Question

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

• A. $B{r}_2$
• B. $Br{O}_4^{-}$
• C. $Br{O}_3^{-}$
• D. HBrO

Answer 1

Answer: (D)

$\stackrel{+1}{\mathrm{HBrO}}\to \stackrel{0}{B{r}_2},E_{HBrO/B{r}_2}^0=1.595V$

$\stackrel{+1}{\mathrm{HBrO}}\to \stackrel{+5}{\mathrm{Br}}{O}_3^{-},E_{Br{O}_3^{-}/HBrO}^0=1.5V$

$E_{cell}^0$ for the disproportionation of HBrO,

$E_{cell}^0=E_{HBrO/B{r}_2}^0-E_{Br{O}_3^{-}/HBrO}^0$

$=1.595-1.5=0.095V=+ve$