The Gibbs energy change (in J) for the given reaction at
Question:

The Gibbs energy change (in J) for the given reaction at $\left[\mathrm{Cu}^{2+}\right]=\left[\mathrm{Sn}^{2+}\right]=1 \mathrm{M}$ and $298 \mathrm{~K}$ is : $\mathrm{Cu}(\mathrm{s})+\mathrm{Sn}^{2+}$ (aq.) $\rightarrow \mathrm{Cu}^{2+}$ (aq.) $+\mathrm{Sn}(\mathrm{s})$ $\left(\mathrm{E}_{\mathrm{Sn}^{2+} \mid \mathrm{Sn}}^{0}=-0.16 \mathrm{~V}, \mathrm{E}_{\mathrm{Cu}^{2+} \mid \mathrm{Cu}}^{0}=0.34 \mathrm{~V}\right.$ Take $\mathrm{F}=96500 \mathrm{C} \mathrm{mol}^{-1}$ )

Solution:

(96500)

$\mathrm{E}_{\text {cell }}^{0}=\mathrm{E}_{\mathrm{Sn}^{2+} / \mathrm{Sn}}^{0}-\mathrm{E}_{\mathrm{Cu}^{2+} / \mathrm{Cu}}^{0}$

$=-0.16-0.34=-0.50 \mathrm{~V}$

$\Delta \mathrm{G}^{0}=-\mathrm{nE}_{\text {cell }}^{0}$

$=-2 \times 96500 \times(-0.5)=96500 \mathrm{~J}$