Question

Consider the change in oxidation state of Bromine corresponding to different emf values as shown in the diagram below : is
Then the species undergoing disproportionation is

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

Answer 1

Answer: (A)

$\stackrel{+1}{\mathrm{HBrO}}⟶B\stackrel{0}{r_2},E_{HBrO/B{r}_2}^0=1.595V$

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

$E_{\text{cell }}^{\circ }$ for the disproportionation of HBrO,

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

$=1.595-1.5$

$=0.095V=+ve$