# Question #6b532

May 22, 2017

$4 M n {O}_{4}^{-} + 12 {H}^{+} + 5 C {H}_{3} C {H}_{2} O H \to 4 M {n}^{2 +} + 5 C {H}_{3} C O O H + 11 {H}_{2} O$

#### Explanation:

Potassium permanganate is an oxidant that may be used to oxidise a primary alcohol to a carboxylic acid. The two half-reactions involved in the oxidation of ethanol to ethanoic acid are

$M n {O}_{4}^{-} + 8 {H}^{+} + 5 {e}^{-} \to M {n}^{2 +} + 4 {H}_{2} O \text{ }$ (1)

$C {H}_{3} C {H}_{2} O H + {H}_{2} O \to C {H}_{3} C O O H + 4 {H}^{+} + 4 {e}^{-} \text{ }$ (2)

To balance the overall equation, each half-reaction must have the same number of electrons. Multiply (1) by 4 and (2) by 5, to get 20 electrons in each half-reaction.

$4 \times$ (1):

$4 M n {O}_{4}^{-} + 32 {H}^{+} + 20 {e}^{-} \to 4 M {n}^{2 +} + 16 {H}_{2} O$

$5 \times$ (2):

$5 C {H}_{3} C {H}_{2} O H + 5 {H}_{2} O \to 5 C {H}_{3} C O O H + 20 {H}^{+} + 20 {e}^{-}$

Next, add the two half-reactions together

$4 M n {O}_{4}^{-} + 32 {H}^{+} + 5 C {H}_{3} C {H}_{2} O H + 5 {H}_{2} O + 20 {e}^{-} \to 4 M {n}^{2 +} + 16 {H}_{2} O + 5 C {H}_{3} C O O H + 20 {H}^{+} + 20 {e}^{-}$

Finally, cancel out the like-species on both sides to give

$4 M n {O}_{4}^{-} + 12 {H}^{+} + 5 C {H}_{3} C {H}_{2} O H \to 4 M {n}^{2 +} + 5 C {H}_{3} C O O H + 11 {H}_{2} O$