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

For $\text{HBrO} \rightarrow \text{Br}_2$, $E^\circ = 1.595 \text{V}$. For $\text{HBrO} \rightarrow \text{BrO}_3^-$, $E^\circ = 1.5 \text{V}$. $E^\circ_{\text{cell}}$ for the disproportionation of $\text{HBrO} = E^\circ_{\text{HBrO/Br}_2} - E^\circ_{\text{BrO}_3^-/\text{HBrO}} = 1.595 - 1.5 = 0.095 \text{V} = +\text{ve}$. Hence, $\text{HBrO}$ undergoes disproportionation.