The cell Pt | H2 (1 bar) | buffer solution || Cu^2+ (1 M) | Cu has EMF 0.537 V at 298 K. Given E deg(Cu^2+/Cu) = 0.34 V and 2.303RT/F = 0.0591 V, the pH of the buffer is (two decimals).

JEE Main levelNumerical answerDifficulty 2 of 3ec102

The cell $Pt ∣ H X_{2} (1 bar) ∣ buffer solution ∣∣ Cu X^{2 +} (1 M) ∣ Cu$ has EMF $0.537 V$ at 298 K. Given $E^{\circ} (Cu X^{2 +} / Cu) = 0.34 V$ and $2.303 RT / F = 0.0591 V$ , the pH of the buffer is (two decimals).

Numerical answer type, tolerance 0.02.

E_{cell} = E_{cathode} - E_{anode} = 0.34 - E (H X^{+} / H X_{2})

, so

E (H X^{+} / H X_{2}) = 0.34 - 0.537 = - 0.197 V

. For the hydrogen electrode at 1 bar,

E = - 0.0591 pH

. Hence

pH = 0.197/0.0591 = 3.33