While studying Redox reaction, I came across this term called n-factor which seems to me as a short-cut trick to do the questions quickly. Since I am preparing for a competitive exam so our chemistry teacher gave us a bunch of formulas to find n-factor of different types of reactions as follows :-

Case 1

Reduction /oxidation reactions in which only one atom of molecule reduce or oxidise. $$ A_{x}^{+p} B_{y} \rightarrow A_{x}^{+q} B_{y_{1}} $$ $$ \begin{array}{l} p>q \Rightarrow \text { Reduction } \\ p<q \Rightarrow \text { oxidation } \end{array} $$ $$ n \text {-factor of } A_{x} B_{y}=x|p-q| $$

Case 2

If more than one atom oxidise or Reduce in molecule. $$ A_{x}^{+P} B_{y}^{+r} \rightarrow A_{x}^{+q}+B_{y_{1}}^{+\omega} $$ $\left.\begin{array}{l}p>q \\ r>\omega\end{array}\right\} \Rightarrow$ Reduction $\left.\begin{array}{l}p<q \\ r<\omega\end{array}\right\} \Rightarrow$ Oxidation

$n$-factor of $A_{x} B_{y}=x|p-q|+y|r-\omega|$

Case 3

Disproportionation reaction $A_{x}^{+p} B_{y} \rightarrow A_{x_{1}}^{+q}+A_{x_{2}}^{+r}$ $$ \begin{array}{l} p>q \Rightarrow \text { Reduction } \\ p<r \Rightarrow \text { oxidation } \end{array} $$ $n$-factor of $A_{x} B_{y}=\frac{(x|p-q|) \times(x|p-r|)}{x|p-q|+x|p-r|}$

Now what I am really interested in is How to derive these formulas ? The first and second cases are pretty easy and logical to think of but the third case really bounced off my head and I am not able to derive the third formula. Please help !

  • $\begingroup$ Hmm, a simplifying tool should not be more complex than tasks it should simplify. $\endgroup$
    – Poutnik
    Commented Sep 14, 2021 at 5:09
  • $\begingroup$ @Poutnik The third formula may look big at first but it indeed is quite easy to apply. I have tried it on several reactions and it gives the correct answer although the normal method of calculation n-factor by balancing was also quite easy but out teacher told us to learn this formula to be used in case of large and scary chemical equations $\endgroup$
    – Tushar
    Commented Sep 14, 2021 at 5:18
  • $\begingroup$ What I mean, I have never needed that, as redox equation enumeration is so simple. $\endgroup$
    – Poutnik
    Commented Sep 14, 2021 at 5:23
  • 1
    $\begingroup$ It should be simplified as $$ \text{n-factor} =x \cdot \frac{|p-q| \times |p-r|}{|p-q|+|p-r|} $$ $\endgroup$
    – Poutnik
    Commented Sep 14, 2021 at 11:05

1 Answer 1


I will try to explain in the third case of disproportionation reaction first after that I believe you would be able to derive the first two cases easily. Let's start

We will begin with the definition of n-factor.

n-factor is the total number of mole of electron transferred for oxidation or reduction of 1 mole of molecule.

Now let's take a general disproportionation chemical reaction as \begin{array}{c} A_{x}^{+p} B_{y} \rightarrow A_{x_{1}}^{+q}+A_{x_{2}}^{+r} \\ p>q \Rightarrow \text { Reduction } \\ p<r \Rightarrow \text { Oxidation } \end{array}

Now we will start balancing this reaction as follows:

Split the reaction in oxidation half and reduction half.

Reduction half $$ A_{x}^{+p} B_{y} \rightarrow A_{x_{1}}^{+q} $$ $$ n_{1}=x|p-q| $$ Oxidation half $$ A_{x}^{+p} B_{y} \rightarrow A_{x_{2}}^{+r} $$ $$ n_{2}=x|p-r| $$

Here $n_{1}$ and $n_{2}$ are n-factor of the the respective half reactions (i.e. the number of mole of electron transferred per mole molecule $A_{x}^{+p} B_{y}$).

Now, in order to balance the equation we must equate the number of electrons transferred in oxidation half and reduction half and then add these half reactions.

Easiest way to equate the electons transferred is as follows:

Reduction half $$ n_{2} A_{x}^{+p} B_{y} \rightarrow n_{2} A_{x_{1}}^{+q} $$ Oxidation half $$ n_{1} A_{x}^{+p} B_{y} \rightarrow n_{1} A_{x_{2}}^{+r} $$ In the above half reaction the number of electrons transferred = $n_{1} n_{2} $

Now we add these two reactions to get the final balanced reaction as $$ (n_{1} + n_{2}) A_{x}^{+p} B_{y} \rightarrow n_{1} A_{x_{2}}^{+r} + n_{2} A_{x_{1}}^{+q} $$

Now in the above reaction we know that the number of electrons transferred is $n_{1} n_{2} $ for $(n_{1} + n_{2})$ mole of $A_{x}^{+p} B_{y} $.

Therefore, $$ \text {n-factor }=\frac{n_{1} n_{2}}{n_{1}+n_{2}} $$ And if we expand this by putting the formulas of $n_{1}$ & $ n_{2} $ we get,

$$ \text{n-factor} =\frac{(x|p-q|) \times(x|p-r|)}{x|p-q|+x|p-r|} $$


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