# Equations for the oxidation reactions of tea teaflavins and EGCG with potassium permanganate

For my chemistry investigation, I used Stamm's potassium permanganate method to investigate tea tannins with potassium permanganate to find the tannin content of teas. The method assumes complete oxidation of tea polyphenols to carbon dioxide.

Assuming that the tea polyphenols consist of only theaflavins, $$\ce{C29H24O12},$$ and Epigallocatechin gallate (EGCG), $$\ce{C22H18O11},$$ could someone help me with finding the equation of the reaction between the listed polyphenols and potassium permanganate respectively, i.e. theaflavin reaction with potassium permamganate and EGCG reaction with potassium permanganate (preferably 2 equations for each redox, but 1 overall equation should be fine too)?

I tried looking for these everywhere, but no source seems to give me the equations I need.

• I upvoted your question and provided a full answer, but you also got two downvotes on the question. And as for half-reactions, your two compounds cannot be oxidized in one fell swoop, so there are no reasonable half-reactions for the highly complicated mess of piecemeal oxidations of degraded fragments of the compounds. Same as for the decay of a dead leaf: its carbon will ultimately be in carbon dioxide. So you are all set. – Ed V Mar 3 '20 at 2:20

The two organic compounds are ultimately being converted to carbon dioxide, $$\ce{CO2}$$, and water, $$\ce{H2O}$$, just as they would be in complete combustion with pure oxygen gas, $$\ce{O2}$$, in a bomb calorimeter. Thus the two balanced combustion equations would be as follows:

$$\ce{C29H24O12 + 29 O2 -> 29 CO2 + 12 H2O} \tag{1}$$

$$\ce{C22H18O11 + 21 O2 -> 22 CO2 + 9 H2O} \tag{2}$$

So now assume that potassium permanganate, $$\ce{KMnO4}$$, is being used as the oxidizing agent, with manganese dioxide, $$\ce{MnO2}$$, as the product. Then, as per Ron's answer (https://chemistry.stackexchange.com/a/85865/79678), and assumed reasonable conditions therein, we have the following reaction:

$$\ce{4 MnO4- + 4 H+ -> 4 MnO2 + 3 O2 + 2 H2O} \tag{3}$$

Showing the spectator potassium ions, and arbitrarily assuming hydrochloric acid as the source of the hydrogen ions, yields

$$\ce{4 KMnO4 + 4 HCl -> 4 MnO2 + 3 O2 + 2 H2O + 4 KCl} \tag{4}$$

Since 29 $$\ce{O2}$$ and 3 $$\ce{O2}$$ have no common factor other than unity, multiply equation 1 by 3, equation 4 by 29, add them, cancel the equal number of $$\ce{O2}$$ molecules on each side and the result is

$$\ce{3 C29H24O12 + 116 KMnO4 + 116 HCl -> 87 CO2 + 116 MnO2 + 94 H2O + 116 KCl} \tag{5}$$

Since 21 $$\ce{O2}$$ is 7 times 3 $$\ce{O2}$$, use equation 2 as is, multiply equation 4 by 7, add them, cancel the equal number of $$\ce{O2}$$ molecules on each side and the result is

$$\ce{C22H18O11 + 28 KMnO4 + 28 HCl -> 22 CO2 + 28 MnO2 + 23 H2O + 28 KCl} \tag{6}$$

It is really surprising that your molecules are transformed to $$\ce{CO2}$$. But if it is the case, if all your theaflavin is oxidized into $$\ce{CO2}$$, the half-equation is :

$$\ce{C29H24O12 + 46 H2O -> 29 CO2 + 116 H+ + 116 e-}$$

The other half-equation depends on the pH of the solution. In highly acidic solution, it produces $$\ce{Mn^2+}$$. In nearly neutral solution, it produces $$\ce{MnO2}.$$