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

1 Answer
Aug 8, 2017

Answer:

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.......