Chemistry - Redox Reactions Question with Solution | TestHub

In acidified $\mathrm{KMn}{\mathrm{O}}_4$, $\mathrm{Mn}{\mathrm{O}}_4^{-}$ is the stronger oxidant and ${\mathrm{H}}_2{\mathrm{O}}_2$ acts as the reductant: ${\mathrm{H}}_2{\mathrm{O}}_2$ $\to$ ${\mathrm{O}}_2$ + $2{\mathrm{H}}^{+}$ + $2{\mathrm{e}}^{-}$, so I is correct. II is also correct: oxygen rises from -1 (peroxide) to 0 (${\mathrm{O}}_2$) - and this loss of electrons is exactly what $"{\mathrm{H}}_2{\mathrm{O}}_2$ is oxidized" means, so II is the correct explanation of I. Hence (a).