Question

For the given reaction:
$2CuBr\left(aq\right)\to CuB{r}_2\left(aq\right)+Cu\left(s\right)$
What will be the equivalent mass of $CuBr$ if molecular weight of $CuBr$ is $M$?

• A. $M$
• B. $0.5M$
• C. $2M$
• D. $0.25M$

Answer 1

The correct option is C $2M$

For the given disproportionation reaction :
$2\stackrel{+1}{C}uBr\left(s\right)⟶\stackrel{+2}{C}uB{r}_2\left(aq\right)+\stackrel{0}{C}u\left(s\right)$
n-factor calculation,
$n_f=(|O.S{.}_{Product}-O.S{.}_{Reactant}|\times \text{number of atoms}$
Oxidation :
$\stackrel{+1}{C}u⟶\stackrel{+2}{C}{u}^{2+}$
$n_f=n_1=|1-\left(2\right)|\times 1=1$
Reduction :
$\stackrel{+1}{C}u⟶\stackrel{0}{C}{u}^{}$
$n_f=n_2=|1-\left(0\right)|\times 1=1$
Valency factor in disproportionation reaction is given by
$\frac{1}{n_f}=\frac{1}{(n_f{)}_{\text{oxidation}}}+\frac{1}{(n_f{)}_{\text{reduction}}}$
$\frac{1}{n_f}=\frac{1}{1}+\frac{1}{1}$
$\frac{1}{n_f}=\frac{1+1}{1\times 1}$
$n_f=\frac{1\times 1}{1+1}=0.5$
$\text{Equivalent mass}=\frac{\text{Molecular mass of}CuBr}{\text{n-factor}}\text{Equivalent mass}=\frac{M}{0.5}=2M$

Equivalent mass :

Equivalent mass or gram equivalent mass =$\frac{Molarmass}{n-factor}$

Earlier it was expressed by taking hydrogen, oxygen, chlorine and silver as the reference elements.

Number of parts by mass of an element which reacts or displaces from a compound 1.008 parts by mass of hydrogen, 8 parts by mass of oxygen and 35.5 parts by mass of chlorine, is known as the equivalent mass of that element.