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

Then the species undergoing disproportionation is

For a reaction to be spontaneous, ${E^o}_{\text{cell }}$ should be positive.

$\begin{array}{l}\mathrm{HBrO}⟶{\mathrm{Br}}_2 \mathrm{E}°=1.595 \mathrm{V},\mathrm{SRP}\left(\text{ cathode }\right)\\ \mathrm{HBrO}⟶{\mathrm{BrO}}_3^{-} \mathrm{E}°=-1.5 \mathrm{V},\mathrm{SOP}\left(\text{ anode }\right)\\ 2\mathrm{HBrO}⟶{\mathrm{Br}}_2+{\mathrm{BrO}}_3^{-}\end{array}$

$E_{\text{cell }}°=\text{ SRP (cathode) - SRP (anode) }$

             $=1.595-1.5 = 0.095\mathrm{V}$

$E_{\text{cell }}°>0\Rightarrow \Delta G°<0\text{ (spontaneous) }$