What is the potential of a hydrogen electrode | vedclass

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(B) The potential of a hydrogen electrode is given by the Nernst equation: $E_{H^{+}/H_2} = E^{\circ}_{H^{+}/H_2} - \frac{0.059}{n} \log \frac{1}{[H^{

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$-0.059$

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(B) The potential of a hydrogen electrode is given by the Nernst equation: $E_{H^{+}/H_2} = E^{\circ}_{H^{+}/H_2} - \frac{0.059}{n} \log \frac{1}{[H^{+}]}$.Given $pH = 1$,we know that $-\log [H^{+}] = 1$,which implies $[H^{+}] = 10^{-1} \ M$.For a standard hydrogen electrode,$E^{\circ}_{H^{+}/H_2} = 0.0 \ V$ and $n = 1$.Substituting the values: $E = 0.0 - \frac{0.059}{1} \log \frac{1}{10^{-1}}$.$E = -0.059 	imes \log(10) = -0.059 	imes 1 = -0.059 \ V$.

What is the potential of a hydrogen electrode in contact with a solution whose $pH$ is $1$ (in $V$)?

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