E deg(MnO4^-, H+/Mn^2+) = 1.51 V. The reduction potential at 298 K when [MnO4^-] = [Mn^2+] = 1 M and pH = 1 is (in V, two decimals). Take (2.303RT)/(F) = 0.0591 V.

JEE Main levelNumerical answerDifficulty 2 of 3ec11

$E^{\circ} (MnO X_{4} X^{-}, H X^{+} / Mn X^{2 +}) = 1.51 V$ . The reduction potential at 298 K when $[MnO X_{4} X^{-}] = [Mn X^{2 +}] = 1 M$ and pH $= 1$ is (in V, two decimals). Take $\frac{2.303 RT}{F} = 0.0591 V$ .

Numerical answer type, answer in V, tolerance 0.01.

MnO X_{4} X^{-} + 8 H X^{+} + 5 e X^{-} Mn X^{2 +} + 4 H X_{2} O

E = 1.51 - \frac{0.0591}{5} lo g \frac{[ Mn X ^{2 +} ]}{[ MnO X _{4} X ^{-} ] [ H X ^{+} ] ^{8}} = 1.51 - \frac{0.0591}{5} \times 8 = 1.51 - 0.095 = 1.415 \approx 1.42 V

. Permanganate is a weaker oxidant at higher pH.