Question

What will be the overall n-factor for the given disproportionation reaction:
$2CuBr\left(aq\right)\to CuB{r}_2\left(aq\right)+Cu\left(s\right)$

• A. 1
• B. 2
• C. 0.5
• D. 2.5

Answer 1

The correct option is C 0.5

For the given reaction :
$2\stackrel{+1}{C}uBr\left(s\right)⟶\stackrel{+2}{C}uB{r}_2\left(aq\right)+\stackrel{0}{C}u\left(s\right)$
Oxidation :
$\stackrel{+1}{C}u⟶\stackrel{+2}{C}{u}^{2+}$
$n_f=n_1=|1-\left(2\right)|\times 1=1$
Reduction :
$\stackrel{+1}{C}u⟶\stackrel{0}{C}{u}^{}$
$n_f=n_2=|1-\left(0\right)|\times 1=1$
$n_f=\frac{n_1\times n_2}{n_1+n_2}$
$n_f=\frac{1\times 1}{1+1}=0.5$