The standard cell potential of the following cell $Zn\ |\ Zn^{2+}(aq)\ ||\ Fe^{2+}(aq)\ |\ Fe$ is 0.32 V. Calculate the standard Gibbs energy change for the reaction: $Zn(s)+Fe^{2+}(aq)\rightarrow Zn^{2+}(aq)+Fe(s)$ (Given: 1 F = 96487 C)

A $-61.75\ kJ\,mol^{-1}$

B $+5.006\ kJ\,mol^{-1}$

C $-5.006\ kJ\,mol^{-1}$

D $+61.75\ kJ\,mol^{-1}$

The answer is A.

Solution

$\Delta_rG^\circ=-nFE^\circ_{cell}$. For this reaction, $n=2$. $\Delta_rG^\circ=-2\times96487\times0.32=-61751.68\ J\,mol^{-1}=-61.75\ kJ\,mol^{-1}$.

Concept Relation between Gibbs energy and standard cell potential

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