Question

In the reaction:
\[3B{{r}_{2}}+6CO_{3}^{2-}+3{{H}_{2}}O\to 5B{{r}^{-}}+BrO_{3}^{-}+6HCO_{3}^{-}\]
(A)- bromine is oxidised and carbonate is reduced
(B)- bromine is reduced and water is oxidised
(C)-bromine is neither reduced nor oxidised
(D)-bromine is both reduced and oxidised

Answer 1

Hint: By calculating the oxidation states of entities in the reaction, in order to find the change in oxidation states, by gaining or losing electrons during the reaction. We can determine the oxidised and reduced substance.

Complete step by step answer:
As we know that in a redox reaction, it involves the oxidation of one substance and the reduction of another substance taking place simultaneously. That is, as one substance (reducing agent) loses electrons, the other (oxidizing agent) accepts the electrons.
But, when both the reduction and the oxidation take place in the same substance, forming two products, that is, when the oxidation state of the substance gets oxidised and reduced into two half-reactions forming two new substances. This is known as the disproportionation reaction.
In the given reaction, it is a disproportionation reaction. It can be shown by taking the oxidation state of the central atoms, that is, bromine and carbon atom on both the sides.
The balanced reaction give as follows:
\[3B{{r}_{2}}+6CO_{3}^{2-}+3{{H}_{2}}O\to 5B{{r}^{-}}+BrO_{3}^{-}+6HCO_{3}^{-}\]
The oxidation state of bromine in $B{{r}_{2}}$, $B{{r}^{-}}$ and $BrO_{3}^{-}$ entities is zero, $(-1)$ and $(+5)$. Whereas, the oxidation state of the carbon atom in $CO_{3}^{2-}$ and $HCO_{3}^{-}$ is $(+4)$ in both the entities. So, it is neither reduced or oxidised.
Therefore, the disproportion takes place in the bromine atom having the two half-reactions as follows:
Oxidation half-reaction: $2B{{r}_{2}}\to BrO_{3}^{-}$ ,as it loses five electrons
Reduction half-reaction: $B{{r}_{2}}\to B{{r}^{-}}$,as it gains one electron

Therefore, in the given reaction option (D)- bromine is both reduced and oxidised.

Note: in this reaction, the $B{{r}_{2}}$molecule acts as both an oxidising and reducing agent, as it gets both reduced into $B{{r}^{-}}$and oxidised into $BrO_{3}^{-}$respectively.