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JEE Main 2025 (Online) 7th April Morning Shift

INTEGER+4 / -12025

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)


JEE Main 2025 (Online) 7th April Morning Shift Chemistry - Electrochemistry Question 26 English


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

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