Question

How do we represent the oxidation of ethanol to acetic acid, accompanied by the reduction of potassium permanganate to manganese(IV) oxide?

Answer 1

Well let us try it..........

`4MnO_4^(-) + 3H_3C-CH_2OH rarr`
`3H_3C-CO_2^(-) +4MnO_2(s) +4H_2O(l) +HO^(-)`

Explanation:

Permanganate is reduced to `MnO_2(s)`

`MnO_4^(-) +4H^(+) + 3e^(-) rarr MnO_2(s) +2H_2O(l)` `(i)`

Ethanol is oxidized to `""^(-)O(O=)C-CH_3`

`H_3C-CH_2OH +H_2O(l) rarr H_3C-CO_2H + 4H^(+) + 4e^(-)` `(ii)`

We add the individual redox reactions in such a way as to eliminate the electrons, i.e. `4xx(i)+3xx(ii)`

`4MnO_4^(-) +4H^(+) + 3H_3C-CH_2OH rarr 3H_3C-CO_2H +4MnO_2(s) +5H_2O(l)`

But basic conditions were specified, and thus we add `4xxHO^-` to each side.......

`4MnO_4^(-) +underbrace(4H^(+) +4HO^(-))_(4H_2O) + 3H_3C-CH_2OH rarr`
`3H_3C-CO_2H +4MnO_2(s) +5H_2O(l) +4HO^(-)`

To give finally.......

`4MnO_4^(-) + 3H_3C-CH_2OH rarr`
`3H_3C-CO_2^(-) +4MnO_2(s) +4H_2O(l) +HO^(-)`

The which I think is balanced with respect to mass and charge; as indeed it must be if we purport to represent chemical reality. What would see in the reaction? The intense, deep purple colour of permanganate ion would dissipate to give a brown precipitate of `MnO_2`.

Note that I am still not happy with this solution. Manganate ion, i.e. `Mn(VI+)`, `MnO_4^(2-)`, a green salt is MORE LIKELY in basic solution........

`MnO_4^(2-) +2H_2O + 2e^(-) rarrMnO_2(s) +4HO^(-)`

I will let you do the adding of this to the oxidation equation.......