Electrochemistry

Given below are half-cell reactions:

$\mathrm{Mn}{\mathrm{O}}_4^{-}+8{\mathrm{H}}^{+}+5{\mathrm{e}}^{-}\to \mathrm{M}{\mathrm{n}}^{2+}+4{\mathrm{H}}_2\mathrm{O};E_{\mathrm{M}{\mathrm{n}}^{2+}/Mn{\mathrm{O}}_4^{-}}^{\circ }=-1.510\mathrm{V}$

$\frac{1}{2}{\mathrm{O}}_2+2{\mathrm{H}}^{+}+2{\mathrm{e}}^{-}\to {\mathrm{H}}_2\mathrm{O};E_{{\mathrm{O}}_2/{\mathrm{H}}_2\mathrm{O}}^{\circ }=+1.223\mathrm{V}$

Will the permanganate ion, $\mathrm{Mn}{\mathrm{O}}_4^{-}$, liberate ${\mathrm{O}}_2$ from water in the presence of an acid?