Sn^2+ is titrated with Fe^3+ according to Sn^2+ + 2Fe^3+ -> Sn^4+ + 2Fe^2+ at 298 K. E deg(Sn^4+/Sn^2+) = 0.15 V and E deg(Fe^3+/Fe^2+) = 0.77 V. The potential of a platinum indicator electrode at the equivalence point, in volt, is (three decimals).

JEE Advanced levelNumerical answerDifficulty 3 of 3ec124

$Sn X^{2 +}$ is titrated with $Fe X^{3 +}$ according to $Sn X^{2 +} + 2 Fe X^{3 +} Sn X^{4 +} + 2 Fe X^{2 +}$ at 298 K. $E^{\circ} (Sn X^{4 +} / Sn X^{2 +}) = 0.15 V$ and $E^{\circ} (Fe X^{3 +} / Fe X^{2 +}) = 0.77 V$ . The potential of a platinum indicator electrode at the equivalence point, in volt, is (three decimals).

Numerical answer type, answer in V, tolerance 0.002.

At equivalence,

E = 0.15 + \frac{0.0591}{2} lo g \frac{[ Sn X ^{4 +} ]}{[ Sn X ^{2 +} ]} = 0.77 + 0.0591 lo g \frac{[ Fe X ^{3 +} ]}{[ Fe X ^{2 +} ]}

. Multiply the first by 2, add to the second:

3 E = 2 (0.15) + 0.77 + 0.0591 lo g \frac{[ Sn X ^{4 +} ] [ Fe X ^{3 +} ]}{[ Sn X ^{2 +} ] [ Fe X ^{2 +} ]}

. From stoichiometry at equivalence,

[Fe X^{2 +}] = 2 [Sn X^{4 +}]

and

[Fe X^{3 +}] = 2 [Sn X^{2 +}]

, so the log term is

lo g 1 = 0

E_{e q} = \frac{2 \times 0.15 + 0.77}{3} = \frac{1.07}{3} = 0.357 V