Consider the change in oxidation state of bromine corresponding to the standard reduction (electrode) potentials shown for each successive step: $\mathrm{BrO}{}_4{}^{-}$ $\stackrel{+1.82\text{V}}{\to }$ $\mathrm{BrO}{}_3{}^{-}$ $\stackrel{+1.50\text{V}}{\to }$ $\mathrm{HBrO}$ $\stackrel{+1.595\text{V}}{\to }$ $\mathrm{Br}{}_2$ $\stackrel{+1.0652\text{V}}{\to }$ $\mathrm{Br}{}^{-}$ Then the species undergoing disproportionation is

Solution & Explanation

\textbf{Answer:} (D) $\mathrm{HBrO}$ \textbf{Solution:} A species disproportionates (simultaneous oxidation and reduction of the same element) only if the resulting cell potential is positive, i.e. the potential of the step to its right (reduction) is greater than the potential of the step to its left (oxidation). For $\mathrm{HBrO}$: Reduction step (to its right): $\mathrm{HBrO}\to \mathrm{Br}{}_2$, $E^{\circ }=+1.595\text{V}$ (cathode). Oxidation step (to its left): $\mathrm{BrO}{}_3{}^{-}\to \mathrm{HBrO}$, $E^{\circ }=+1.50\text{V}$ (anode, reversed). $E_{cell}^{\circ }=E_{cathode}^{\circ }-E_{anode}^{\circ }=1.595-1.50=+0.095\text{V}>0$ Since $E_{cell}^{\circ }$ is positive, $\mathrm{HBrO}$ spontaneously disproportionates: $5\mathrm{HBrO}\to 2\mathrm{Br}{}_2+\mathrm{BrO}{}_3{}^{-}+\dots$ giving $\mathrm{Br}{}_2$ and $\mathrm{BrO}{}_3{}^{-}$. Hence the answer is $\mathrm{HBrO}$.