What pressure (bar) of H₂ would be required to make emf of hydrogen electrode zero in pure water at 25°C?

We need to find the pressure of $$H_2$$ that makes the EMF of a hydrogen electrode zero in pure water at 25 degrees C. The half-cell reaction for the hydrogen electrode is $$2H^+(aq) + 2e^- \rightarrow H_2(g), \quad E^\circ = 0 \text{ V}$$ and the Nernst equation for this half-cell is $$E = E^\circ - \frac{RT}{2F}\ln\frac{P_{H_2}}{[H^+]^2}$$.

Setting $$E = 0$$ gives $$0 = 0 - \frac{RT}{2F}\ln\frac{P_{H_2}}{[H^+]^2}$$ which implies $$\ln\frac{P_{H_2}}{[H^+]^2} = 0$$ and hence $$\frac{P_{H_2}}{[H^+]^2} = 1$$. Therefore $$P_{H_2} = [H^+]^2$$.

In pure water at 25 degrees C, $$[H^+] = 10^{-7}$$ M and thus $$P_{H_2} = (10^{-7})^2 = 10^{-14}$$ bar. The correct answer is Option D: $$10^{-14}$$.