For the following reaction, 2H^+(aq) + 2e^- longrightarrow H_2(g) whi…

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 electrode potential at any concentration is given by the Nernst equation. For the hydrogen electrode, 2H⁺(aq) + 2e⁻ → H₂(g), with E° = 0:

    \( E = E^{\circ} - \displaystyle\frac{0.0591}{2}\log\displaystyle\frac{p_{H_2}}{[H^+]^2} = -\displaystyle\frac{0.0591}{2}\log\displaystyle\frac{p_{H_2}}{[H^+]^2} \)

The potential becomes more negative whenever the ratio p_{\( \ce{H2} \)}/[H⁺]² becomes larger, that is, when the partial pressure of H₂ increases and when [H⁺] decreases. Both changes in option (3) push E the same way.

Why the other options are incorrect

• Option (1) E_cell becomes more negative with an increase in partial pressure of H₂ and [H⁺] ions. — Incorrect. A rise in \( \ce{[H^{+}]} \) makes E more positive, against the effect of a rise in p_{\( \ce{H2} \)}; the two changes oppose each other, so E does not necessarily become more negative.

• Option (2) E_cell becomes more positive with an increase in partial pressure of H₂ and [H⁺] ions. — Incorrect. A rise in p_{\( \ce{H2} \)} makes E more negative, not more positive; again the two changes oppose each other.

• Option (4) E_cell becomes more positive with an increase in partial pressure of H₂ and decrease in [H⁺] ions. — Incorrect. Both changes make the ratio larger, so E becomes more negative, not more positive.

NCERT (2026-27 edition):
Nernst Equation – “Nernst showed that for the electrode reaction:
    Mⁿ⁺(aq) + ne^– → M(s)
the electrode potential at any concentration measured with respect to standard hydrogen electrode can be represented by:
    E_(Mⁿ⁺/M) = E^o_(Mⁿ⁺/M) – (RT/nF) ln ([M]/[Mⁿ⁺])”