Given below are half-cell reactions:
\(\mathrm{MnO}_{4}^{-}+8 \mathrm{H}^{+}+5 \mathrm{e}^{-} \rightarrow \mathrm{Mn}^{2+}+4 \mathrm{H}_{2} \mathrm{O}\)
\(\mathrm{E}^{°} _{\mathrm{Mn}^{2+} / \mathrm{MnO}_{4}^{-}}=-1.510 \mathrm{~V}\)
\(\frac{1}{2} \mathrm{O}_{2}+2 \mathrm{H}^{+}+2 \mathrm{e}^{-} \rightarrow \mathrm{H}_{2} \mathrm{O}\)
\(\mathrm{E}_{\mathrm{O}_{2} / \mathrm{H}_{2} \mathrm{O}}^{\circ}=+1.223 \mathrm{~V}\)
Will the permanganate ion, \(\mathrm{MnO}_{4}^{-}\) liberate O2 from water in the presence of an acid ?
1
No, because \(E_{\text {cell }}^{\circ}\)= -2.733 V
2
Yes, because \(E_{\text {cell }}^{\circ}\)= +0.287 V
3
No, because \(E_{\text {cell }}^{\circ}\)= -0.287 V
4
Yes, because \(E_{\text {cell }}^{\circ}\)= +2.733 V