Question

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

$Br{O}_4^{-}\stackrel{1.82V}{\to }Br{O}_3^{-}\stackrel{1.5V}{\to }HBrO\stackrel{1.595V}{\to }B{r}_2\stackrel{1.0652V}{\to }B{r}^{-}$

Then the species undergoing disproportionation is

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

Answer 1

The correct option is A $HBrO$

As, $Br{O}_4^{-}$ is present in its highest oxidation state $(+7),$ so its will not undergo disproportionation,

For, $Br{O}_3^{-}$
$\therefore \left(1\right)+\left(2\right)=\left(3\right)$
$2H_{2}O+2Br{O}_3^{-}\to 2Br{O}_4^{-}+4H^{+}+4e^{-}....\left(1\right),$
$E_1^0=-1.82$
$4e^{-}+5H^{+}+Br{O}_3^{-}HBrO+2H_{2}O...\left(2\right),E_2^0=1.5V$
$H^{-}+3Br{O}_3^{-}\to 2Br{O}_4^{-}+HBrO-\left(3\right)$

$E_3^0=\frac{-182\times 4+1.5\times 4}{4}$
$=-ve$

Also for $HBrO$

$4e^{-}+6H^{+}+2HBrO\to 2B{r}_2+4H_{2}O....\left(1\right),E_1^0=1.595$
$2H_{2}O+HBrO\to Br{O}_3^{-}+5H^{-}+4e^{-}....\left(2\right),E_2^0=-1.5V$
$\left(1\right)+\left(2\right)=3$
$H^{+}+3HBrO\to 2B{r}_2+2H_{2}O+Br{O}_3^{-}-\left(3\right)$

$\therefore E_3^0=\frac{1.595\times 4-1.5\times 4}{4}$
$=+ve$

So, $HBrO$ will disproportionate.