Chemistry - Redox Reactions Question with Solution | TestHub

Acidified $\mathrm{KMn}{\mathrm{O}}_4$ is decolourized only by reducing agents. ${\mathrm{H}}_2\mathrm{S}$ (S -2 $\to$ 0 or +6), ${\mathrm{H}}_2{\mathrm{O}}_2$ (O -1 $\to$ 0) and $\mathrm{FeC}{\mathrm{l}}_2$ ($\mathrm{F}{\mathrm{e}}^{2+}$ $\to$ $\mathrm{F}{\mathrm{e}}^{3+}$) all reduce $\mathrm{Mn}{\mathrm{O}}_4^{-}$. In $\mathrm{CuS}{\mathrm{O}}_4$, copper is already in its common stable +2 state and $\mathrm{S}{\mathrm{O}}_4^{2-}$ has S at +6, so nothing is oxidized and the colour stays. Hence (b).