of film.

Example

Lin seed oil, Castor oil.

3. Thinner

It is the volatile portion of paint. It is added to reduce the viscosity of the paint

for easy application on the surface. It easily evaporates after paint is applied.

Functions

To reduce viscosity of paint.

To dissolve vehicle and other additives.

To suspend the pigments.

To increase elasticity of film.

To increase penetration of vehicle.

To improve drying of film.

Example

Turpentine, Dipentine, Xylol.

4. Drier

It is a substance used to speed up drying of the paint.

Functions

To act as oxygen carrier or catalyst.

To provide oxygen essential for oxidation and polymerization of drying oil.

Example

Metallic soap, linoleate and resinate of Co, Mn etc.







5. Extender or Filler

These are white pigments that form bulk of the paint.

Functions

To reduce cost of paint To prevent shrinkage and cracking of film

To modify shades of pigment

To retard settling of pigments in paint.

Example

Talc gypsum, china-day.

6. Plasticizer

It is added to the paint to provide elasticity to the film and prevent its cracking.

Example

Triphenyl phosphate, Tricresyl phosphate

7. Antiskinning agent

It is a chemical added to the paint to prevent gelling and peeling of the paint.

Example

Polyhydroxy phenols.

Pigment Volume Concentration (P.V.C.)

The P.V.C. of a paint is calculated using the equation.

P.V.C. =

Volume of pigment in the paint

Volumeof pigment in the paint + Volume of non-volatile vehicle in the paint

If P.V.C. is high, durability, adhesion and consistency of the paint will be low.

Failure of Paints

A paint may fail due to any one of the following reasons:

i) Chalking : It is the gradual powdering of the paint film on the painted

surface. This happens due to improper dispersion of pigment in vehicle.

ii) Cracking : A paint film cracks due to unequal expansion or contraction of

paint coats.






iii) Evasion : This is very

iv) Blistering : It is du

sunshine.

Metallic Coating

Electroplating or Electro-depo

It is the deposition of co

through an electrolytic solution o

The base metal to be ele

electrode forms anode. The elect

Objectives or uses or applicati

i) To enhance resistanc

ii) To give a decorative

iii) To enhance resistanc

iv) To improve hardness

v) To obtain polished su

Theory

If the coating metal itse

bath does not change during ele

replenished continuously by diss

Example

Electroplating of Gold

II

quick chalking.

to improper surface exposure of paint to st

ition

at metal on the base metal by passing direct cu

f a soluble salt of the coat metal.

troplated acts as cathode. The coat metal or an

olyte is a soluble salt of coat metal.

ns of Electroplating:

to corrosion of base metals.

ppearance.

to chemical attack.

and wearing resistance.

rface.

lf forms the anode, the concentration of electr

trolysis. The metal ions deposited on the cathod

lution of the anode.

rong

rent

inert

lyte

are







It is the deposition of a noble metal (from its salt solution) on a catalytically

active metal surface using a reducing agent without use of electric current.

The reducing agent reduces the metal ions. The metal atoms get deposited

over the surface to give a thin uniform coating.

Metal ions + reducing agent metal (deposited) + oxidation product

Example

Electroless nickel plating

The various steps are:

Step I : Pretreatment and activation of the surface:

The surface to be plated is degreased by using organic solvents or alkali and

then accompanied by acid treatment.

i) The surface of stainless steel is activated by dipping in hot solution of 50%

H2SO4.

ii) Mg alloy surface is activated by giving a thin coating of zinc and copper over

it.

iii) Al, Cu, Fe, brass etc, do not require activation.

iv) Plastic, glass etc, are activated by dipping in a solution of SnCl2 /HCl and then

in PdCl2 solution. On drying a thin layer of palladium is formed on the

surface.

Step II : Preparation of plating bath:

The plating bath consists of:

i) Coating Metal : A solution of NiCl2 20g/lit.

ii) Reducing agent : Sodium hypophosphite 20g/lit.

iii) Exaltant to accelerate coating rate and complexing agent : Sodium succinate

15g/lit.

iv) Buffer to maintain pH at 4.5 : Sodium acetate 10g/lit.

v) Temperature 93oC

The pretreated object is immersed in the plating bath for required time. The

following reactions occur and nickel is coated on the object.

Cathode : Ni2+ + 2e- Ni

Anode : H2PO2- + H2O H2PO3

- + 2H+ + 2e-







Overall Reaction : Ni2+ + H2PO2- + H2O Ni + H2PO3

- + 2H+

Uses of Nickel Plating

i) For decorative coating of jewellery, decorative items and automobile spares

ii) For coating of polymers for decorative purpose.iii) For electronic appliances.

Advantages of electroless plating over electro plating

i) Electricity is not necessary

ii) Complicated parts are uniformly coated

iii) Plastics, glass etc, are easily coated

iv) Good mechanical, chemical and magnetic properties are obtained.

Differences between Electroplating and Electroless plating:

Electroplating Electroless Plating

1. It is done by passing current. It is done by auto catalytic redox

reaction.

2. Separate anode is required. Catalytic surface of the object acts as

anode.

3. Anode reaction: M Mn+ + ne- Anode reaction : R O + ne-

4. Cathode reaction: Mn+ + ne- M Cathode reaction: Mn+ + ne- M

5. Irregular objects are not satisfactorily

plated

All objects are satisfactorily plated.

6. Object to be coated forms the cathode Object to be coated forms catalytically

active surface.

7. It is carried out on conducting

materials

It is carried out even on insulators.

Control of corrosion by modifying the environment:There are five methods

1. Deaeration :

Presence of oxygen increases corrosion rate. Deaeration involves removal of

dissolved oxygen by increasing the temperature together with the mechanical

agitation. This also removes dissolved oxygen.







12) Define Galvanic corrosion.

13) What are the differences between dry and wet corrosions?

14) Explain the terms with examples.

Ans: a) Pipeline corrosion, b) Pitting corrosion, c) Crevice corrosion

15) What is concentration cell corrosion?Ans : It is the corrosion occurring when a metal is exposed to an electrolyte with

varying amount of oxygen. (E.g.) pitting corrosion.

16) Name some factors that affect corrosion.

17) A steel screw in a brass marine hardware corrodes. Explain.

18) Wire mesh corrodes faster at the joints. Why?

Ans : The joints are welded and stressed. The joints become anodic with respect to

wires. At the anodic part oxidation takes place and the metal is corroded.

19) Iron corrodes faster than aluminium. Explain.

Ans : During corrosion, aluminium forms a non-porous tightly adhering protective

oxide film (Al2O3) on the surface. So further corrosion is prevented by the film.

20) How does the nature of the metal affect corrosion?

21) Impure metal corrodes faster than pure metal explain.

Ans : The impurity in the metal causes heterogeneity. Tiny electrochemical cells

are formed at the exposed parts, with impurity acting as cathode. The anodic part

corrodes.

22) Bolt and nut made of the same metal is preferred. Explain.

Ans : Such a combination does not permit local galvanic cell formation. So

corrosion cannot take place.

23) How does the nature of environment affect corrosion?

24) Using chemical equations explain the corrosion of iron in weakly alkaline

solutions.

25) Corrosion of metal is the highest at the metal junction in a galvanic couple. Why?

Ans : At the metal junction, air cannot diffuse easily. So the junction is less

aerated and acts as anode. The more aerated remaining part acts as cathode. Thus

a galvanic cell is formed at the junction and the anodic part corrodes.

26) State the basic design rules in controlling corrosion.

27) What is cathodic protection?







Ans : The metal to be protected from corrosion is connected to more active metal

like Mg, Al. The Mg or Al metal forms anode and undergoes corrosion, protecting

the parent metal, which acts as cathode.

28) Why does a drop of oil, water, dust or salt solution resting on iron surface lead to

corrosion?Ans : The area covered by water or solution is less exposed to air. So the covered

area acts as anode and corrodes. The uncovered area is exposed to air and acts as

cathode.

29) Which of the following metals provide cathodic protection of iron? Al,Zn,Cu,Ni.

Ans : Al and Zn.

30) What is a sacrificial anode? How does it protect a submerged pipeline?

31) How is impressed current cathodic protection carried out?

32) What are (corrosion) inhibitors? Give example.

33) How do inhibitors resist corrosion?

Ans : Anodic inhibitors form a sparingly soluble salt with newly formed metal

ions and get adsorbed on the metal surface. Thus further corrosion is prevented.

Cathodic inhibitors decrease the rate of corrosion, by blocking active metal

sites due to adsorption.

34) What are vapour phase inhibitors? How do they function?

35) State some requirements of a good paint.

36) Describe the role of pigment in a paint. Give an example.

37) Explain the function of drier in a paint.

38) What is the role of extender in paints?

39) Explain the following terms : (a) Plasticizer; (b) Thinner; (c) Vehicle or drying oil.

40) Define electroplating (or) electrodeposition.

41) Mention some factors that affect the quality of electroplating.

42) What is meant by electroless plating?

43) Mention some uses or applications or advantage of (i) Ni plating; (ii) Au plating.

44) What is galvanization?

Ans : The process of giving a zinc coating on a base metal is galvanization.

45) What are the reasons for failure of a paint?

PART - B : QUESTIONS







1) What are the differences between chemical and electrochemical corrosion?

Explain any four factors that affect electrochemical corrosion.

2) Discuss the mechanism of electrochemical corrosion.

3) Explain the various types of dry or chemical corrosion.

4) Discuss the different types of oxide films formed in chemical corrosion.5) Write notes on a) Hydrogen embrittlement corrosion; (b) Decarburisation;

(c)Liquid metal corrosion.

6) Explain hydrogen evolution type corrosion.

7) Discuss the following types of electrochemical corrosion.

a. Galvanic corrosion

b. Differential aeration or concentration cell corrosion

8) Write briefly about a) Pitting corrosion; b) Pipeline corrosion; c) Crevice

corrosion.

9) What are the factors affecting corrosion?

10) Explain water line corrosion.

11) Proper designing controls corrosion - Elucidate.

12) What is sacrificial anode? Discuss its role in the prevention of corrosion.

13) Write notes on

a. Impressed current cathodic protection

b. Corrosion inhibitors

14) Discuss the effect of pH on corrosion.

15) Explain Galvanic corrosion.

16) What are the important constituents of a paint? Explain the function of various

constituents.

17) What are the main objectives of electroplating? Explain the electroplating of gold.

18) Write a note on nickel electroplating.

19) Give a brief account of electroless plating.







UNIT-III

FUELS & COMBUSTION

IntroductionMost of the industries depend largely upon the power generated by the

combustion of fuels. Any source of heat energy is called as fuel. Fuel is a substancethat combines with oxygen and produce large amount of heat.

