chapter 6.3 redox equilibrium

8/10/2019 Chapter 6.3 Redox Equilibrium

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6.3 Redox Equilibria

A) Redox Reactions

(a) Redox reactions in terms of electron transfer

Oxidation loss of electron(s)

Reduction gain of electron(s)

e.g.(1) loss of 2 e-(oxidation)

Cu + Cl2 CuCl2

gain of e-(reduction)

(2) loss of 2 e-(oxidation)

Zn + Cu2+ Zn2+ + Cu

gain of 2 e-(reduction)

(b) Oxidation State (Number)

Rules for Assigning Oxidation State

1. Oxidation state of an atom in an element = 0

e.g. Na OS = 0 H2 OS = 0 O2 OS = 0

2. The more electronegativeatom in a compound has the

ve oxidation state


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e.g. NH3 OS of N = -3 NF3 OS of N = +3

3. Sum of oxidation state of a neutral compound = 0

e.g. NaCl (+1) + (-1) = 0 H2O (+1 x 2) + (-2) = 0

4. Sum of oxidation state of a simple / complex ion = the

charge on the ion.

e.g. SO42- (-6) + 4(-2) = -2 NH4+ (-3) + 4(+1) = +1

5. Some elements always have the same oxidation state in their

compounds

(i) Group I metals: +1

(ii) Group II metals: +2

(iii) Al: +3

(iv) H: +1 in covalent compounds with non-metals

-1 with group I & II metals (e.g. NaH, MgH2)

(v) F: -1

(vi) O: -2 except

1. +2 in OF2

2. 1 in peroxides (e.g. Na2O2, H2O2)

3. 1/2 in superoxides (e.g. KO2)

e.g. POCl3 P + (-2) + 3(-1) = 0 P = +5


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HPO3- (+1) + P + 3 (-2) = -2 P = +3

6. Assigning oxidation state to elements in covalent compounds

Oxidation state = formal charge on the atom when the

shared electrons are assigned to the more electronegative

one of the bonded atoms.

e.g. CH4, EN of C > H

OS of C = -4

OS of H = +1

N2H4 EN of N > H

OS of N = -2

OS of H = +1

S2O82

S

O

O

O

O S

O

O

O

O

-1

-1

-2

-2

-2

-2

-2

-2

+6

+6

C

H

H H

H

-1 -1 +1

+1

+1

+1

-1-1

N

H

N

H

H H

+1 +1

-1

-1+1 -1 +1

-10 0


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S2O32- 2(S) + 3(-2) = -2 S = +2

+2 is only the average OS of S

or

Ex 6.3-1

Try to assign the OS of the underlined elements.

OH ClO S4O62-

CO32 ClO3

NH4+ ClO4

CrO42 MnO4

Cr2O72 MnO42

(c)Redox reactions in terms of oxidation number

Oxidation increase in oxidation state

Reduction decrease in oxidation state

S

S

OO

-

O-

0

+4

-2

-2 -2

S

S

OO

-

O-

-2

-2 -2

+6

-2 more electropositive

owing to its direct

bonding to more

electronegative oxygen

atom

SO S

S

S

O

O-

OO

-

O


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e.g. 0 +1 oxidation

2Na(s) + Cl2(g)2NaCl(s)

0 1 reduction

Disproportionation a chemical reaction in which a particular

chemical species is simultaneously oxidized and reduced

e.g. reduction

3ClO- 2Cl- + ClO3-

+1 -1 +5

oxidation

Oxidizing agent- substance which oxidize others but itself being

reduced

Reducing agent substance which reduce others but itself being

oxidized

Oxidizing agent and reducing agent are relative only. A

compound may be an oxidizing agent in one reaction but it may

be a reducing agent in another reaction.


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iii) In an acidic solution, balance Hby adding the appropriate

no. of H+to the side deficient in H.

