$Fe X^{2 +}$ is titrated with $Ce X^{4 +}$ at 298 K; $E^{\circ} (Fe X^{3 +} / Fe X^{2 +}) = 0.68 V$ , $E^{\circ} (Ce X^{4 +} / Ce X^{3 +}) = 1.44 V$ , $2.303 RT / F = 0.0591 V$ . The change in indicator electrode potential between the point where 99.9% of the $Fe X^{2 +}$ has been oxidised and the point where 0.1% excess $Ce X^{4 +}$ (relative to the initial $Fe X^{2 +}$ ) has been added is closest to

Answer: A

At 99.9% titrated,

[Fe X^{3 +}] / [Fe X^{2 +}] = 999 \approx 1 0^{3}

, so

E = 0.68 + 3 \times 0.0591 = 0.857 V

. At 0.1% excess,

[Ce X^{4 +}] / [Ce X^{3 +}] = 0.001

, so

E = 1.44 - 3 \times 0.0591 = 1.263 V

. The jump across the equivalence region is

1.263 - 0.857 = 0.405 V

, i.e.

0.76 - 6 \times 0.0591

. The full difference of formal potentials (0.760 V) is never reached in the titration itself.