DefinitionFuel may be defined as any substance which undergoes combustion with

oxygen to supply heat energy without producing any objectionable gases.Classification of Fuels

On the basis of physical state fuels are broadly classified into three types.FUELSolid Liquid Gaseous

Characteristics of a good fuelA good fuel should have the following characteristics

1. It should have high calorific value.2. It should have moderate ignition temperature.3. It should contain low moisture content.

4. It should not give any harmful gases during combustion.5. It should have low% of non-combustible matter.6. It should be cheap and readily available.7. It should not undergo spontaneous combustion.8. Transportation should be easy.

Calorific valueIt may be defined as the amount of heat liberated when unit mass of fuel

undergoes complete combustion.Solid Fuels

CoalCoal is formed by the decay of vegetable and woody materials under high

pressure and heat.The process of conversion of woody material to coal is known as coalificationor metamorphism.

Classification of coal by rankVarious types of coal can be recognized on the basis of rank from the parent

material. Various types of coals are;






The progressive transformation of coal by rank results ini. Decrease in the moisture content

ii. Decrease in H2,O2 ,N2 and S contentiii. Decrease in volatile matter contentiv. Increase in 'C' contentv. Increase in calorific value

vi. Increase in hardness

Analysis of coalIn order to confirm the quality or rank of the coal the following two methods arecarried out. They are,

1) Proximate analysis.2) Ultimate analysis.

1. Proximate analysis

This method is used for the determination of i) Moisture contentii) Volatile matteriii) Ash contentiv) Fixed carbon

i) Determination of moistureA known weight of coal sample is taken in silica crucible and it is heated in an

electric hot air oven at 100-110oC for 1 hour. After 1 hour, the crucible is cooled andweighed. The process of heating, cooling and weighing is repeated until the weight of the crucible becomes constant.

Loss in weight of coal due to moisture content is calculated as follows.% of moisture = Loss in weight x 100

Weight of coal taken

ii) Determination of volatile matterThe dried sample of coal from the first experiment is covered with a lid and

heated is a muffle furnace at 950oc for 7 minutes. Then, the crucible is cooled andweighed. The process of heating, cooling and weighing is repeated until the weight of the crucible becomes constant.

Loss in weight of coal due to volatile matter is calculated as follows.% of volatile matter = Loss in weight x 100

Weight of dried coaliii) Determination of ash

After the removal of volatile matter, the coal left in the crucible is heated inthe muffle furnace without lid at 700-750oC for 30 minutes. Then, the crucible iscooled andweighed. The process of heating, cooling and weighing is repeated until the weight of the crucible becomes constant. The residual weight of coal due to formation of ashcan becalculated as follows







% of ash = Weight of ash formed x100

Weight of dried coal after removed of volatile matter

iv) Fixed carbon

It is determined by deducing the moisture, volatile and ash contents from 100.

% of fixed carbon = [100-( % of moisture + % of volatile matter + % of ash)]

Significance (or) Importance of proximate analysis

S.No.

Coal

Composition MeritsDemerits

1.Moisturecontent

i) It reduces the dry ash

ii) It makes the coal bed inuniform

i) It increases the transportcharges.ii) It reduces the calorificvalue.

2.Volatilematter

Low volatile mattercontaining coal is suitable forthe manufacture of coal gasand metallurgical coke.

i) It reduces the calorificvalueii) The coal with largeamount of volatile matterburns with a smoky andlong flame.

3.

Ash

Content

The ash consists of silica,alumina, magnesia, lime etc.,which are used in various

smallscale industries.

i) It reduces calorificvalueii) It increases the

transport charges.iii) It is a useless matterand its disposal is a bigproblem

4.Fixed

Carbon

i) It has high calorific valueii) It helps to design thefurnace and the shape of thefire box

--------

2) Ultimate analysisThis method is used for the determination of

i) Carbon and hydrogenii) Nitrogeniii)Sulphuriv) Oxygen

i) Determination of carbon and hydrogen







A known weight of coal sample is heated in the presence of cupric oxide incombustion apparatus. The carbon and hydrogen in the coal sample are oxidised toCO2 and H2O respectively.

C + O2 CO2

(12) (44)

H2 + ½ O2 H2O(2) (18)

The liberated CO2 and H2O vapour are passed through the tubes containingknown weight of KOH solution and anhydrous CaCl2 respectively. When the reactionis over, the tubes are disconnected and weighed. The increase in weight of KOH tuberepresent the weight of CO2 while increase in weight of CaCl2 tube represents theweight of H2O respectively.

Knowing the weights of CO2 and H2O, the % of C and H can be calculated asfollows:

% of carbon =(12/44) x (Weight of CO2 formed / Weight of coal sample) x100

% of Hydrogen = (2/18) x (Weight of H2O formed / Weight of coal sample) x100

ii) Determination of nitrogenNitrogen can be determined by Kjeldal's method. In this method, a known

weight of coal sample is heated with con. H2SO4 is the presence of K2SO4and CuSO4

.Nitrogen present in the coal sample is converted into ammonium sulphate and a clear

solution is obtained.The % of N2 present is the coal sample can be determined by volumetric

analysis.

2N +3H2 + H2SO4 (NH4)2 SO4

The solution is then treated with excess NaOH, the ammonia gas is liberated.

(NH4)2 SO4 +2NaOH 2NH3 + Na2SO4 +2H2O

The liberated ammonia is distilled over and absorbed by known volume of 0.1N

H2SO4.The volume of unused 0.1N H2SO4 is determined by titrating it against 0.1NNaOH . Thus, the amount of acid neutralized by liberated ammonia is determined.From this the % of nitrogen can be calculated as follows.

% of N = (V1-V2 ) m1 x Normality x 14 x 100

Weight of coal sample x 1000

% of N = Volume of acid (ml) x Normality x 1.4







value. SO2 , SO3 etc. which makeair pollution.

4. OxygenLower the % of O 2 higher isits calorific value.

If the coal has high % of O2 ,it absorbs moisture and willreduce the calorificvalue.

CarbonisationThe process of conversion of coal in to coke by strong heating in the absence

of air is known as carbonization. This is also called as destructive distillation.

Caking and coking coalsThe product obtained from the carbonisation is soft, plastic and coherent mass

is called coking coal. On the other hand if the product is hard and strong then the cokeis called coking coal.

Metallurgical cokeA coke which is used in metallurgical purposes is called as metallurgical coke.

They are hard, strong, porous and coherent.

Requisites or Characteristics of metallurgical coke1. It should have minimum percentage of moisture, ash, sulphur and

phosphorous.2. It should have pores in nature, so that the oxygen can easily contact the carbon

which ensures the complete combustion3. It should have high mechanical strength to with stand high pressure.4. It should have low rate of combustion5. It must be uniform and medium in size.6. It should have high calorific value

7. It must be cheap and readily available8. It should burn easily9. It should have very low reactivity

Manufacture of metallurgical coke - Otto Hoffmann by product ovenOtto Hoffmann oven consists of a number of series of narrow silica chambers.

Each chamber is provided with a charging hole at the top. The oven is provided withiron door at each end for discharging the coke.







Coal is charged into chambers and the doors are closed. The chambers areheated by a mixture of preheated air and producer gas at 1200oC. The coal undergoescombustion and liberate waste gases. The heat of the waste gases are used for heat theregeneration of oven at about 1000oC before leaving the chamber.

Therefore, heating of the oven is continued till the liberation of volatile gasesis completely. After 24 hours, the coke is removed from the oven and quenched withwater.The yield of coke is about 75%.

The valuable by products like coal gas, tar, ammonia, H2S benzol etc., arerecovered.

Advantages

1.The carbonization time is less2.Heating can be done externally by producer gas.3.Variable by products can be recovered.

Liquid Fuels

Types of PetrolDepending upon the type and nature of hydrocarbons (paraffin) present in the

crude oil, petrol can be classified into the following three types.1. Straight run petrol2. Cracked Petrol3. Polymer petrol

1. Straight run petrolThe crude oil on fractional distillation yields only about 15 - 20% gasoline.

This is known as straight run gasoline. The quality of straight run gasoline is not sogood. It containsmainly straight chain paraffin, which ignite readily and more rapidly than any otherhydrocarbons and hence it produces knocking (unwanted sound) in IC engines.

2. Cracked petrol.







Cracking is defined as “the decomposition of high boiling hydrocarbons of

high molecular weight into simpler, low boiling hydrocarbons of low molecularweight.”

C10H22 C5H12 + C5H10

Decane n-Pentane Pentene

B.Pt : 174oC B.Pt : 36oC

Thus the petrol obtained by cracking is known as cracked petrol.

Types of CrackingThere are two kinds of cracking

1. Thermal cracking2. Catalytic cracking

1. Thermal CrackingIf cracking is carried out at higher temperature and pressure without any

catalyst, it is called Thermal Cracking. The petrol so obtained is called cracked petrol.There are two types of thermal cracking.

(i) Liquid Phase Thermal CrackingIn this method, the heavy oil is cracked at a temperature of 475 - 530oC under

high pressure of 100 kg/cm2 to keep the reaction product in liquid state. The crackedproducts are then separated into various fractions in a fractionating column. The yieldof gasoline is about 50-60% and the octane number is 65-70.

(ii) Vapour phase thermal crackingIn this method, the heavy oil is first vapourised and then cracked at a

temperature of 600 - 650oC under a lower pressure of 10 - 20 kg/cm2. The yield of

gasoline is about, 70%. This process is suitable only for those oils which are readilyvapourised.

2. Catatytic CrackingWhen cracking is carried out at lower temperature and pressure in the presence

of suitable catalyst, it is called Catalytic Cracking. The catalyst used is aluminiumsilicate or alumina.

3. Polymer petrolThe gaseous by-products, obtained during cracking, contain olefins (like

ethylene, propene and butene) and alkanes (like methane, ethane, propane). Thesegases undergo polymerization at high temperature and pressure with or withoutcatalyst to give petrol rich in branched alkanes. Thus the gasoline obtained bypolymerisation is called polymer petrol.

Types of polymerisationThere are two types of polymerisation.

(a) Thermal polymerisation







It is carried out at 500 - 600oC and 70 - 350 kg / cm2 pressure. The productsare gasoline and gas oil mixture, from which gasoline is separated by fractionaldistillation.

(b) Catalytic polymerisation.It is carried out at lower temperature of 150 - 200oC in presence of catalyst

likeH3PO4 .

Synthetic petrolThe synthetic petrol can be obtained by hydrogenation of coal. It is nothing

but heating of coal with hydrogen at high temperature and pressure, gasoline or petrolis obtained. The following two methods are available for the hydrogenation of coal.

1.Bergius process (or) direct hydrogenation.2.Fischer- Tropsch process (or) indirect hydrogenation.