In an alkaline solution, balance H by adding the

appropriate no. H2Oto the side deficient in H and then add

an equal no.of OH-on the opposite side.

iv) Then balance the charge by adding e(s) to the side deficient

in negative charge.

3. Multiply the two balanced half-equations by appropriate no. so

that the no. of egained in one half equation is equal to that lost

in the other.

4. Add the 2 half-equations and eliminate the e. Collect like terms

and cancel any duplications on both sides if necessary.

Example

(1) MnO4-+ Fe2+ Mn2++ Fe3+

MnO4- Mn2+

MnO4- Mn2+ + 4H2O

MnO4- + 8H+ Mn2+ + 4H2O

MnO4- + 8H+ + 5e Mn2+ + 4H2O(1)


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Fe2+ Fe3+

Fe2+ Fe3+ + e ..(2)

(1) + (2) x 5

MnO4- + 8H+ + 5Fe2+ Mn2+ + 4H2O + 5Fe3+

(2) Cr2O72-+ SO2 Cr3++ SO42-

Cr2O72- Cr3+

Cr2O72- 2Cr3+

Cr2O72- 2Cr3+ + 7H2O

Cr2O72- + 14 H+ 2Cr3+ + 7H2O

Cr2O72- + 14 H+ + 6 e 2Cr3+ + 7H2O..(1)

SO2+ 2H2O SO42-

SO2+ 2H2O SO42- + 4H+

SO2+ 2H2O SO42- + 4H++ 2e..(2)

(1)+ (2) x 3

Cr2O72- + 14 H+ + 3SO2+ 6H2O2Cr3+ + 7H2O + 3SO42-+ 12H+

Cr2O72- + 2H+ + 3SO2 2Cr3+ + H2O + 3SO42-


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(3) MnO4-+ H2O2MnO2+ O2 (basic)

MnO4- MnO2

MnO4- MnO2+ 2H2O

MnO4-+ 4H2O MnO2+ 2H2O + 4OH-

MnO4-+ 2H2O + 3e MnO2+ 4OH-.(1)

H2O2 O2

H2O2 + 2OH- O2+ 2H2O

H2O2 + 2OH- O2+ 2H2O + 2e (2)

(1)x 2 + (2) x 3

2MnO4-+ 4H2O + 3H2O2 + 6OH-2MnO2+ 8OH-+ 3O2+ 6H2O

2MnO4-+ 3H2O2 2MnO2+ 2OH-+ 3O2+ 2H2O

(4) SnO22- Sn + SnO32- (basic)

SnO22- Sn + 2H2O

SnO22-+ 4H2O Sn + 2H2O + 4OH-

SnO22-+ 2H2O + 2e Sn + 4OH-(1)

SnO22- SnO32-

SnO22-+ H2O SnO32-

SnO22-+ H2O + 2OH- SnO32-+ 2H2O

SnO22-+ 2OH- SnO32-+ H2O + 2e (2)


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(1)+ (2)

2SnO22- + 2H2O + 2OH- Sn + 4OH-+ SnO32-+ H2O

2SnO22- + H2O Sn + 2OH-+ SnO32-

Ex 6.3-3

Balance the following equations

i) Cu + NO3Cu2++ NO

ii) MnO4+ IMn2++ I2

iii) SO32+ Br2SO42+ Br(acidic)

iv) Mn3+MnO2+ Mn2+(acidic)

v) Sn2++ H2O2Sn4++ H2O (acidic)

vi) MnO2+ PbO2MnO4+ Pb2+(acidic)

vii) Cl2Cl+ ClO3(basic)

viii)Cl2Cl+ ClO(basic)

ix) H2O2+ Cr(OH)4-CrO42-+ H2O (basic)

B) Electrochemical Cells

(a) Introduction

Redox reaction can be split into two half reversible


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Mn+(aq) + ne M(s)

At equilibrium: rate of oxidation = rate of reduction

An electrode potential is set up at the interface of the

solid and solution phases single electrode

potential

The position of equilibrium depends on

1. nature of metal / metal ion system,

2. concentration of metal ions,

3. temperature.

