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:

Official Ans. by NTA $(96500.00)$

Solution:

$\Delta \mathrm{G}=\Delta \mathrm{G}^{\circ}+\mathrm{RT} \ln \left[\frac{\mathrm{Sn}^{+2}}{\mathrm{Cu}^{+2}}\right]$

$=-2 \times 96500[(-0.16)-0.34]+\mathrm{RT} \ln \left(\frac{1}{1}\right)$

$=96500 \mathrm{~J}$

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