Consider the change in oxidation state of Bromine corresponding to different emf values:${\text{BrO}}_4^{-}\stackrel{1.82\text{V}}{\to }{\text{BrO}}_3^{-}\stackrel{1.5\text{V}}{\to }\text{HBrO}\stackrel{1.595\text{V}}{\to }{\text{Br}}_2\stackrel{1.0652\text{V}}{\to }{\text{Br}}^{-}$ The species undergoing disproportionation is:

For a species to undergo disproportionation, the standard potential for reduction ($E_{right}^{\circ }$) must be greater than the standard potential for oxidation ($E_{left}^{\circ }$).For $\text{HBrO}$:$E_{red}^{\circ }(\text{HBrO}\to {\text{Br}}_2)=1.595\text{V}$.$E_{ox}^{\circ }(\text{HBrO}\to {\text{BrO}}_3^{-})=1.5\text{V}$.Since $1.595>1.5$, $\text{HBrO}$ will disproportionate.For ${\text{BrO}}_3^{-}$: $E_{right}^{\circ }(1.5)<E_{left}^{\circ }(1.82)$, so it won't.