[aux]

Hello, I've been trying to write the KMno4 decomposition as a redox reaction but with no success. The reaction is:
 
2 KMnO4 -> K2MnO4 + MnO2 + O2

I know that Mn reduces from +7 to +5 and +4, and O oxydizes from -2 to 0. But I assume I can't just use Mn and O as standlone atoms in the half-reactions?

Anyway, here's what I was thinking the reactions should be:

O: 2O^2- -> O₂ + 4e^-

R: MnO₄^- + e^- -> MnO₄^2-

R: MnO₄^- + 8H⁺ + 3e^- -> Mn⁴⁺ + 4H₂O

Adding it all up, I'd get:

2MnO₄^- +  8H⁺ + 2O^2- -> MnO₄^2- + Mn⁴⁺ + 4H₂O + O₂

Which is almost correct except for the fact that my hydrogen and oxygen ions on the left side don't cancel out with the water on the right. Where did I go wrong?

[chenbeier]

1. Build redox pairs
Which are the ones?
Potassium is spectator leave it out of the reaction.

[Borek]

I know that Mn reduces from +7 to +5 and +4

+6 and +4, not that it matters much.

R: MnO₄^- + e^- -> MnO₄^2-

R: MnO₄^- + 8H⁺ + 3e^- -> Mn⁴⁺ + 4H₂O

Lest's call them R1 and R2 respectively.

This is tricky. Why do you try O + R1 + R2 and not some other combination? Just to balance number of electrons you can as well do O + 4R1, or 3O + 4R2, or some more elaborate version like 11O + 23R1 + 7R2. As these are not unique it is obvious that something is wrong with this approach. Most often it means the process as written is actually sum of several separate processes (in your case O+R1 and O+R2). You introduced the ambiguity by trying to move the process into the realm of reactions taking place in water solutions and splitting the molecules into arbitrary parts.

[chenbeier]

@aux Develop the third equation without H⁺ and water.

Use only permanganete , Manganesedioxide and oxide ions.

After that make an addition of all 3 equations.

[aux]

@Borek

Impossible to balance the three half (third?) reactions. I've tried setting up a system of two equations based on the number of electrons they've got and the fact that I need the O2- and H+ to cancel out with water on the other side, and I get a chemically impossible solution (I'd need to multiply R1 or R2 by a negative number).

So yeah you could be right. I just randomly took a reaction from my textbook and tried solving it as a redox for practice.



@chenbeier

Aight I'll give it a shot, so...


O: 2O^2- -> O₂ + 4e^-

R: MnO₄^- + e^- -> MnO₄^2-

R: MnO₄^- + 3e^- -> MnO₂ + 2O^2-

Adding it all up, I get:

2MnO₄^- + 2O^2- -> O₂ + MnO₄^2- + MnO₂ + 2O^2-

And now my O2- cancel out...

2MnO₄^- -> O₂ + MnO₄^2- + MnO₂

And that does seem to be correct? Damn, thanks dude!

[Borek]

R: MnO₄^- + e^- -> MnO₄^2-

R: MnO₄^- + 3e^- -> MnO₂ + 2O^2-

Trick is, if these reactions were independent, there would be no single reason why they have to be added up in the total equation on a 1:1 basis, and not in some other ratio, after all it is only total number of electrons exchanged between reduction and oxidation that matters. However, no other ratio works here: R2 and O need to be 1:1, otherwise O^2- doesn't cancel out. So in fact these reductions are not two separate and independent processes, this is one combined process:

2MnO₄^- + 4e^- MnO₄^2- + MnO₂ + 2O^2-