Reduction- Oxidation Reactions

2nd lecture

1

Learning ObjectivesWhat are some of the key things we learned from this lecture?

• Types of electrochemical cells. √

• How to measure Standard electrode potential. √

• reference electrode. √

• How to measure the electrode potential from the Nernst

equation. √

• Factors affecting the oxidation potential. √

2

Electrochemical cell

An electrochemical cell is a device used for generating an electromotive force (voltage) and current from chemical reactions, (galvanic, voltaic cell).

or the reverse, inducing a chemical reaction by a flow of current, (electrolytic).

The current is caused by the reactions releasing and accepting electrons at the different ends of a conductor.

There are two types of electrochemical cells:

• Electrolytic cell, (electrical Energy chem. Energy)

• Galvanic cell (voltaic), (chemical energy) electrical energy)

In an electrochemical cell, an electric potential is created

between two dissimilar metals. This potential is a measure

of the energy per unit charge which is available from the

oxidation/reduction reactions to drive the reaction. It is

customary to visualize the cell reaction in terms of two half-

reactions, an oxidation half-reaction and a reduction half-

reaction.

Reduced species oxidized species + ne-

Oxidation at anode

Oxidized species + ne- reduced species

Reduction at cathode

Electrolytic cell

When driven by an external voltage applied to the

electrodes, the electrolyte provides ions that flow to

and from the electrodes, where charge-transferring, or

redox, reactions can take place.

Only for an external electrical potential (i.e. voltage) of

the correct polarity and large enough magnitude can

an electrolytic cell decompose a normally stable, or

inert chemical compound in the solution.

At anode: 2Cl- Cl2 +2e- (oxidation)

At cathode: 2Na+ + 2e- 2Na (reduction)

Cell reaction: 2Na+ + 2Cl- 2Na + Cl2 (redox reaction)

2- Galvanic Cells• In redox reaction, electrons are transferred from the oxidized species to the reduced species.

• Imagine separating the two 1/2 cells physically, then providing a conduit through which the electrons travel from one cell to the other.

8H+ + MnO4- + 5e Mn2+ + 4H2O

Fe2+ Fe3+ + e- x 5

We need to “complete the circuit” by allowing positive ions to flow as well.

We do this using a “salt bridge” which will allow charge neutrality in each cell to be maintained.

Salt bridge/porous disk: allows for ion migration suchthat the solutions will remain neutral.

Galvanic CellGalvanic Cell

Electrochemical cell in which chemical reactions are used to

create spontaneous current (electron) flow

Standard electrode potential (Standard electrode potential (EE0 0 ))• If the concentration of the components of the cell

are unity (1M), measured at 250c• The potential is called standard potential.

• It is not possible to measure the absolute potential of an electrode, but it could be measured relative to a standard reference electrode.

How to measure E0 standard electrode potential?

Suppose we have Cu rod immersed in its ions (half cells).

In order to measure potential of an electrode, it must be connected to another electrode (half cell) having known constant potential called reference electrode and we use

“Normal (standard) Hydrogen Electrode“.

So an electric current will flow from that having a higher potential to the other one, this current is called e.m.f. Which is the algebric difference of the electrode potential of the two half cells.

If the reference electrode has a potential equal to zero, therefore, so the e.m.f. measured will be equal to E0

Reference Electrodes

What is Normal Hydrogen Electrode (NHE)? It consists of a piece of platinium foil coated electrolytically

with platinum black and immersed in a solution of HCl (normal solu).

Hydrogen gas at pressure of 1 atmosphere is passed. Platinum black layer absorbs a large amount of H2

and can be considered as a bar of H2.

Under fixed conditions: 1 atm pressure Normal solu of HCl The potential is assumed to be zero. By connecting the electrode of unknown

potential to “NHE“, the E0 can be measuredby a potentiometer.

(1 atm)

(1 N)

Pt wire

Glass tube to contain H2 (g)

Bubbles of H2 (g)

H+ + e- ½ H2

Standard electrode potential E0: Is the e.m.f. produced when a half cell consisiting of an element immersed in one molar solution of its ions is coupled with NHE.

By measuring E0 for all elements and arrange into a series known as “Electrochemical Series “:Starting from

E0 = -ve value, to

hydrogen where E0 = zero, then to

E0 = +ve value

From this series we can conclude that:

1- The sign of E0 is similar to the charge on the metal electrode.

