The reduction potential of the hydrogen electrode Pt | H2 (4 bar) | H+ in a solution of pH 4 at 298 K is (take 2.303RT/F = 0.0591 V, log 2 = 0.301)

JEE Main levelSingle correct optionDifficulty 2 of 3ec087

The reduction potential of the hydrogen electrode $Pt ∣ H X_{2} (4 bar) ∣ H X^{+}$ in a solution of pH 4 at 298 K is (take $2.303 RT / F = 0.0591 V$ , $lo g 2 = 0.301$ )

For

2 H X^{+} + 2 e X^{-} H X_{2}

E = 0 - \frac{0.0591}{2} lo g \frac{p _{H X_{2}}}{[ H X ^{+} ] ^{2}}

lo g \frac{4}{( 1 0 ^{- 4} ) ^{2}} = lo g 4 + 8 = 0.602 + 8 = 8.602

E = - 0.02955 \times 8.602 = - 0.254 V

. Ignoring the pressure gives

- 0.0591 \times 4 = - 0.236 V