$E^{\circ} (Fe X^{3 +} / Fe X^{2 +}) = 0.77 V$ . The reduction potential of a platinum electrode dipped in a solution with $[Fe X^{3 +}] = 0.1 M$ and $[Fe X^{2 +}] = 1.0 M$ at 298 K is (take $2.303 RT / F = 0.0591 V$ )

Answer: A

For

Fe X^{3 +} + e X^{-} Fe X^{2 +}

n = 1

and

E = E^{\circ} + 0.0591 lo g \frac{[ Fe X ^{3 +} ]}{[ Fe X ^{2 +} ]}

. Here

lo g (0.1/1.0) = - 1

, so

E = 0.77 - 0.0591 = 0.7109 \approx 0.711 V

. Lowering the oxidised form lowers the potential.