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Question:
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Given
$$
\mathrm{E}_{\mathrm{Cu}^{2+} / \mathrm{Cu}}^{\circ}=0.34 \mathrm{~V}, \mathrm{E}_{\mathrm{Cu}^{2+} / \mathrm{Cu}}^{\circ}=0.15 \mathrm{~V}
$$
Standard electrode potential for the half cell $\mathrm{Cu}^{+} / \mathrm{Cu}$ is
Options:
$$
\mathrm{E}_{\mathrm{Cu}^{2+} / \mathrm{Cu}}^{\circ}=0.34 \mathrm{~V}, \mathrm{E}_{\mathrm{Cu}^{2+} / \mathrm{Cu}}^{\circ}=0.15 \mathrm{~V}
$$
Standard electrode potential for the half cell $\mathrm{Cu}^{+} / \mathrm{Cu}$ is
Solution:
1263 Upvotes
Verified Answer
The correct answer is:
$0.53 \mathrm{~V}$
$0.53 \mathrm{~V}$
$$
\begin{aligned}
& \text { } \mathrm{Cu}^{++}+e^{-} \longrightarrow \mathrm{Cu}^{+} \text {; } \\
& E_1^0=0.15 \mathrm{~V} ; \Delta G_1^0=-n_1 E_1^0 F \\
& \mathrm{Cu}^{2+}+2 e \longrightarrow \mathrm{Cu} \quad \text {; } \\
& E_2^0=0.34 \mathrm{~V} ; \Delta G_2^0=-n_2 E_2^0 F \\
&
\end{aligned}
$$
On subracting eq.(i) from eq. (ii) we get
$$
\begin{aligned}
& \mathrm{Cu}^{+}+e^{-} \longrightarrow \mathrm{Cu} ; \Delta G^0=\Delta G_2^0-\Delta G_1^0 \\
&-n E^0 F=-\left(n_2 E^0 F-n_1 E_1^0 F\right) \\
& E^0=\frac{n_2 E_2^0 F-n_1 E_1^0 F}{n F} \\
&=\frac{2 \times 0.34-0.15}{1} \\
&=0.53 \mathrm{~V}
\end{aligned}
$$
\begin{aligned}
& \text { } \mathrm{Cu}^{++}+e^{-} \longrightarrow \mathrm{Cu}^{+} \text {; } \\
& E_1^0=0.15 \mathrm{~V} ; \Delta G_1^0=-n_1 E_1^0 F \\
& \mathrm{Cu}^{2+}+2 e \longrightarrow \mathrm{Cu} \quad \text {; } \\
& E_2^0=0.34 \mathrm{~V} ; \Delta G_2^0=-n_2 E_2^0 F \\
&
\end{aligned}
$$
On subracting eq.(i) from eq. (ii) we get
$$
\begin{aligned}
& \mathrm{Cu}^{+}+e^{-} \longrightarrow \mathrm{Cu} ; \Delta G^0=\Delta G_2^0-\Delta G_1^0 \\
&-n E^0 F=-\left(n_2 E^0 F-n_1 E_1^0 F\right) \\
& E^0=\frac{n_2 E_2^0 F-n_1 E_1^0 F}{n F} \\
&=\frac{2 \times 0.34-0.15}{1} \\
&=0.53 \mathrm{~V}
\end{aligned}
$$
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