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Types of Redox Reactions

3MnO2- + 4H+⟶...

Question

${3 M n O}^{2 -} + 4 H^{+} ⟶ {2 M n O_{4}}^{-} + M n O_{2} + 2 H_{2} O$

The reaction is :

Combustion

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Redox

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Double displacement

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Disproportionation

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Solution

The correct options are
B Redox
D Disproportionation
The term "redox" comes from two concepts involved in electron transfer: reduction and oxidation.
A disproportionation reaction is a reaction in which a substance is simultaneously oxidized and reduced, giving two different products.
$M n O_{4}^{2 -} (V I)$ is oxidised to ${M n O}_{4}^{-} (V I I)$ and reduced to $M n O_{2} (I V)$ .
So in the above reaction, both oxidation and reduction were done, so it is redox and disproportionation reaction.

Hence, option B & D are correct.

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Similar questions

2 C u^{+}

\to

C u^{2 +}

C u^{0}

3 M n O_{4}^{2 -}

4 H^{+}

\to

2 M n O_{4}^{-}

M n O_{2}

2 H_{2} O

2 K M n O_{4}

△ - \to

K_{2} M n O_{4}

M n O_{2}

O_{2}

2 M n O_{4}^{-}

3 M n^{2 +}

2 H_{2} O

⟶

5 M n O_{2}

4 H^{+}

C u^{+} \to C u^{2 +} + C u

3 M n O_{4}^{- 2} + 4 H^{+} \to 2 M n O_{4}^{-} + M n O_{2} + 2 H_{2} O

2 K M n O_{4} \to K_{2} M n O_{4} + M n O_{2} + O_{2}

2 M n O_{4}^{-} + 3 M n^{+ 2} + 2 H_{2} O \to 5 M n O_{2} + 4 H^{+}

(a) 2 C u^{+} \to C u^{2 +} + C u (b) 3 M n O_{4}^{2 -} + 4 H^{+} \to 2 M n O_{4}^{-} + M n O_{2} + 2 H_{2} O (c) 2 K M n O_{4} \to K_{2} M n O_{4} + M n O_{2} + O_{2} (d) 2 M n O_{4}^{-} + 3 M n^{2 +} + 2 H_{2} O \to 5 M n O_{2} + 4 H^{+}

3 M n O_{4}^{2 -} + 4 H^{+} ⟶ M n O_{2} + 2 M n O_{4}^{-} + 2 H_{2} O

2 C u^{+}

\to

C u^{2 +}

C u^{0}

3 M n O_{4}^{2 -}

4 H^{+}

\to

2 M n O_{4}^{-}

M n O_{2}

2 H_{2} O

2 K M n O_{4}

△ - \to

K_{2} M n O_{4}

M n O_{2}

O_{2}

2 M n O_{4}^{-}

3 M n^{2 +}

2 H_{2} O

⟶

5 M n O_{2}

4 H^{+}

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