Tell also disproportion             

Consider the change in oxidation state of Bromine corresponding to different emf values as shown in the diagram below.       

Then the species undergoing disproportionation is                 
^{$$\begin{gathered} \left(\mathrm{i}\right) {\mathrm{BrO}}_3^{–} \\ \left(\mathrm{ii}\right) {\mathrm{BrO}}_4^{–} \\ \left(\mathrm{iii}\right) \mathrm{HBrO} \\ \left(\mathrm{iv}\right) {\mathrm{Br}}_2 \end{gathered}$$}
 

Dear Student,

From the above equations it is only HBrO which undergoes both oxidation as well reduction. Other species undergo only reduction.

​​​​​The reactions are - HBrO → BrO₃^-  E = 1.5 (reverse) ,HBrO → Br₂  E = 1.595
                                       +1        +5                                        +1             0

             E^∘cell = E^∘_HBro/Br2 -  E^∘_BrO3/HBrO = 1.595 - 1.5 = 0.095 V

Regards.