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