1.Bergius process (or) Direct hydrogenation of coal

In this process, the finely powdered coal is made in to a paste with heavy oiland nickel oleate catalyst. The coal paste and hydrogen gas mixture is pumped intothe converter. The mixture is heated to 400-5000C under 200-250 atmosphere. Duringthis process, the unsaturated coal becomes saturated hydrocarbons.

Coal dust + H2Catalyst

Saturated hydrocarbon 2

200 - 250 atm400-4500C

The saturated hydrocarbon further undergoes decomposition to yield a mixtureof lower hydrocarbons.

DecompositionSaturated hydrocarbon Crude oil (or) lower

hydrocarbons






2

The mixture is passed through a condenser, where the crude oil is obtained.The crude oil is then fractioned to get

i) Gasolineii) Middle oil andiii) Heavy oil.

The middle oil is further hydrogenated in the presence of catalyst to give moregasoline. The heavy oil is again used for making a paste with fresh coal dust.

2. Fisher Tropsch Process or Indirect hydrogenation of coal

In this process, coal is first converted to coke. Then, coke is heated and steamis passed over it. Water gas is produced.

C + H2O1200 C CO + H2

The water gas is mixed with hydrogen and the mixture is passed throughchamber containing Fe2O3 to remove H2S. Then it is passed through a chambercontaining mixture of Fe2O3 + Na2O3 to remove organic sulphur compounds.






The purified gas is compressed to 5 to 25 atmospheres over a cobalt, thoriumand MgO catalyst on kieselghur at 200oC. A mixture of straight chain paraffins andolefins areproduced.

nCO + 2nH2 CnH2n + nH2Oolefin

nCO + (2n+1)H2 CnH2n+2 + nH2Oparafin

The mixture of olefins and straight chains are passed through a condenser,where the crude oil is obtained.

The crude oil is then sent through fractionating column to yields,i) Gasoline andii) Heavy oil.

The heavy oil is used for cracking to get more gasoline.

Knocking

DefinitionKnocking is defined as a sharp metallic sound produced in the internal

combustion engine by immature ignition of air and gasoline mixture.Causes of knocking

In an internal combustion engine (Petrol engine), a mixture of gasoline vapourand air is used as fuel. The mixture is ignited by an electric spark. But in some cases,the rate of combustion will not be uniform due to unwanted impurities present ingasoline. Therefore, the rate of ignition increases gradually and the final portion of theair fuel mixture gets ignited instantaneously, producing an explosive sound called asknocking. Knocking reduces the efficiency of the engine.Chemical structure and knocking

The tendency of knocking of a fuel depends upon the molecular structure,design of engine, fuel-air ratio etc., The knocking tendency decreases as follows.

Straight chain alkanes > Monosubstituted alkanes > Cyclo alkanes > alkanes >aromatics.

Octane number or octane ratingThe knocking tendency of gasoline is usually expressed in terms of octane

number. Isooctane has better resistance to knocking and its antiknock value has beengiven as 100. On the other hand n-heptane has very poor resistance to knocking andits antiknock value has been given as zero.

DefinitionOctane number is defined as the percentage by volume of isooctane present in

a mixture of isooctane and n-heptane.







Isooctane (octane No = 100)

CH3-CH2 – CH2-CH2-CH2-CH2-CH3

n- heptanes (Octane Number = 0)Anti-knocking agents/leaded petrol

Knocking can be minimized or prevented by addition of suitable additives

such as tetraethyl lead (CH2)4Pb to petrol. This petrol is called leaded petrol and theprocess is called as sweetening of petrol.Mechanism of prevention of knocking

Proper combustion causes knocking due to formation of free radicalmechanism. When addition of TEL to petrol, it under goes thermal decomposition toform ethyl free radical. It combines with the free radicals in knocking process andthus the chain growth is stopped.Disadvantages of TEL

During thermal decomposition of TEL, certain amount of lead oxide andmetallic lead may deposit on the plug and cylinder wall. In order to avoid this,ethylene dibromide is added to dilute the lead pollution.

CH – Br2

Pb+ CH2 = CH2 + PbBr2

CH – Br2

Diesel oilIt is obtained from fractional distillation of petroleum between 250 - 320oC.It

is a mixture of C15H32 to C18H38 hydrocarbons. Its calorific value is about 11000kcal/kg. It is used as a very good diesel engine fuel.

Ignition lag or Ignition delay

The combustion of a fuel in a diesel engine is not instantaneous and the timebetween injection of the fuel and its ignition is called as ignition lag or ignition delay.

Diesel indexThe quality of a diesel oil is indicated by diesel index number using the

following formula.Diesel index number

Specific gravity (API) x Aniline point in Fo 100Aniline point and specific gravity are noted from API. (American Petroleum Institute)scale.

Cetane number (or) Cetane ratingThe knocking tendency of diesel is usually expressed in terms of cetane

number. Cetane has better resistance to knocking and its antiknock value has beengiven as 100. On the other hand 2-methyl naphthalene has poor resistance to knockingand its antiknock value has been given as zero.

CH3- (CH2)14- CH3

Cetane (or) hexa decans(Cetene number = 100)







2 Methyl naphthalene (Cetane number = 0)

The cetane number is defined as “the percentage by volume of hexa decane

present in a mixture of hexa decane and - methyl naphthalene, which has the sameignition lag as the fuel under test “

The cetane number decreases in the following order

Straight chain alkanes > naphthalene > alkenes > branched alkanes > aromatic.The cetane number of diesel can be improved by addition of additives such as

ethyl nitrate or isoamyl nitrate to diesel. This process is called as doping.

Gaseous fuels

Producer gas

The average composition of the producer gas is as follows.CO 30%N 52-55% 2

H 8-12% 2

CO 3% 2

It's calorific value is 1300 Kcal/kg

ManufactureThe furnace consists of a tall steel vessel lined with silica bricks inside. It is

provided with cup and cone arrangement to feed coal into furnace. It is also providedwith inlet and outlet for passing air and steam and removes the ash respectively.

The low grade coal is fed in to the furnace as shown in the figure. Then amixture of air and steam is passed over a red hot coke at 11000C; the producer gas isproduced. The following reactions take place in different zones.

Manufacture of producer gas

i) Ash zone






}

The lowest zone consists mainly of ash. It is used to preheat the incoming airand steam.

ii) Oxidation zoneIn this zone, the coal undergoes oxidation to form CO and CO2. It is an

exothermic reaction. The temperature of this zone is 11000C.

C + ½ O2 CO + N2(Air) + N2

ExothermicC + O2 CO2

iii) Distillation zoneThis is the upper part of the fuel bed. The incoming coal is heated by outgoing

gases to remove volatile matter. The temperature of this zone is 400-5000C.

Uses1. It is used for heating open hearth furnace, glass furnace, muffle furnace etc.,

2. It can be used as reducing agent in metallurgical operation.

Water gasThe average composition of water gas is as follows.

CO = 40 - 42 %H = 48 - 51% 2

CO = 3 - 5 % 2

N = 3-6 % 2

It's calorific value is about 2800 Kcal/kg.

Manufacture

The furnace consists of a tall steel vessel lined with silica bricks inside. It isprovided with cup and cone arrangement to feed the coke into furnace .At the bottom,it is provided with two inlets for passing air and steam.

The coke is fed into the furnace as shown in the figure. When steam and airare passed alternatively over a red hot coke at 900 - 10000C, water gas is produced.







ReactionThe reaction of water gas production involves the following steps.

i. Air is passed through the coke bed, carbon burns to CO2 and the temperature of thefuel bed increase to over 10000C.

C + O2 CO2 (Exothermic)

ii. Steam is passed through the red hot coke; CO and H2 are produced. The reaction isbeing endothermic, the temperature of the fuel and falls below 10000c.

C + H2O CO + H2 ( Exothermic)

In order to maintain the temperature of the fuel bed above 10000C, the air andsteam are passed alternatively.

Uses1. It is used as a source of hydrogen.2. A mixture of water gas and producer gas is used for the manufacture of ammonia.3. It is mixed with hydrogen to produce methanol.4. It is used for the preparation of carbureted water gas (Water gas + gaseoushydrocarbon) which can be used for lighting and heating purposes.

Liquefied Petroleum Gas (LPG)It is obtained as a by - Product during fractional distillation of crude petroleum

oil or by cracking of heavy oil. It consists of propane and butane. It can be readilyliquefied under pressure, so it can be economically stored and transported incylinders. The average composition of LPG is as follows.Constituents

n - Butane = 38.5 %Iso butane = 37 %Propane = 24.5 %

Its calorific value is about 25,000 kcal /kg

Uses1. It is used as a domestic and industrial fuel.2. It is also used as a motor fuel.

Anna University Questions1. What is water gas? How is it manufactured? Write the chemical reaction takingplace in

different zones. [June 05, Dec 06]2. How will you obtain coke by Otto Hottmann method.[June 05, Dec 05, Dec06]3. Explain the proximate analysis of coal. [June 05, Dec 05, Dec06]4. How is ultimate analysis carried out? [Dec 06].5. Describe the different methods by which synthetic petrol can be obtained [Dec 05].6. How will you obtain synthetic petrol by Bergius process [June 05].7. Describe the manufacture of gasoline by Fischer Troprch method [ June 06 May07].8 .Explain the mechanism of petrol knocks [Dec 06].







pound of water through 10C.

Higher and Lower calorific values

a) Higher (or) Gross calorific values (or) GCV (or) HCVIn general, all fuels contain hydrogen whose calorific value can be determined

by experimentally. During combustion, H is converted to H O, the latent heat 2of

steam also gets included in the measured heat. It is called as higher (or) gross calorificvalue.Definition

It may be defined as the total amount of heat liberated when a unit mass fuelundergoes complete combustion and the products of combustion are cooled to roomtemperature.

b) Lower or Net calorific value (or) NCV (or) LCVIn most of the combustion process, the steam and moisture are not condensed,

but they escape along with combustion gases. Hence, a lesser amount of heat isavailable. This is known as lower or net calorific value.

DefinitionIt may be defined as the total amount of heat liberated when a unit mass of

fuel undergoes complete combustion and the products of combustion are allowed toescape.

[NCV = GCV - Latent heat of steam]

H2 + ½ O2 H2O2gms 16gms 18gms1 8 9

'1' part of H2 produces '9' parts of steam. The latent heat of steam is 587 cal/gm

NCV = {GCV - Mass of hydrogen x 9 x Latent heat of steam}NCV =[ GCV – (9H/100) x 587]

Where 'H' is % of hydrogen in fuel

Theoretical calculation of calorific value by Dulong formulaThe formula based on the following assumptions

i. Most of the fuels have C, H, S and Oii. Calorific value is the sum of the calorific value of each element.

iii. The calorific values of C, H and S are found to be 8080, 34500 and 2240 k cal / kg respectively.

