Question

Consider the change in oxidation state of Bromine corresponding to different EMF values as shown in the diagram below.
Then the species undergo disproportionation is _________________.
$Br{O_4}^ - \xrightarrow{{1.82V}}Br{O_3}^ - \xrightarrow{{1.5V}}HBrO\xrightarrow{{1.595V}}B{r_2}\xrightarrow{{1.0652V}}B{r^ - }$
A.$B{r_2}$
B.$HOBr$
C. $Br{O_3}^ - $
D.$Br{O_4}^ - $

Answer 1

Hint: A chemical reaction in which a species is both oxidized as well as reduced is called disproportionation reaction. Consider an example:
$2{H_2}{O_2} \to {H_2}O + {O_2}$
In the above example, the oxidation state of oxygen in hydrogen peroxide ${H_2}{O_2}$is $\left( { - 1} \right)$ while that in water is $\left( { - 2} \right)$ and in oxygen it is 0. So there oxygen is both reduced to water by gaining electrons as well as oxidized to oxygen by losing electrons.

Complete step by step answer:
Now, in the given species, the different oxidation states of bromine are:
$Br{O_4}^ - \to \left( { + 7} \right)$ oxidation state, i.e. its highest oxidation state.
$Br{O_3}^ - \to \left( { + 5} \right)$ oxidation state, an intermediate oxidation state but its oxidation potential is too low for it to undergo oxidation.
$HOBr \to \left( { + 1} \right)$ oxidation state, an intermediate oxidation state. Oxidation and reduction potentials comparable.
$B{r_2} \to \left( 0 \right)$ oxidation state, i.e. its elementary form.
$B{r^ - } \to \left( { - 1} \right)$ oxidation state, i.e. its lowest oxidation state with a complete octet.
The bromine in $HOBr$ is present in $\left( { + 1} \right)$ oxidation state. It can oxidise further into $Br{O_3}^ - $ and can also reduce to form$B{r_2}$ . The reduction and oxidation potential values of $HOBr$ can be represented as:
$HOBr\xrightarrow{{ - 1.5V}}Br{O_3}^ - $
$HOBr\xrightarrow{{1.595V}}B{r_2}$
When $HOBr$ is converted to, $Br{O_3}^ - $ the EMF value is $ - 1.5{\text{ V}}$ and when $HOBr$ is converted to $B{r_2}$ , the value of the EMF is $1.595{\text{ V}}$.
Hence, the resultant EMF from both is $0.095{\text{ V}}$ which is a positive EMF value, suggesting a favourable reaction.

Hence, the right answer is option (B) $HOBr$.

Note:
If a reduction diagram is given in the question, we can use a shortcut trick to find out the species which will undergo a disproportionation reaction. The species whose reduction potential on the right is more than the reduction potential on the left, that species will be unstable and undergo disproportionation reaction.