Single electrode potential cannot be measured absolutely

c) Electrochemical Cells Made up of Metal / Metal Ions Systems

Made up of 2 different metal / metal ion half cells

Different metal / metal ion half cells will have different

electrode potentials


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If we connect an electrical wire between the two different metal

electrodes.

The potential difference will drive electrons through the

conducting wire from one cell to another (but only

instantaneously).

Current will stop after a while as charges would be built up in

the two half cells which prevent further electrons movement.

The problem can be solved by adding a salt bridge or a

porous partition. It enables electrical contact while

preventing the mixing of solutions which leads to the direct

redox reaction.

Salt bridge contains a strong electrolyte (e.g. saturated KCl,

KNO3 or NH4NO3) held in a jelly like matrix in a inverted


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Utube or just a piece of filter paper soaked with strong

electrolyte

Porous partition contains tiny passage that allows hindered

flow of ions.

e.g. Daniell cell (composed of Cu and Zn half cell)

Zn electrode: Zn(s)Zn2+(aq) + 2e anode (-ve pole)

Cu electrode: Cu2+(aq) + 2eCu(s) cathode (+ve pole)

Electrons flow from Zn electrode through external circuit

to Cu electrode.

Overall reaction: Zn(s) + Cu2+

(aq) Zn2+(aq) + Cu(s)

d) Electromotive Force (e.m.f.) of Electrochemical Cell

Absolute potentialof a half cell CANNOTbe measured.


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Only thepotential differencebetween 2 different half cells can be

measured.

The potential difference of 2 half cells can be measured by the

following circuit.

Potential difference (measured by the voltmeter) increases with

the increasing resistance (decreasing current).

NO currentflowmaximum potential differencee.m.f. of the

cell


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e) E.m.f. Measurement of Electrochemical Cell

A voltmeter cannot measure the e.m.f. of an electrochemical cell

accurately because current is taken by voltmeter.

E.m.f. can be measured by:

(1) Use a voltmeter of high resistance (faster method)

e.g. a transistor voltmeter, impedance voltmeter or

a digital multimeter.

(2) A potentiometer

When no current flows (i.e. no deflection on

galvanometer),

e.m.f. of the cell under measurement =

known e.m.f. of the standard cellACdistance

ADdistance


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(f) Cell Diagrams (IUPAC Convention)

Commonly used notations

1. Solid line () boundaries between electrodes and

solution (phase boundaries).

2. Dotted line (M)porous partition

(MM

)salt bridge3. E or Ecell e.m.f. of a cell in volt.

4. Sign (+ or) polarity of theRIGHTHANDelectrode.

e.g. Daniell cell

Anode Cathode

Zn(s)Zn2+(aq)MCu2+(aq)Cu(s) E= + 1.1 V

Zn(s)Zn2+(aq)Cu2+(aq)Cu(s)

Reaction represented:

Zn(s)+ Cu2+(aq)Zn2+(aq)+ Cu(s)

If the cell diagram is represented as:

Cu(s)Cu2+(aq)MZn2+(aq)Zn(s) E=1.1 V

Cu(s)Cu2+(aq)Zn2+(aq)Zn(s)

Reaction represented:

Zn2+(aq)+ Cu(s)Zn(s)+ Cu2+(aq)


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C) Standard Electrode Potential and Electrochemical Series

a) Standard Hydrogen Electrode

Since only potential difference between 2 electrodes

can be measured standard electrodemust be

chosen so as to compare electrode potential of

different system.Standard hydrogen electrodeis one of the standard

electrodes and its electrode potential is arbitrarily

assigned as 0.

