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

Aug 8, 2017

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

$4 M n {O}_{4}^{-} + 3 {H}_{3} C - C {H}_{2} O H \rightarrow$
$3 {H}_{3} C - C {O}_{2}^{-} + 4 M n {O}_{2} \left(s\right) + 4 {H}_{2} O \left(l\right) + H {O}^{-}$

#### Explanation:

Permanganate is reduced to $M n {O}_{2} \left(s\right)$

$M n {O}_{4}^{-} + 4 {H}^{+} + 3 {e}^{-} \rightarrow M n {O}_{2} \left(s\right) + 2 {H}_{2} O \left(l\right)$ $\left(i\right)$

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

${H}_{3} C - C {H}_{2} O H + {H}_{2} O \left(l\right) \rightarrow {H}_{3} C - C {O}_{2} H + 4 {H}^{+} + 4 {e}^{-}$ $\left(i i\right)$

We add the individual redox reactions in such a way as to eliminate the electrons, i.e. $4 \times \left(i\right) + 3 \times \left(i i\right)$

$4 M n {O}_{4}^{-} + 4 {H}^{+} + 3 {H}_{3} C - C {H}_{2} O H \rightarrow 3 {H}_{3} C - C {O}_{2} H + 4 M n {O}_{2} \left(s\right) + 5 {H}_{2} O \left(l\right)$

But basic conditions were specified, and thus we add $4 \times H {O}^{-}$ to each side.......

$4 M n {O}_{4}^{-} + {\underbrace{4 {H}^{+} + 4 H {O}^{-}}}_{4 {H}_{2} O} + 3 {H}_{3} C - C {H}_{2} O H \rightarrow$
$3 {H}_{3} C - C {O}_{2} H + 4 M n {O}_{2} \left(s\right) + 5 {H}_{2} O \left(l\right) + 4 H {O}^{-}$

To give finally.......

$4 M n {O}_{4}^{-} + 3 {H}_{3} C - C {H}_{2} O H \rightarrow$
$3 {H}_{3} C - C {O}_{2}^{-} + 4 M n {O}_{2} \left(s\right) + 4 {H}_{2} O \left(l\right) + H {O}^{-}$

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 $M n {O}_{2}$.

Note that I am still not happy with this solution. Manganate ion, i.e. $M n \left(V I +\right)$, $M n {O}_{4}^{2 -}$, a green salt is MORE LIKELY in basic solution........

$M n {O}_{4}^{2 -} + 2 {H}_{2} O + 2 {e}^{-} \rightarrow M n {O}_{2} \left(s\right) + 4 H {O}^{-}$

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