# Equivalent mass in case of disproportionation reaction

I came across a question where the equivalent mass of hypophosphorus acid is to be found in the given reaction: $$\ce{2H3\overset{+1}{P}O2 -> H3\overset{+5}{P}O4 + \overset{-3}{P}H3}$$

If $$M$$ is the molecular mass of $$\ce{H3PO2}$$ then $$E_\text{oxidation}=M/4$$ (for change in oxidation state from $$+1$$ to $$+5$$) and $$E_\text{reduction}=M/4$$ (for change in oxidation state from $$+1$$ to $$-3$$)

Equivalent mass of $$\ce{H3PO2}$$ in this disproportionation reaction

$$= E_\text{oxidation}+E_\text{reduction}= M/4+M/4= M/2$$

Now, I am not able to comprehend that, why these two types of equivalent mass has been added to arrive the equivalent mass of $$\ce{H3PO2}$$.

My take: The concept of equivalent mass collapses in this situation!

• in my understanding, there is no sense in the eq. mass addition. Consider formally both original molecules as different compounds. – Poutnik Sep 10 '19 at 12:41

In your case, you will have two equivalent weights one with respect to oxidation, and one respect to reduction. Solve the disporptionation reaction below and see what you get as an equivalent weight for $$\ce{Br2}$$ in each case.
$$\ce{3Br2 + 6OH^- -> 5Br^- + BrO3^- + 3H2O}$$