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NEET Hard

Consider the change in oxidation state of Bromine corresponding to different emf values as shown in the diagram below: BrO4−→1.82VBrO3−→1.5VHBrO\text{BrO}_4^- \xrightarrow{1.82 \text{V}} \text{BrO}_3^- \xrightarrow{1.5 \text{V}} \text{HBrO} and Br−←1.0652VBr2←1.595VHBrO\text{Br}^- \xleftarrow{1.0652 \text{V}} \text{Br}_2 \xleftarrow{1.595 \text{V}} \text{HBrO}. Then the species undergoing disproportionation is

1

Br2\text{Br}_2

2

BrO4−\text{BrO}_4^-

3

BrO3−\text{BrO}_3^-

4

HBrO\text{HBrO}

Step-by-Step Solution

For HBrO→Br2\text{HBrO} \rightarrow \text{Br}_2, E∘=1.595VE^\circ = 1.595 \text{V}. For HBrO→BrO3−\text{HBrO} \rightarrow \text{BrO}_3^-, E∘=1.5VE^\circ = 1.5 \text{V}. Ecell∘E^\circ_{\text{cell}} for the disproportionation of HBrO=EHBrO/Br2∘−EBrO3−/HBrO∘=1.595−1.5=0.095V=+ve\text{HBrO} = E^\circ_{\text{HBrO/Br}_2} - E^\circ_{\text{BrO}_3^-/\text{HBrO}} = 1.595 - 1.5 = 0.095 \text{V} = +\text{ve}. Hence, HBrO\text{HBrO} undergoes disproportionation.

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