Question

Consider the following reaction and select the correct statements:
$B{r_2} + O{H^ - }\xrightarrow{{}}Br{O_3}^ - + B{r^ - } + {H_2}O$
A.Equivalent weight of $B{r_2},$ when it is reduced to $B{r^ - }$ is $80\,g/equiv.$
B.Equivalent weight of $B{r_2},$ when it is oxidized to $B{r^ - }$ is $96\,g/equiv.$
C.Net equivalent weight of $B{r_2}$ is $96\,g/equiv.$
D.It is a disproportionation reaction.

Answer 1

We can calculate the equivalent mass by using formula mass of oxidizing/reducing agent divided by the total change in oxidation number of an element per molecule, which undergoes reduction/oxidation.

Complete step by step answer:
We can write the formula for the equivalent mass as,
$Equivalent\,mass = \dfrac{{Formula\,mass\,of\,oxidizing/reducing\,agent}}{{Total\,change\,in\,oxidation\,number\,per\,molecule,which\,undergoes\,reduction/oxidation}}$
The equation is,
$\dfrac{1}{2}B{r_{_02}} \to B{r_{ - 1}}^ -$
On reducing bromine, the oxidation number of bromine goes from ${\text{0}}$ to ${\text{ - 1}}{\text{.}}$ The change in the oxidation number of bromine atom is ${\text{ + 1}}{\text{.}}$ Therefore, the total change in oxidation number in bromine molecule is ${\text{2}}{\text{.}}$ We can give the equivalent mass of reduced bromine as $\dfrac{M}{2} = 80.$ Therefore, the option (A) is correct.
On oxidizing bromine, the oxidation number of bromine goes from ${\text{0}}$ to ${\text{ + 5}}{\text{.}}$ The change in the oxidation number of bromine atom is ${\text{ + 5}}{\text{.}}$ Therefore, the total change in oxidation number in bromine molecule is ${\text{10}}{\text{.}}$ We can give the equivalent mass of oxidized bromine as $\dfrac{M}{2} = 80.$ So, the option (B) is correct.
The sum of the equivalent weight in oxidation half and the equivalent weight is reduction half is equal to the net equivalent weight.
We can give the net equivalent as $16 + 80 = 96.$ Therefore, the option (C) is correct.
In the given reaction, bromine is reduced to bromide ion and bromine is oxidized to $Br{O_3}^ - .$ Hence, it is a disproportionation reaction.
$\therefore$Option (D) is correct.

Note:
We know that disproportion reactions are otherwise called redox reactions. In redox reactions, the compound goes oxidation and reduction simultaneously. Some of the common examples of redox reactions are fruit browning, respiration, rusting of metals, and photosynthesis. Oxidizing agents transfer oxygen to another species, and reducing agents remove oxygen from another species.