Q58 / 200 · NEET 2011 (Screening) · Chemistry
The electrode potentials for
$\begin{aligned}
& \mathrm{Cu}^{2+}(a q)+e^{-} \longrightarrow \mathrm{Cu}^{+}(a q) \\
& \text { and } \mathrm{Cu}^{+}(a q)+e^{-} \longrightarrow \mathrm{Cu}(s) \\
& \text { are }+0.15 \mathrm{~V} \text { and }+0.50 \mathrm{~V} \text { respectively. The value } \\
& \text { of } E_{\mathrm{Cu}^{2+} / \mathrm{Cu}}^{\circ} \text { will be }
\end{aligned}$
- Correct answerA$0.325 \mathrm{~V}$
- B$0.650 \mathrm{~V}$
- C$0.150 \mathrm{~V}$
- D$0.500 \mathrm{~V}$
Solution
$$
\begin{aligned}
& \mathrm{Cu}^{2+}+e^{-} \quad \rightarrow \mathrm{Cu}^{+} ; \\
& E_1^{\circ}=0.15 \mathrm{~V} ; \Delta G_1^{\circ}=-n_1 E_1^{\circ} F \\
& \mathrm{Cu}^{+}+e^{-} \rightarrow \mathrm{Cu} ; \\
& E_2^{\circ}=0.50 \mathrm{~V} ; \Delta G_2^{\circ}=-n_2 E_2^{\circ} F \\
& \hline \mathrm{Cu}^{2+}+2 e^{-} \quad \rightarrow \mathrm{Cu} ; E^{\circ}=? ; \Delta G^{\circ}=-n E^{\circ} F \\
& \Delta G^{\circ}=\Delta G_1^{\circ}+\Delta G_2^{\circ} \\
& \text { or }-2 E^{\circ} F=-1 F \times 0.15+(-1 F \times 0.50) \\
& \text { or }-2 E^{\circ} \mathrm{F}=-0.15 \mathrm{~F}-0.50 \mathrm{~F} \\
& \text { or }-2 F E^{\circ}=-F(0.15+0.50) \\
& \therefore \quad E^{\circ}= \frac{0.65}{2}=0.325 \mathrm{~V}
\end{aligned}
$$
Hence, the correct answer is option: 'A'.