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We all know that

$\text{Equivalent weight} = \frac{\text{Molecular weight}}{\text{n-factor}}$

However consider a reaction like:

$\ce{Br2 + OH- -> Br- + BrO3- + H2O}$ where oxidation numbers of $\ce{Br}$ are $0, -1, +5$ respectively.

What should here be the equivalent weight of $\ce{Br2}$?

PS: I understand that I should also show what I've attempted. But, this is just a question that came up into my head and I've made it as descriptive as possible. I don't really see anything else that I must add to make this question complete.

Thank you!

UPDATE: I just came across a formula for calculating the n-factor for a dispropportionation reaction $n_f=\frac{n_1\cdot n_2}{n_1+n_2}$ (sources, 1 and 2), for $n_1$ the magnitude of n-factor in oxidation reaction and $n_2$ the magnitude of n-factor in reduction reaction. Can anyone prove how to arrive at this formula?

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  • $\begingroup$ @IvanNeretin Thanks! But isn't there only one reaction here that we're seeing? Or is it the fact that you're considering oxidation and reduction reactions separately, as if they're the basic parts of the given reaction? $\endgroup$ – Gaurang Tandon Feb 16 '17 at 13:17
  • $\begingroup$ True, it's one reaction, so what? Like I said, either use the idea of equivalent the two-pronged way, or consider it unusable in this case. $\endgroup$ – Ivan Neretin Feb 16 '17 at 13:32
  • $\begingroup$ @Gaurang Tandon... To your update...The result can be derived by writing the two half reactions, expressing their equivalent weight as M/n1 and M/n2 and equating their sum with M/N, where N is the resultant n factor. Thus $N=\frac{n_1\cdot n_2}{n_1+n_2}$ .........Yet I do not understand why n1 and n2 aren't simply added and divided by two.......because n factor is the no of electrons which one mole of the compound reacts with or supplies in a redox reaction. $\endgroup$ – Pranoy De May 18 '18 at 9:30
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You may consider this formalism inapplicable here altogether. Alternatively, you may assume you have two different kinds of bromine (one that gets oxidized and another that gets reduced), each with its own factor.

True, in this way, the same bromine molecule would have two different equivalent weights in the same reaction. There's nothing strange or unusual about that. After all, the same molecule surely can have different equivalent weights in different reactions.

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The n-factor of a compound in a reaction gives the number of moles of electron lost or gained by it (during oxidation and reduction respectively) per mole of the given compound. In a disproportionation, the compound undergoing both the oxidation and reduction is the same.

It is important to understand that the n-factor depends not just on the change in oxidation state of the element but also the number of moles of the element undergoing that change. This is the reason why we cannot simply take the average of the two n-factors as the n-factor of the disproportionate. It is necessary to also consider how much of the reactant is getting oxidised and how much of it is getting reduced, something along the lines of a weighted mean.

It is quite simple to derive the expression you have found for the n-factor of a disproportionated compound. Consider the following general reaction:

$$\ce{nA -> xB + n-xC}$$

Where x moles of A (for which n-factor is to be calculated) oxidises to B and the rest of it reduces to give C. Let the n-factors for the oxidation and reduction respectively be $n1$ and $n2$. Which means $n1x$ moles of electrons go into oxidation, and $(n-x)n2$ moles into reduction.

We have 2 equations, one by conservation of electrons, $$n1x=n2(n-x)$$ $$\implies x =\frac{(n2).n}{n1+n2}$$ and the other from the definition of n-factor: $$ n_f = \frac{moles ~of~ electrons ~transferred}{moles ~of ~reactant} $$ $$= \frac{n1x}{n}~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~$$ $$= \frac{n1 ~. ~ n2}{n1+n2}~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~$$

An interesting observation here, is that upon simplifying the obtained equation you get the formula for the the equivalent resistance of resistors connected parallelly in an electric circuit. The correspondence between them is intuitive. $$ \frac{1}{n_f} = \frac{1}{n1} + \frac{1}{n2} $$

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  • $\begingroup$ It looks like you are partially answering the question, and at the same attempting to ask a new question. Stack Exchange sites are sticking with strict Q&A format meaning a question should receive a short, but complete and a well-defined answer without discussion or explicitly raising new questions. If you have a new question, please ask a new question instead of attaching one to your answer. $\endgroup$ – andselisk Jun 16 at 17:54

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