Therefore, the Dulong's formula can be written as follows.

HCV = 1/100 { 8080 C + 34500 (H- O/8) + 2240S }kcal/kg

LCV = HCV- { (9/100) H x 587} kcal/kg

Problems based on calorific values

Problem 1.






Calculation of minimum air requirements1. Theoretical oxygen demand Substances always combine in definite proportionswhich are determined by their molecular weights ie, stoichiometric equation.Ex:Combustion of carbon

C + O2 CO2

12 32 44 (by weight)1 1 1 (by volume)

12 parts by weight of C requires 32 parts by weight of O2 of produce 44 partsby weight of CO2

2. Nitrogen, ash, carbondioxide etc are present in the fuel are incombustible.

3. Air contains 21% oxygen (79% being nitrogen) by volume and 23% oxygen (77%being nitrogen) by weight. Therefore,

Minimum weight of air required = (100/23) x Minimum O2

Minimum volume of air required = (100/21) x Minimum O2

4. At STP (273 K and 760mm. Pressure ) 22.4 liters of any gas weights equal to itsone gram molecule.

Example: 22.4 litres of CO2 at STP will weight 44 gm.

5. At NTP, all perfect gases obey the following gas equation

PV = n RT

WhereP - PressureV - Volumen - Number of moles of gas







R - gas constantT - absolute temperature.

6. Net amount of O2 required can be calculated as follows.Net amount of O2 required = Total amount of O2 required - O2 present in the

fuel.

7. The amount of air required if any excess air is supplied

Theoretical amount of air

= ------------------------------- x {100 + % Excess air}

100

Problems based on combustion reaction

Problem 1.On analysis, an oil was found to have the following percentage composition C

= 50%, H = 10% 0 = 5% and S = 1.8% calculate the weight of oxygen and requiredfor the complete burning of 1 kg of this oil. (Air contains 23% oxygen by weight).Also calculate GCV & NCV.

Solution1. 1 kg of the fuel contains

50/100 = 0.5 kg of carbon10/100 = 0.1 kg of Hydrogen5/100 = 0.5 kg of Oxygen1.8/100 = 0.18 kg of Sulphur

2. The combustion equations of the various elements present in the fuel are as

follows:a. C + O2 CO2

12kg 32kg 44kgb. H2 + ½ O2 H2O

2kg 16kg 18kgc. S + O2 SO2

32kg 32kg 64kg

a) 12 kgs of carbon requires 32 kgs of oxygen0.5 kg of carbon requires = (32 x 0.5) / 12

= 1.33 kg of oxygen

b) 2 kgs of hydrogen requires 16 kgs of oxygen0.1 kg of hydrogen requires = (16 x 0.1)/2

= 0.8 kg of oxygen

c) 32 kgs of sulphur requires 32 kgs of oxygen0.018 kg of sulphur requires = (32 x 0.018) / 32

= 0.018 kg of oxygen







Total amount of O2 required = 1.33 + 0.8 + 0.018= 2.148 kgs

But, the amount of O2 already present in the fuel = 0.05 kg of O2

Net amount of O2 required = Total amount of O2 - O2 already present in thefuel

= 2.148 -0.05= 2.098 kgs

We know that, 23 kgs of O2 is supplied by 100 kgs of air

2.098 kg of O2 is supplied by = (100x 2.098) / 23= 9.12 kgs of air

The minimum amount of air required for complete combustion of 1 kg of a fuel =9.12 kgs

GCV = (1/100) {8080 C + 34500 (H- O/8) + 2240 S } kcal/kg= (1/100) {8080 x 50 + 34500 (10 - 5/ 8 ) + 2240 x 1.8 }

kcal/kg= (1/100) {404000 + 323437.5 + 4032} kcal/kg= 4040 + 3234.375 + 40.32 kcal/kg

GCV = 7314.695 kcal/kg

NCV = GCV - { (9/100) H x 587} kcal/kg= 7314.695- { (9/100) x 10 x 587} kcal/kg= 7314.695 528.3 kcal/kg

NCV = 6786.39 kcal/kg

Problem 2.Calculate the volume of air required for the complete combustion of 100 m3 of

the gaseous fuel having following composition by volumeH2 = 50% CH4 = 36% N2 = 1.5% CO = 6% C2H4 = 4% CO2 = 4%.Air contains 21% oxygen by volume.

1.1 m3 of fuel contains50/100 = 0.5 m3 of H2

36/100 = 0.63 m3 of CH4

1.5/100 = 0.015 m3 of N2

6/100 = 0.06 m3 of CO

4/100 = 0.04 m

3

of C2H44/100 = 0.04 m3 CO2

2. N2 and CO2 are non combustible constituents they do not undergo combustion.3. The combustion equation of the remaining constituents is written as follows.

a) H2 + ½ O2 H2O1 vol 0.5 vol

1 m3 of H2 requires 0.5 m3 of O2

0.5m3 of H2 requires = 0.5 x 0.5







Analysis of these gases give an idea about the complete combustion of fuel ornot.

i. If CO is high in the flue gas shows incomplete combustion of the fuel andshort supply of oxygen.

ii. If CO2 and O2 are high in the flue gas shows complete combustions of the fueland excess supply of oxygen.

Description of Orsat's apparatusIt consists of a horizontal tube having three way stop cock at one end. The end

of three way stop cock is connected to a U tube containing fused CaCl2 to removemoisture in the gas. The another end of the tube is connected with a graduated burette.The burette is surrounded by a water jacket in order to keep the temperature of gasconstantly. The lower end of the burette is connected by a water reservoir by means of rubber tube. The level of the water in burette can be raised or loweredby raising orlowering the reservoir.

The middle of the horizontal tube is connected with 3 bulbs(A, B and C) forabsorbing flue gases as follows:

i. Bulb 'A' containing KOH solution and it absorbs only CO2

ii. Bulb 'B' containing alkaline pyrogallol solution and it absorbs only O2

iii. Bulb 'C' containing ammoniacal cuprous chloride solution and it absorbsCO.

Working of Orsat apparatusThe bulbs A, B and C are cleaned and filled by respective solutions. Now, the

three way stop cock is opened and the burette is filled with water, by raising the waterreservoir to remove air from the burette. Then, the flue gas is taken in the burette upto 100 cc by raising and lowering the reservoir. The 3 way stop cock is now closed.

Absorption of gases in bulbs

i) Absorption of CO2

The stopper of the bulb 'A' is opened and the flue gas is allowed to pass byraising the water reservoir. CO2 present in flue gas is absorbed by KOH. This processis repeated several times by raising and lowering the water reservoir until the volumeof burette becomes constant. The decrease in volume of burette indicates the volumeof CO2 in 100 cc of the flue gas. Now the stopper of the bulb ‘A’ is closed.







Component

Component is the minimum number of independently variable constituents by

means of which the composition of each phase can be expressed in the form of a

chemical equation.

Let us consider the equilibrium :

ice(s)i water(l) vapour(g)

There are three phases. All have the same chemical composition namely H2O.

The component of the system is one.

Thermal decomposition of CaCO3: It is a two component system.

CaCO3(s) CaO(s) + CO2(g)

The composition of each phase can be expressed in the form of a chemical

equation using two constituents. When CaCO3 and CO2 are components,Phase Component

CaCO3 CaCO3+0 CO2

CaO CaCO3 - CO2

CO2 0 CaCO3 + CO2

PCl5(s) PCl3(l) + Cl2(g) is a two component system.

While NH4Cl dissociates, the following equilibrium exists.

NH4Cl(s) NH3(g) + HCl(g)

The system has two phases namely solid and gas. Each can be expressed using

one chemical constituent namely NH4Cl, as long as NH3 and HCl exists in equal

amounts. It is a one component system.

Degree of Freedom

It is the minimum number of independent variables such as pressure,

temperature and concentration that must be fixed to define the system completely.

Let us consider the equilibrium :

ice(s) iq(s) water(l) vapour(g)

All the three phases will be in equilibrium only at a particular temperature and

pressure. So we need not specify any variable. The degree of freedom of the system is

zero. The system is non-variant.







For a gas or mixture of gases, we have to specify both temperature and

pressure. So the degree of freedom is two.

Let us consider the equilibrium

water(l) vapour(g)

To define the system we have to mention either pressure or temperature. The

degree of freedom is one. The system is univariant.

Phase Diagram

Phase diagram is the graph obtained by plotting one degree of freedom against

another.

When a phase diagram is plotted between temperature and pressure, it is called

P-T diagram. It is used for one component system.When a phase diagram is plotted between temperature and composition, it is

called T-C diagram. It is used for two component system.

Uses of Phase Diagram

i) We can predict from phase diagram whether a eutectic mixture or solid

solution or compound is formed on cooling a liquid mixture of two metals.

ii) We can understand the properties of materials in a heterogeneous equilibrium

system.

iii) A study of low melting eutectic alloys used in soldering can be made using

phase diagram.

Application of Phase rule to Water system

Water system is a one component system. There are three equilibria:

ice(s) water(l)

water(l) vapour(g)

ice(s) vapour(g)

The phase diagram contains curves, areas and a triple point.






Curves

Along the vapourisation curve OA, water and vapour are in equilibrium.

water(l) vapour(g)

Along the sublimation curve OB, ice and vapour are in equilibrium.

ice(s) vapour(g)

Along the melting point curve OC, ice and water are in equilibrium.

ice(s) water(l)

OC is slightly inclined towards pressure axis. This shows that melting point of ice decreases with increase of pressure.

Along all the curves, two phases are in equilibrium. Applying phase rule,

F = C - P + 2 = 1 - 2 + 2 = 1

The degree of freedom is one or univariant. To define any point along the

curve, it is enough to mention either pressure or temperature.

Areas

Areas AOC, BOC, AOB represent water, ice and vapour respectively. In any

area, only one phase is present. Applying phase rule,

F = C - P + 2 = 1 - 1 + 2 = 2

To define any point in an area, we have to mention both pressure and

temperature.

P r e s s u r e

Temperature







Triple Point

The three curves OA, OB and OC meet at the tripe point ‘O. At ‘O ice, water

and vapour are in equilibrium,

ice(s) water(l) vapour(g)

Applying phase rule, F = C P+2 = 1 3 + 2 = 0

The degree of freedom is zero. The triple point is self-defined. Temperature

and pressure at ‘O are 0.0075C and 4.58mm respectively.

Metastable curve OA’

Sometimes water can be cooled carefully below 0C without ice formation.

Thus we realize the super cooled water - vapour pressure curve OA'. The metastable

equilibrium is,

Super-cooled water vapour

The equilibrium is unstable. Water can be converted into ice by slight

disturbance.

