Redox Reactions- MnO4- and H2O2

Started by Pratham Jaiswal, September 09, 2020, 06:09:02 AM

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Pratham Jaiswal

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question

uma

You need to write down the balanced equation of KMnO4 with H2O2
KMnO4 is reduced to Mn 2+
And H2O2 is oxidized to O2
If you look at the half reactions then in the reduction of one mole of MnO4 - we need 5 electrons
In the oxidation of 1 mole of H2O2 to O2 we get only two electrons.


5 electrons + MnO4-  -------> Mn2+

H2O2 ---->  O2 + 2 electrons


Multiply above equation by 2 and below equation by 5
This will equalize the number of electrons transferred and from here we can work out the mole ratio and you can check for every one mole MnO 4 - we need 5 / 2 moles of H2O2

10 electrons + 2MnO4-  -------> 2Mn2+

5H2O2 ---->  5O2 + 10 electrons

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