JEE Main 2025 — Electrochemistry Question with Solution

1 Faraday electricity was passed through ${\mathrm{Cu}}^{2+}(1.5\mathrm{M},1\mathrm{L})/\mathrm{Cu}$ and 0.1 Faraday was passed through ${\mathrm{Ag}}^{+}(0.2\mathrm{M},1\mathrm{L})/\mathrm{Ag}$ electrolytic cells. After this the two cells were connected as shown below to make an electrochemical cell. The emf of the cell thus formed at 298 K is __________ mV (nearest integer)

$$\begin{array}{rl}\text{ Given : }{\mathrm{E}}^{\circ }{\mathrm{Cu}}^{2+}/\mathrm{Cu}& =0.34\mathrm{V}\\ & \\ {\mathrm{E}}^{\circ }{\mathrm{Ag}}^{+}/\mathrm{Ag}& =0.8\mathrm{V}\\ & \\ \frac{2\cdot 303\mathrm{RT}}{\mathrm{F}}& =0.06\mathrm{V}\end{array}$$