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The standard reduction potential for $\mathrm{Mg}^{2+} / \mathrm{Mg}$ is $-2.37 \mathrm{~V}$ and for $\mathrm{Cu}^{2+} / \mathrm{Cu}$ is $0.337$. The $E_{\text {cell }}^{\circ}$ for the following reaction is $\mathrm{Mg}+\mathrm{Cu}^{2+} \longrightarrow \mathrm{Mg}^{2+}+\mathrm{Cu}$
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Verified Answer
The correct answer is:
$+2.7 \mathrm{~V}$
$E_{\text {cell }}^{\circ}=E_{\text {cathode }}^{\circ}-E_{\text {anode }}^{\circ}$
$$
\begin{aligned}
=E_{\left(\mathrm{Cu}^{2+} / \mathrm{Cu}\right)}^{\circ}-E_{\left(\mathrm{Mg}^{2+} / \mathrm{Mg}\right)}^{\circ} &=0.337-(-2.37) \mathrm{V} \\
&=2.7 \mathrm{~V}
\end{aligned}
$$
$$
\begin{aligned}
=E_{\left(\mathrm{Cu}^{2+} / \mathrm{Cu}\right)}^{\circ}-E_{\left(\mathrm{Mg}^{2+} / \mathrm{Mg}\right)}^{\circ} &=0.337-(-2.37) \mathrm{V} \\
&=2.7 \mathrm{~V}
\end{aligned}
$$
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