Given below are half cell reactions : MnO 4 - + 8 H + + 5 e - → Mn 2 + + 4 H 2 O , E Mn 2 + / MnO 4 - o = -…

The reaction between ${\mathrm{MnO}}_4^{-}$ and ${\mathrm{H}}_2\mathrm{O}$ is

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

$\mathrm{E}°={\mathrm{E}}_{{\mathrm{MnO}}_4^{-}/{\mathrm{Mn}}^{+2}}^{\mathrm{o}} + {\mathrm{E}}_{{\mathrm{H}}_2\mathrm{O}/{\mathrm{O}}_2}^{\mathrm{o}}$

$=1.510 \mathrm{V}-1.223 \mathrm{V}$

${\mathrm{E}}^{\mathrm{o}}=+0.287 \mathrm{V}$

As $\mathrm{E}°$ of the cell reaction is positive, so this reaction is feasible. Therefore permanganate ion ${\mathrm{MnO}}_4^{-}$ liberate ${\mathrm{O}}_2$ from water in presence of an acid.

Hence, option D is correct.