# Writing equation of reaction in stoichiometry of redox titrations

While solving some questions on the stoichiometry on redox titrations, I come across this question:

Calculate the volume of $0.05\ \mathrm{M} \ \ce{KMnO4}$ solution required to oxidize completely $2.70\ \mathrm{g}$ of oxalic acid in acidic medium.

I know how to solve this question. The only problem I am facing is that the reaction equation is not given. How do I know what the products of this reaction would be? Can someone help me in how to write the equation of the reaction by the information given in the question.

• You can assume that oxalic acid will be oxidized to $\ce{CO2}$ and $\ce{H2O}$ while $\ce{MnO4^-}$ will be reduced to $\ce{Mn^{2+}}$. – aventurin Aug 28 '16 at 10:58
• @aventurin assume...? Here I can see the solution to find this out but in an exam how am I supposed to know? – Osheen Sachdev Aug 28 '16 at 11:01
• Permanganate in acidic solution is a well-known oxidant (permanganometry, en.wikipedia.org/wiki/Permanganometry). So either by intuition or by looking at the standard electrode potentials en.wikipedia.org/wiki/… and the Nernst equation. – aventurin Aug 28 '16 at 11:57
• en.wikipedia.org/wiki/Sodium_oxalate – Mithoron Aug 28 '16 at 14:02

See the following steps:

1. We know what $\ce{KMnO4}$ will do by reading the question:

Calculate the volume of […] $\ce{KMnO4}$ solution required to oxidize completely […] oxalic acid.

2. We need the structure of oxalic acid. (This is knowledge required to answer the question, no way around.) Structure of oxalic acid. Image taken from Wikipedia, where a full list of authors is available.

3. Determine oxidation numbers of all the atoms involved. Shortcut: hydrogen has $\mathrm{+I}$, oxygen has $\mathrm{-II}$ and carbon has $\mathrm{+III}$.

4. Determine what can be oxidised. Only carbon really can.

5. Determine what it can be oxidised to. The final product must have $\ce{C^{IV}}$, which is most common as carbon dioxide or carbonate.

6. Solve the redox equation using either $\ce{CO2}$ or $\ce{CO3^2-}$. Both will work.

• Well this works here as carbon can only be oxidized to 4+ from 3+ but in the reduction of potassium dichromate by FeSO4 how would we know the change in the oxidation state of chromium...? Would we have to memorize it or is there a way to figure it out? – Osheen Sachdev Aug 29 '16 at 15:17
• @OsheenSachdev You did ask about oxalic acid, so I gave you that answer. For chromium, it’s basically either remember that chromium(III) is a stable oxidation state or correctly interpret the green colour at the end of the reaction to mean chromium(III). – Jan Aug 30 '16 at 9:35

Firstly, oxalic acid with both carbon atoms in +3 oxidation state, which undergoes oxidation has only to form CO2 Secondly, the formation of products from the reduction of KMnO4 is dependent heavily on pH of the solution. But like in the question, if the pH of the solution is not mentioned, then it's assumed to be acidic, and the product is Mn(II), since most of the volumetric titrations are done in an acidic solution provided by acids which will not participate in the redox reaction whatsoever.