# How much KMnO4 is required for the complete oxidation of ferric/ferrous oxalate in acidic medium?

Find out the amount of $\ce{KMnO4}$ required for the complete oxidation of $\pu{5 mol}$ of an equimolar mixture of ferric oxalate and ferrous oxalate.

I understand that $\ce{MnO4-}$ will be oxidised to $\ce{Mn^{2+}}$ and ferrous ions to ferric ions. I wrote the following reaction: $$\ce{FeC2O4 + Fe2(C2O4)3 + MnO4^- + 8H+ + e- -> 2Fe^3+ + 4CO2 + Mn^2+ + 4H2O}$$

Then, I applied equivalence concept but could not get a correct answer. Any ideas?

• $\ce {Fe^{ +2}}$converts to $\ce {Fe^{ +3}}$ and and carbon from +3 to +4 , calculate required equivalents for them and then equate then with equivalents of $\ce {KMnO4}$ May 9, 2017 at 10:53
• Can you show more effort on how you applied the equivalence concept ? May 9, 2017 at 10:54

I'd like the question to be tackled in two parts. In each part, we're going to deal with one of ferric oxalate and ferrous oxalate.

Since the mixture ($\pu{5 mol}$) given to us is equimolar, we have $\pu{2.5 mol}$ of ferric oxalate and $\pu{2.5 mol}$ of ferrous oxalate. I second your understanding that $\ce{MnO4-}$ will be oxidised to $\ce{Mn^{2+}}$, because the medium is acidic, and ferrous ions to ferric ions.

Ferric oxalate doesn't contain any $\ce{Fe^3+}$ ions, so our main focus here is the conversion of oxalate ($\ce{C2O4^2-}$) into carbon dioxide. The oxidation state of carbon in oxalate is $+3$, while it's $+4$ in carbon dioxide. So there's a net change of $+1$ per carbon atom. Oxalate contains two carbon atoms, so that gives us a valency factor of $2$ per mole of oxalate ion. However, for $1$ mole of $\ce{Fe2(C2O4)3}$, you'll have $3$ moles of oxalate ions that are undergoing the aforementioned change. This gives us an overall valency factor for $\ce{Fe2(C2O4)3}$ as $2 \times 3 = 6$. The valency factor for the $\ce{MnO4- -> Mn^2+}$ change is $5$. Now, we're going to equate the equivalents as per the Law of Equivalence.

\begin{align*} n_{\ce{KMnO_4}} \times V_{f_1} &= n_{\ce{Fe2(C2O4)3}} \times V_{f_1'}\\ x \times 5 &= 2.5 \times 6 \\ x &= 3 \end{align*}

For the part of the mixture that contains $\ce{FeC2O4}$, we need to change our valency factor. This is because here we're taking into account the conversion of ferrous ions into ferric ions; this conversion has a valency factor of $1$. So the overall valency factor, in this case, would be $2+1=3$. Now, since this has been done, we proceed to apply the same equation as in the previous case.

\begin{align*} n_{\ce{KMnO_4}} \times V_{f_2} &= n_{\ce{FeC2O4}} \times V_{f_2'}\\ x \times 5 &= 2.5 \times 3 \\ x &= 1.5 \end{align*}

Thus, overall, it can be concluded that you'll need $\pu{3+1.5=4.5 mol}$ of $\ce{KMnO4}$ for the complete oxidation. The beauty of this? I didn't use any reaction-equilibrium equations.

This is the reason we use equivalent concept, to save ourselves from the hassle of writing (and perhaps balancing) chemical reactions.