For $MnO_4^-$ to liberate $O_2$ from water, $MnO_4^-$ must undergo reduction and $H_2O$ must undergo oxidation.

Cathode (Reduction): $MnO_4^- + 8H^+ + 5e^- \rightarrow Mn^{2+} + 4H_2O$ The given potential is for the reverse process (oxidation of $Mn^{2+}$), so the standard reduction potential is: $E^\circ_{cathode} = -E^\circ_{Mn^{2+}/MnO_4^-} = -(-1.510\text{ V}) = +1.510\text{ V}$

Anode (Oxidation): $H_2O \rightarrow \frac{1}{2}O_2 + 2H^+ + 2e^-$ The standard reduction potential for oxygen is given as: $E^\circ_{anode} = E^\circ_{O_2/H_2O} = +1.223\text{ V}$

The standard cell potential is: $E^\circ_{cell} = E^\circ_{cathode} - E^\circ_{anode} = 1.510\text{ V} - 1.223\text{ V} = +0.287\text{ V}$

Since $E^\circ_{cell}$ is positive, the reaction is spontaneous. Therefore, the permanganate ion will liberate $O_2$ from water.