Let us say that we have the following two reactions:
$$\ce{A + O2 -> B}$$ where $\ce A \text{ and } \ce B $ are different oxides of metal $\ce M$.
Weights of $\ce A$ and $\ce B$ are $\pu{5.72 g}$ and $\pu{6.36 g}$.
$$\ce{B + H2->M + H2O}$$
The weights of $\ce B$ and $\ce M$ are $\pu{4.77 g}$ and $3.81$ g, respectively.

So I was trying to infer some information from the given data and this is what I did:

In the first reaction we can calculate the equivalent weight of $\ce A$ and $\ce B$ to be $\pu{71.5 g}$ and $\pu{79.5 g}$ respectively.
In the second reaction we can calculate the equivalent weight of $\ce B$ and $\ce M$ to be $\pu{39.75 g}$ and $\pu{31.7 g}$ respectively.

If we compare the equivalent weight of $\ce B$ in the two reactions, the equivalent weight of $\ce B$ in the first reaction is twice the equivalent weight in second reaction. What exactly does this mean?

  • $\begingroup$ Should the second equation be $\ce{B + 2nH+ -> M + nH2O}$? Otherwise, the second equation will have mass imbalance. $\endgroup$
    – Ben Norris
    May 14, 2014 at 13:31
  • $\begingroup$ @BenNorris Since I'm working with equivalents, I don't need to balance the equation. Although the edited question has H2 instead of H2O. $\endgroup$ May 14, 2014 at 13:40

1 Answer 1


Equivalent weights are considering a redox equation with one mole equivalent of electrons. But some redox reactions require more than one mole equivalent of electrons to be equilibrated.

Thus, if you have the second reaction that leads to a weight equivalent of $1/2$ of that for the first reaction, it means that the reaction 2 has consumed $1/2$ of $B$ compared to reaction 1 for one mole equivalent of electrons.

You still cannot proceed to the molecular weights on these basis, but you can now try to consider that the most likely case is that the first one involved only one electron and the second one two electrons. Thus you can write that this is consistent with:

First equation:

$$\ce{A \rightarrow B + e-}$$

(equilibrated by $\ce{\frac 14 O_{2} + e- \rightarrow {1\over 2} O^2-}$).

And for the second equation:

$$\ce{B + 2 e- -> M}$$

(equilibrated with $\ce{H2 + O^2- -> H2O + 2 e-}$).

One this is said, one should consider the molar weight of $\ce M$ to be twice the equivalent weight of $\ce M$ calculated from the second equation, leading to a molar weight of 63.5 g/mol. Quite unsurprisingly, this is the molar weight of copper, so $M = Cu$. And going backwards, $\ce{B = CuO}$ with copper in its second oxidation state. We can check this, as the molecular weight of CuO is 76.5 g/mol, exactly the equivalent weight calculated for reaction 1. And thus $\ce{A = Cu2O}$ with copper in its first oxidation state.


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