To determine if the permanganate ion $$MnO$_4^-$ can liberate $$O$_2$ from water, we need to calculate the $E^\circ_{cell}$ for the overall reaction.$\\$The given half-cell reactions are:$\\$1. $$MnO$_4^- + 8H^+ + 5e^- \rightarrow$Mn$^{2+} + 4H_2O, \quad E^\circ_{Mn^{2+}/MnO_4^-} = -1.510 \,$V$\\$2. $$O$_2 + 2H^+ + 2e^- \rightarrow H_2O, \quad E^\circ_{O_2/H_2O} = +1.223 \,$V$\\$To find the overall cell potential $E^\circ_{cell},$ we need to reverse the second reaction because we want $$O$_2$ to be produced.$\\$Reversed reaction:$\\$3. $2H_2O \rightarrow$O$_2 + 4H^+ + 4e^-, \quad E^\circ_{H_2O/O_2} = -1.223 \,$V$\\$Now, balance the number of electrons transferred in both reactions.$\\$Multiply the first reaction by 2 and the reversed second reaction by 5:$\\$4. $2($MnO$_4^- + 8H^+ + 5e^- \rightarrow$Mn$^{2+} + 4H_2O) \\$5. $5(2H_2O \rightarrow$O$_2 + 4H^+ + 4e^-) \\$This gives:$\\$6. $2$MnO$_4^- + 16H^+ + 10e^- \rightarrow 2$Mn$^{2+} + 8H_2O \\$7. $10H_2O \rightarrow 5$O$_2 + 20H^+ + 20e^- \\$Combine the reactions:$\\$8. $2$MnO$_4^- + 16H^+ + 10H_2O \rightarrow 2$Mn$^{2+} + 8H_2O + 5$O$_2 + 20H^+ \\$Simplify the equation:$\\$9. $2$MnO$_4^- + 2H_2O \rightarrow 2$Mn$^{2+} + 5$O$_2 + 4H^+ \\$Calculate $E^\circ_{cell}: \\ E^\circ_{cell} = E^\circ_{cathode} - E^\circ_{anode} \\ E^\circ_{cell} = (-1.510 \,$V$) - (-1.223 \,$V$) \\ E^\circ_{cell} = -1.510 \,$V$+ 1.223 \,$V$\\ E^\circ_{cell} = -0.287 \,$V$\\$Since $E^\circ_{cell}$ is negative, the reaction is not spontaneous.$\\$Therefore, the permanganate ion $$MnO$_4^-$ will not liberate $$O$_2$ from water.$\\$This corresponds to Option 2: No, because $E^\circ_{cell} = -0.287 \,$V$.$