At equilibrium: 2H+(aq) +2e H2(g)

Cell diagram: Pt(s)H2(g, 1 atm)2H+(aq, 1.0 M)MM

Platinum electrode is used as an inert electrode by which

electrons can leave or enter the electrode system. Platinum

black acts as a catalyst, being porous, it retains a comparatively


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large quantity of hydrogen.

b) Standard Electrode Potential E

Electrode potential depends on [ion] and temp. Thus we have to

standardize the potential measurement conditions.

Eis a value of an electrode potential relative to the standard

hydrogen electrode under the conditions:

1.[ion] = 1.0 M

2.temp. = 298 K

3.pressure of gas = 1 atm.

c)StandardElectrode Potential from e.m.f. Measurements

Standard electrode potential of a metal / metal ion system can be

measured by using standard hydrogen electrode under standard

conditions.

Pt(s)H2(g, 1 atm)2H+(aq, 1.0 M)MMMn+(aq, 1.0M)M(s)

E= Ecell = E

R.H.S.EL.H.S.= ER.H.S.

where ER.H.S..= the electrode to be measured

EL.H.S= the standard Hydrogen electrode = 0


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Electrode potential to be measured is always written on R.H.S.

and thus it always refers to the Reduct ion Potential.

Mn+(aq)+ ne M(s)

The more ve the E, the higher the tendency to release eand

vice versa.

Oxidized form Reduced form E/ VoltsK+(aq)+ e

K(s) 2.92Ca2+(aq)+ 2e

Ca(s) 2.87Na+(aq)+ e

Na(s) 2.71Mg2+(aq)+ 2e

Mg(s) 2.37Al3+(aq)+ 3e

Al(s) 1.66Zn2+(aq)+ 2e

Zn(s) 0.76Fe2+(aq)+ 2e Fe(s) 0.44Pb2+(aq)+ 2e

Pb(s) 0.13

d) Redox Equilibria Extended to Other Systems

Reduction potential of a redox system other than metal / metal

ion system can also be measured. Pt is used as an inert

electrode.

e.g. Br2(aq)+ 2e 2Br(aq) Non-metal / non-metal ion

MMBr2(aq), 2Br(aq)Pt(s)


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Fe3+(aq) + e Fe2+(aq) Ion / ion

MMFe3+(aq), Fe2+(aq)Pt(s)

MnO4(aq)+ 8H+(aq)+ 5e

Mn2+(aq)+ 4H2O(l)

MM[MnO4(aq)+ 8H+(aq)], [Mn2+(aq)+ 4H2O(l)Pt(s)

2IO3-(aq) + 12H+(aq) + 10 e I2(aq) + 6H2O(l)

MM[2IO3-

(aq, 1M)+ 12H

+

(aq, 1M)], [I2(aq, 1M)+ 6H2O(l)]Pt(s)

Rules for constructing cell diagram of system other than metal

/ metal ion.

1. The reduced form of ions is put nearest to the inert electrode.

2. Reduced form is separated oxidized form by comma

3. For oxidized & reduced forms containing more than one

chemical species, they must be included in the oxidized &

reduced forms of the half cell diagram by square brackets.


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e) Electrochemical Series

E.C.S. is formed when the electrode potentials of various

redox systems are tabulated together.

The more ve electrode potential are at the TOP

Greater tendency to lose e

Stronger Reducing Agent


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D) Uses of Standard Reduction Potentials

(a) To construct a working electrochemical cell and calculate its

e.m.f.

Consider the following standard reduction potential at 298 K.

Cu2+(aq) + 2e Cu(s) E

= +0.337V

Ag+(aq)+ e Ag(s) E

= +0.799V

Zn2+(aq)+ 2e Zn(s) E=0.763V

As to construct a working cell, i.e. a cell which runs

spontaneously, the e.m.f. of the cell MUST BE +ve .

