Question

The formula for calculation of n-factor $\left(n_f\right)$ for a disproportionation reaction is given as:

• A. $n_f=(n_f{)}_{\text{oxidation}}+(n_f{)}_{\text{reduction}}$
• B. $\frac{1}{n_f}=\frac{1}{(n_f{)}_{\text{oxidation}}}+\frac{1}{(n_f{)}_{\text{reduction}}}$
• C. $n_f=(n_f{)}_{\text{oxidation}}^2+(n_f{)}_{\text{reduction}}^2$
• D. $n_f=(n_f{)}_{\text{oxidation}}-(n_f{)}_{\text{reduction}}$

Answer 1

The correct option is B $\frac{1}{n_f}=\frac{1}{(n_f{)}_{\text{oxidation}}}+\frac{1}{(n_f{)}_{\text{reduction}}}$

The formula for calculation of n-factor$\left(n_f\right)$ for disproportionation reaction:
$n_f=\frac{(n_f{)}_{\text{oxidation}}\times (n_f{)}_{\text{reduction}}}{(n_f{)}_{\text{oxidation}}+(n_f{)}_{\text{reduction}}}$

Taking reciprocal both sides

$\frac{1}{n_f}=\frac{(n_f{)}_{\text{oxidation}}+(n_f{)}_{\text{reduction}}}{(n_f{)}_{\text{oxidation}}\times (n_f{)}_{\text{reduction}}}$
$\frac{1}{n_f}=\frac{(n_f{)}_{\text{oxidation}}}{(n_f{)}_{\text{oxidation}}\times (n_f{)}_{\text{reduction}}}+\frac{(n_f{)}_{\text{reduction}}}{(n_f{)}_{\text{oxidation}}\times (n_f{)}_{\text{reduction}}}$
Hence,

$\frac{1}{n_f}=\frac{1}{(n_f{)}_{\text{oxidation}}}+\frac{1}{(n_f{)}_{\text{reduction}}}$