Question

$BrO_{3}^{-}$ ion reacts with $B{{r}^{-}}$ to form $B{{r}_{2}}$, in acid medium. The equivalent mass of $B{{r}_{2}}$ in this reaction is:
A. $\dfrac{4M.w.}{6}$
B. $\dfrac{3M.w.}{5}$
C. $\dfrac{5M.w.}{3}$
D. $\dfrac{5M.w.}{8}$

Answer 1

Hint: We will find the equivalent mass of $B{{r}_{2}}$ in this reaction by the formula of equivalent weight, which is the molar mass divided by the n-factor. n-factor is basically the number of electrons lost and gained in the disproportionation reaction, here the number of these two should be the same.

Complete Solution :
 -As we are being provided with the information that $BrO_{3}^{-}$ ion reacts with $B{{r}^{-}}$ to form $B{{r}_{2}}$, in acid medium. We can write the reaction as:
\[2BrO_{3}^{-}+12{{H}^{+}}+10{{e}^{-}}\to B{{r}_{2}}+6{{H}_{2}}O\]
Here, ${{n}_{1}}$ = 10
\[2B{{r}^{-}}\to B{{r}_{2}}+6{{e}^{-}}\]
Here, ${{n}_{2}}$ = 2

- We can see here that this reaction is a disproportionation reaction or we can say redox reaction, where both the oxidation as well as reduction reaction takes place.
- Now, we will find out the n-factor for the above reaction by the formula:
$\begin{align}
 & n-factor=\dfrac{{{n}_{1}}\times {{n}_{2}}}{{{n}_{1}}+{{n}_{2}}} \\
 & =\dfrac{10\times 2}{10+2} \\
 & =\dfrac{20}{12} \\
 & =\dfrac{5}{3} \\
\end{align}$

- Now, we will find the equivalent weight, which is the molar mass divided by the n-factor.
$\begin{align}
  & =\dfrac{M.w.}{n-factor} \\
 & =\dfrac{M.w.}{5/3} \\
 & =\dfrac{3M.w.}{5} \\
\end{align}$

- Hence, we can conclude that the correct option is (b), that is when $BrO_{3}^{-}$ ion reacts with $B{{r}^{-}}$ to form $B{{r}_{2}}$, in acid medium. The equivalent mass of $B{{r}_{2}}$ in this reaction is $\dfrac{3M.w.}{5}$
So, the correct answer is “Option B”.

Note: - As we know that equivalent weight of any compound depends upon the chemical reaction in which it takes place. Basically, equivalent weight is a relative quantity, hence it does not have a unit.
- Whenever, we express the equivalent weight of substance in a unit of grams, then we call it the Gram Equivalent Weight.