Ecell = Eright - Eleft

Case 1:

Cell diagram:

Cu(s)Cu2+(aq 1.0 M)MMAg+(aq 1.0 M)Ag(s) E= +0.462V

Ecell= -0.799 (0.337) = +0.462 V

2Ag+(aq)+ Cu(s) 2Ag(s)+ Cu2+(aq) E= +0.462V


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Case 2:

Cell diagram:

Zn(s)Zn2+(aq 1.0 M)MMCu2+(aq 1.0 M)Cu(s) E= +1.1V

Ecell= (+0.337) (- 0.763) = + 1.1 V

Cu2+(aq)+ Zn(s) Cu(s)+ Zn2+(aq) E= +1.100V

Ex 6.3-4

When a standard Ni2+(aq)Ni(s)half cell is connected to a standard Cu2+(aq)Cu(s)

half cell, the cell voltage is 0.59 V. The copper electrode is the cathode.

Determine the standard reduction potential for Ni2+(aq)Ni(s)half cell. (Given:

Eof Cu2+(aq)Cu(s)= +0.34 V)


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Ex 6.3-5

A cell is based on the reaction of

5Fe(s)+ 2MnO4(aq)+ 16H+(aq)5Fe2+(aq)+ 2Mn2+(aq)+ 8H2O(l)

Given that: Fe2+(aq)Fe E=0.44 V

MnO4(aq)Mn2+(aq) E= +1.51 V

a) Write balance equations for the reactions occurring at the cathode

and the anode.

b) State the direction in which electrons move when the electrodes

are connected externally.

c) Calculating the e.m.f. of the cell

d) Write the cell diagram.


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b) To Predict the Feasibility of Redox Reactions

E = +ve energetically feasible

= 0 in equilibr ium

= -ve not energetically feasible

Evalues predicts whether a reaction is energetically feasible or

not, it does not tell the rate at which the reaction takes place.

Eis an intensive property, i.e. independent of the amount of

the no. of etransferred.

In using Eto predict feasibility of a reaction, care must be taken

that prediction apply only to standard conditionsonly.

In general, ifEcell> 0.4 V, we can predict with confidence.

Ex 6.3-6

Given that: Cu2+(aq)+ 2e Cu(s) E

= +0.34 V

Ag+(aq)+ e Ag(s) E= +0.80 V

i) Write an ionic equation for the spontaneous reaction

ii) Calculate Ecell

iii) Write the cell diagram for this reaction.

iv) Explain why the e.m.f. measured by a potentiometer differs from that


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calculated in (iii).

Ex 6.3-7

Given that: Ag+(aq) + e Ag(s) E

= + 0.80 V

AgCl(s)+ e Ag(s)+ Cl

(aq) E

= + 0.22 V

i) Write the equation for the spontaneous reaction.

ii) Calculate the cell e.m.f.

iii) Write the cell diagram for the spontaneous reaction.


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Ex 6.3-8

Given that: Zn2+(aq) + 2e Zn(s) E

=0.76 V

Hg2Cl2(aq) + 2e 2Hg(l)+ 2Cl

(aq) E

= +0.28 V

i) Write an ionic equation for the spontaneous reaction.

ii) Calculate Ecell.

iii) Write the cell diagram for this reaction.

Ex 6.3-9

(a) Given that

Br2(aq) + 2e 2Br-(aq) E= +1.09 V

O2(g) + 2H2O(l) + 4e 4OH-(aq) E= +0.40 V

Predict whether Br2or O2is a stronger oxidizing agent.


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(b) Given that:

MnO42(aq) + 4H+(aq) +2e

MnO2(s)+ 2H2O(l) E= +2.26 V

MnO4(aq) + e

MnO42(aq) E

= +0.56 V

Show that MnO42(aq)disproportionate in acidic medium.

Ex 6.3-10

Given that: S4O62(aq), 2S2O32(aq)Pt(s) E= +0.10 V

Br2(aq), 2Br(aq)Pt(s) E= +1.10 V

I2(aq), 2I(aq)Pt(s) E= +0.55 V

i) Write the cell diagram for the cell with highest e.m.f. and the equation for the

chemical change that takes place.


