Question

In the reaction, ${\text{P}}_{\text{2}}{\text{H}}_{\text{4}}\to {\text{PH}}_{\text{3}}+{\text{P}}_{\text{4}}{\text{H}}_{\text{2}}$, calculate the given equivalent mass of ${\text{P}}_{\text{2}}{\text{H}}_{\text{4}}$. (at mass P = 31)

Answer 1

$P_2{H}_4⟶P{H}_3+P_4{H}_2$

Oxidation state of $P$ in $P_2{H}_4\Rightarrow 2x+4=0$

$\Rightarrow x=-2$

Oxidation state of $P$ in $P_4{H}_2\Rightarrow 4x+2=0$ (Oxidation)

$\Rightarrow x=-\frac{1}{2}$

Oxidation state of $P$ in $P{H}_3\Rightarrow -3$ (Reduction)

This is a disproportionation reaction.

$P_2{H}_4$ $n_{factor}$ in oxidation $=2|(-2+\frac{1}{2})|=3$

$P_2{H}_4$ $n_{factor}$ in reduction $=2|(-2+3)|=2$

Net $n_{factor}=\frac{n_{red}\times n_{oxid}}{n_{red}+n_{oxid}}=\frac{3\times 2}{3+2}=\frac{6}{5}$

$E_{equ}=n_{factor}\times m$

$=\frac{31\times 5}{6}$

$=25.833$