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Consider the change in oxidation state of Bromine corresponding to different emf values as shown in the diagram below: $BrO_4^- \xrightarrow{1.82\ V} BrO_3^- \xrightarrow{1.5\ V} HBrO \xrightarrow{1.595\ V} Br_2 \xrightarrow{1.0652\ V} Br^-$. Then the species undergoing disproportionation is
Explanation
A species disproportionates when $E^\circ$ to its right exceeds $E^\circ$ to its left.
Detailed Solution
$\overset{+1}{HBrO} \rightarrow \overset{0}{Br_2}$, $E^\circ_{HBrO/Br_2} = 1.595$ V
$\overset{+1}{HBrO} \rightarrow \overset{+5}{BrO_3^-}$, $E^\circ_{BrO_3^-/HBrO} = 1.5$ V
$E^\circ_{cell}$ for the disproportionation of HBrO $= E^\circ_{HBrO/Br_2} - E^\circ_{BrO_3^-/HBrO}$
$= 1.595 - 1.5 = 0.095$ V (positive)
Hence HBrO undergoes disproportionation.