To solve this problem, we need to determine the pressure of $H_2$ required to make the potential of a hydrogen electrode zero in pure water at $298$ K.$\\$The potential of a hydrogen electrode is given by the Nernst equation:$\\ E = E^\circ - \frac{RT}{nF} \ln \left( \frac{[H^+]^2}{P_{H_2}} \right) \\$Where:$\\$• $E$ is the electrode potential.$\\$• $E^\circ$ is the standard electrode potential, which is $0$ V for the hydrogen electrode.$\\$• $R$ is the universal gas constant, $8.314 \, \mathrm{J \, mol^{-1} \, K^{-1}}. \\$• $T$ is the temperature in Kelvin, $298 \, \mathrm{K}. \\$• $n$ is the number of moles of electrons transferred, $2$ for $H_2. \\$• $F$ is Faraday's constant, $96485 \, \mathrm{C \, mol^{-1}}. \\$• $[H^+]$ is the concentration of hydrogen ions.$\\$• $P_{H_2}$ is the pressure of hydrogen gas.$\\$For the potential to be zero, $E = 0.$ Thus, the equation becomes:$\\ 0 = 0 - \frac{RT}{2F} \ln \left( \frac{[H^+]^2}{P_{H_2}} \right) \\$Simplifying, we get:$\\ \frac{RT}{2F} \ln \left( \frac{[H^+]^2}{P_{H_2}} \right) = 0 \\$This implies:$\\ \ln \left( \frac{[H^+]^2}{P_{H_2}} \right) = 0 \\$Taking the exponential on both sides, we have:$\\ \frac{[H^+]^2}{P_{H_2}} = 1 \\$Thus, $P_{H_2} = [H^+]^2 \\$In pure water at $298$ K, the concentration of $H^+$ ions is $10^{-7} \, \mathrm{M}. \\$Therefore, $P_{H_2} = (10^{-7})^2 = 10^{-14} \, \mathrm{atm}. \\$Thus, the pressure of $H_2$ required is $10^{-14} \, \mathrm{atm}. \\$This corresponds to Option 1.