Question

A sample of $FeS{O}_4$ and $Fe{C}_2{O}_4$ was dissolved in dilute sulphuric acid. The complete oxidation of the reaction mixture required 40 mL of N/15 $KMn{O}_4.$ after the oxidation, the reaction mixture was reduced by Zn and $H_{2}S{O}_4.$ On using the same solution of $KMn{O}_4$ to oxidize the resulting solution, 25 mL was required. Calculate the ratio of equivalents of Fe in oxalate and Fe(II) sulphate. (Fe: 56)

Answer 1

The oxidation reactions are as follows:
For $FeS{O}_4$,
$F{e}^{2+}\to F{e}^{3+}+e^{-}$ ..(1)
For $Fe{C}_2{O}_4$,
$F{e}^{2+}\to F{e}^{3+}+e^{-}$ ...(2)
$C_2{O}_4^{2-}\to 2C{O}_2+2e^{-}$ ..(3)
The reduction reactions are as follows
$F{e}^{3+}+e^{-}\to F{e}^{2+}..\left(4\right)Zn\to Z{n}^{2+}+2e^{-}..\left(5\right)Or2F{e}^{2+}+Zn\to Z{n}^{2+}+2F{e}^{2+}..\left(6\right)$
Finally, on being oxidized once again
$F{e}^{2+}\to F{e}^{3+}+e^{-}$ ..(7)
Let the equivalents of Fe in $FeS{O}_4$ be 'a' and in $Fe{C}_2{O}_4$ be 'b'.
In the first oxidation, we know the equivalents of $KMn{O}_4$ used (acidic medium).
It is clear that the $KMn{O}_4$ is oxidizing Fe in (1), Fe in (2) and $C_2{O}_4^2-$ in (3).
Therefore equivalents of $KMn{O}_4=a+b+b..\left(8\right)$
This gives us one equation in a and b.

In the second oxidation, $KMn{O}_4$ oxidizies only all the Fe as the $C{O}_2$ from the oxidation of $C_2{O}_4^{2-}$ escapes as it is a gas. This is a crucial thing to note.
$\therefore$ equivalents of $KMn{O}_4$ in the second oxidation = a+b
This gives us a second equation in a and b.
Equivalents of $KMn{O}_4$ in first oxidation = $40\times \frac{1}{15}=2.67$ meq

Equivalent of $KMn{O}_4$ in second oxidation $=25\times \frac{1}{15}=1.67$ meq
$\therefore a+2b=2.67$
$a+b=1.67$
Solving for a and b
a = $\frac{2}{3}$, b = 1
Hence $\frac{b}{a}=\frac{3}{2}=1.5$