2- The greater the –ve value of E0, the greater the tendency of the metal to pass into ionic state (oxidized) i.e. E0 is a quantitative measure of the ability of the element to lose electrons giving its ions.

3- Metal with more negative potential will displace any other metal below it in the series from its solution.

System E0

Zn/Zn2+ -0.76Fe/Fe2+ -0.44Sn/Sn2+ -0.13H2(pt)/H+ 0Cu/Cu2+ +0.34

Reduction PotentialsElectrode Couple "E0, V"Na+ + e- --> Na -2.7144Mg2+ + 2e- --> Mg -2.3568Al3+ + 3e- --> Al -1.676Zn2+ + 2e- --> Zn -0.7621Fe2+ + 2e- --> Fe -0.4089Cd2+ + 2e- --> Cd -0.4022Tl+ + e- --> Tl -0.3358Sn2+ + 2e- --> Sn -0.141Pb2+ + 2e- --> Pb -0.12662H+ + 2e- --> H2(SHE) 0S4O62- + 2e- --> 2S2O32- 0.0238Sn4+ + 2e- --> Sn2+ 0.1539SO42- + 4H+ + 2e- --> H2O + H2SO3(aq) 0.1576Cu2+ + e- --> Cu+ 0.1607S + 2H+ + 2e- --> H2S 0.1739AgCl + e- --> Ag + Cl- 0.2221Saturated Calomel (SCE) 0.2412UO22+ + 4H+ + 2e- --> U4+ + 4H2O 0.2682

Reduction PotentialsHg2Cl2 + 2e- --> 2Cl- + 2Hg 0.268Bi3+ + 3e- --> Bi 0.286Cu2+ + 2e- --> Cu 0.3394Fe(CN)63- + e- --> Fe(CN)64- 0.3557Cu+ + e- --> Cu 0.518I2 + 2e- --> 2I- 0.5345I3- + 2e- --> 3I- 0.5354H3AsO4(aq) + 2H+ + 2e- -->H3AsO3(aq) + H2O 0.57482HgCl2 + 4H+ + 2e- -->Hg2Cl2 + 2Cl- 0.6011Hg2SO4 + 2e- --> 2Hg + SO42- 0.6152I2(aq) + 2e- --> 2I- 0.6195O2 + 2H+ + 2e- --> H2O2(l) 0.6237O2 + 2H+ + 2e- --> H2O2(aq) 0.6945Fe3+ + e- --> Fe2+ 0.769Hg22+ + 2e- --> Hg 0.7955Ag+ + e- --> Ag 0.7991Hg2+ + 2e- --> Hg 0.85192Hg2+ + 2e- --> Hg22+ 0.9083NO3- + 3H+ + 2e- -->HNO2(aq) + H2O 0.9275

If an inert electrode such as platinum is placed in a redox system e.g. Fe3+/Fe2+ it will assume a definite potential (oxidation potential) indicative of the position of equilibrium.Oxidation potential is the pot. difference arround an inert electrode immersed in solu. of oxidized and reduced form.

If the oxidizing tendency predominate,

the system will take electrons from platinum

leaving it positively charged

If the system has reducing properties,

electrons will be given up to platinum and it

will acquire a negative charge.

The magnitude of the potential will be a measure of oxidizing or reducing character of the system. If +ve oxidizing and if –ve reducing.

To measure the oxidation potential on pt wire, the system should be connected to NHE.

Standard Oxidation potential (SOP): Is the e.m.f. produced when a half cell consisting of an inert electrode (pt) dipped in equal concentration of both oxidized and reduced form of a system is connected with NHE.

Depending on the SOP the systems were arranged in a series from this series we can conclude that:

1- The higher the SOP of the system, the stronger the oxidizing power of its oxidized form and the weaker the reducing power of its reduced form.

2- the most powerful oxidizing agents are those at the top of the series with higher +ve potential, and the most powerful reducing agents occupy the bottom with higher –ve potential.

There should be adifference in oxidation potential between the two systems of about 0.4 v for oxidation to occur.

