Reduction of the metal centre in aqueous permanganate ion involves\

Question

Professor PC Thomas Classes · Answer

(a, c, d) : In alkaline medium. $$ \mathrm{MnO}_{4}^{-}+e^{-} \longrightarrow \mathrm{MnO}_{4}^{2-} $$ But $\mathrm{MnO}_{4}^{-}$is further reduced to $\mathrm{MnO}_{2}$\ (in case of aqueous $\mathrm{KMnO}_{4}$ )\ $\mathrm{MnO}_{4}^{2-}+2 \mathrm{H}_{2} \mathrm{O}+2 e^{-} ightarrow \mathrm{MnO}_{2}+4 \mathrm{OH}^{-}$ Complete reaction :\ $2 \mathrm{KMnO}_{4}+\mathrm{H}_{2} \mathrm{O} ightarrow 2 \mathrm{MnO}_{2}+2 \mathrm{KOH}+3 \mathrm{O}$\ $\mathrm{MnO}_{4}^{-}+2 \mathrm{H}_{2} \mathrm{O}+3 e^{-} ightarrow \mathrm{MnO}_{2}+4 \mathrm{OH}^{-}$\ In acidic medium :\ $\mathrm{MnO}_{4}^{-}+8 \mathrm{H}^{+}+5 e^{-} ightarrow \mathrm{Mn}^{2+}+4 \mathrm{H}_{2} \mathrm{O}$\ In neutral medium :\ $\mathrm{MnO}_{4}^{-}+2 \mathrm{H}_{2} \mathrm{O}+3 e^{-} ightarrow \mathrm{MnO}_{2}+4 \mathrm{OH}^{-}$\ Hence, number of electrons lose in acidic and neutral medium are 5 and 3 electrons respectively.\