1. What is the equivalent weight of white phosphorus () in the following disproportionation reaction — Electrochemistry Chemistry Question

💡 Solution & Explanation

In this reaction, $P_4$ (0) oxidises to $H_2PO_2^-$ (+1) losing 4 electrons ($n_1=4$) and reduces to $PH_3$ (-3) gaining 12 electrons ($n_2=12$). The net n-factor is $\frac{n_1 \times n_2}{n_1 + n_2} = \frac{48}{16} = 3$. Hence, Equivalent weight = M/3.