Nernst Equation FOR OXIDATION POTENTIAL• Compensated for non unit activity (not 1 M)• Relationship between cell potential and activities• aA + bB +ne- <--> cC + dD

• At 298K, 2.3RT/F = 0.0592

• What is potential of an electrode of Zn(s) and 0.01 M Zn2+

• Zn2+ +2e- <--> Zn °= -0.763 V• activity of metal is 1

2.30RT

nFlog

[C]c[D]d

[A]a[B]b

0.763 0.0592

2log

10.01

0.822V

• Eind.electrode = E0 - 0.059 log [products] n [reactants]

• Eind.electrode = E0M/M+

- 0.059 log 1

n [M+]

Factors affecting the oxidation potential:

1- Common ion effect:

If we determine a sample of Fe2+ with KMnO4 in presence of Cl- ions we have 3 systems: MnO4

-/Mn2+ (1.52)

Cl2/Cl- (1.36)

Fe3+/Fe2+ (0.77)

KMnO4 will oxidize both Cl- and Fe2+ so more KMnO4 will be consumed.

To overcome this problem the oxidation potential of MnO4-/Mn2+

should be reduced below Cl2/Cl- HOW?

By adding Zimmermann´s reagent

It contains [MnSO4, H3PO4, H2SO4]

E25 = E0 - 0.059/n log [Mn2+] / [MnO4-]

Presence of MnSO4 increases the conc of Mn2+ ion.

So the oxidation potential will decrease below that Cl2/Cl-

2-Effect of H+ ion concentration:[H+] has a deciding effect on the oxidation potential of oxidizing

agents containing oxygen and so [H+] should be included in Nernest equation.

E increases with increasing [H+] i.e.increasing acidity.

MnO4- + 8H+ + 5e ↔ Mn2++ 4H2O

EMnO4-/Mn2+ = E0 - 0.059/5 log [Mn2+] / [MnO4

-] [H+]8

At pH 5: KMnO4 oxidizes I- but not Br- and Cl-

At pH 3: KMnO4 oxidizes I- and Br- but not Cl-

At lower pH : KMnO4 oxidizes I- , Br- and Cl-

AsO43- + 2H+ + 2e AsO3

3- + H2O

EAsO43 -/ AsO3

3- = E0 - 0.059/2 log [ASO33-] / [AsO4

3-] [H+]2

If [H+] increases: The oxidation potential of AsO43- increases.

If [H+] decreases: The oxidation potential of AsO43-decreases and the

reducing power of AsO33- increases.

Therefore: AsO43- oxidizes I- to I2 in acid medium.

While: I2 oxidizes AsO3-3 in alkaline medium with NaHCO3

AsO4-3 + 2 I- + 2H+ AsO3

-3 + I2 + H2Oacid

alkaline

3-Effect of complexing agents:

E.g.1- Oxidation potential of I2/2I- system increases in presence of HgCl2 WHY?

Because HgCl2 forms a complex with the I- ions [HgI4]2-

I2 + 2 e 2I-

EI2 /2I-= E0 -

removing the I- ions from the share of the reaction. minimizing its concentration. decreasing the ratio of [I-]2 / I2 . increasing the oxidation potential of I2 /2I- system .

][

][log

2

059.0

2

2

I

I

E.g.2- Oxidation potential of Fe3+/ Fe2+ system decreases on addition of F- or PO4

3- WHY?

Because F- and PO43- form stable complexs with Fe3+

which are [FeF6]3- or [Fe (PO4)2]-3

Fe3+ + e Fe2+

EFe3+/Fe2+ = E0 -

removing the Fe3+ ions from the share of the reaction. minimizing its concentration. Increasing the ratio of Fe2+/ Fe3+ . lowering the oxidation potential of Fe3+/ Fe2+ system .

][

][log

1

059.03

2

Fe

Fe

4-Effect of precipitating agents:

Oxidation potential of Ferricyanide/ Ferrocyanide system increases on addition of Zn2+ WHY?

Because Zn2+ will precipitate Zn2[Fe(CN)6]

Fe (CN)63- + e Fe(CN)6

4-

E = Eo -

removing the Ferrocyanide ions from the share of the reaction.

minimizing conc of ferrocyanide. decreasing the ratio of [Fe(CN)6]

4-/ [Fe (CN)6]3- .

increasing the oxidation potential of [Fe(CN)6]3-/ [Fe (CN)6]4-

][Fe(CN)

] (CN) Fe[log

1

059.0-3

6

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