$25 mL$ 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 $30 mL$ of titrant has been added, the indicator electrode potential in mV is (nearest integer).

Numerical answer type, answer in mV, tolerance 1.

Answer: 1399 mV

Equivalence is at 25 mL, so 5 mL of

Ce X^{4 +}

is in excess while 25 mL-equivalents of

Ce X^{3 +}

have formed. After equivalence the potential is set by the

Ce X^{4 +} / Ce X^{3 +}

couple:

[Ce X^{4 +}] / [Ce X^{3 +}] = 5/25 = 0.2

E = 1.44 + 0.0591 lo g 0.2 = 1.44 - 0.0591 \times 0.699 = 1.44 - 0.0413 = 1.3987 V \approx 1399 mV