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For the following disproportion reaction $3Br_2+6OH^{-}\rightarrow 5Br^-+BrO_3^-+3H_2O$. Correct statement below is (Given $Br_2$ molecular mass is 160)

$\begin{array}{1 1}(a)\;\text{equivalent weight of }Br_2\;\text{in the net reaction is 96}\\(b)\;\text{equivalent weight of }Br_2\;\text{when it is reduced to Br is 80}\\(c)\;\text{equivalent weight of }Br_2\;\text{when it is oxidized to }BrO_3^-\text{ is 16}\\(d)\;\text{all the above are correct}\end{array}$

1 Answer

To obtain $2Br^-$ from $Br_2$
$Br_2+2e^-\rightarrow 2Br^-$
$\therefore$ Equivalent weight=$\large\frac{160}{2}$
$\Rightarrow 80$
To obtain $BrO_3^-$ from $Br_2$
$Br_2\rightarrow 2Br^{+5}+10e^-$
Net oxidation state changed '5'.Total number of electrons involved in the net reaction $(n)=10$
$\therefore$ Equivalent weight=$\large\frac{m.wt}{n}$
$\Rightarrow \large\frac{160}{10}$
$\Rightarrow 16$
Hence (d) is the correct answer.
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