Question: n factor of \[\text{Fe}{{\text{C}}_{\text{2}}}{{\text{O}}_{\text{4}}}\]during its oxidation by acidi...

Question

n factor of $\text{Fe}{{\text{C}}_{\text{2}}}{{\text{O}}_{\text{4}}}$ during its oxidation by acidified $\text{KMn}{{\text{O}}_{\text{4}}}$ is:

Answer 1

Solution

Hint: Solve this question using the concept of law of equivalents, writing the chemical equation and balancing it. Then, write the change in oxidation state of compounds to find out the n factor.

Complete step by step answer:
Let us start this question by understanding n-factor.
n-factor – It is defined separately for acids and bases. For acids, n-factor is defined as “the number of ${{\text{H}}^{\text{+}}}$ ions replaced by 1 mole of acid in a reaction”. Whereas, for bases, n-factor is defined as “the number of $\text{O}{{\text{H}}^{\text{-}}}$ ions replaced by 1 mole of base in a reaction”.
Now, let us write the equation for the oxidation of $\text{Fe}{{\text{C}}_{\text{2}}}{{\text{O}}_{\text{4}}}$ by acidified $\text{KMn}{{\text{O}}_{\text{4}}}$ . (also let us take sulphuric acid as the acidic medium).
$10Fe{{C}_{2}}{{O}_{4}}+6KMn{{O}_{4}}+24{{H}_{2}}S{{O}_{4}}\to 5F{{e}_{2}}{{(S{{O}_{4}})}_{3}}+20C{{O}_{2}}+6MnS{{O}_{4}}+3{{K}_{2}}S{{O}_{4}}+24{{H}_{2}}O$
Let us see the effect of potassium permanganate - $\text{KMn}{{\text{O}}_{\text{4}}}$ on ferrous oxalate $\text{Fe}{{\text{C}}_{\text{2}}}{{\text{O}}_{\text{4}}}$ .
Ferrous oxalate is composed of iron (II) and oxalate. Therefore, we will see that net n factor for both of them.
As we can see –
$F{{e}^{+2}}$ gets converted to $F{{e}^{+3}}$
Change in oxidation number = +1
${{C}_{2}}{{O}_{4}}^{-2}$ gets converted to $C{{O}_{2}}$
Change in oxidation number = +2
So, the total oxidation number = 1+2 = 3.
Therefore, the answer is –
n factor of $\text{Fe}{{\text{C}}_{\text{2}}}{{\text{O}}_{\text{4}}}$ during its oxidation by acidified $\text{KMn}{{\text{O}}_{\text{4}}}$ is 3.

Note: Law of equivalents – “According to the law of equivalence, whenever two substances react, the equivalents of one will be equal to the equivalents of other and the equivalents of any product will also be equal to that of the reactant”.