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Potassium Permanganate

MnO2-4 1 mole...

Question

$M n O_{4}^{2 -}$ (1 mole) in neutral aqueous medium is disproportionate to:

2/3 mole of

M n O_{4}^{-}

and 1/3 mole of

M n O_{2}

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1/3 mole of

M n O_{4}^{-}

and 2/3 mole of

M n O_{2}

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1/3 mole of

M n_{2} O_{7}

and 1/3 mole of

M n O_{2}

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2/3 mole of

M n_{2} O_{7}

and 1/3 mole of

M n O_{2}

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Solution

The correct option is A 2/3 mole of $M n O_{4}^{-}$ and 1/3 mole of $M n O_{2}$
${3 M n O}_{4}^{2 -} n e u t r a l a q e o u s ⟶ m e d i u m {M n O}_{2} + 2 {M n O}_{4}^{-} + O_{2}$
${M n O}_{4}^{2 -} n e u t r a l a q e o u s ⟶ m e d i u m \frac{1}{3} {M n O}_{2} + \frac{2}{3} {M n O}_{4}^{-} + \frac{1}{3} O_{2}$
$\Rightarrow$ In neutral aqeous medium, ${M n O}_{4}^{2 -}$ disproportionates to give $\frac{1}{3}$ mole ${M n O}_{2}$ and $\frac{2}{3}$ mole ${M n O}_{4}^{-}$ .

Suggest Corrections

Similar questions

1

M n O_{4}^{2 -}

M n O_{2} + 4 H C l ⟶ M n C l_{2} + 2 H_{2} O + C l_{2}

2

M n O_{2}

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11.2

C l_{2}

C l_{2}

A^{n +}

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2.68 \times 10^{- 3}

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M n O^{-}

How much charge is required for the following reductions :

(i) 1 mole of $A l^{3 +}$ to Al ?

(ii) 1 mole of $C u^{2 +}$ to Cu ?

(iii) 1 mole of $M n O_{4}^{-} t o M n^{2 +}$ ?

B a (O H)_{2}

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