Reaction of KMnO 4 in neutral or faintly alkaline solution is given by the equation:

Question

Reaction of KMnO 4 in neutral or faintly alkaline solution is given by the equation: (A) MnO 4-+3 e -+2 H 2 O → MnO 2+4 OH - (B) MnO 4-+8 H ++5 e - → Mn 2++4 H 2 O (C) MnO 4-+ e - → MnO 42- (D) 2 MnO 4-+2 OH - → 2 MnO 42-+1 / 2 O 2+ H 2 O

Tardigrade Admin · Accepted Answer

In neutral or faintly alkaline medium, KMnO 4 itself gets changed to MnO 2.

Q. Reaction of $K M n O_{4}$ in neutral or faintly alkaline solution is given by the equation:

$M n O_{4}^{-} + 3 e^{-} + 2 H_{2} O \to M n O_{2} + 4 O H^{-}$

$M n O_{4}^{-} + 8 H^{+} + 5 e^{-} \to M n^{2 +} + 4 H_{2} O$

$M n O_{4}^{-} + e^{-} \to M n O_{4}^{2 -}$

$2 M n O_{4}^{-} + 2 O H^{-} \to 2 M n O_{4}^{2 -} + 1/2 O_{2} + H_{2} O$

In neutral or faintly alkaline medium, $K M n O_{4}$ itself gets changed to $M n O_{2}$ .

Q. Reaction of KMnO4​ in neutral or faintly alkaline solution is given by the equation:

MnO4−​+3e−+2H2​O→MnO2​+4OH−

MnO4−​+8H++5e−→Mn2++4H2​O

MnO4−​+e−→MnO42−​

2MnO4−​+2OH−→2MnO42−​+1/2O2​+H2​O