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Types of Redox Reactions

MnO42- 1 mole...

Question

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

2 / 3

mole

M n O_{4}^{-}

and

1 / 3

mole

M n O_{2}

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1 / 3

mole

M n O_{4}^{-}

and

2 / 3

mole

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 $M n O_{4}^{-}$ and $1 / 3$ mole $M n O_{2}$
Option (A) is correct.
$M n O_{4}^{2 -}$ (1 mole) in neutral aqueous medium disproportionate to $2 / 3$ mole $M n O_{4}^{-}$ and $1 / 3$ mole $M n O_{2}$ .
$3 K_{2} M n O_{4} + 2 H_{2} O ⟶ 2 K M n O_{4} + M n O_{2} + 4 K O H$

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Similar questions

M n O_{4}^{2 -}

(1 mole) in neutral aqueous medium is disproportionate to:

1

mole of

M n O_{4}^{2 -}

in neutral aqueous medium disproportionates to:

Q. Assertion :In the following reaction:

S O_{2} + 2 H_{2} S ⟶ 3 S + 2 H_{2} O

1

mole

S O_{2}

and

1

mole

H_{2} S

form

3

moles sulphur. Reason:

1

mole

S O_{2}

and

2

moles

H_{2} S

form

3

moles sulphur and

2

moles

H_{2} O

Q. 1 mole of

B a (O H)_{2}

will exactly neutralize

Q. How much charge is required to reduce:

(a)

1 m o l e

A l^{3 +}

A l

and

(b)

1 m o l e

M n O_{4}^{-}

M n^{2 +}

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MnO2−4 (1 mole) in neutral aqueous medium disproportionate to :

The correct option is A 2/3 mole MnO−4 and 1/3 mole MnO2Option (A) is correct.MnO2−4 (1 mole) in neutral aqueous medium disproportionate to 2/3 mole MnO−4 and 1/3 mole MnO2.3K2MnO4+2H2O⟶2KMnO4+MnO2+4KOH

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