Question

In the reaction, $VO + F{e_2}{O_3} \to FeO + {V_2}{O_5}$, what is the n – factor for ${V_2}{O_5}$?

Answer 1

We need to know that the n – factor is the conversion factor or a valency factor. The n – factor is different for both redox reaction and non – redox reaction. Because, in the case of redox reaction, for a substance, the n – factor in the redox reaction is equal to the number of moles of gain of the electrons or number of moles of loss of the electron present in a molecule. But in the case of a non – redox reaction, the n – factor is equal to the product of displaced moles and the charge of displaced moles.

Complete answer:
In the given reaction, the vanadium oxide is reacted with ferric oxide and there is a formation of ferrous oxide with vanadium pentoxide. And this is a redox reaction. The oxidation and redox reaction can be written as,
$2{V^{II}} - 6{e^ - } \to 2{V^v}$ ------ (Oxidation)
$F{e^{III}} + 6{e^ - } \to 6F{e^{II}}$ ------ (Reduction)
Here, the oxidation state of vanadium oxide is equal to $+ 2$ and the oxidation state of vanadium pentoxide is equal to $+ 5$.
Therefore, the n – factor is equal to the product of the number of vanadium in vanadium pentoxide and the magnitude of change in the oxidation number.
Hence,
$n - factor = 2 \times \left( {5 - 2} \right) = 2 \times 3 = 6$
n- Factor of vanadium pentoxide is equal to six.

Note:
For a redox reaction, the n – factor is equal to the number of moles of gain of the electron or loss of the electron. But it is different for both redox and non – redox reactions. For an acid, the n – factor is equal to the number of acidic hydrogen ions which a molecule of the acid would give when it is dissolved in the solvent. And in the case of base, it is equal to the number of hydroxide ions dissolved in the solvent.