In acidic medium MnO2-4-

The correct option is A disproportionates to MnO_2 and MnO^ _4 In acidic medium, #160;Reaction: 2MnO^2 _4 H^+→ MnO_2 + MnO^ _4 #160; #16 ...

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Byju's Answer

Standard XII

Chemistry

Potassium Permanganate

In acidic med...

Question

In acidic medium $M n O_{4}^{2 -}$ :

disproportionates to

M n O_{2}

and

M n O_{4}^{-}

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is oxidised to

M n O_{4}^{-}

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is reduced to

M n O_{2}

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is reduced to

M n^{2 +}

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Solution

The correct option is A disproportionates to $M n O_{2}$ and $M n O_{4}^{-}$
In acidic medium,
Reaction:
$2 M n O_{4}^{2 -}^{+} M n O_{2} + M n O_{4}^{-}$
+6 +4 +7

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Similar questions

100

mL of

H_{2} O_{2}

is oxidised by

100

mL of

0.01 M

K M n O_{4}

in acidic medium (

M n O_{4}^{⊝}

reduced to

M n^{2 +}

100

mL of the same

H_{2} O_{2}

is oxidised by

V

mL of

0.01 M

K M n O_{4}

in basic medium (

M n O_{4}^{⊝}

reduced to

M n O_{2}

). Hence,

V

is :

Q. Pyrolusite in

M n O_{2}

is used to prepare

K M n O_{4}

.
Steps are :

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

V O^{2 +}

is oxidised to

V O_{2}^{+}

M n O_{4}^{-}

in acidic medium, which in turn is reduced to

M n^{2 +}

. What is balanced equation for this reaction ?

100

mL of

1 M K M n O_{4}

oxidised

100

mL of

H_{2} O_{2}

in acidic medium (when

{M n O_{4}}^{-}

is reduced to

M n^{2 +}

). Volume of same

K M n O_{4}

required to oxidise

100

mL of

H_{2} O_{2}

in neutral medium (when

{M n O_{4}}^{-}

is reduced to

M n O_{2}

) will be:

M n O_{4}^{2 -}

can be converted to

M n O_{4}^{-}

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In acidic medium MnO2−4:

The correct option is A disproportionates to MnO2 and MnO−4In acidic medium, Reaction:2MnO2−4H+→MnO2+MnO−4 +6 +4 +7

In acidic medium $M n O_{4}^{2 -}$ :

The correct option is A disproportionates to $M n O_{2}$ and $M n O_{4}^{-}$
In acidic medium,
Reaction:
$2 M n O_{4}^{2 -}^{+} M n O_{2} + M n O_{4}^{-}$
+6 +4 +7