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!

  • 2
    $\begingroup$ in my understanding, there is no sense in the eq. mass addition. Consider formally both original molecules as different compounds. $\endgroup$ – Poutnik Sep 10 '19 at 12:41

I like and dislike the question. I assume this is from your Indian university entrance exams. The good part is that it is tricky, the sad part is that the concept of equivalents is so outdated that any North American student will barely recognize it. They should abolish teaching equivalents exactly because this problem you are encountering.

Fact number no. 1, unlike molar equivalents of a given molecule equivalent weights are reaction dependent. Hence addition of equivalent weights is irrelevant. Equivalent weights are useful in analytical chemistry because they simplify calculations. The case of disproportionation reactions and their equivalents is rarely discussed because nobody uses disproportionation reactions in titrations, gravimetry etc.

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.

$\ce{3Br2 + 6OH^- -> 5Br^- + BrO3^- + 3H2O}$


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