The reduction potential of a hydrogen half-ce | vedclass

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Question

(C) The reduction half-reaction for the hydrogen electrode is: $H^{+} + e^{-} \longrightarrow \frac{1}{2} H_{2}$.Using the Nernst equation: $E = E^{\c

Vedclass · Accepted Answer

$p(H_2) = 2\,atm$ and $[H^{+}] = 1.0\,M$

Vedclass · Answer

(C) The reduction half-reaction for the hydrogen electrode is: $H^{+} + e^{-} \longrightarrow \frac{1}{2} H_{2}$.Using the Nernst equation: $E = E^{\circ} - \frac{0.0591}{1} \log \frac{(P_{H_{2}})^{1/2}}{[H^{+}]}$.Since $E^{\circ} = 0 \, V$ for the standard hydrogen electrode,the equation becomes $E = -0.0591 \log \frac{(P_{H_{2}})^{1/2}}{[H^{+}]}$.For $E$ to be negative,$\log \frac{(P_{H_{2}})^{1/2}}{[H^{+}]}$ must be positive,which means $\frac{(P_{H_{2}})^{1/2}}{[H^{+}]} > 1$.In option $C$,$P_{H_{2}} = 2 \, atm$ and $[H^{+}] = 1.0 \, M$,so $\frac{(2)^{1/2}}{1} = \sqrt{2} \approx 1.414 > 1$. Thus,$E < 0$.

The reduction potential of a hydrogen half-cell will be negative if:

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