The product of oxidation of I - with MnO 4 - in alkaline medium is:-
Solution
Reaction: \(6 \mathrm{MnO}_4^{-}+\mathrm{I}^{-}+6 \mathrm{OH}^{-} ightarrow \mathrm{IO}_3^{-}+6 \mathrm{MnO}_4^{2-}+3 \mathrm{H}_2 \mathrm{O}\)
Here the oxidation state of \(\mathrm{Mn}\) reduced from +7 to +6 and the oxidation state of I oxidised from -1 to +5.
\(6 \mathrm{e}^{-}\)is involved in the reaction.
Oxidation product form is \(1 \mathrm{O}_3^{-}\). ,
Asked in: JEE-TOPICTESTS-CHEMISTRY