The reduction potential of a hydrogen electro | vedclass

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(C) The reduction potential of the hydrogen electrode is calculated using the Nernst equation for the reaction: $2H^{+} + 2e^{-} \rightarrow H_2$. $E_

Vedclass · Accepted Answer

$-0.413$

Vedclass · Answer

(C) The reduction potential of the hydrogen electrode is calculated using the Nernst equation for the reaction: $2H^{+} + 2e^{-} ightarrow H_2$. $E_{red} = E_{red}^{0} - \frac{0.0591}{n} \log \frac{P_{H_2}}{[H^{+}]^2}$. For a neutral solution,the $H^{+}$ ion concentration is $10^{-7} \ M$ because the $pH = 7$. Substituting the values: $E_{H^{+}|H_2} = 0 - \frac{0.0591}{2} \log \frac{1}{(10^{-7})^2}$. $E_{H^{+}|H_2} = -\frac{0.0591}{2} \log 10^{14}$. $E_{H^{+}|H_2} = -\frac{0.0591 	imes 14}{2} \log 10$. $E_{H^{+}|H_2} = -0.4137 \ V \approx -0.413 \ V$.

The reduction potential of a hydrogen electrode at $25^{\circ} C$ in a neutral solution is $(P_{H_2} = 1 \ atm)$. (in $V$)

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