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The Gibbs free-energy change of a galvanic cell reaction is given by:
A$\Delta G = -E/(nF)$
B$\Delta G = +nFE$
C$\Delta G = -nFE$
D$\Delta G = nF/E$
Answer & Solution
Correct answer: C. $\Delta G = -nFE$
$\Delta G = -nFE$, where $n$ = number of electrons transferred, $F$ = Faraday constant (96 485 C/mol), $E$ = cell EMF. A positive $E$ ⇒ negative $\Delta G$ ⇒ spontaneous reaction.
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