Consider the change in oxidation state of Bromine corresponding to different emf values as shown in the diagram below:

Then the species undergoing disproportionation is

$$\begin{gathered} \mathrm{Calculate} {\mathrm{E}}_{\mathrm{cell}}^0 \mathrm{corresponding} \mathrm{to} \mathrm{each} \mathrm{compound} \mathrm{undergoing} \mathrm{disproportionation} \mathrm{reaction}. \\ \mathrm{The} \mathrm{reaction} \mathrm{for} \mathrm{which} {\mathrm{E}}_{\mathrm{cell}}^0 \mathrm{comes} \mathrm{out} \mathrm{to} \mathrm{be} +\mathrm{ve} \mathrm{is} \mathrm{spontaneous}. \\ \mathrm{HBrO}\stackrel{ }{\to }{\mathrm{Br}}_2; \mathrm{E}º=1.595,\mathrm{SRP}\left(\mathrm{cathode}\right) \\ \mathrm{HBrO}\stackrel{ }{\to }{\mathrm{BrO}}_3^{-}; \mathrm{E}º=-1.5\mathrm{V},\mathrm{SOP}\left(\mathrm{Anode}\right) \\ 2\mathrm{HBrO}\stackrel{ }{\to }{\mathrm{Br}}_2+{\mathrm{BrO}}_3^{-} \\ {\mathrm{E}}_{\mathrm{cell}}^0=\mathrm{SRP}\left(\mathrm{cathode}\right)-\mathrm{SRP}\left(\mathrm{Anode}\right) \\ =1.595-1.5 \\ =0.095\mathrm{V} \\ {\mathrm{E}}_{\mathrm{cell}}^0>0\Rightarrow \mathrm{\Delta G}º<0\left(\text{spontaneous}\right) \end{gathered}$$