The given reaction is $$\ce{NH4NO2 \rightarrow N2 + 2H2O}$$

The way I did it was $$\ce{NH4+ \rightarrow \frac{1}{2} N2 + 4H+ + 3e-}$$ and $$\ce{NO2- +3e- +4H+ \rightarrow \frac{1}{2} N2 + 2H2O}$$ Therefore n factor is 3

But the book I am referring to says 6. Can someone please help?


"n factor" of oxidizing or reducing agent = total number of electrons transferred per a mole of the reactant.

When more than one element in a substance is oxidized/reduced, the n-factor of the substance is taken as the sum of the n-factors of the individual elements of the substance.

This is the case here.You have ammonium ion that undergoes oxidation reaction (n-factor= 3) and nitrite ion that undergoes reduction reaction (n-factor= 3). So, the n-factor of ammonium nitrite is 6.

  • $\begingroup$ Even though one is getting reduced and the other oxidised? Because if we usually add half reactions, the net e- transfer is not the sum. Am i right? $\endgroup$
    – user165253
    Feb 10 '15 at 16:46
  • $\begingroup$ It's a matter of definitions of terms. Let's take the case of separate oxidizing and reducing agents. Permanganate ions reactiong with fer(II) ions. "n factor" of KMnO4 when it's reduced to ion Mn2+ is five. Of course, when permangante ion is reduced, there is another chemical substance, ion Fe (II) that is oxidized to ion Fe(III), and the n factor of it is 1. $\endgroup$ Feb 10 '15 at 18:24

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