# Equivalent weight in case of disproportionation reaction

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?

• @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? – Gaurang Tandon Feb 16 '17 at 13:17
• 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. – Ivan Neretin Feb 16 '17 at 13:32
• @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. – Pranoy De May 18 '18 at 9:30