Reduced or Condensed phase rule

A solid-liquid equilibrium of an alloy has practically no vapour phase and the

effect of pressure is negligible. The experiment is regarded as being conducted atconstant atmospheric pressure. Since the vapour phase is neglected, the system is

called a condensed system. As pressure is kept constant, the phase rule becomes F' =

C - P + 1. This is called reduced phase rule or condensed phase rule.

Classification of a two component system

Two component system is classified into three types based on solubility and

reactivity.

i) Simple eutectic formation

ii) a) Compound formation with congruent melting point

b) Compound formation with incongruent melting point

iii) Solid solution







i) Simple eutectic formation

It is a two component system of two solids which are completely miscible in

the molten state. But the two solids are completely immiscible in the solid state. They

do not form any solid solution or react chemically. Of the different mixtures of the

two solids, the mixture with the lowest melting point is called eutectic mixture.

ii) a) Compound formation with congruent melting point

Here the two component system of two solids form one or more compounds

with definite composition. If the compound melts at a constant temperature, with the

same composition in the solid and molten state, the compound has a congruent

melting point.

b) Compound formation with incongruent melting point

If the compound formed in a two component solid system, decomposes at a

temperature below its melting point, the compound has incongruent melting point.

Also there is formation of a new solid phase with a different composition from that of

the original.

iii) Formation of solid solution

It is a two component system of two solids, which are completely miscible

both in the solid and molten state. They form solid solution with atomic level mixing.

The condition for formation of solid solution is that the two solids must not differ in

atomic radius by more than 15%.

Thermal analysis or Cooling curve

It is the study of the cooling curves of various compositions of a system during

solidification.

Example 1 :







A pure substance in the molten state is cooled slowly. The temperature is

noted at different intervals. Temperature is plotted against time.

The cooling from A to B is continuous. At the freezing point B, solid begins to

separate. The temperature remains constant, until all the liquid is solidified from B to

C . Then temperature of the solid begins to decrease gradually.

Example 2 :

A molten mixture of A and B is cooled slowly. The cooling from E to F is

continuous. At F one solid begins to separate. There is a break in the rate of cooling

from F to G. At G the second solid also begins to separate. Now the temperature

remains constant until all the molten liquid is completely solidified at H. From H

cooling of the solid mass starts. The temperature along the horizontal line GH is theeutectic temperature.







For different compositions of A and B, the experiment is repeated and cooling

curves are drawn. From these curves, a phase diagram is drawn with temperature in Y

axis and composition in X axis.

Use of Cooling curves

1) From cooling curves, melting points and eutectic points are obtained.

2) Behaviour of compounds is understood.

3) Percentage purity of compounds is determined.

4) The composition corresponding to the freezing point gives the composition of

an alloy.

Simple eutectic system or lead-silver system

The lead-silver system is a two component system with the formation of asimple eutectic. The phase diagram has areas, curves and a eutectic point.

i) Curve AO:

A is the freezing point of pure silver. As lead is gradually added to silver,

melting point of silver decreases along AO. Solid Ag and liquid melt are in

equilibrium along AO.

Solid Ag melt

Applying reduced phase rule,

F' = C - P + 1 = 2 - 2 + 1; F' = 1

Along AO the system is univariant.







At any point above AOB, the system is bivariant.

v) Applying Pattinson’s process in the desilverisation of argentiferrous lead

Argentiferrous lead contains about 0.1% silver. It is melted to liquid state. Let

us consider the melt at the point p as in the figure. If the melt is allowed to cool, Pb

begins to crystallize at q. The solution becomes richer and richer in silver. If cooled

further, more lead separates along BO. The melt becomes richer in Ag. When the

eutectic point is reached the percentage of Ag rises to 2.6%. The process of raising the

percentage of lead in the alloy is Pattinsons process.

Uses of eutectic systems

i) Study of eutectic system helps to predict suitable alloy composition.ii) Eutectic system is useful to prepare solders, used to join two metal pieces.

Differences between melting point, eutectic point and triple point

Melting point : It is the temperature at which a solid and its liquid phase, having the

same composition are at equilibrium.

Solid A liquid A

Eutectic point : It is the temperature at which two solids A and B and a liquid phase

are at equilibrium.

Solid A + Solid B Liquid

Triple point : It is the temperature at which three phases namely, solid, liquid and

vapour are in equilibrium.

Solid A liquid A vapour A

Eutectic point is a melting point. But melting point need not be eutectic point.

Merits of phase rule

1) The rule is applied to physical as well as chemical equilibrium.







2) It shows that different systems with the same degree of freedom have similar

behavior.

3) We can decide whether the given number of substances can remain in

equilibrium or not.

4) We can classify equilibrium states in terms of phases, components and degrees

of freedom.

Limitations of phase rule

i) The rule is applied to only systems in equilibrium.

ii) All the phases must be at the same temperature and pressure.

iii) Solid and liquid states must not be in finely divided state. Otherwise there may

be deviations.

iv) External forces like electrical, magnetic and gravitational forces should be

absent. Only P, T, C variables are to be considered.

ALLOYS

Alloy

Alloy is a homogeneous solid solution of two or more elements, one of which

is a metal. Alloy with mercury as a constituent is called amalgam.

Properties of alloys

1) Alloys are harder and less malleable. They have lower melting point than the

component metals.

2) Alloys have low electrical conductivity.

3) They resist corrosion and action of acids.

Importance or need or purpose or use or advantages of making alloy

i) Increasing the hardness of metalsIn general pure metals are soft, but their alloys are hard. Gold and silver are

usually soft. When alloyed with copper they become hard. On addition of 0.5%

Arsenic, lead becomes so hard that we can make bullets.

ii) Lowering the melting point







The melting point of a metal decreases on alloying. (e.g.) Woods metal (alloy

of lead bismuth, tin and cadmium) melts at 60.5C. This is below the melting point of

any constituent metal.

iii) Increasing corrosion resistance

Alloying a metal makes it less reactive and helps resist corrosion. (e.g.) Pure

iron rusts. When alloyed with chromium, it opposes corrosion.

iv) Change in chemical activity

The chemical reactivity of a metal can be increased or decreased by alloying.

v) Change in colour

Alloying improves the colour of metal. (e.g.) Brass with 90% Cu and 10% Znis golden yellow in colour.

vi) Casting property

Alloying makes a soft, brittle metal into hard, fusible and easily castable.

Alloy of lead with 5% tin and 2% antimony is used for casting printing type.

Function (or) Effect of alloying elements

If we add small amounts of Ni, Cr, Mn, V, Mo, W etc., to steel, there is adrastic change in properties like tensile strength, resistance to corrosion, coefficient of

expansion etc. The resulting products are called alloy steels or special steels.

Element Effect on properties Use

1. Nickel Fine grain formation, lowcoefficient of expansion, highcorrosion resistance.

Making balance wheels.

2. Chromium High tensible strength,hardening, high corrosionresistance.

Making surgicalinstruments, cutlery

3. Manganese High abrasion resistance Making grinding wheelsteering, spindles, rails4. Vanadium Production of reversible stress,

increase in tensile strength andabrasion resistance

Making axles, crank pin,piston and locomotiveforgings

5. Molybdenum Cutting hardness at hightemperature

Making high speed stools

6. Tungsten Increase in magnetic retentionand cutting hardness

Making cutting tools,permanent magnets







Heat treatment of alloys (steel)

Definition : It is the process of heating and cooling steel under carefully controlled

conditions.

Purpose or objectives of heat treatment

1. To improve corrosion resistance

2. To remove internal stress and strain

3. To improve electrical and magnetic properties

4. To remove trapped gases

5. To refine grain structure

6. To reduce brittleness and increase toughness and ductility.

Various types (methods) of Heat treatment of alloys (steel).

There are several types, 1. Annealing, 2. Hardening, 3. Tempering,

4.Normalizing, 5.Carburizing, 6. Nitriding, 7.Cyaniding.

1. Annealing

The alloy or steel is heated to a high temperature and then slowly cooled in a

furnace.

Purpose

To increase machinability

To remove trapped gases

Types of annealing

Low temperature annealing

High temperature annealing.

Low temperature annealing

The steel is heated to a temperature below the lower critical temperature and

then cooled slowly.

Purpose

To increase machinability

To reduce hardness







To increase ductility and shock-resistance

To remove internal stress and strain.

High temperature annealing

The steel is heated to 30 to 50C above the higher critical temperature. It is

held at that temperature for sufficient time to allow for internal changes. Then it is

cooled to room temperature.

Purpose

To make steel soft with increase in toughness.

To increase ductility and machinability

2. Hardening (Quenching)

The steel is heated beyond the critical temperature and then cooled suddenly in

oil or brine-water. This process increases the hardness of steel.

Purpose

To increase abrasion resistance for making cutting tools.

To increase resistance to wear.

3. Tempering

Already hardened steel is heated to a temperature below its hardening

temperature and cooled slowly. To retain strength and hardness, reheating temperature

should not exceed 400C. To increase ductility and toughness, reheating temperature

must be in the range 400 to 600C.

Purpose

To increase ductility and toughness and to reduce brittleness.

To make cutting tools, blades etc.

To remove internal stress and strain.

4. Normalising

Steel is heated to a temperature above its higher critical temperature and

cooled gradually in air.







Purpose

To make steel suitable for engineering works.

To have homogeneity in structure with refined grains.

To remove internal strain and stress

To increase toughness.

5. Carburising

Mild steel article is placed in a cast iron box containing pieces of charcoal. It

is heated to 900-950C and kept at the temperature for sufficient time. Carbon is

absorbed to required depth. The article is cooled slowly within the iron box. The outer

part of the of the article becomes a high-carbon steel with 0.8 to 1.2% carbon.

6. Nitriding

Steel or alloy is heated to about 550C in pressure of ammonia. Ammonia

decomposes to give N2 which combines with the surface of alloy to form hard nitride.

Purpose

To form super hard surface.

7. Cyaniding:Steel is immersed in molten salt bath containing KCN at 1120K and then

quenched in oil or water. Purpose to get hard surface.

Classification (or) types of alloys

Based on the type of base metal, alloys are classified into ferrous and non-

ferrous alloys.

Properties of Ferrous alloys

They have high strength and yield point.

The possess ductility and weldability.

They are corrosion and abrasion resistant.

They have less cracks and distortion.

Important ferrous alloys







Nichrome

It is an alloy with 60% nickel, 12% chromium, 25% iron and 2% manganese.

Properties

It has high resistance to oxidation and heat.

It has high melting point.

It possesses high electrical resistance.

It withstands heat upto 1100C.

Use

To make heating elements and resistance coils.

To make machineries exposed to high temperature.

Used in electric irons other household electrical equipments.

To manufacture boiler parts, steam lines stills, retort, aero-engine valves, gas-turbines.

