E deg(Ce^4+/Ce^3+) = 1.61 V and E deg(Fe^3+/Fe^2+) = 0.77 V. For the reaction Fe^2+ + Ce^4+ -> Fe^3+ + Ce^3+ at 298 K, log_10 K (nearest integer) is. Take (2.303RT)/(F) = 0.0591 V.

JEE Main levelNumerical answerDifficulty 2 of 3ec064

$E^{\circ} (Ce X^{4 +} / Ce X^{3 +}) = 1.61 V$ and $E^{\circ} (Fe X^{3 +} / Fe X^{2 +}) = 0.77 V$ . For the reaction $Fe X^{2 +} + Ce X^{4 +} Fe X^{3 +} + Ce X^{3 +}$ at 298 K, $lo g_{10} K$ (nearest integer) is. Take $\frac{2.303 RT}{F} = 0.0591 V$ .

Numerical answer type.

E_{cell}^{\circ} = 1.61 - 0.77 = 0.84 V

with

n = 1

lo g K = \frac{n E ^{\circ}}{0.0591} = \frac{0.84}{0.0591} = 14.2 \approx 14

. The very large

K

(

\sim 1 0^{14}

) is why

Ce X^{4 +}

titration of

Fe X^{2 +}

goes essentially to completion.

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