Fe^2+ is titrated with KMnO4 in a solution maintained at pH 1 ([H+] = 0.1 M) at 298 K: MnO4- + 5Fe^2+ + 8H+ -> Mn^2+ + 5Fe^3+ + 4H2O. E deg(MnO4-, H+/Mn^2+) = 1.51 V, E deg(Fe^3+/Fe^2+) = 0.77 V, 2.303RT/F = 0.0591 V. The platinum electrode potential at the equivalence point, in mV, is (nearest integer).

JEE Advanced levelNumerical answerDifficulty 3 of 3ec125

$Fe X^{2 +}$ is titrated with $KMnO X_{4}$ in a solution maintained at pH 1 ( $[H X^{+}] = 0.1 M$ ) at 298 K: $MnO X_{4} X^{-} + 5 Fe X^{2 +} + 8 H X^{+} Mn X^{2 +} + 5 Fe X^{3 +} + 4 H X_{2} O$ . $E^{\circ} (MnO X_{4} X^{-}, H X^{+} / Mn X^{2 +}) = 1.51 V$ , $E^{\circ} (Fe X^{3 +} / Fe X^{2 +}) = 0.77 V$ , $2.303 RT / F = 0.0591 V$ . The platinum electrode potential at the equivalence point, in mV, is (nearest integer).

Numerical answer type, answer in mV, tolerance 1.

At equivalence:

E = 0.77 + 0.0591 lo g \frac{[ Fe X ^{3 +} ]}{[ Fe X ^{2 +} ]}

and

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

. Multiply the second by 5, add the first:

6 E = 0.77 + 5 (1.51) + 0.0591 lo g \frac{[ Fe X ^{3 +} ] [ MnO X _{4} X ^{-} ]}{[ Fe X ^{2 +} ] [ Mn X ^{2 +} ]} + 8 \times 0.0591 lo g [H X^{+}]

. At equivalence

[Fe X^{3 +}] = 5 [Mn X^{2 +}]

and

[Fe X^{2 +}] = 5 [MnO X_{4} X^{-}]

, so the concentration log term vanishes.

E_{e q} = \frac{0.77 + 7.55}{6} - \frac{8 \times 0.0591}{6} pH = 1.3867 - 0.0788 = 1.3079 V \approx 1308 mV