Question

Given a balanced redox reaction:
$2Mn{O}_4^{⊝}+8H^{⨁}+B{r}_2\to 2M{n}^{2+}+2Br{O}_3^{⊝}+2H_{2}O$
If the molecular weights of $Mn{O}_4^{⊝}$ and $B{r}_2$ be $M_1$ and $M_2$ respectively, then:

• A. equivalent weight of $Mn{O}_4^{⊝}$ $=\frac{M_1}{5}$
• B. equivalent weight of $B{r}_2$ = $\frac{M_2}{10}$
• C. the $n$-factor ratio of $Mn{O}_4^{⊝}:B{r}_2$ is $1:1$
• D. none of the above

Answer 1

Answer: (A), (B)

equivalent weight of $Mn{O}_4^{⊝}$ $=\frac{M_1}{5}$
equivalent weight of $B{r}_2$ = $\frac{M_2}{10}$

Equivalent weight $=\frac{Molarmass}{n-factor}$

$Mn{O}_4^{-}+5e^{-}+8H^{+}\to M{n}^{2+}+4H_{2}O$

As in this reaction, $1$ mol of $Mn{O}_4^{-}$ gains $5$ electrons to get converted to $M{n}^{2+}$.
So, $n-$ factor $=5$.
Hence, the equivalent weight of $Mn{O}_4^{-}=\frac{M_1}{5}$. ( if $M_1=$ molecular weight of $Mn{O}_4^{-}$)

In the same way , we can see that $B{r}_2$ has $0$ oxidation state of $Br$, whereas in $Br{O}_3^{-}$ the oxidation state of $Br$ is $+5$.

$B{r}_2\to 2Br{O}_3^{-}+10e^{-}$
Hence, $n-$ factor $=10$ .

So, equivalent weight of $B{r}_2=\frac{M_2}{10}$. ($M_2=$ Molecular weight of $B{r}_2$)