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 species to undergo disproportionation,
$E_{cell}=+ve;\mathrm{\Delta }G=-ve$
Disproportionation of HBrO :
$\begin{array}{l}HBrO\to {BrO}_3^{-}+{Br}_2\\ E_{cell}=E_c-E_a\\ =E_{HBro/{Br}_2}-E_{HBrO/{BrO}_3^{-}}\\ =1.595-1.5\\ =0.095V\\ E_{\text{cell }}=+ve\text{ and }\Delta G=-ve\end{array}$
HBrO disproportionation is Spontaneous
(1) Disproportionation of ${\mathrm{BrO}}_3^{-}$:
$\begin{array}{l}{\mathrm{BrO}}_3^{-}\to {\mathrm{BrO}}_4^{-}+\mathrm{HBrO}\\ {\mathrm{E}}_{\mathrm{cell}}={\mathrm{E}}_{\mathrm{c}}-{\mathrm{E}}_{\mathrm{a}}={\mathrm{E}}_{{\mathrm{BrO}}_3^{-}/\mathrm{HBrO}}-{\mathrm{E}}_{{\mathrm{Bro}}_4^{-}/{\mathrm{BrO}}_3^{-}}=1.5-1.82=-0.32\mathrm{V}\end{array}$
Therefore ${\mathrm{BrO}}_3^{-}$ does not undergo disproportionation reaction.
(2) ${\mathrm{BrO}}_4^{-}$ does not undergo disproportionation as the maximum oxidation state of Br = +7
(3) Disproportionationof Br₂
$\begin{array}{l}{Br}_2\to HBrO+{Br}^{-}\\ E_{\text{cell }}=E_c-E_a\\ =E_{{Br}_2/{Br}^{-}}-E_{HBro/{Br}_2}\\ =1.0653-1.595\\ =-0.5298V\end{array}$
$E_{cell}=-ve$ no disproportionation reaction.