For a hydrogen electrode, the reduction half-reaction is: $2\text{H}^+(aq) + 2e^- \rightarrow \text{H}_2(g)$ According to the Nernst equation at $298 \text{ K}$: $E = E^\circ - \frac{0.0591}{2} \log \frac{P_{\text{H}_2}}{[\text{H}^+]^2}$ We are given that the potential $E = 0$, and the standard potential of the hydrogen electrode $E^\circ = 0$. In pure water at $298 \text{ K}$, the concentration of hydrogen ions is $[\text{H}^+] = 10^{-7} \text{ M}$. Substituting these values into the Nernst equation: $0 = 0 - \frac{0.0591}{2} \log \frac{P_{\text{H}_2}}{(10^{-7})^2}$ $0 = \log \frac{P_{\text{H}_2}}{10^{-14}}$ $\frac{P_{\text{H}_2}}{10^{-14}} = 10^0 = 1$ $P_{\text{H}_2} = 10^{-14} \text{ atm}$