How to derive formulas for n factor of different types of reactions?

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 !




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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