Question

When $Br{O}_3^{-}$ ion reacts with ${Br}^{-}$ in acid medium, ${Br}_2$ is liberated. The equivalent weight of ${Br}_2$ in this reaction is:

• A. $\frac{5M}{8}$
• B. $\frac{5M}{3}$
• C. $\frac{3M}{5}$
• D. $\frac{4M}{6}$

Answer 1

The correct option is C $\frac{3M}{5}$

Given,

$Br{O}_3^{-}$ reacts with $B{r}^{-}$ in acidic medium

The reaction involved is

$Br{O}_3^{-}+5B{r}^{-}+6H^{+}\leftrightharpoons 3B{r}_2+3H_{2}O$

Here, in $Br{O}_3^{-}$, $Br$ has $+5$ oxidation state & in $Br{O}_3^{-}⟶3B{r}_2$ reduction is $5$

Also, in $5B{r}^{-}⟶3B{r}_2$ oxidation is $5$

$\therefore$ n factor=$5$.

Now, we know, Equivalent weight=$\frac{Molecularweight}{nfactor}$

Equivalent weight of $3B{r}_2=\frac{3\times M}{5}=\frac{3M}{5}$