How to calculate valence factor (n-factor) for a element?

Suppose I am given some reaction in which 
$$\ce{C6H12O6 -> CO2}$$
and I want to calculate n-factor for this reaction to ultimately calculate equivalent weight of carbon for this reaction. Since $$E=\frac{M}{n}$$
I know that n-factor is no of electron gained/lost by one atom of a compound.


So Initial Oxidation state of carbon is $$6x + 12-12=0$$ $$x=0$$ meaning 
and final oxidation state of carbon is $$-4$$ Hence, n-factor of this reaction should be $4$ but it's given to be $28$. How is this possible? Can someone explain how to correctly calculate n-factor with few more examples?




Hint : 


n-factor of a molecule/compound is defined as the change in oxidation state **per molecule**.  

You have correctly calculated the change of **one** carbon atom as 4.  

But how many carbon atoms are there in the glucose molecule? 


Note: 


* The **average** oxidation state of carbon in glucose is zero while in reality the different carbons have different OS. ([Reference](http://butane.chem.uiuc.edu/pshapley/GenChem2/B7/1.html))
* n-factor of a reaction is not defined but it is defined for a single species participating in that reaction.
* n-factor of glucose is not equal to the n-factor of the product formed in that reaction, that is, $\ce{CO\_2}$.


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Your reaction, $\ce{C6H12O6 -> CO2}$ is a redox half-reaction. The *n*-factor of a molecule/compound in a redox reaction is defined as the change in the oxidation state per molecule (as defined in Del Pate's answer). The easy to visualize definition is the amount of electrons (in $\pu{mol}$) donate or accepted per $\pu{1 mol}$ of the compound in the given redox half-reaction.



> 
> Can someone explain how to correctly calculate n-factor with few more examples?
> 
> 
> 


Let's look at following redox half-reactions and find the *n*-factor by these two different definitions:


$$\text{Example 1:} \qquad \qquad \ce{MnO4- ->[acid] Mn^2+ }$$


The oxidation number of $\ce{Mn}$ of the reactant is $+7$ while that of the product is $+2$. The change in the oxidation state per $\ce{MnO4-}$ is $+5$. Thus, the *n*-factor of $\ce{MnO4-}$ is $5$ in acid medium.


If we balance this redox half-reaction, we get:


$$\ce{MnO4- + 8 H+ + 5 e- -> Mn^2+ + 4 H2O}$$


Thus, the amount of electrons accepted per $\pu{1 mol}$ of $\ce{MnO4-}$ in the given redox half-reaction is $5$. Hence, the *n*-factor of $\ce{MnO4-}$ is $5$ in acid medium.


$$\ce{MnO4- ->[base/neutral] Mn^4+ \ (\text{as} MnO2)}$$


The oxidation number of $\ce{Mn}$ of the reactant is $+7$ while that of the product is $+4$. The change in the oxidation state per $\ce{MnO4-}$ is $+3$. Thus, the *n*-factor of $\ce{MnO4-}$ is $3$ in base/neutral medium.


$$\ce{MnO4- + 2 H2O + 3 e- -> MnO2 + 4 OH-}$$


Thus, the amount of electrons accepted per $\pu{1 mol}$ of $\ce{MnO4-}$ in the given redox half-reaction is $3$. Hence, the *n*-factor of $\ce{MnO4-}$ is $3$ in base/neutral medium.


$$\text{Example 2:} \qquad \qquad \ce{CuS -> Cu^2+ + SO2 }$$


The oxidation number of $\ce{S}$ of the reactant is $-2$ while that of the product is $+4$. The change in the oxidation state per $\ce{CuS}$ is $+6$. Thus, the *n*-factor of $\ce{CuS}$ is $5$ in this redox reaction.


If we balance this redox half-reaction, we get:


$$\ce{CuS + 2 H2O -> Cu^2+ + SO2 + 4 H+ + 6 e- }$$


Thus, the amount of electrons accepted per $\pu{1 mol}$ of $\ce{CuS}$ in the given redox half-reaction is $6$. Hence, the *n*-factor of $\ce{CuS}$ is $6$ in this redox half-reaction.


Now, let's look at the given reaction in the question:


$$\ce{C6H12O6 -> CO2}$$


Suppose we don't know the oxidation number of $\ce{C}$ of glucose $(\ce{C6H12O6})$ since there are $\ce{6C}$s in the molecule. Thus, we can approach to solve the problem by the second method. If we balance this redox half-reaction, we get:


$$\ce{C6H12O6 + 6 H2O -> 6CO2 + 24 H+ + 24 e- }$$


Thus, the amount of electrons accepted per $\pu{1 mol}$ of $\ce{C6H12O6}$ in the given redox half-reaction is $24$. Hence, the *n*-factor of $\ce{C6H12O6}$ is $24$ in this redox half-reaction. This demonstration reveals that even without knowing the change in oxidation number during the reaction, you can find the *n*-factor.


**Note:** This finding shows that the given answer for the question is incorrect.


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