xConsider the following reduction potentials: Cu2+ + 2e– Cu E° = 0.339 V Pb2+ + 2e– Pb E° = –0.130 V For a galvanic cell employing the Cu, Cu2+ and Pb, Pb2+ couples, calculate the maximum amount of work that would accompany the reaction of one mole of lead under standard conditions.

Cathode is where reduction takes place and anode is where oxidation takes place. The potential of a electrochemical reaction ([tex]E^{\circ}(\text{cell})[/tex]) is equal to

[tex]E^{\circ}(\text{cell}) = E^{\circ}(\text{cathode}) - E^{\circ}(\text{anode})[/tex].

There are two half-reactions in this question. [tex]\rm Cu^{2+} + 2\,e^{-} \rightleftharpoons Cu[/tex] and [tex]\rm Pb^{2+} + 2\,e^{-} \rightleftharpoons Pb[/tex]. Either could be the cathode (while the other acts as the anode.) However, for the reaction to be spontaneous, the value of [tex]E^{\circ}(\text{cell})[/tex] should be positive.

In this case, [tex]E^{\circ}(\text{cell})[/tex] is positive only if [tex]\rm Cu^{2+} + 2\,e^{-} \rightleftharpoons Cu[/tex] is the reaction takes place at the cathode. The net reaction would be

[tex]\rm Cu^{2+} + Pb \to Cu + Pb^{2+}[/tex].

Its cell potential would be equal to [tex]0.339 - (-0.130) = \rm 0.469\; V[/tex].

The maximum amount of electrical energy possible (under standard conditions) is equal to the free energy of this reaction:

[tex]\Delta G^{\circ} = n \cdot F \cdot E^{\circ} (\text{cell})[/tex],

where

[tex]n[/tex] is the number moles of electrons transferred for each mole of the reaction. In this case the value of [tex]n[/tex] is [tex]2[/tex] as in the half-reactions.
[tex]F[/tex] is Faraday's Constant (approximately [tex]96485.33212\; \rm C \cdot mol^{-1}[/tex].)

[tex]\begin{aligned}\Delta G^{\circ} &= n \cdot F \cdot E^{\circ} (\text{cell})\cr &= 2\times 96485.33212 \times (0.339 - (-0.130)) \cr &\approx 9.0 \times 10^{4} \; \rm J \cr &= 90\; \rm kJ\end{aligned}[/tex].