Given below are half-cell reactions:
\(MnO_4^- + 8H^+ + 5e^- → Mn^{2+} + 4H_2O\),
\(E°_{(Mn^{+2}/MnO_4^-)} = –1.510 V\)
\((\frac{1}{2})O_2 + 2H^+ + 2e^- → H_2O,\)
\(E°(O_2/H_2O) = +1.223 V\)
Will the permanganate ion, \(MnO_4^-\) liberate \(O_2\) from water in the presence of an acid?