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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.

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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.

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  • $\begingroup$ It makes sense, though where is the potassium left? $\endgroup$ – ChairOTP Mar 5 '13 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 '13 at 22:17
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    $\begingroup$ 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 '13 at 23:04
  • $\begingroup$ @BenNorris Thanks!! I have edited the equation using \ce{} $\endgroup$ – kaliaden Mar 6 '13 at 7:54
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    $\begingroup$ @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 '13 at 8:11

protected by Community Oct 6 '14 at 17:50

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