Question

Consider a voltaic cell constructed from the following redox couples and make the following predictions about what would occur if you ran this cell under standard conditions.

couple

Pb2+ | Pb Ce4+, Ce3+ | Pt

reduction half reaction

Pb2+ + 2 e- <==> Pb Ce4+ + e- <==> Ce3+

reduction potential E ?re

- 0.13 volts + 1.44 volts

(a) Write the anode, cathode and overall cell reactions.

(b) Calculate the ideal cell voltage, E ?cell, for this voltaic cell.

Answer 1

The given reduction half-cell reactions are

Pb2+ + 2 e- <==> Pb , E = - 0.13V

Ce4+ + e- <==> Ce3+ , E = +1.44V

Since the reduction potential for the reduction of Ce4+ to Ce3+ is positive, it is feasibe and hence it will be reduced. Hence Ce4+ acts ascathode. (answer)

Reduction half cell reaction: Ce4+ + e- <==> Ce3+ , E = +1.44V

Since the reduction potential for the reduction of Pb2+ to Pb is negative, it is not feasibe. However the reverse of this reaction (i.e oxidation) will be feasible. Hence Pb wil be oxidised and acts as anode (answer)

Oxidation half cell reaction: Pb <======> Pb2+ + 2e- , E = +0.13 V

Now the overall cell reaction can be written as

Pb + 2Ce4+ <========> Pb2+ + 2Ce3+ (answer)

and the cell can be represented as

Pb|Pb2+ ||Pt||Ce4+|Ce3+

Now the over all cell voltage voltage can be calculated as

Ecell = E(oxidation) +E(reduction) = 0.13+1.44 = 1.57 V (answer)

Since the reduction potential