Use $E^{\circ} (MnO X_{4} X^{-} / Mn X^{2 +}) = + 1.51 V$ and $E^{\circ} (Fe X^{3 +} / Fe X^{2 +}) = + 0.77 V$ (acidic medium, $2.303 RT / F = 0.0591 V$ at 298 K). For the balanced reaction of $MnO X_{4} X^{-}$ with $Fe X^{2 +}$ in acid to give $Mn X^{2 +}$ and $Fe X^{3 +}$ , $lo g K$ is closest to

Answer: D

Balanced equation:

MnO X_{4} X^{-} + 5 Fe X^{2 +} + 8 H X^{+} \to Mn X^{2 +} + 5 Fe X^{3 +} + 4 H X_{2} O

, so

n = 5

E_{cell}^{\circ} = 1.51 - 0.77 = 0.74 V

lo g K = n E^{\circ} /0.0591 = 5 \times 0.74/0.0591 = 62.6