Consider the change in oxidation state of Bromine corredponding to different emf values as shown in the diagram below :
.
The species undergoing disproportionation is .

Options

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

Answer

D

Explanation

Key Concept The reaction in which same species is oxidised as well as reduced is called disproportionation reaction. Firstly, calculate the value of `E_("cell")^(@)` of each species undergoing disproportionation reaction. The reaction whose `E_("cell")^(@)` value is positive will be feasible (spontaneous).
(i) Given, `BrO_(3)^(-) to HBrO, E_(BrO_(3)^(-)//HBrO)=1.5V`
`BrO_(3)^(-) to BrO_(4)^(-), E_(BrO_(3)^(-)//BrO_(4)^(-))=-1.82V`
`therefore 2overset(+5)(BrO_(3)^(-)) to overset(+1)(HBrO)+overset(+7)(B)rO_(4)^(-)`
`E_("cell")^(@)=E_("red")^(@)+E_(ox)^(@)`
`E_(BrO_(3)^(-)//HBrO)^(@)+E_(BrO_(3)^(-)//BrO_(4)^(-))^(@)`
`=1.5-1.82=-0.32 V`
`(ii) overset(+1)(HBrO) to overset(0)(Br_(2)), E_(HBrO//Br_(2))^(@)=1.595 V`
`overset(+1)(HBrO)to overset(+5)(BrO_(3)^(-)), E_(HBrO//BrO_(3)^(-))^(@)=-1.5V`
`overset(+1)(2HBrO)to overset(0)(Br_(2))+overset(+5)(BrO_(3)^(-))`
`E_("cell")^(@)=E_(Br_(2)//Br^(-))+^(@)+E_(Br_(2)//HBrO)^(@)`
`=1.0652-1.595=-0.5298V`
`therefore` Among the given options, only HBrO undergoes disproportionation.