For the following reaction,

\(2H^+(aq) + 2e^- \longrightarrow H_2(g)\)

which is the correct statement under non-standard conditions.

\(E_{cell}\) becomes more negative with an increase in partial pressure of \(H_2\) and decrease in \([H^+]\) ions.

The correct answer is option 3. \(E_{cell}\) becomes more negative with an increase in partial pressure of \(H_2\) and decrease in \([H^+]\) ions.

The Nernst equation relates the standard cell potential \((E^o)\), cell potential \((E_{cell})\), temperature, gas concentrations, and ion concentrations:

\(E_{cell} = E^o - \frac{RT}{nF} × ln(Q)\)

where:

\(R\) is the gas constant

\(T\) is the temperature

\(n\) is the number of electrons transferred in the reaction

\(F\) is Faraday's constant

\(Q\) is the reaction quotient (ratio of product concentrations to reactant concentrations raised to their stoichiometric coefficients)

In this case:

Reaction: \(2H^+(aq) + 2e^- \longrightarrow H_2(g)\)

\(n = 2\) (number of electrons transferred)

\(Q = \frac{[H_2]}{[H^+]^2}\) (since the reaction quotient uses concentrations raised to their stoichiometric coefficients)

According to the Nernst equation, the cell potential \((E_{cell})\) is inversely proportional to the natural logarithm of the reaction quotient \((ln(Q))\).

When the partial pressure of H₂ increases (higher \([H_2]\)), \(Q\) increases, and \(ln(Q)\) also increases (positive value). This term becomes more negative, resulting in a more negative \(E_{cell}\).

When the concentration of \(H^+\) ions decreases (lower \([H^+]\)), \(Q\) decreases, and \(ln(Q)\) becomes more negative (larger negative value). This term again makes \(E_{cell}\) more negative.

Therefore, a combination of increasing H₂ pressure and decreasing \(H^+\) concentration will make the cell potential \((E_{cell})\) more negative. This signifies a tendency for the reaction to proceed in the forward direction (consuming \(H^+\) and producing \(H_2\)) to reach equilibrium.