Question

The equivalent weight for underlined molecule in the following reaction is M/n (M = molar mass of $P_4)$.
Find n.
${\underset{–}{P}}_4\to P{H}_3+H_{2}P{O}_2^{-}$

Answer 1

$P_4^0⟶\stackrel{-3}{P{H}_3}+3H_2\stackrel{+1}{P{O}_2^{-}}$

Reduction half reaction:

$P_4^0+12e^{-}⟶\stackrel{-3}{4P{H}_3}$

Therefore, n-factor is 12.

Oxidation half reaction:

$P_4^0⟶3H_2\stackrel{+1}{P{O}_2^{-}}+4e^{-}$

Therefore, n-factor is 4.

$\frac{M}{12}$ + $\frac{M}{4}$ = $\frac{M}{3}$

$\therefore$ n-factor in this reaction = 3