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.

  • $\begingroup$ 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

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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.