Calculate ${\mathrm{E}}_{\mathrm{cell}}^{\mathrm{o}}$ corresponding to each compound undergoing disproportionation reaction. The reaction for which ${\mathrm{E}}_{\mathrm{cell}}^{\mathrm{o}}$ comes out +ve is spontaneous.

$\text{HBrO}\to \text{B}{\text{r}}_2 {\text{E}}^{\text{o}}=1.595, \text{S}\text{R}\text{P} \left(\text{Cathode}\right)$
$2\text{HBrO}\to \text{B}{\text{r}}_2+\text{B}\text{r}{\text{O}}_3^{-}$
${\text{E}}_{\text{c}\text{e}\text{l}\text{l}}^{\text{o}}=$ SRP (cathode) - SRP (Anode)

$=1.595-1.5$

$=0.095\mathrm{ }\mathrm{V}$

${\text{E}}_{\text{cell}}^{\text{o}}>0\Rightarrow \mathrm{\Delta }{\text{G}}^o<0$ [spontaneous]