25 mL of 0.1 M Fe^2+ is titrated with 0.1 M Ce^4+ at 298 K. E deg(Fe^3+/Fe^2+) = 0.68 V, E deg(Ce^4+/Ce^3+) = 1.44 V, 2.303RT/F = 0.0591 V, log 2 = 0.301. After 20 mL of Ce^4+ has been added, the indicator electrode potential is

JEE Main levelSingle correct optionDifficulty 2 of 3ec121

$25 mL$ of $0.1 M$ $Fe X^{2 +}$ is titrated with $0.1 M$ $Ce X^{4 +}$ at 298 K. $E^{\circ} (Fe X^{3 +} / Fe X^{2 +}) = 0.68 V$ , $E^{\circ} (Ce X^{4 +} / Ce X^{3 +}) = 1.44 V$ , $2.303 RT / F = 0.0591 V$ , $lo g 2 = 0.301$ . After $20 mL$ of $Ce X^{4 +}$ has been added, the indicator electrode potential is

Before equivalence, 20 mL of

Ce X^{4 +}

converts 20 mL-equivalents of

Fe X^{2 +}

Fe X^{3 +}

, leaving 5 mL-equivalents of

Fe X^{2 +}

[Fe X^{3 +}] / [Fe X^{2 +}] = 20/5 = 4

(volume cancels).

E = 0.68 + 0.0591 lo g 4 = 0.68 + 0.0591 \times 0.602 = 0.68 + 0.0356 = 0.716 V

. Inverting the ratio gives 0.644 V.