Hydrogen electrode reaction

The standard hydrogen electrode is based on: $2H^+(aq) + 2e^- \rightleftharpoons H_2(g)$

For this electrode, the Nernst equation at $25^\circ\mathrm{C}$ is: $E = E^\circ - \frac{0.0591}{2} \log \frac{P_{H_2}}{[H^+]^2}$

Since for hydrogen electrode, $E^\circ = 0$, $E = - \frac{0.0591}{2} \log \frac{P_{H_2}}{[H^+]^2}$

Condition for emf to be zero

We are asked for the pressure of $H_2$ required so that the electrode emf becomes zero in pure water.

So set: $E = 0$

Then, $0 = - \frac{0.0591}{2} \log \frac{P_{H_2}}{[H^+]^2}$

Hence, $\log \frac{P_{H_2}}{[H^+]^2} = 0$ $\frac{P_{H_2}}{[H^+]^2} = 1$ $P_{H_2} = [H^+]^2$

Use $[H^+]$ for pure water

In pure water at $25^\circ\mathrm{C}$, $[H^+] = 10^{-7} \, \text{M}$

Therefore, $P_{H_2} = (10^{-7})^2 = 10^{-14} \text{ bar}$

Match with options

$P_{H_2} = 10^{-14} \text{ bar}$

This corresponds to Option B.

Verification with stored answer

Stored correct answer: B

Our derived answer: B

So they agree.