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ii) Which is the positive electrode?

iii) An aqueous solution of I2oxidizes S2O32(aq)to S4O62

(aq), but an aqueous

Br2(aq)oxidizes it further to SO32(aq).

a) What does this tell about the relative oxidizing power of Br2and I2?

b) What conclusion can you draw about the Eof the of the following half

cell?

[2SO32(aq)+ 6H+(aq)], [2S2O32(aq)+ 3 H2O(l)]Pt(s)


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E) Effect of Concentration Change on Electrode Potential

e.g. Cu(s) Cu2+(aq,1M)MAg+(aq)Ag(s)

Measure Ecellfor different concentration of Ag+(aq)

Plot Ecellvs log [Ag+(aq)]

Cell reaction Cu(s) + 2Ag+(aq) Cu

2+(aq) + 2Ag(s)

(a) E

cell= EAg - E

Cu= (+0.8) (+ 0.34) = + 0.46 V

Ag+(aq) + e Ag(s) E= + 0.8 V

By Le Chatelier Principle

[Ag+] increase in tendency to form Ag+

equilibrium shifts to LHS

electrode potential of Ag becomesless +ve

Ecellbecomes less +ve

+0.46 Vlog [Ag+] = -8.8

Log [Ag+]

Ecell


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(b) From the graph Ecellas [Ag+]

consistent with the prediction

And Ecelllog [Ag+] EAglog [Ag

+]

(c) When Ecell= 0 log [Ag+(aq)] = -8.8

Ecell= 0 equilibrium is established

[Ag+]eqn= 1.6 x 10-9M

29-2

2

c)10x(1.6

1

]Ag[

][CuK ==

+

+

Kc= 4 x 1017dm3mol-1

(d) log [Ag+] < -8.8 Ecell= -ve

(e) Nernst Equation

E = E

+[ion]ln

Fz

RT

R - gas constant = 8.31 JK-1 T temp

F Faraday constant z charge of metal ion

E = E+ [ion]log

F

RT2.310

z


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E = E+ [ion]log

z

0.0610

Ex 6.3-11

Al3+

(aq) + 3e Al(s) E= -1.66 V

Cu2+(aq) + 2e Cu(s) E= +0.337 V

Calculate Ecellif [Cu

2+

] = 0.1 M & [Al

3+

] = 0.01M

F) Practical Cell for Electricity Production through Chemical

Reactions

There are basically 3 types of cells and batteries

i) Primary cells

ii) Secondary cells

iii) Fuel cells


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i) Primary Cells

CANNOT be recharged.

Most common, convenient and cheapest to use

Produce environmental problems with their disposal after use

e.g.Leclanche (Zinc Carbon) cell

Anode: Zn container

Cathode: Graphite

Electrolyte: NH4Cl and ZnCl2 +

MnO2(in paste form)

Anode: Zn(s)Zn2+(aq)+ 2e

Cathode:

2NH4+(aq)+2MnO2(s)+2eMn2O3(s)+2NH3(aq)+H2O(l)

Overall: Zn(s) + 2NH4+(aq)+ 2MnO2(s)

Zn2+(aq)+ Mn2O3(s)+ 2NH3(aq)+ H2O(l)

Cell diagram: Zn(s)Zn2+(aq)MMnO2(s)Mn2O3(s)C(s) E

= +1.50 V

When a current is drawn for sometimes

NH3(g)& Zn2+accumulates

Equilibrium shifts to LHS

Ecelldrops


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When the cell is stored for some time without use

Zn + NH4+Zn2++ NH3+ H2occurs

[NH4+] decreases & eqn shifts to LHS

Ecelldrops

e.g. Silver oxide battery

Silver oxide cells look like buttons.

The () electrode (anode) is zinc powder.

The (+) electrode (cathode) is silver oxide.

