Given below are half cell reactions : MnO _ 4 ^ - + 8 H ^ + + 5 e ^...

The reaction between ${MnO}_{4}^{-}$ and ${H}_{2}O$ is

${E}^{o}$ for the cell reaction:-

$E ° = {E}_{{MnO}_{4}^{-} / {Mn}^{+ 2}}^{o} + {E}_{{H}_{2}O / {O}_{2}}^{o}$

$= 1 . 510V - 1 . 223V$

${E}^{o} = + 0 . 287V$

As $E °$ of the cell reaction is positive, so this reaction is feasible. Therefore permanganate ion ${MnO}_{4}^{-}$ liberate ${O}_{2}$ from water in presence of an acid.

Hence, option D is correct.