Stainless steel

It is an alloy steel containing chromium with other elements like Nickel,

Molybdenum. It is corrosion resistant due to the formation of a tough film of

chromium oxide at the surface.

There are two types of stainless steels.

a. Heat treatable stainless steel

b. Non-heat treatable stainless steel

a. Heat treatable stainless steel

This steel contains upto 1.2% carbon, less than 12-16% of chromium.

Properties

It is magnetic, tough and resistant to weather and water. It can be heated upto

800C.

Use

Making surgical instruments, scissors, blades etc.







b. Non-heat treatable stainless steel

It has less strength at high temperature. Corrosion resistance is more. There

are two types.

i) Magnetic type

Composition

12-20% chromium and 0.35% carbon.

Properties

It has very high corrosion resistance.

It can be forged and rolled.

Use

To make chemical equipments and automobile parts.

ii) Non-magnetic type

Composition

0.15% carbon, 18-25% chromium, 8 to 20% of nickel. Total percentage of

nickel and chromium is above 23%.

18/8 stainless steel

It has 18% chromium and 8% nickel and 0.15% carbon.

Properties

It has extreme resistance to corrosion. The resistance is increased by the

addition of a small amount of molybdenum.

Use

To make household utensils, sinks and surgical instruments.

Non-ferrous alloys

These do not have iron as a constituent. The main constituents of non-ferrous

alloys are copper, zinc, lead, tin etc. They have low melting points compared to

ferrous alloys.

Properties

Non-ferrous alloys are characterized by







i) Softness and ductility

ii) Attractive colours

iii) Low density and coefficient of friction

iv) Corrosion resistance

Important Non-ferrous alloys

Copper alloys

Brass and Bronze are alloys of copper.

Brass

Brass is copper alloy with copper and zinc as main constituents.

Some important brasses, their properties and uses are tabulated below.

Type of Brass Composition Property UseCommercialbrass

Cu = 90%Zn = 10%

i. Golden colourii. Stronger and harder

than Cu

Forging, rivethardwares jewellery

Dutch metal Cu = 80%Zn = 20%

i. Golden colourii. Suitable for

drawing andforming operation

Tube, flexible hoses jewellery

Cartridge brass Cu = 70%Zn = 30%

Soft and ductile cold,-deformed

Condenser tubes,sheets householdarticles

German Silver Cu = 25 to 50%Zn = 10 to 35%Sn = 5 to 30%

Silver colour,malleable, corrosionresistance

Ornament, bolt nut,utensil, coins

Jobin brass Cu = 59 to 62%Zn = 35 to 38%Sn = 0.5 to 1.5%

High corrosion andabrasion resistance

Propeller and marineworks

Bronze

Bronze is a copper alloy with copper and tin as main constituents. Bronze has

lower melting point than steel. It has better heat and electricity conducting propertythan steel. It has good corrosion resistance.

Some important bronzes, their properties and uses are tabulated below:

Type of Bronze Composition Property Use

Coinage Bronze Cu = 89-92%Sn = 8-11%

Soft, ductile Pump, valve, coin,statue wire, utensils

Gun metal Cu = 85% Hard and tough, Foundry work, heavy







Zn = 4%Sn = 8%Pb = 3%

resistant to explosiveforces.

load bearing, steamplant

Aluminiumbronze

Cu = 90-93%Al = 7-10%

Yellow colour,fusible, resistant tocorrosion and

abrasion.

Jewellery, utensil,coin bushes

PhosphorusBronze

Cu = 87-90%Sn = 10-13%P = 0.4 - 1%

Hard, brittle, lowcoefficient of friction

Bearing, gear, tap,spring turbine, blade.

Nickel bronze Cu = 90%Ni = 9%Fe = 1%

Hard, good tensilestrength, corrosionresistant

Valves, semi-hardbearings.

QUESTIONS

PART - A

1. Define phase rule.

2. Define the terms : phase, component, degrees of freedom.

3. Comment on degrees of freedom.

Ans : It is the minimum number of independent variables such as pressure,

temperature, concentration that must be fixed to define the system completely.

(e.g.) In water system, at the triple point, three phases are in equilibrium.Applying phase rule,

F = C - P + 2 = 1 - 3 + 2 = 0

The degree of freedom is zero; the triple point is self-defined and non-variant.

If F=1, the system is called univariant. We have to mention one variable to

define the system.

4. Calculate the number of components in the following:

KCl(s) + Water(l) KCl hydrate(s)

MgCO3(s) MgO(s) + CO2(g)

PCl5(s) PCl3(l) + Cl2(g)

5. Calculate the number of phases in the following:

A gaseous mixture of N2, H2, NH3.

H2(g) + Cl2(g) 2HCl(g)








Rhombic sulphur(s) Monoclinic sulphur(s)

6. Calculate the number of phases, degree of freedom and components for the

equilibrium:


7. What is a phase diagram?

8. What is triple point?

9. Explain metastable equilibrium.

10. Explain invariant point.

11. How does the melting point of ice change with pressure?

12. What is meant by reduced phase rule?

13. Define eutectic point.14. Eutectic mixture is a mixture and not a compound. Explain.

Ans : In the solid eutectic, the two components are completely immiscible, though

they are miscible in the liquid state. Also the solid eutectic shows the properties of

both the components. So it is a mixture and not a compound.

15. What are the uses of eutectic mixture?

16. What are the applications of phase rule?

17. State the merits of phase rule.

18. Explain the limitations of phase rule.

19. Define an alloy.

20. Name some properties of alloys.

21. What are the advantages of alloys?

22. State some ferrous and non-ferrous alloys.

23. What is the composition of nichrome? What are its uses?

24. Write the composition and uses of the following alloys :

German silver, Dutch metal, Gun metal, brass (commercial), solder, phosphor

bronze, wood metal.

25. Explain the following terms. What are their purposes?

Hardening, tempering, annealing, normalizing, carburising, cyaniding, nitriding.

26. What is quenching?







27. What is 18/8 steel? What are its uses?

PART - B

1. Apply phase rule to a one component system.

2. Discuss in detail lead-silver system with the help of the phase diagram.

3. What is thermal analysis? Draw the cooling curves of a pure substance and its

mixture with another solid and discuss.

4. Why is condensed phase rule applied to a two component alloy system?

5. Apply phase rule to a two component system forming eutectic mixture.

6. Define alloy. What is the effect of alloying a metal? Explain with examples.

7. What are the properties of alloys?8. Mention the purposes of making alloys.

9. Name some ferrous and non-ferrous alloys. Give their composition, properties and

uses.

10. Write a note on stainless steels.

11. Discuss the various methods of heat treatment of alloys and steel.

12. What is Bronze? Name some important bronzes, their composition, properties and

uses.

13. Define Brass. Write briefly about different types of brass.

14. Explain with examples how the property of steel changes with the addition of

small amounts of nickel, chromium, manganese vanadium, molybdenum and

tungsten.







UNIT - V

Analytical techniques

Spectroscopy

The study of interaction of electromagnetic radiation with atoms or molecules

is spectroscopy. During this interaction, radiation is absorbed or emitted by themolecules or atoms. The measurement of the intensity of the emitted or absorbed

radiation forms the basis of spectroscopy.

Absorption spectrum or absorption spectroscopy

When electromagnetic radiation is passed through a solution or vapour, certain

wavelengths are absorbed by the solution or the vapour. The atoms or molecules

undergo transition to a higher energy level. The measurement of the decrease in the

intensity of the radiation is absorption spectroscopy. The spectrum obtained is

absorption spectrum.

Emission spectrum (emission spectroscopy)

Atoms or molecules are excited to higher energy levels by heat or electric

discharge. The excited atoms or molecules come to the ground state by emitting

radiation. The study of the emitted radiation is emission spectroscopy. The spectrum

obtained is emission spectrum.

Atomic spectra

It is the study of interaction of electromagnetic radiation with atoms. Atoms

are excited to higher electronic energy levels. The spectrum contains discrete lines.

Molecular spectra

It is the study of interaction of electromagnetic radiation with molecules.

Molecules are excited to higher rotational, vibrational and electronic energy levels.

The spectrum contains bands.

Differences between atomic and molecular spectra







Atomic spectra Molecular spectra

1. It arises due to the interaction of

electromagnetic radiation with atoms

It arise due to the interaction of

electromagnetic radiation with

molecules

2. The spectrum has discrete lines The spectrum has complicated bands.3. It is due to electronic transitions It is due to rotational vibrational and

electronic transitions

Absorption laws

Lambert’s law:

Let a beam of monochromatic radiation pass through a homogeneous

absorbing medium. Now the rate of decrease in intensity of the radiation ‘dI with the

thickness of the absorbing medium ‘dx, is proportional to intensity ‘I of the incident

radiation.

dI dIkI (or) kdx

dx I

k= absorption coefficient - (1)

Let Io and I be the intensities before entering the medium (x=0) and after

passing through any thickness ‘x respectively.

Integrating equation (1)between these limits.

o

I x

I o

dIkdx

I

o

Iln kx

I - (2)

This is Lamberts Law.

Beer’s law or Beer-Lambert’s law

Let a beam of monochromatic radiation pass through a homogeneous

absorbing medium. Now the rate of decrease in intensity of the radiation ‘dI with the

thickness of the absorbing medium ‘dx is proportional to the intensity of the incident

radiation ‘I as well as the concentration ‘C of the solution.

dIkI C

dx







k = molar absorption coefficient (or)

dII

= kCdx -(1)

Let Io and I be the intensities before entering the medium (x=0) and after

passing through thickness ‘x respectively. Integrating equation (1) between theselimits,

o

I x

I o

dIkC dx

I

o

Iln kCx

I (or) oI

ln kCxI

2.303 =

oI klog CxI 2.303

(2)

A = Cx

oIA log

I absorbance or optical density

k2.303

molar extinction coefficient.

Equation (2) is Beers law (or) Beer - Lamberts law.

Application of Beer Lambert’s law

Absorbance As of solution of known concentration Cs is measured. Now,

As = Csx

x = As /Cs -(1)

Then the absorbance Au of unknown solution is measured.

Au = Cux

u

u

Ax

C - (2)

From equations (1) and (2)

s u

s u

A A

C C (or) u

u s

s

AC C

A

knowing Au, As and Cs, unknown concentration Cu can be calculated.







Limitation of Beer - Lambert’s law

1. Applicable only for dilute solutions.

2. Monochromatic radiation must be used.

3. The solute should not dissociate or associate in solution.4. Not applicable to suspensions.

5. Solution should not contain impurities.

6. Temperature should not change to a large extent.

Colorimetry (or) Colorimetric analysis

Colorimetry is an analytical technique to measure the concentration of

coloured solutions. The absorbance ‘A of the solution is measured using a

photoelectric colorimeter. Then the concentration is determined using Beer-Lamberts

law. If the substance is colourless, a suitable complexing agent is added to convert it

into a coloured compound.

