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Topic 4 : Disproportionation Reaction

1. Which of the following reactions are disproportionation reactions?⇒ (2019)

(I) $C u^{+} \to C u^{2 +} + C u$
(II) $3 M n O_{4}^{-} + 4 H^{+} \to 2 M n O_{4}^{-} + M n O_{2} + 2 H_{2} O$
(III) $2 K M n O_{4} \to K_{2} M n O_{4} + M n O_{2} + O_{2}$
(IV) $2 M n O_{4}^{-} + 3 M n^{2 +} + 2 H_{2} O \to 5 M n O_{2} + 4 H^{+}$

A. (I), (II)

B. (I), (II), (III)

C. (II), (III), (IV)

D. (I), (IV)

Correct option is (A)

Disproportionation reaction are the reactions in which the same element/compound get oxidized and reduced simultaneously.

Since I and II reactions are as follows

$2 C u^{+} \to C u^{2 +} + C u^{0}$

And

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

Other two reactions are correct and belong to comproportionation reaction.

Thus option A is correct.

2. Consider the change in oxidation state of Bromine corredponding to different emf values as shown in the diagram below :
Consider the change in oxidation state of Bromine corredponding to different emf values as shown in the diagram below : . The the species undergoing dispropprtionation is . .
The the species undergoing dispropprtionation is .⇒ ()

A. $B r_{2}$

B. $B r O_{4}^{-}$

C. $B r O_{3}^{-}$

D. HBrO

The correct option is (D)

$H B r O$
$+ 1 H B r O \to 0 B r_{2}, E_{\frac{H B r O}{B r_{2}}}^{0} = 1.595 V$
$+ 1 H B r O \to + 5 B r O_{3}^{-}, E_{\frac{B r O_{3}^{-}}{H B r O}}^{0} = 1.5 V$
$E_{c e l l}^{0}$ for the disproportionation of $H B r O$ .
$E_{c e l l}^{0} = E_{\frac{H B r O}{B r_{2}}}^{0} - E_{\frac{B r O_{3}^{-}}{H B r O}}^{0}$
=1.595 -1.5
=0.095 V = +ve

⇒Redox

Topic 4 : Disproportionation Reaction