What is the change in the reduction potential | vedclass

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(C) The reduction reaction for a hydrogen electrode is: $2H^+ (aq) + 2e^- \rightarrow H_2 (g)$.Using the Nernst equation at $298 \, K$: $E_{H^+/H_2} =

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Increases by $0.413 \, V$

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(C) The reduction reaction for a hydrogen electrode is: $2H^+ (aq) + 2e^- ightarrow H_2 (g)$.Using the Nernst equation at $298 \, K$: $E_{H^+/H_2} = E^0_{H^+/H_2} - \frac{0.0591}{1} \log \frac{1}{[H^+]}$.Since $E^0_{H^+/H_2} = 0 \, V$ and $pH = -\log[H^+]$,the equation becomes: $E = 0.0591 	imes pH$.For $pH_1 = 0$: $E_1 = 0.0591 	imes 0 = 0 \, V$.For $pH_2 = 7$: $E_2 = 0.0591 	imes 7 = 0.4137 \, V$.Change in potential $\Delta E = E_2 - E_1 = 0.4137 - 0 = 0.4137 \, V$.Since the potential increases from $0 \, V$ to $0.4137 \, V$,it increases by $0.413 \, V$.

What is the change in the reduction potential of a hydrogen electrode when the $pH$ of the initial solution changes from $0$ to $7$ by neutralization?

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