Photoelectric colorimeter - Instrumentation

I. Components

1. Radiation Source: Tungsten filament lamp is used to get visible light between

400-750nm.

2. Monochromator: It allows the radiation of required wavelength to pass through

the solution. It absorbs other wavelengths.

3. Slit i) Entrance slit: It selects a narrow source of light.

ii) Exit slit : It provides narrow band of dispersed spectrum for observation by

the detector.

4. Cell: It holds the test solution. It is transparent. It is made of colour-corrected fused

glass or quartz.

5. Detector : It measures the intensity of the transmitted radiation from the test

solution. Photosensitive devices are used. They produce a current which is

proportional to the transmitted radiation.

6. Recorder (or) Meter: It measures the fraction of radiation abosorbed.







II. Working

A narrow beam of light enters through the monochromator and passes through

the test solution in the cell. The light then passes into the detector. The instrument is

provided with filters and slits to select desired wavelength. The detector generates a

current proportional to the intensity of light transmitted from the test solution. The

current or intensity of transmitted light is more if the solution absorbs less light, i.e.,

the concentration is low.Current transmitted light 1 / Concentration

Now a days, the transmitted light passes into a meter which shows the fraction

of light absorbed, which is proportional to the concentration.

Estimation of Iron by colorimeter

Principle

Iron (III) forms an intense blood-red coloured complex with thiocyanate in

solution.Fe3+ + 6NH4CNS [Fe(CNS)6]

3- + 6 +

4NH

Blood-red

coloured complex

Reagents

i) Standard iron solution : 0.865 g of FAS is dissolved in minimum amount

of distilled water. 10 ml of conc.HNO3 is added and the solution is diluted

to one litre. 1ml of this solution conatins 0.1 mg of Fe.

ii) Ammonium thiocyanate solution : A 10% solution in distilled water is

prepared.

iii) 1:1 HCl acid: 50ml conc.HCl is added to 50ml distilled water.







Procedure

A series of standard solutions of Fe3+ are prepared by adding NH4CNS

solution and a little 1:1 HCl.

A blank is prepared by adding same amount of reagents.

The colorimeter is set at zero absorbance using the blank solution with asuitable filter. The absorbance of each standard solution is measured using the same

filter. Absorbance is plotted against concentration in a graph. A straight line passing

through the origin is obtained. This is calibration curve.

According to Beer-Lamberts law,

A = Cx

Since the thickness x is a constant for the given cell,

A C.

The absorbance of the test solution is measured under the same conditions.

The concentration of the test solution is read from the calibration curve.

Applications of Colorimetry

1. Very small concentrations of many metal ions can be determined.

2. Stability constants of complexes can be evaluated.

3. Geometry of inorganic complexes, cis-trans isomers can be examined.

4. Dissociation constants of acid-base indicators can be determined.

5. Molar extinction coefficients of complexes can be evaluated.

Flame photometry

Atomised metal is introduced into a flame. The metal atoms are excited and

then come to the ground state. The intensity of the emitted light is measured. The







wavelength of the light emitted identifies the metal. Its intensity measures the

concentration of the metal.

Theory or Principle

A metal salt solution is introduced into the flame.i) The solvent evaporates leaving behind the solid salt.

ii) The solid salt vaporises and dissociates into atoms.

iii) Some of the metal atoms are excited to higher energy levels by absorbing

heat energy.

iv) The excited atoms return to the ground state by emitting radiation.

v) The emitted radiation passes through the filter and falls on a detector. The

detector generates a current which is amplified and recorded.

Evaporation vapourization dissociaton

M+ X- =========== MX============MX==========M + X

(solution) (solid) (gas) (gas) (gas)

(gas) M*------------- M

excited state ground

state

Instrumentation

Components

1. Burner

Propane-air or acetylene-air flame is used. The flame should,

i) vapourise the solvent from the test solution.

ii) dissociate the solid into atoms.

iii) excite atoms and make them emit radiation.







2. Mirror

Radiation from the flame is emitted in all directions. To increase the radiation

reaching the detector, a concave mirror is kept behind the burner.

3. Slit

i) Entrance slit : It is kept between flame and monochromator. It allows only radiation

from the flame and mirror.

ii) Exit slit : It is placed between detector and monochromator. It prevents entry of

interfering radiations.

4. Monochromator

It permits only radiation of required wavelength to pass through. It absorbs

other radiations.

5. Detector

Radiation from the monochromator falls on a detector. It generates a current

which is proportional to the intensity of radiation.

6. Amplifier and Recorder

The current from the detector is amplified and recorded.

Working

At a constant pressure, air is sent into an atomiser. This suction draws some

test solution into the atomiser. The mixture of air and sample solution is mixed with

fuel and burnt in the burner. The radiation emitted from the burner flame is sent

successively through lense, filter, detector, amplifier and recorder. A graph, called

calibration curve, is drawn between recorder reading (intensity of emitted radiation)







and concentration of sample solution. A similar experiment is performed for the test

solution. Its concentration is read from the graph.

Applications of flame photometry1. Estimation of sodium

The flame photometer is switched on. Air and gas supplies are regulated.

Initially double distilled water is sent in and ignition is started. After the instrument is

warmed up for nearly 10 minutes, it is adjusted for zero reading in the display.

Sodium gives a characteristic yellow emission at 589nm. The instrument is set

at this wavelength.

Series of standard solutions of sodium chloride (1,2,3,4,....10ppm) are

prepared. They are introduced one by one into the flame and readings (intensity of emitted light) are noted. A calibration graph is drawn between intensity of emitted

light and concentration. A straight line passing through origin is obtained.

Now the test solution is introduced into the flame and the reading is noted.

The concentration of sodium in the (water) sample is read from the graph.

2. Quantitative analysis

Alkali and alkaline earth metals, transition metals like Fe, Mn, Cu can be

estimated.

3. Qualitative analysis

Alkali and alkaline earth metals can be detected from the colour of the flame.

Element max (nm) Colour of the flame







Ca 422 nm Brick red

Li 670 nm Scarlet red

K 766 nm Red

4. Measurement of alkali and alkaline earth metals, Fe, Cu, Mn etc are useful inmedicine, agriculture and botany.

5. It is highly useful in the analysis of biological fluids and tissues.

6. Industrial and natural waters, petroleum products, glass, cement are analysed

by this technique.

Interferences in flame photometry

i) Spectral interference : When two elements show different spectra but both emit at

the same wavelength, this interference occurs.

ii) Ionic interference : It occurs due to ionisation of metal atoms at high temperature.

Ionisation reduces atomic emission.

iii) Cation-anion interference : Anions like oxalate, phosphate form stable complexes

with cations and so interfere.

Cation-Cation Interference: This reduces intensity of the signal.

Limitations of flame photometry

i) Inert gases and all metal ions cannot be determined.

ii) Molecular nature of the metal present in the original sample cannot be ascertained.

iii) Liquid samples alone are to be used.

Atomic absorption spectroscopy (AAS)

Principle

It is based on the atomisation of sample accompanied by absorption of

characteristic radiation by the ground state gaseous atoms. When a radiation of







suitable wavelength is passed through the flame containing metal atoms, part of the

radiation is absorbed. The extent of absorption is proportional to the concentration of

metal atoms in the flame. The amount of radiation absorbed is determined in AAS.

Components

i. Radiation source

It is a hollow cathode lamp.It has a glass tube containing a noble gas like

argon (anode) and a hollow cathode made of the analyte metal. The lamp emits stable

intense characteristic radiation of the metal to be estimated.

ii. Chopper

It is a rotating wheel placed between the hollow cathode lamp and the flame. It

breaks the steady light from the lamp into a pulsating light. This is because the

recorder recognises only a pulsating current.

iii. Burner

Total combustion burners or premixed burners are used to convert liquid

sample into gas and the molecules into atomic state.

iv. Nebuliser

Before the liquid sample enters the burner, it is converted into small droplets

by the nebuliser.

v. Monochromator

It selects a particular wavelength from the radiation emitted by the hollow

cathode lamp. Prisoms and gratings are used as monochromators.

vi. Detector

Photomultiplier tubes are used as detectors. They convert the light falling on

them into electric current.

vii. Recorder

The signal from the amplifier is recorded in a chart paper or digital read out

device.







Working

The characteristic radiation from the radiation source passes through the flame

to which the sample solution is aspirated. The metallic compounds are decomposed

and reduced to the elemental state forming a cloud of atoms. The atoms absorb a

fraction of radiation in the flame. The unabsorbed radiation from the flame passesthrough the monochromator, and then to the detector, amplifier and recorder.

Quantitative estimation of nickel

Requirements

i) Wavelength : 232 nm

ii) Light source : Hollow cathode lamp (nickel)

iii) Flame : Nitrous oxide - acetylene flame-reducing (rich, red)

Step I : Preparation of standard stock solution

1.0g of nickel nitrate is dissolved in minimum amount of dil. 1:1 HNO3 and

made up to one lit. with 1% HNO3. This standard nickel solution has a strength of

1000ppm.

Step II : Construction of calibration curve

The instrument is switched on and allowed to warm up for ten minutes. 232nm

wavelength is selected. A blank solution is aspirated into the flame and the assembly

is adjusted for zero absorbance.

From the standard nickel stock solution, a series of standard solutions are

prepared. They are aspirated into the flame one by one and the absorbance is

measured. A calibration graph is plotted between absorbance and concentration of

nickel (ppm).







Step III : Estimation of unknown concentration

The test solution is aspirated into the flame under the same conditions and its

absorbance is measured. The corresponding concentration is read from the graph.

Applications of AAS

i) Nearly seventy to eighty metals present in any sample can be accurately and

rapidly determined.

ii) Trace amount of lead in alloy can be estimated.

iii) Vanadium in lubricating oil can be determined.

iv) AAS is used to analyse clinical samples like blood, urine, hair, teeth and tissues

etc.

v) AAS is applied in food industry, soil analysis and pharmaceutical industry.

Limitations of AAS

i) Sample preparation is tedious and time consuming.

ii) Chemical form of metal cannot be ascertained.

iii) Only metals can be determined.

UV - Visible Spectroscopy

UV-Visible spectroscopy involves transition of valence electrons of molecules

or ions from the ground electronic state E0 to a higher electronic state E1. The ultra

violet region is 100 to 400 nm. The visible region is 400 to 800 nm. The amount of

energy required for the transition is,

E = hυ = E1 - E0

Type of transitions involved in organic molecules







There are three types of electrons in organic molecules - bonded electrons,

bonded electrons and non-bonded electrons (n). All the three types of electrons are

present in formaldehyde :

H

H

C

Ö : n

The types of transitions and their order of energeis are:

n * < * < n * << *

1. n * transitions

These are shown by unsaturated molecules with heteroatoms like N, O, S andhalogen. They are due to the excitation of an electron from non-bonding lone pair (n)

to an empty anti-bonding * orbital.

