Consider the change in oxidation state of bromine corresponding to different EMF values as shown in the diagram below:
`BrO_(4)^(-) overset(1.82V)rarr BrO_(3)^(-) overset(1.5V)rarr HBrO overset(1.595V)rarr Br_(2) overset(1.0652V)rarr Br^(-)`
Then the species undergoing disproportionation is -

Options

A. `HBrO`
B. `BrO_(3)^(-)`
C. `Br_(2)`
D. `BrO_(4)^(-)`

Answer

A

Explanation

Both `BrO_(3)^(-) & HBrO` can undergo disproportionation reaction (in `BrO_(4)^(-) & Br^(-)`, bromine is in its highest and lowest oxidation states, respectively.)
For `BrO_(3)^(-),overset(+5)(BrO_(3)^(-))rarr overset(+7)(BrO_(4)^(-))+overset(+1)(HBrO)`
`E_("cell")^(@)=(1.5-1.82)=-0.32V lt0`
For `HBrO^(-), overset(+1)(HBrO)rarr overset(+5)(BrO_(3)^(-))+overset(0)(Br_(2))`
`E_("cell")^(@)=(1.595-1.5)=0.095V gt 0`
Thus, `HBrO` can undergo disproportionation reaction.