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(D) The half-cell reaction for the hydrogen electrode is: $2 H^{+}_{(aq)} + 2 e^{-} \rightarrow H_{2(g)}$.Using the Nernst equation: $E = E^{\circ} -

Vedclass · Accepted Answer

$-17.7$

Vedclass · Answer

(D) The half-cell reaction for the hydrogen electrode is: $2 H^{+}_{(aq)} + 2 e^{-} ightarrow H_{2(g)}$.Using the Nernst equation: $E = E^{\circ} - \frac{0.059}{n} \log \frac{1}{[H^{+}]}$.For the standard hydrogen electrode,$E^{\circ} = 0 \ V$ and $n = 1$ (for the reduction of $H^{+}$ to $\frac{1}{2} H_2$).$E = 0 - 0.059 	imes \log \frac{1}{[H^{+}]}$.Since $pH = -\log[H^{+}]$,we have $\log \frac{1}{[H^{+}]} = pH$.Therefore,$E = -0.059 	imes pH$.Given $pH = 3$,$E = -0.059 	imes 3 = -0.177 \ V$.Converting to the required format: $-0.177 \ V = -17.7 	imes 10^{-2} \ V$.

The hydrogen electrode is dipped in a solution of $pH=3$ at $25^{\circ} C$. The potential of the electrode will be . . . . . . $\times 10^{-2} \ V$. $\left(\frac{2.303 RT}{F}=0.059 \ V\right)$

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