Question

Consider following reaction and select the correct statements:
${Br}_2+{OH}^{-}⟶Br{O}_3^{-}+{Br}^{-}+H_{2}O$ (unbalanced)

• A. Equivalent weight of ${Br}_2$, when it is reduced to ${Br}^{-}$ is $80$ g/equiv
• B. Equivalent weight of ${Br}_2$, when it is oxidised to $Br{O}_3$ is $96$ g/equiv
• C. Net equivalent weight of ${Br}_2$ is $96$ g/equiv
• D. It is disproportionation reaction

Answer 1

Answer: (A), (C), (D)

Equivalent weight of ${Br}_2$, when it is reduced to ${Br}^{-}$ is $80$ g/equiv
It is disproportionation reaction
Net equivalent weight of ${Br}_2$ is $96$ g/equiv

(A) The expression for the equivalent mass is : $\text{equivalent mass}=\frac{\text{Formula mass of oxidizing/reducing agent}}{\text{Total change in oxidation number of an element per molecule that undergoes reduction/oxidation}}$
$\frac{1}{2}\underset{0}{{Br}_2}⟶\underset{-1}{{Br}^{-}}$
When bromine is reduced, the oxidation number of bromine changes from 0 to -1. The change in the oxidation number of bromine atom is 1. Total change in oxidation number in bromine molecule is 2.
The equivalent mass of bromine when it is reduced is $=\frac{M}{2}=80$.
Thus, the option A is correct.

(B) When bromine is oxidized, the oxidation number of bromine changes from 0 to +5. The change in the oxidation number of bromine atom is 5. Total change in oxidation number in bromine molecule is 10.
The equivalent mass of bromine when it is oxidized is $=\frac{M}{2}=80$.
Thus, the option B is incorrect.

(C) Net equivalent weight is equal to the sum of the equivalent weight in oxidation part and the equivalent weight in reduction part.
It is equal to $16+80=96$.
Thus, the option C is correct.

(D) In this reaction bromine is reduced to bromide ion and oxidized to $Br{O}_3^{-}$. Hence, it is an disproportionation reaction.
Hence, the option D is correct.