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Aluminium displaces hydrogen from dilute $\mathrm{HCl}$, whereas silver does not. The emf of a cell prepared by combining $\mathrm{Al} / \mathrm{Al}^{3+}$ and $\mathrm{Ag} / \mathrm{Ag}^{+}$ is $2.46 \mathrm{~V}$. The reduction potential of silver electrode is $+0.80 \mathrm{~V}$. The reduction potential of aluminium electrode is
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The correct answer is:
$-1.66 \mathrm{~V}$
$E^{\circ}=E_{\left(\mathrm{Ag}^{+} / \mathrm{Ag}\right)}^{\circ}-E_{(\mathrm{Al}}$
$\begin{aligned}
&2.46=0.80-E_{\left(\mathrm{Al}^{3+} / \mathrm{Al}\right)}^{0} \\
&\therefore \quad E_{\left(\mathrm{Al}^{3+} / \mathrm{Al}\right)}^{0}=-1.66 \mathrm{~V}
\end{aligned}$
$\therefore$ Reduction potential $=-1.66 \mathrm{~V} .$
$\begin{aligned}
&2.46=0.80-E_{\left(\mathrm{Al}^{3+} / \mathrm{Al}\right)}^{0} \\
&\therefore \quad E_{\left(\mathrm{Al}^{3+} / \mathrm{Al}\right)}^{0}=-1.66 \mathrm{~V}
\end{aligned}$
$\therefore$ Reduction potential $=-1.66 \mathrm{~V} .$
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