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I need to balance the equation for the oxidation of primary alcohol. The media is neutral to start out with, but the product will be basic. I assigned oxidation numbers and added stoichiometric coefficients accordingly:

$$\ce{3 R-\overset{-1}{C}H2OH(aq) + 4 \overset{+7}{Mn}O4^-(aq) -> 3 R-\overset{+3}{C}OO^-(aq) + 4 \overset{+4}{Mn}O2(s)}$$

The next step is balancing the charge(?): currently, net charge on the left is −4, and −3 on the right. From what I know, a reaction like the above will be balanced in a neutral media by adding $\ce{H+}$ or $\ce{OH-}$ on the right side of the equation. To balance that out, you will instead add $\ce{H2O}$ on the left side.

We, therefore, need to add $\ce{OH-}$ on the left side of the equation:

$$\ce{3 R-\overset{-1}{C}H2OH(aq) + 4 \overset{+7}{Mn}O4^-(aq) -> 3 R-\overset{+3}{C}OO^-(aq) + 4 \overset{+4}{Mn}O2(s) + OH^-(aq)}$$

Now that it is balanced, we add $\ce{H2O}$ to equal… oh wait, that's not possible. We are missing $\ce{H}$ on the right side, and we can't add more $\ce{OH-}$ since that would make the charges(?) not be balanced.

I need an advice on this with an explanation.

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  • $\begingroup$ It might be worth invoking acid on one side and water on the other $\endgroup$
    – Beerhunter
    Jan 19, 2018 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, 2018 at 19:54
  • $\begingroup$ 3CH3CH2OH + 2MnO4- ---> 3CH3COOH + 2MnO2 + 4H+ + H2O $\endgroup$ Apr 18, 2018 at 10:20
  • $\begingroup$ @TheSmartestNoo please consult Chemistry Meta for some information on how to apply chemistry markup with MathJax. We are using the mhchem extension here. $\endgroup$ Jul 3, 2022 at 23:15

2 Answers 2

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The equation proposed by deLange is interesting, but it is not an answer, because it requires $\ce{H+}$ ions to proceed, and Oliver, the question's author, states that the reaction should not occur in an acidic solution.

So the $\ce{H+}$ ions have to disappear from the deLange's equation, which I repeat here. $$\ce{4H+ + 4 MnO4^- + 3 RCH2OH -> 4 MnO2 + 5 H2O + 3 RCOOH}$$ The only way of getting rid of these $\ce{H+}$ ions is to add $\ce{4 OH-}$ ions on both sides of deLange's final equation. This will transform the $\ce{4 H+}$ into $\ce{4 H2O}$ on the left-hand-side. On the right-hand side, this will transform $\ce{3 RCH2COOH}$ into $\ce{3 RCH2COO- + 3 H2O}$ and the fourth $\ce{OH-}$ ion will remain in the solution, which becomes basic. The obtained equation will be$$\ce{4 H2O + 4 MnO4^- + 3 RCH2OH -> 4 MnO2 + 5 H2O + 3 RCOO- + 3 H2O + OH-}$$which can be shortened according to $$\ce{4 MnO4^- + 3 RCH2OH -> 4 MnO2 + 3 RCOO- + 4 H2O + OH-}$$If the initial solution was neutral, it becomes basic at the end.

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The half reaction for the reduction of permanganate is $\ce{3e- + 4H+ + MnO4^- -> MnO2 + 2H2O}$

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

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

$\ce{16H+ + 4MnO4- + 3RCH2OH + 3H2O -> 4MnO2 + 8H2O + 3RCOOH + 12H+}$

Cancelling things that appear on both sides

$\ce{4H+ + 4MnO4- + 3RCH2OH -> 4MnO2 + 5H2O + 3RCOOH}$

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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