# For the given cell;

Question:

For the given cell;

$\mathrm{Cu}(\mathrm{s})\left|\mathrm{Cu}^{2+}\left(\mathrm{C}_{1} \mathrm{M}\right)\right|\left|\mathrm{Cu}^{2+}\left(\mathrm{C}_{2} \mathrm{M}\right)\right| \mathrm{Cu}(\mathrm{s})$

change in Gibbs energy $(\Delta \mathrm{G})$ is negative, if:

1. $\mathrm{C}_{1}=\mathrm{C}_{2}$

2. $\mathrm{C}_{2}=\frac{\mathrm{C}_{1}}{\sqrt{2}}$

3. $\mathrm{C}_{1}=2 \mathrm{C}_{2}$

4. $C_{2}=\sqrt{2} C_{1}$

Correct Option: , 4

Solution:

For the concentration cell, $E_{\text {cell }}^{0}=0$

As $\Delta G=-n F E$

If $\Delta G=-\mathrm{ve}$, then $E_{\text {cell }}$ is $+\mathrm{v}_{\mathrm{e}}$.

$E_{\text {cell }}=E_{\text {cell }}^{\mathrm{o}}-\frac{R T}{2 F} \ln \frac{C_{1}}{C_{2}}$

$E_{\text {cell }}=0-\frac{R T}{2 F} \ln \frac{C_{1}}{C_{2}}$

$E_{\text {cell }}=\frac{R T}{2 F} \ln \frac{C_{2}}{C_{1}}$

So, $C_{2}>C_{1}$

Thus, $C_{2}=\sqrt{2} C_{1}$ relation is correct.