What pressure (bar) of $\mathrm{H}_2$ would be required to make emf of hydrogen electrode zero in pure water at $25^{\circ} \mathrm{C}$ ?

<p>The electromotive force (emf) of a hydrogen electrode can be derived from the Nernst equation, which for the hydrogen half-cell reaction $\mathrm{H_2(g)} \rightarrow 2\mathrm{H^+(aq)} + 2\mathrm{e^-}$ is given by:</p> <p>$$ E = E^\circ + \frac{RT}{2F} \ln \left( \frac{[\mathrm{H^+}]^2}{P_{\mathrm{H_2}}} \right) $$</p> <p>where:</p> <ul> <li><em>E</em> is the electrode potential.</li> <li><em>E</em>^∘ is the standard electrode potential, which is 0 V for the hydrogen electrode.</li> <li><em>R</em> is the gas constant, $8.314 \, \mathrm{J \cdot mol^{-1} \cdot K^{-1}}$.</li> <li><em>T</em> is the temperature in Kelvin: $25^{\circ} \mathrm{C} = 298 \, \mathrm{K}$.</li> <li><em>F</em> is the Faraday constant, $96485 \, \mathrm{C \cdot mol^{-1}}$.</li> <li>$[\mathrm{H^+}]$ is the concentration of hydrogen ions.</li> <li>$P_{\mathrm{H_2}}$ is the pressure of hydrogen gas.</li> </ul> <p>In pure water at $25^{\circ} \mathrm{C}$, $[\mathrm{H^+}] = 10^{-7} \, \mathrm{M}$. To make the emf zero, we set <em>E</em> to 0 in the Nernst equation:</p> <p>$$ 0 = 0 + \frac{8.314 \times 298}{2 \times 96485} \ln \left( \frac{(10^{-7})^2}{P_{\mathrm{H_2}}} \right) $$</p> <p>Simplifying the constants:</p> <p>$\frac{8.314 \times 298}{2 \times 96485} \approx 0.0128$</p> <p>Thus, the equation becomes:</p> <p>$0 = 0.0128 \ln \left( \frac{(10^{-7})^2}{P_{\mathrm{H_2}}} \right)$</p> <p>Since <em>ln</em> term must be zero for this equation to hold true (as 0 divided by any number is still 0), the expression inside the logarithm must equal 1:</p> <p>$\frac{(10^{-7})^2}{P_{\mathrm{H_2}}} = 1$</p> <p>Simplifying this gives:</p> <p>$(10^{-7})^2 = P_{\mathrm{H_2}}$</p> <p>$10^{-14} = P_{\mathrm{H_2}}$</p> <p>Therefore, the required pressure of $\mathrm{H_2}$ to make the emf of the hydrogen electrode zero in pure water at $25^{\circ} \mathrm{C}$ is $10^{-14}$ bar.</p> <p>The correct answer is <strong>Option B: $10^{-14}$</strong>.</p>