Question

1 mole of equimolar mixture of ferric oxalate and ferrous oxalate will require x mole of $KMn{O_4}$ in acidic medium for complete oxidation, the value of x is:
A) 0.5 mol
B) 0.9 mol
C) 1.2 mol
D) 4.5 mol

Answer 1

Hint: The chemical formula for Ferric Oxalate is $F{e_2}{({C_2}{O_4})_3}$ ; The Iron is in $ + 3$ oxidation state. Ferrous Oxalate is given by the formula $Fe({C_2}{O_4})$ ; in here the iron is in $ + 2$ oxidation state.
In redox reactions the no. of equivalents of Oxidising agent and reducing agent will be equal. We can write it as:
 $no.of{\text{ }}equivalent{\text{ of O}}{\text{.A}} = no.of{\text{ equivalent of R}}{\text{.A}}$
$no.of{\text{ }}equivalent = moles \times {n_f}$
(Where O.A is the Oxidizing agent and R.A is the reducing agent and ${n_f}$ is the n-factor)

Complete answer: It has been given that the mixture is equimolar, which means that in 1 molar mixture 0.5 mole will be of Ferric Oxalate and 0.5 mole will be of Ferrous Oxalate. In acidic Medium, on reaction with $KMn{O_4}$ it will give us two reactions.
The first reaction of Ferric Oxalate with $KMn{O_4}$ can be given as:
 $F{e_2}{({C_2}{O_4})_3} + MnO_4^ - + {H^ + } \to 2F{e^{ + 3}} + 6C{O_2} + {H_2}O + M{n^{ + 2}}$-- (1)
In here we can say that the Oxidation state of Carbon changes from $ + 3 \to + 4$ . And Mn changes from $ + 7 \to + 2$ ; i.e., it gets reduced.
The second reaction of ferrous oxalate can be given as:
 $Fe{C_2}{O_4} + MnO_4^ - + {H^ + } \to F{e^{ + 3}} + 2C{O_2} + {H_2}O + M{n^{ + 2}}$-- (2)
In here the oxidation state of C changes from $ + 3 \to + 4$ and Fe changes from $ + 2 \to + 3$ . Both the atoms get oxidized simultaneously.
Now note this formula: The n-factor for any reaction change can be given as: ${n_f} = |change{\text{ in O}}{\text{.S|}} \times no{\text{ }}{\text{. of atoms}}$
${n_f}$ value of equation (1) can be given as: ${n_f} = | + 1| \times 6 = 6$
${n_f}$ value of equation (2) can be given as: ${n_f} = {n_1} + {n_2} = | + 1| \times 2 + | + 1| \times 1 = 3$
As said earlier the no. of equivalent of O.A agent will be equal to reducing agent.
No. of equivalent of $KMn{O_4} = $ no. of equivalent of $F{e_2}{({C_2}{O_4})_3} + Fe{C_2}{O_4}$
 $x \times 5 = 6 \times \dfrac{1}{2} + 3 \times \dfrac{1}{2}$ (Consider the no. of moles of $KMn{O_4}$ to be x)
 $5x = 4.5$
$\Rightarrow x = 0.9$
The correct answer is Option (B).

Note:
The n-factor of certain Oxidizing agents can be easily remembered in different mediums. The n-factor of ${K_2}C{r_2}{O_6}$ is always 6 in the acidic medium. The n-factor of $KMn{O_4}$ in Basic, Acidic and Neutral Medium is 1,5 and 3 respectively. You can remember this as $BAN - 153$
Ferric Oxalate is a dicarboxylic acid occurring in plants and animals. It is also produced by the metabolism of ascorbic acid in the body. Ferrous Oxalate is a weak inorganic oxidising agent. It is used for manufacturing silver bromide gelatine plates and decorative glassware.