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The reduction potential of hydrogen half-cell will be negative if
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$\mathrm{p}\left(\mathrm{H}_{2}\right)=2 \mathrm{~atm}$ and $\left[\mathrm{H}^{+}\right]=1.0 \mathrm{M}$
Reaction $2 \mathrm{H}^{+}+2 \mathrm{e} \rightarrow \mathrm{H}_{2}$
$\therefore \mathrm{n}=2, \quad \mathrm{Q}=\frac{\mathrm{P}_{\mathrm{H}_{2}}}{\left[\mathrm{H}^{+}\right]^{2}}$
As $\mathrm{E}_{\mathrm{H}^{+} / \mathrm{H}_{2}}=\mathrm{E}_{\mathrm{H}^{+} / \mathrm{H}_{2}}^{0} \frac{-0.059}{\mathrm{n}} \log \mathrm{Q}=-\frac{0.059}{2} \log \mathrm{Q}$
$\therefore$ if $\mathrm{Q}>1 \quad \mathrm{E}_{\mathrm{H}^{+} / \mathrm{H}_{2}} < 0$
(A) $Q=\frac{1}{1}=1 \quad X$
(B) $Q=\frac{1}{2^{2}} < 1 \quad X$
(C) $Q=\frac{2}{12}>1$
(D) $Q=\frac{2}{2^{2}} < 1 \quad X$
$\therefore \mathrm{n}=2, \quad \mathrm{Q}=\frac{\mathrm{P}_{\mathrm{H}_{2}}}{\left[\mathrm{H}^{+}\right]^{2}}$
As $\mathrm{E}_{\mathrm{H}^{+} / \mathrm{H}_{2}}=\mathrm{E}_{\mathrm{H}^{+} / \mathrm{H}_{2}}^{0} \frac{-0.059}{\mathrm{n}} \log \mathrm{Q}=-\frac{0.059}{2} \log \mathrm{Q}$
$\therefore$ if $\mathrm{Q}>1 \quad \mathrm{E}_{\mathrm{H}^{+} / \mathrm{H}_{2}} < 0$
(A) $Q=\frac{1}{1}=1 \quad X$
(B) $Q=\frac{1}{2^{2}} < 1 \quad X$
(C) $Q=\frac{2}{12}>1$
(D) $Q=\frac{2}{2^{2}} < 1 \quad X$
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