Annoying Redox Rxn

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[dumples92]

For chemistry, I have to balance this redox reaction. That means I have to balance all the charges, and make the charges of both sides equal zero. Here is the equation:

MnO₄^-(aq) + C₂H₅OH(aq) → Mn²⁺(aq) + CH₃COOH(aq)

[kyle1990]

Please show that you've attempted the problem before you post homework questions. Are you having trouble getting started?

[dumples92]

Well, sort of.
I tried working it out, starting by assigning oxidation numbers. The problem is that in
CH3COOH, each of the carbons have a different oxidation number.
I just can't figure out how i am supposed to figure out how many electrons it gains.

[kyle1990]

If you can't figure out one of the half reactions, try the other one first.

MnO4-  -------> Mn2+

I won't show the work here, but if you need me to, just ask and i will:

8H+   +   5e-    +   MnO4-   ------>  Mn2+   +   4H2O

So there's the first half reaction, now onto the second. The carbon in the CH3 group in the acetic acid formed in this reaction has the same oxidation state as it did in the starting material: -3. There is a change in the oxidation number of the other carbon atom, however, from -1 to +3. The oxidation half-reaction therefore formally corresponds to the loss of four electrons by one of the carbon atoms.

 CH3CH2OH +  H2O ------> CH3CO2H  + 4 e- + 4 H+



Now balance the final overall reaction. Don't forget that the number of electrons lost is the oxidation reaction MUST equal the number of electrons gained in the reduction reaction