# 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?

• How did you get the 2 of 2x+... ? – Del Pate Mar 28 '15 at 6:10
• Sorry for typo . – Heisenberg Mar 28 '15 at 6:19
• Are you sure its 28 and not 24? – Del Pate Mar 28 '15 at 6:28
• Not sure , I could be wrong. – Heisenberg Mar 28 '15 at 6:28
• You say it's given to be 28. Is it given 28 on some website/text? – Del Pate Mar 28 '15 at 6:30

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)

• 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, $\ce{CO_2}$.

• And I differ from the answer provided by your notes. If I am wrong somewhere, feel free to correct me. – Del Pate Mar 28 '15 at 6:42
• Okay what if we had 2 carbon atom on product side? – Heisenberg Mar 28 '15 at 6:52
• Hypothetically let's suppose c2o4 – Heisenberg Mar 28 '15 at 6:53
• @Heisenberg The OS of Carbon in $C_2O_4$ is the same as in $CO_2$. Hence the n-factor with respect to glucose shall remain the same. – Del Pate Mar 29 '15 at 5:40
• Yes @Heisenberg. I have edited the answer. – Del Pate Mar 29 '15 at 16:50

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.