25 mL of 0.1 M Fe^2+ is titrated potentiometrically with 0.1 M Ce^4+ (E deg(Fe^3+/Fe^2+) = 0.68 V, E deg(Ce^4+/Ce^3+) = 1.44 V). After 12.5 mL of titrant has been added, the indicator electrode potential is

JEE Main levelSingle correct optionDifficulty 2 of 3ec119

$25 mL$ of $0.1 M$ $Fe X^{2 +}$ is titrated potentiometrically with $0.1 M$ $Ce X^{4 +}$ ( $E^{\circ} (Fe X^{3 +} / Fe X^{2 +}) = 0.68 V$ , $E^{\circ} (Ce X^{4 +} / Ce X^{3 +}) = 1.44 V$ ). After $12.5 mL$ of titrant has been added, the indicator electrode potential is

12.5 mL is half the equivalence volume (25 mL), so exactly half of the

Fe X^{2 +}

has been oxidised and

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

. Before equivalence the potential is set by the

Fe X^{3 +} / Fe X^{2 +}

couple:

E = 0.68 + 0.0591 lo g 1 = 0.68 V

. The half-equivalence potential equals the formal potential of the analyte couple.