These materials are tightly compacted powders separated by

a moist paste of silver oxide containing some potassium

hydroxide. Moistened paper serves as the salt bridge.

The cell produces 1.5V.


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Anode: Zn(s)+ 2OHZnO(s)+ H2O(l)+ 2e

Cathode: Ag2O(s)+ H2O(l)+2e2Ag(s)+ 2OH(aq)

Overall: Zn(s)+ Ag2O(s)ZnO(s)+ 2Ag(s)

The silver oxide cell is smalland lasts for a long time. It

also gives a steady current. However, it is more

expensivethan other types of dry cells.

Cell diagram:

Zn(s)ZnO(s)KOH(aq)Ag2O(s)Ag(s)

ii) Secondary Cells

Must be charged up first before it can be used.

Usually rechargeable

e.g. Lead acid accumulator


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Cathode: Pb coated with PbO2

Anode: Pb metal

WhenDISCHARGING

Anode: Pb(s)+ SO42(aq)PbSO4(s)+ 2e

Cathode: PbO2(s)+4H+(aq)+SO42(aq)+2e

PbSO4(s)+2H2O(l)

discharge

Cell rxn: Pb(s)+PbO2(s)+4H+(aq)+2SO42(aq) 2PbSO4(s)+2H2O(l)

recharge

Cell diagram:

Pb(s) PbSO4(s) H2SO4(aq) PbO2PbSO4(s) Pb(s)

E= -0.35 V E

= +1.69 V

E

cell= +1.69 (-0.35) = + 2.05 V

When the battery is recharged by connecting to an external source,

electrons are forced in the opposite direction

Reverse the cell reaction

It is mainly used in motor cars. Usually 6 cells are joined in

series to give a total of 12 V

Life time: 3-5 years


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Ex 6.3-12

The rechargeable nickel-cadmium battery has electrodes

of solid cadmium and solid nickel (IV) oxide, NiO2,

coated on a conductor. When this battery discharges,

solid Cd(OH)2and solid Ni(OH)2are formed. The

electrolyte is a basic medium of KOH. Given

Cd(OH)2 Cd E= -0.76 V

NiO2 Ni(OH)2 E= +0.49 V

(a) Write balanced equations for the reactions occurring

at cathode and anode when the battery discharges.

(b) Write the overall cell reaction and find Ecell.

(c) What is the direction of electrons flow when the

battery discharges?

(d) Write the cell diagram by IUPAC convention.

(e) Why is the battery re-chargeable and how, in principle,

this is carried out?


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iii) Fuel Cells

Differences between fuel cells and electrochemical

cells:

Fuel cell is a primary cell which efficiently converts the chemical

energy into electrical energy without losing to heat, mechanical

linkages

Environmentally friendly

Continuous supply of electrical energy

Open system

Flexible: micro to mega Watts

e.g. Hydrogen-oxygen fuel cell

Fuel (H2) and oxidizing agent (O2) are supplied in a continuous

flow into the anode and cathode compartments respectively.


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Porous Ni electrodes act as the electrical conductor and also a

catalyst for the reactions.

Anode: 2H2(g)+ 4OH(aq)4H2O(l)+ 4e

Cathode: O2(g)+ 2H2O(l)+ 4e4OH(aq)

Overall reaction: 2H2(g)+ O2(g)2H2O(l)

Cell diagram:

Ni(s)H2(g)H2O(l)KOH(aq)O2(g)2OH(aq)Ni(s)

It is used as the energy source in spacecraft and the crew can

drink the water produced during the reaction.

