JEE Main 2025 — 7 April, Morning Shift — Question 19

$\mathrm{CulCu}^{2+} \| \mathrm{Ag}^{+} / \mathrm{Ag}$

$\mathrm{E}_{\text {cell }}=\mathrm{E}^{\circ}-\frac{0.0591}{2} \log \frac{\left[\mathrm{Cu}^{2+}\right]}{\left[\mathrm{Ag}^{+}\right]^{2}}$

1 F deposits 1 equivalent of $\mathrm{Cu}^{2+}$

$=0.5 \mathrm{~mol}$

Initial moles of $\mathrm{Cu}^{2+}=1 \times 1.5=1.5$

Final moles $=1.5-0.5=1$

$$\left[\mathrm{Cu}^{2+}\right]_{\text {final }}=\frac{1}{1}=1 \mathrm{M}\left[\mathrm{Ag}^{+}\right]_{\text {final }}=0.1 \mathrm{M}$$

Putting values $E_{\text {cell }}=[0.80-0.34]-\frac{0.06}{2} \log \frac{\left[\mathrm{Cu}^{2+}\right]}{\left[\mathrm{Ag}^{+}\right]^{2}}$

$\simeq 400 \mathrm{mV}$