Example :

Acetaldehyde and acetone

n * transition occurs at 270 to 300 nm. An electron from the non-bonding

orbital of oxygen atom is excited to a * orbital.

2. * transitions

These are shown by molecules with electrons. (e.g.) olefins, aromatic

compounds. They are due to the excitation of an electron from bonding orbital to

anti-bonding * orbital.

Example :







Ethylene CH2 = CH2

A * transition occurs at 200 nm.

3. n * transitions

These are shown by saturated compounds containing a lone pair of electrons.

They are due to the excitation of an electron from non-bonding orbital (n) to a *

anti-bonding orbital.

Example :

Trimethyl amine (CH3)3 N:

n * transition occurs at 227 nm. An electron from the non-bonding orbital

of ‘N atom is excited to a * orbital.

3. * transitions

These are shown by all saturated compounds. They are due to the excitation of

an electron from bonding orbital to a * anti-bonding orbital.

Example : CH4

* transition occurs at 122 nm. * are high energy transitions.

They are weak, and usually not observed.

Chromophores

These are groups with multiple bonds and are responsible for the colour of acompound.

N O

nitroso azo nitro carbonyl

N N N

O

OOC

Auxochromes

These are groups that are not themselves responsible for the colour of the

compound. But they enhance the colour of chromophore by conjugation.

(e.g.) - OH - NH2

Chromophore Auxochrome

1. This group is responsible for

colour

This group is not responsible for colour,

but enhances the colour of chromophore







2. Contains multiple bond. Contains lone pair of electrons

3. (e.g.) - N = O - N = N - - NH2 - OH

Instrumentation of UV- Visible SpectrophotometerComponents

i) Radiation source

Hydrogen discharge and deuterium lamps as well as mercury lamps are the

common sources. They must supply a steady radiation of sufficient intensity.

ii) Monochromator

It disperses the radiation according to the desired wavelength. It comprises an

entrance slit, a dispersing prism or grating and an exit slit.

iii) Cells

They hold the sample and reference solutions. The cell must be uniform in

construction, inert to solvents and transparent.

iv) Detector

Barrier layer cell, photomultiplier tube, photocell are used as detectors. The

detector converts the radiation falling on it into electric current.

v) Recorder

The signal from the detector is recorded on a chart paper by the recorder.

Working

The radiation from the source passes through the monochromator, which

selects a narrow range of wavelength to pass through an exit slit. The radiation from

the monochromator is split into two equal beams. One half (reference beam) passes

through cell containing the blank solution. The other half (sample beam) passes

through the cell containing test solution. The detector compares the intensities of the







two beams. If sample absorbs, the intensity of the sample beamI will be less than

that of reference beam (I0). The recorder plots absorbance versus wavelength.

Applications

i) Determination of structure of compound

Hartley established that compounds of similar constitution have similar

absorption spectra. This is Hartley rule. Using this, we can characterise organic

compounds by comparing the UV-visible of spectrum unknown compound with that

of known compounds available in the reference books.

ii) Study of coordinate complexes

Visible spectroscopy is used to study the structure of coordination complexes.

We can identify the cis-trans isomers, octahedral, tetrahedral and square planar

structures.

iii) Determination of formation constants

Visible spectroscopy is used to determine the composition and the formation

constants of coordination complexes.

iv) Kinetic studies

UV-visible spectroscopy is used to study the kinetics of reactions involving

coloured complexes. The absorbance is directly proportional to the concentration. If

the concentration changes, the absorbance also changes. The change in absorption

(concentration) with time is used as a tool in studying the kinetics of ligand

displacement reactions and hydrolysis of coordination complexes.

v) Determination of concentration of solutions

Beer Lamberts law is,

0Ilog cx

I

This law is used to determine molar extinction coefficient of a compound.

Also we can determine concentration of solutions, since absorbance 0Ilog

Iis directly

proportional to concentration. Absorbances of solutions of known concentrations are







measured. A graph is plotted between absorbance and concentration. The absorbance

of unknown solution is measured. Its concentration is read from the graph.

vi) Study of Tautomerism

The percentage of keto and enol forms in a tautomeric equilibrium is estimatedfrom the relative intensities of absorbance bands of the UV spectrum.

(e.g.) Acetoacetic ester.

CH3 C CH2 COOC2H5

O

Keto form

max= 275nm

CH3 C COOC2H5

OH

Enol form

max= 244nm

vii) Determination of calcium in blood

The amount of calcium in blood is estimated. The Ca present in 1 ml of serum

is converted into its oxalate. It is dissolved in dil.H2SO4 and treated with ceric

sulphate solution. The absorbance is measured at 315nm. The amount of calcium in

the blood is determined.

viii) Molecular weight determination

The molecular weight of a compound can be determined by converting it into

a derivative which gives an absorption band. (e.g.) The molecular weight of an amine

is determined by converting it into a picrate derivative.

Infrared (IR) Spectroscopy

If infra red radiation (400-4000 cm-1) is passed through molecules, vibrational-

rotational transitions occur in them.

There are two types of fundamental vibrations in a molecule.

1) Stretching vibration

In this vibration, the bonds are elongated and compressed. But the bond angle

remains unchanged. There are two kinds of stretching vibrations:







i) Symmetric stretching

ii) Asymmetric stretching

2) Bending or deformation vibrations

In this vibration, the bond angle increases and decreases but the bond length

does not change. Bending vibrations are of two kinds.

i) In plane bending vibrations: Here all the atoms are in the same plane during

vibration.

ii) Out of plane bending vibration : Here the atoms go out of plane during vibration.

In plane bending Out of plane bending

Number of modes of vibrations

i) A linear molecule has (3n-5) fundamental vibrational modes. n= number of atoms

in a molecule.







(e.g.) CO2 is a linear molecule.

It can exhibit 3 x 3 - 5 = 4 vibrations. They are shown above.

ii) A non-linear molecule has (3n-6) fundamental vibrational modes. n= number of atoms in the molecule.

(e.g.) Water (H2O) is a non-linear molecule. It has (3n-6) = 3 x 3 - 6 = 3 vibrational

modes. They are shown below.

O

H H

O

H H

O

H H

symmetricstretching

Asymmetricstretching

Bending

Instrumentation of IR Spectrometer

Components

i) Radiation Source

Nichrome wire and Nernst glower are the main sources of IR radiation. Nernst

glower is filament made of oxides of Zr, Th and Ce held together using a binder.

ii) Monochromator

It permits only radiation of desired wavelength to pass through and absorbs

other wavelengths.

iii) Cell

The cell holding the sample solution should be transparent to IR radiation.

NaCl cells are used.

iv) Detector

The IR radiation falling on the detector is converted into an electric signal.

Some detectors are:

i) Thermocouple






ii) Pyroelectric detec

v) Recorder

The electrical signal fro

against wavelength.

Working

The radiation from the s

beam passes through the sample

When the sample abso

intensities. This produces an os

signal from detector gets amplifi

display.

Applications of IR spectroscop

1. Identification of Compounds

Each molecule gives its

same IR spectra, they contain i

can be established by comparing

spectra available in the literature.

2. Detection of functional grou

Different functional gro

absorption frequency, the nature

identified.

Examples

II

tors

the detector is recorded as percentage transmitt

urce is split into two beams of equal intensity.

and the other through the reference.

bs, the two emerging beams will have diff

illating signal to be measured by the detector.

d and goes to the recorder which gives a print o

own unique IR spectrum. If two samples giv

entical molecules. Thus the identity of a comp

the IR spectrum of a compound with the standar

s

ups absorb at different IR regions. So from

of functional group present in the molecule ca

ance

One

rent

The

t or

the

und

d IR

the

n be







Functional group IR Frequency (cm-1

)

C O aldehyde & ketones 1680 - 1720

N H amines 3300 - 3400

O H alcohols, phenols 3500 - 3650

C N cyanides 2200 - 2250

NO2 nitro compound 1350 - 1380

Ar C H aromatic ring 3000 - 3100

R O R ether 1100 - 1200

3. The progress of a reaction can be studied by observing the decrease or

increase in the intensity of a particular peak at different time intervals.

4. Structure determination

We can easily find out whether a molecule is linear or not.Example :

Let us consider NO2. If it is linear, it will show (3n-5) = 3 x 3 - 5 = 4 bands. If

it is non-linear, it will show (3n-6) = 3 x 3 - 6 = 3 bands. The actual IR spectrum of

NO2 showed 3 bands. So we conclude that NO2 is non-linear.

N

O O5. Cis-trans isomers can be identified.

6. Purity of samples

Pure sample gives a sharp and well-resolved IR bands. Impure sample gives

not only broad and poorly resolved bands but also additional bands. So from the IR

spectrum of the sample we can ascertain the purity.

7. Molecular weight of a compound can be determined by measuring end group

concentrations in the IR spectrum.

8. The crystallinity of a solid can be ascertained by changes in the IR spectrum.

PART - A QUESTIONS







Ans : The region 700-1400 cm-1 is finger-print region. It has rich and complicated

intense IR bands. The region is used to identify a molecule just as we use finger

prints of people to identify them.

23. Calculate the number of vibrational modes (or) bands for H2, H2O2, CH4, C6H6,

CO2, H2O, C2H2, C2H6.24. Name some applications of IR spectroscopy.

25. What is the principle of atomic absorption spectroscopy?

26. Explain the principle of flame photometry?

27. What is the frequency of a radiation of wavelength 5000Å.

Ans :10

14 18

C 3 x 106 x10 s

5000 x 10

PART B QUESTIONS

1. Derive Beer-Lamberts law. What are its limitations?

2. What is the principle of colorimetry? Explain the various parts and working of a

colorimeter. Write about its applications.

3. What is flame photometry? Discuss its principle.

4. Write briefly about the instrumentation, working and applications of flame

photometry.

5. Explain the interferences in flame photometry. What are the limitations of flame

photometry?

6. State the principle of atomic absorption spectroscopy. Write a note on the

instrumentation and working and applications of the spectrometer.

7. How is nickel quantitatively estimated using AAS?

8. Narrate the applications and limitations of AAS.

9. What is UV-visible spectroscopy?

10. Discuss the various types of transitions in UV-visible spectroscopy.

11. Explain the instrumentation, working and applications of UV-visible

spectroscopy.

12. Write about the various types of fundamental vibrations in IR spectroscopy.

13. Sketch the possible vibrational modes of Water and CO2.

14. Explain the instrumentation, working and applications of IR spectroscopy.





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