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Chemistry Forums for Students => High School Chemistry Forum => Topic started by: makonia on September 09, 2014, 11:25:22 AM

Title: How to balance redox
Post by: makonia on September 09, 2014, 11:25:22 AM
How do I balance this reaction:
KMnO4 → K2MnO4 + MnO2 + O2
Mn is redused to both +4 and +6, also they both come from the same molecule.
Title: Re: How to balance redox
Post by: sjb on September 09, 2014, 12:23:50 PM
Quote from: makonia on September 09, 2014, 11:25:22 AM
How do I balance this reaction:
KMnO4 → K2MnO4 + MnO2 + O2
Mn is redused to both +4 and +6, also they both come from the same molecule.

So manganese is reduced - you can write two half equations for that. Is anything oxidised?
Title: Re: How to balance redox
Post by: AdiDex on September 09, 2014, 01:12:01 PM
Is this redox reaction in acidic medium or basic medium or nothing is given to you...?
Title: Re: How to balance redox
Post by: makonia on September 09, 2014, 01:27:37 PM
The only thing given is that the process is done by heating up KMnO4 in order to produce oxygen.

I suppose i could write two half equations, but they specifically ask that i solve it by oxidation numbers which confuses me because I don't know how to do that when it is redused into two different oxidation numbers. The one that is oxidised is oxygen, which goes from -2 to 0
Title: Re: How to balance redox
Post by: makonia on September 09, 2014, 01:58:39 PM
The way I think Mn starts as +7 and is then reduced to +6 in K2MnO4 and +4 in the other one. that way it receives 4 electrons combined and since oxygen only gives two I have to dobble the amount of oxygen on both sides of the equation. Does this make any sense or am I completely lost right now?
Title: Re: How to balance redox
Post by: Borek on September 09, 2014, 02:10:42 PM
I would not bother with treating it as redox. Start balancing potassium.
Title: Re: How to balance redox
Post by: makonia on September 09, 2014, 02:17:15 PM
Thank you! figured it out now :)