Question

Consider the change in oxidation state of Bromine corresponding to different emf values as shown in the diagram below :

Then the species undergoing disproportionation is

• A. \(\text{BrO}^{-}_3\)
• B. \(\text{Br}_2\)
• C. \(\text{BrO}_4^{-}\)/div>
• D. HBrO

Answer 1

The Correct Option is D

First, let's understand the concept of disproportionation. Disproportionation is a specific type of redox reaction in which a single substance is simultaneously oxidized and reduced, forming two different products.

In the context of Bromine compounds, we need to analyze the given emf values associated with the changes in oxidation states shown in the diagram:

• \(\text{BrO}_4^{-} \rightarrow \text{BrO}_3^{-}\) has an emf of \(1.82 \, \text{V}\).
• \(\text{BrO}_3^{-} \rightarrow \text{HBrO}\) has an emf of \(1.5 \, \text{V}\).
• \(\text{Br}_2 \rightarrow \text{Br}^-\) has an emf of \(1.0652 \, \text{V}\).
• \(\text{HBrO} \rightarrow \text{Br}_2\) has an emf of \(1.595 \, \text{V}\).

For a species to undergo disproportionation, an appropriate potential difference must exist. The species must be able to both oxidize and reduce itself into other substances.

Why HBrO undergoes disproportionation:

\(\text{HBrO}\) is an intermediate in the above reactions, with the potential of being reduced to \(\text{Br}_2\) (emf \(1.595 \, \text{V}\)) and also oxidized to \(\text{BrO}_3^{-}\) (since it forms in the same process via \(\text{BrO}_3^{-}\) with an emf of \(1.5 \, \text{V}\)). It can split itself into two different reactions forming separate compounds:

• Oxidation: \(\text{HBrO} \rightarrow \text{BrO}_3^{-}\)
• Reduction: \(\text{HBrO} \rightarrow \text{Br}_2 \rightarrow \text{Br}^-\)

Conclusion:

The values and the sequence of reactions imply that \(\text{HBrO}\) is capable of reacting with itself to produce different oxidation states of Bromine, thus undergoing disproportionation.

Correct answer: HBrO