The pressure of \(H_2\) required to make the potential of \(H_2\) electrode zero in pure water at \(298K\) is (Given \(E_{cell} = 0.0 V\))

\(10^{-14} atm\)

The correct answer is option 2. \(10^{-14} atm\).

The potential of a hydrogen electrode in pure water at \(298 \, \text{K}\) (\(25 \, \text{°C}\)) is \(0.0 \, \text{V}\) when the pressure of \(H_2\) gas at the electrode is equal to the standard hydrogen electrode (SHE) potential, which is \(0.0 \, \text{V}\) by definition.

The standard pressure of \(H_2\) gas is \(1 \, \text{atm}\), and the standard concentration of \(H^+\) ions in pure water at \(25 \, \text{°C}\) is \(10^{-7} \, \text{M}\) (corresponding to \(pH = 7\)).

Given that the \(E_{\text{cell}} = 0.0 \, \text{V}\), the Nernst equation can be written as:

\(E_{\text{cell}} = E^{\circ}_{\text{cell}} - \frac{0.0592}{n} \log Q \)

For a hydrogen electrode in pure water, the half-reaction is:

\(2H^+ + 2e^- \rightleftharpoons H_2 \)

Since the concentration of \(H^+\) ions in pure water is \(10^{-7} \, \text{M}\), the reaction quotient (\(Q\)) can be written as:

\(Q = \frac{[H_2]}{[H^+]^2} = \frac{P_{H_2}}{(10^{-7})^2} \)

Given that \(E_{\text{cell}} = 0.0 \, \text{V}\) and \(E^{\circ}_{\text{cell}} = 0.0 \, \text{V}\), we can rearrange the Nernst equation to solve for \(Q\):

\(0.0 \, \text{V} = 0.0 \, \text{V} - \frac{0.0592}{2} \log Q \)

\( \log Q = 0 \)

\(Q = 1 \)

Substituting \(Q = 1\) into the expression for \(Q\), we get:

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

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

\( P_{H_2} = 10^{-14} \, \text{atm} \)

Therefore, the pressure of \(H_2\) required to make the potential of the \(H_2\) electrode zero in pure water at \(298 \, \text{K}\) is \(10^{-14} \, \text{atm}\).

So, the correct answer is Option 2: \(10^{-14} \, \text{atm}\).