A galvanic cell is made of two hydrogen electrodes, one of which is t…

The correct answer is Option (2) → 0.1 M CH₃COOH

By the Nernst equation, the potential of a hydrogen electrode (H₂ at 1 bar, 298 K) depends on the \( \ce{H^+} \) concentration of its solution:

    \( \displaystyle \ce{H^+(aq) + e^- -> 1/2 H2(g)} \)

    \( \displaystyle E_{\mathrm{H^+/H_2}}=0-0.059\log\frac{1}{[\mathrm{H^+}]}\,\mathrm{V} \)

For [H⁺] below 1 M this potential is negative, so this electrode is the anode and the standard hydrogen electrode (E° = 0) is the cathode:

    \( \displaystyle E_{\mathrm{cell}}=0-E_{\mathrm{H^+/H_2}}=0.059\log\frac{1}{[\mathrm{H^+}]}\,\mathrm{V} \)

So the emf is largest when [H⁺] is smallest.

• 0.1 M HCl — a strong acid, fully ionised: [H⁺] = 0.1 M.

• 0.1 M CH₃COOH — a weak acid, only partly ionised: [H⁺] is much less than 0.1 M, the lowest of the four.

• 0.1 M HNO₃ — a strong acid, fully ionised: [H⁺] = 0.1 M.

• 0.1 M H₂SO₄ — a strong acid that gives two \( \ce{H^+} \) per molecule: [H⁺] is more than 0.1 M.

The acetic acid solution has the lowest [H⁺], so it gives the maximum emf.

NCERT (2026-27 edition):
2.3 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: … concentration of solid M is taken as unity and we have

    \( \displaystyle E_{(\mathrm{M^{n+}/M})}=E^{\circ}_{(\mathrm{M^{n+}/M})}-\frac{RT}{nF}\ln\frac{1}{[\mathrm{M^{n+}}]} \qquad (2.8) \)”

Weak Electrolytes – “Weak electrolytes like acetic acid have lower degree of dissociation at higher concentrations …”