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R-CH2OH(aq) + MnO4{-}(aq) --> R-COO{-}(aq) + MnO2(s)

The above equation is the oxidation of a primary alcohol. The media is neutral to start out with, but the product will be basic*.

From what I know, a reaction like the above will, when in a neutral media, be balanced by adding H+ or OH- on the right side of the equation, not the left. To balance that out, you will instead add H2O on the left side of things.

My research is as follows:
R-CH2OH(aq) + MnO4{-}(aq) --> R-COO{-}(aq) + MnO2(s)
Oxidation number for C:
Left: -I ... right: +III
Oxidation number for Mn
Left: +VII ... right: +IV
Result: 3R-CH2OH(aq) + 4MnO4{-}(aq) --> 3R-COO{-}(aq) + 4MnO2(s)

Next step is balancing the charge(?) of the equation. The same amount of charge on left as right. Right now, it is as follows:
Left: -4 right: -3

We therefore need to add OH- on the left side of the equation. 3R-CH2OH(aq) + 4MnO4{-}(aq) --> 3R-COO{-}(aq) + 4MnO2(s) + OH-(aq) Now that it is balanced, we add H2O to equal ... oh wait, that's not possible. We are missing H on the right side, and we can't add more OH- since that would make the charges(?) not be balanced.

I need advice on this with explanation.

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  • $\begingroup$ It might be worth invoking acid on one side and water on the other $\endgroup$ – Beerhunter Jan 19 '18 at 19:50
  • $\begingroup$ Not sure what you mean here. The equation should be solvable as above. Anyhow, I figured it out on my own. I didn't know you could balance with OH- and H2O on the same side; which means in the above example the answar would be: 3R-CHOH-R + 2MnO4- → 3R-CO-R + 2MnO2 + 2OH- + 2H2O $\endgroup$ – Oliver Jan 19 '18 at 19:54
  • $\begingroup$ 3CH3CH2OH + 2MnO4- ---> 3CH3COOH + 2MnO2 + 4H+ + H2O $\endgroup$ – nagender penthala Apr 18 '18 at 10:20
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The half reaction for the reduction of permanganate is 3e- + 4H+ + MnO4- --> MnO2 + 2H2O

The half reaction for the oxidation of an alcohol to an acid is RCH2OH + H2O --> RCOOH + 4e- + 4H+

Add 4 times the reduction half reaction to 3 times the oxidation

16H+ + 4MNO4- + 3RCH2OH + 3H2O --> 4MnO2 + 8H2O + 3RCOOH + 12H+

Cancelling things that appear on both sides

4H+ + 4MNO4- + 3RCH2OH --> 4MnO2 + 5H2O + 3RCOOH

and Bob's your uncle. Writing it this way also shows that adding acid pushes it to the right, IOW this reaction should be carried out in acid solution.

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