The electrode potentials for Cu^2+(aq)+e^- longrightarrow Cu^+(aq) and Cu^+(aq)+e^- longrightarrow Cu(s) are +0.15 ~V and +0.50 ~V , respectively. The value of E_Cu^2+^° / Cu will be:

Correct answer: (B) 0.325 ~V. . Cu^2++1 e^- arrow Cu^+ ; Δ G_1^0=-n_1 E_1^0 ~F ] fracCu^++1 e^- arrow Cu ; Δ G_2^0=-n_2 E_2^0 ~FCu^2++2 e^- arrow Cu Δ G^°=Δ G_1^0+Δ G_2^0 -n E^° F=-1 n_1 E_1^0 ~F+(-1) n_2 E_2^0 ~F -n E^° F=-F (n_1 E_1^0+n_2 E_2^0 ~F ) E^°=fracn_1 E_1^0+n_2 E_2^0n=(0.15 × 1+0.50 × 1)/2=0.325

Q8 / 120 · VITEEE 2019 · Chemistry

The electrode potentials for $\mathrm{Cu}^{2+}(\mathrm{aq})+\mathrm{e}^{-} \longrightarrow \mathrm{Cu}^{+}(\mathrm{aq})$ and $\mathrm{Cu}^{+}(\mathrm{aq})+\mathrm{e}^{-} \longrightarrow \mathrm{Cu}(\mathrm{s})$ are $+0.15 \mathrm{~V}$ and $+0.50 \mathrm{~V}$, respectively. The value of $\mathrm{E}_{\mathrm{Cu}^{2+}}^{\circ} / \mathrm{Cu}$ will be:

A
$0.500 \mathrm{~V}$
Correct answer
B
$0.325 \mathrm{~V}$
C
$0.650 \mathrm{~V}$
D
$0.150 \mathrm{~V}$

Solution

$\left.\quad \mathrm{Cu}^{2+}+1 \mathrm{e}^{-} \rightarrow \mathrm{Cu}^{+} \quad ; \Delta \mathrm{G}_{1}^{0}=-\mathrm{n}_{1} \mathrm{E}_{1}^{0} \mathrm{~F}\right]$ $\frac{\mathrm{Cu}^{+}+1 \mathrm{e}^{-} \rightarrow \mathrm{Cu} \quad ; \Delta \mathrm{G}_{2}^{0}=-\mathrm{n}_{2} \mathrm{E}_{2}^{0} \mathrm{~F}}{\mathrm{Cu}^{2+}+2 \mathrm{e}^{-} \rightarrow \mathrm{Cu} \quad \Delta \mathrm{G}^{\circ}=\Delta \mathrm{G}_{1}^{0}+\Delta \mathrm{G}_{2}^{0}}$ $-\mathrm{n} \mathrm{E}^{\circ} \mathrm{F}=-1 \mathrm{n}_{1} \quad \mathrm{E}_{1}^{0} \mathrm{~F}+(-1) \mathrm{n}_{2} \mathrm{E}_{2}^{0} \mathrm{~F}$ $-\mathrm{n} \mathrm{E}^{\circ} \mathrm{F}=-\mathrm{F}\left(\mathrm{n}_{1} \mathrm{E}_{1}^{0}+\mathrm{n}_{2} \mathrm{E}_{2}^{0} \mathrm{~F}\right)$ $\mathrm{E}^{\circ}=\frac{\mathrm{n}_{1} \mathrm{E}_{1}^{0}+\mathrm{n}_{2} \mathrm{E}_{2}^{0}}{\mathrm{n}}=\frac{0.15 \times 1+0.50 \times 1}{2}=0.325$

Hence, the correct answer is option: 'B'.

Solution

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