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Potassium Permanganate

Reduction of ...

Question

Reduction of the metal centre in aqueous permaganate ion involves:

3

electrons in neutral medium

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5

electrons in neutral medium

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3

electrons in alkaline medium

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5

electrons in acidic medium

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Solution

The correct options are
A $3$ electrons in neutral medium
D $5$ electrons in acidic medium
Acidic medium: $M n O_{4}^{-} + 8 H^{+} + 5 e^{-} \to M n^{2 +} + 4 H_{2} O$
Neutral medium: $M n O_{4}^{-} + 4 H^{+} + 3 e^{-} \to M n O_{2} + 2 H_{2} O$ .

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M n O_{4}^{-}

M n O_{3}^{-}

C O_{2}

How many mole of electrons are involved in the reduction of one mole of $M n O_{4}^{-}$ ion in alkaline medium to $M n O_{4}^{- 2}$ ?

C O_{2}

K M n O_{4}

K = 39.09

M n = 54.94

O = 16.0

2 K M n O_{4} + 2 K O H \to 2 K_{2} {M n O}_{4} + H_{2} O + [O]

K M n O_{4}

K M n O_{4}

K M n O_{4}

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