JEE MAIN - Chemistry (2025 - 7th April Morning Shift - No. 23)
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)
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$$\begin{aligned} \text { Given : } \mathrm{E}^{\circ} \mathrm{Cu}^{2+} / \mathrm{Cu} & =0.34 \mathrm{~V} \\\\ \mathrm{E}^{\circ} \mathrm{Ag}^{+} / \mathrm{Ag} & =0.8 \mathrm{~V} \\\\ \frac{2 \cdot 303 \mathrm{RT}}{\mathrm{~F}} & =0.06 \mathrm{~V} \end{aligned}$$
Explanation
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reaction
$$\begin{aligned} & \mathrm{E}=\mathrm{E}^{\circ}-\frac{0.06}{\mathrm{n}} \log \frac{\left[\mathrm{Cu}^{+2}\right]}{\left[\mathrm{Ag}^{+}\right]^2} \\ & \mathrm{E}=(0.8-0.34)-\frac{0.06}{2} \log \frac{1}{(0.1)^2}=0.4 \mathrm{~V} \end{aligned}$$
Correct answer $=400 \mathrm{mV}$
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