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For a cell with $E^0 = 1.10$ V and $n = 2$ electrons transferred, the standard Gibbs energy change $\Delta G^0$ is approximately ($F = 96 485$ C/mol):
A$-106\ \text{kJ/mol}$
B$+212\ \text{kJ/mol}$
C$-1.10\ \text{kJ/mol}$
D$-212\ \text{kJ/mol}$
Answer & Solution
Correct answer: D. $-212\ \text{kJ/mol}$
$\Delta G^0 = -nFE^0 = -2 \times 96 485 \times 1.10 \approx -2.12 \times 10^5$ J/mol $\approx -212$ kJ/mol. Large negative $\Delta G$ ⇒ thermodynamically very favourable.
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