Question

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.

Answer 1

Answer:

Approximately $\rm 90\; kJ$.

Explanation:

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

$E^{\circ}(\text{cell}) = E^{\circ}(\text{cathode}) - E^{\circ}(\text{anode})$.

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

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

$\rm Cu^{2+} + Pb \to Cu + Pb^{2+}$.

Its cell potential would be equal to $0.339 - (-0.130) = \rm 0.469\; V$.

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

$\Delta G^{\circ} = n \cdot F \cdot E^{\circ} (\text{cell})$,

where

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

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