Reduction of the metal centre in aqueous permangenate 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
$K M n O_{4}$ converts to $M n O_{2}$ in neutral medium. So, change in oxidation state here is $3$

$K M n O_{4}$ converts to $M n^{2 +}$ . Change in oxidation state of $M n$ is $5$ .(i.e. from $+ 7$ to $+ 2$ )

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M n O_{4}^{-}

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

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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]

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K M n O_{4}

K M n O_{4}

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