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:

$$ E = E^\circ + \frac{RT}{2F} \ln \left( \frac{[\mathrm{H^+}]^2}{P_{\mathrm{H_2}}} \right) $$

where:

• E is the electrode potential.
• E^∘ is the standard electrode potential, which is 0 V for the hydrogen electrode.
• R is the gas constant, $$8.314 \, \mathrm{J \cdot mol^{-1} \cdot K^{-1}}$$.
• T is the temperature in Kelvin: $$25^{\circ} \mathrm{C} = 298 \, \mathrm{K}$$.
• F is the Faraday constant, $$96485 \, \mathrm{C \cdot mol^{-1}}$$.
• $$[\mathrm{H^+}]$$ is the concentration of hydrogen ions.
• $$P_{\mathrm{H_2}}$$ is the pressure of hydrogen gas.

In pure water at $$25^{\circ} \mathrm{C}$$, $$[\mathrm{H^+}] = 10^{-7} \, \mathrm{M}$$. To make the emf zero, we set E to 0 in the Nernst equation:

$$ 0 = 0 + \frac{8.314 \times 298}{2 \times 96485} \ln \left( \frac{(10^{-7})^2}{P_{\mathrm{H_2}}} \right) $$

Simplifying the constants:

$$ \frac{8.314 \times 298}{2 \times 96485} \approx 0.0128 $$

Thus, the equation becomes:

$$ 0 = 0.0128 \ln \left( \frac{(10^{-7})^2}{P_{\mathrm{H_2}}} \right) $$

Since ln 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:

$$ \frac{(10^{-7})^2}{P_{\mathrm{H_2}}} = 1 $$

Simplifying this gives:

$$ (10^{-7})^2 = P_{\mathrm{H_2}} $$

$$ 10^{-14} = P_{\mathrm{H_2}} $$

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.

The correct answer is Option B: $$10^{-14}$$.