Question

Potassium permanganate acts as an oxidant in alkaline and acidic medium. The final products formed from $ KMn{{O}_{4}} $ in the two conditions are respectively

• A. $ Mn{{O}^{2-}} $ and $ M{{n}^{3+}} $
• B. $ M{{n}^{3+}} $ and $ M{{n}^{2+}} $
• C. $ M{{n}^{2+}} $ and $ M{{n}^{3+}} $
• D. $ Mn{{O}_{2}} $ and $ M{{n}^{2+}} $

Answer 1

Answer: (D)

In dilute alkaline solution, $ Mn{{O}_{2}} $ is formed. $ \overset{+7}{\mathop{Mn}}\,O_{4}^{-}+2{{H}_{2}}O+3{{e}^{-}}\xrightarrow[{}]{{}}\overset{+4}{\mathop{Mn}}\,{{O}_{2}}+4O{{H}^{-}} $
But in very strong base and with an excess of however manganate ion $ (MnO_{4}^{2-}) $ is formed.
$ MnO_{4}^{2-}+{{e}^{-}}\xrightarrow[{}]{{}}MnO_{4}^{2-} $
In the acidic medium MnS04 (i.e., $ M{{n}^{2+}} $ )is formed. $ 2\overset{+7}{\mathop{KM}}\,n{{O}_{4}}+3{{H}_{2}}S{{O}_{4}}\xrightarrow[{}]{{}}{{K}_{2}}S{{O}_{4}}+\overset{+2}{\mathop{2Mn}}\,S{{O}_{4}} $
$ +3{{H}_{2}}O+5(O) $