Given below are half cell reactions: MnO_4^-+8 H^++5 e^- → Mn^2++4 H_2 O; E_Mn^2+ / MnO_4^-^°=-1.510 V; 1/2 O_2+2 H^++2 e^- → H_2 O; E_O_2 / H_2…

Given below are half cell reactions : $$\begin{aligned} & \mathrm{MnO}_4^{-}+8 \mathrm{H}^{+}+5 \mathrm{e}^{-} \rightarrow \mathrm{Mn}^{2+}+4 \mathrm{H}_2 \mathrm{O} \\ & \mathrm{E}_{\mathrm{Mn}^{2+} / \mathrm{MnO}_4^{-}}^{\circ}=-1.510 \mathrm{~V} \\ & \frac{1}{2} \mathrm{O}_2+2 \mathrm{H}^{+}+2 \mathrm{e}^{-} \rightarrow \mathrm{H}_2 \mathrm{O} \\ & \mathrm{E}_{\mathrm{O}_2 / \mathrm{H}_2 \mathrm{O}}^{\circ}=+1.223 \mathrm{~V} \end{aligned} $$ Will the permanganate ion, $\displaystyle \mathrm{MnO}_4^{-}$liberate $\displaystyle \mathrm{O}_2$ from water in the presence of an acid ?