I've already searched online some information about this equation, but everywhere I read, there's also sulfuric acid, which we did not use in the laboratory. We only used potassium permanganate and oxalic acid and water, there was no sulfuric acid. The purpose of the laboratory was to calculate experimentally the rate of the reaction at different concentrations, but in order to do so, I need the balanced equation. Can anyone help me out?

By the way, the oxalic acid was dihydrate.

If anyone is wondering, one of the experiments was as follows:

First we took $1.5~\pu{mL}$ of $\ce{H2C2O4}$ and added $1.5~\pu{mL}$ of $\ce{H2O}$, and finally, we added $1~\pu{mL}$ of $\ce{KMnO4}$ and then we shook the test tube until the colour changed from purple to ochre. Because from what I read, they said the $\ce{H2SO4}$ was necessary so the reaction takes place, but in the experiment, the reaction did occur and we didn't use it.

  • $\begingroup$ Does the reaction take place in any anhydrous conditions, for instance, in dmf? $\endgroup$
    – user3785
    Nov 25, 2013 at 4:10
  • $\begingroup$ related chemistry.stackexchange.com/questions/59905/… $\endgroup$ Sep 29, 2016 at 18:43
  • $\begingroup$ One can think of potassium permanganate as a substitute for H2O2 in REDOX reactions, in which case, we have this reference: osti.gov/biblio/6229619 . " The rate of oxidation of oxalate by hydrogen peroxide in acid solution at the reflux temperature was adequate for process application; reaction half-times at 100/sup 0/C were less than one hour when hydrogen peroxide concentration was greater than 0.5 M. The reaction was first order... to both oxalate and hydrogen peroxide concentrations and had an activation energy 58.7 kJ/g-mol...rate increased with hydrogen ion concentration" $\endgroup$
    – AJKOER
    May 20, 2020 at 12:46

2 Answers 2


We had a titration of oxalic acid vs. potassium permanganate for an experiment in which we used concentrated $\ce{H2SO4}$.

Actually, the reaction requires an acidic medium i.e. $\ce{H+}$ is involved as a reactant... I suppose the $\ce{H+}$ released by the oxalic acid would be enough since you are studying the rate of the reaction though I am not sure.

The balanced ionic equation is: $\ce{2MnO4- + 5H2C2O4 + 6H+ -> 2 Mn^2+ + 10CO2 + 8H2O}$

As you can see, $\ce{H+}$ is involved as a reactant and that is why you might have read that conc.$\ce{H2SO4}$ is necessary.

  • $\begingroup$ It makes sense, though where is the potassium left? $\endgroup$
    – ChairOTP
    Mar 5, 2013 at 21:47
  • $\begingroup$ @ChairOTP, kaliaden left out K bacause it is a spectator ion, meaning it does not participate in the reaction, and so it is often left out. $\endgroup$
    – user467
    Mar 5, 2013 at 22:17
  • 1
    $\begingroup$ If you wrap your chemical equations in \ce{ }, you will get prettier formulae and equations. \ce{ } converts 2MnO4^- +5H2C2O5 + 6H+ -> 2Mn^{2+} + 10CO2 + 8H2O into $\ce{2MnO4^- +5H2C2O5 + 6H+ -> 2Mn^{2+} + 10CO2 + 8H2O}$ $\endgroup$
    – Ben Norris
    Mar 5, 2013 at 23:04
  • $\begingroup$ @BenNorris Thanks!! I have edited the equation using \ce{} $\endgroup$
    – kaliaden
    Mar 6, 2013 at 7:54
  • 2
    $\begingroup$ @ChairOTP Like trb456 mentioned, The $K^+$ is a spectator ion and is not actually involved in the reaction. The same goes for the $SO_4^{2-}$ when $H_2SO_4$ is used, only the $H^+$ is actually involved in the reaction while the $SO_4^{2-}$ is a spectator ion. So you could actually use any other acid instead of $H_2SO_4$ as long as it doesn't react in any other way with any of the reactants. $\endgroup$
    – kaliaden
    Mar 6, 2013 at 8:11

Oxalic acid is a relatively strong acid for a carboxylic acid, and according to my sources below, can auto-catalyze the reaction with Potassium Permanganate. So, the reaction you performed was likely just the same mechanism that you have seen everywhere else. My source is as follows:

Kovacs K.A.; Grof P.; Burai L.; Riedel M. (2004). "Revising the mechanism of the permanganate/oxalate reaction". J. Phys. Chem. A. 108 (50): 11026–11031.


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