The reaction between MnO4− and H2O is

Eo for the cell reaction:-
E∘=EMnO4−/Mn+2o+EH2O/O2o
=1.510V−1.223V
Eo=+0.287V
As E∘ of the cell reaction is positive, so this reaction is feasible. Therefore permanganate ion MnO4− liberate O2 from water in presence of an acid.
Hence, option D is correct.