Given below are half-cell reactions: \[ \begin{array}{l} \mathrm{MnO}_{4}^{-}+8 \mathrm{H}^{+}+5...
0Given below are half-cell reactions:
\[
\begin{array}{l}
\mathrm{MnO}_{4}^{-}+8 \mathrm{H}^{+}+5 e^{-} \rightarrow \mathrm{Mn}^{2+}+4 \mathrm{H}_{2} \mathrm{O} \\
\mathrm{E}_{\mathrm{Mn}^{2+} / \mathrm{MnO}_{4}^{-}}^{\circ}=-1.510 \mathrm{~V} \\
\frac{1}{2} \mathrm{O}_{2}+2 \mathrm{H}^{+}+2 e^{-} \rightarrow \mathrm{H}_{2} \mathrm{O} \\
\mathrm{E}_{\mathrm{O}_{2} / \mathrm{H}_{2} \mathrm{O}}^{\circ}=+1.223 \mathrm{~V}
\end{array}
\]
Will the permanganate ion, \( \mathrm{MnO}_{4}^{-} \)liberate \( \mathrm{O}_{2} \) from water in the presence of an acid?
1. No, because \( E_{\text {cell }}^{\circ}=-2.733 \mathrm{~V} \)
2. Yes, because \( E_{\text {cell }}^{\circ}=+0.287 \mathrm{~V} \)
-3. No, because \( E_{\text {cell }}^{\circ}=-0.287 \mathrm{~V} \)
4. Yes, because \( E_{\text {cell }}^{\circ}=+2.733 \mathrm{~V} \)
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