It cannot be used as a source of portable power because of heavy

weight (500 lb)

G) Corrosion of Iron and Its Prevention

(a) Chemistry of Rusting

(i) Electrochemical Process Involving in Rusting


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Around the edge of water droplet higher [O2]

act as CATHODE

O2(g)+ 2H2O(l)+ 4e4OH(aq) E= +0.40 V

(turn phenophthalein pink)

At the centre of water droplet lower [O2]

act as ANODE

Fe(s)Fe2+(aq)+ 2e E=0.44 V

(turn K3Fe(CN)6blue)

O2(g) + 2H2O(l) + 2Fe(s)4OH-(aq) + 2Fe2+(aq)E= +0.84V

E

cellis +ve spontaneous reaction

Fe2+(aq)formed in the anodic regions travel to the cathodic regions

where it reacts with the OHproduced in the cathodic regions.

Fe2+(aq)+ 2OH(aq)Fe(OH)2(s)

Fe(OH)2(s)further react with O2(g)in air

4Fe(OH)2(s)+ O2(g)+ 2 H2O(l)4Fe(OH)3(s)

Fe(OH)3(s)Fe2O3nH2O(s) (rust)


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ii) Essential Conditions for Rusting

1. Moisture (water)

2.Air (O2)

iii) Factors Increasing the Rusting Rate

1. Low pH because it favours FeFe2++ 2e

2. High temperature

3. Present of dissolved electrolytes e.g. NaCl because

conductivity

b) Prevention of Rusting

i) Application of a protective layer (coating)

Nonmoving objects: Painting, varnishing or coating with

plastic

Moving objects: Greasing or oiling

ii) Cover with a thin layer of another metal which is resistant to

corrosion

1. Tinplating: By immersing Fe in molten Sn (used to make


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Zn protects Fe from rusting by sacrificial protection. Galvanized

iron is not used for canning food because Zn2+is poisonous.

3. Chromiumplating, Nickelplating: By electrolysis

iii) Alloying

Stainless steel: An alloy of Fe with Cr (min. 12%), Ni and

Mn. Cr forms a layer of impermeable and

insoluble oxide layer on surface of alloy

grains.

iv) Sacrificial protection

Use more reactive metals, e.g. Mg as sacrificial anode which

supply eto the iron (cathode) to prevent it from oxidation, e.g.

attach large Mg blocks to the steel of the hull of ship, offshore

oil drilling platform or underground oil tank.


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v) Cathodic protection

Iron is treated as the cathode and connected to the ve

terminal of a battery. This can inhibit the formation of Fe2+.

iv) Socioeconomic Implications of Corrosion and Prevention

Corrosion of iron can cost much to the society

i) Money spent on:

1. Replacement of the corroded articles

2. Prevention of corrosion

3. Indirect cost, e.g. those for the maintenance of machines,

those due to lost production when machines fail down or

where they are shut down for maintenance. (In 1980 in US,

70 billions US dollars were lost annually because of

corrosion)

ii) Wastage of natural resources.

It is estimated that 1 tonne of steel is converted into rust very

90 sec in Britain and about 40 % of the steel made in US is

used to replace steel lost by rusting.

iii) Inconvenience to human beings and even loss of life.


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Corrosion results in the formation of cracks and crevices

which weaken the strength of metals

concrete in buildings may fall off

aged vehicles may cause more road accidents

Disadvantages of corrosions are outweighed by the relative

low cost, abundance and ease of extraction

The society should assume a greater responsibility for

corrosion prevention in order to have a greater improvement

of the situation.

HKAL Past Paper Questions

86 IIA 2(b) 87 IIA 2(b)88 IA 1(c) 89 IIA 1(b)90 IA 2(b), IIA 1(b) 91 IA 2(a), IIA 2(b)92 IIA 1(c) 93 IA 1(d), IIA 1(c), 3(a)94 IA 1(d)(e), IIA 3(a)(i) 95 IIA 3(c)96 IIB 6(d)(e) 97 IA 2(c), IIA 4(a)(b)98 IIA 4(a) 99 IIA 1(b)00 IIA 3(d) 01 IA 4(a)(b), IIA 4(a)02 IA 1(b), IIA 3(b)

chapter 6.3 redox equilibrium

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