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Q.

Reaction of KMnO₄ in neutral or very weakly alkaline solution can be represented as

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a

${MnO}_{4}^{-} + 2 H_{2} O + 3 e^{-} ⟶ {MnO}_{2} + 4 {OH}^{-}$

b

${MnO}_{4}^{-} + 8 H^{+} + 5 e^{-} ⟶ {Mn}^{2 +} + 4 H_{2} O$

c

${MnO}_{4}^{-} + e^{-} ⟶ {MnO}_{4}^{2 -}$

d

$2 {MnO}_{4}^{-} + 2 {OH}^{-} ⟶ 2 {MnO}_{4}^{2 -} + 1 / 2 O_{2} + H_{2} O$

answer is A.

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Detailed Solution

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${MnO}_{4}^{-} + 2 H_{2} O + 3 e^{-} ⟶ {MnO}_{2} + 4 {OH}^{-}$

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