What happens to a redox reaction when it is shown as half-reactions?

1 Answer
Jul 28, 2017

Answer:

Well, not much.........

Explanation:

Our strategy of separating out oxidation and reduction equations to represent the overall chemical reaction is a means to an end. And the end is to give a stoichiometrically balanced equation that accurately represents the macroscopic chemical change that occurs.

I grant that it may seem a little abstract to assign oxidation numbers, and represent the loss and gain and electrons, and both of these are highly conceptual entities. But it does finally give us an equation that represents gross chemical transfer. And as all chemical reactions conserve mass and charge, our use of electrons helps us to represent this conservation of charge:

#MnO_4^(-) +8H^(+) + 5e^(-) rarr Mn^(2+) + 4H_2O# #"; reduction (i)"#
#Mn(+VII)rarrMn(+II)#

#RCH_2OH+H_2Orarr RCO_2H+ +4H^(+) + 4e^(-)# #"; oxidation (ii)"#
#C(-I)rarrC(+III)#

And we cross-multiply the reactions, so that electrons do not appear in the final reaction, i.e. #4xx(i)+5xx(ii):#

#4MnO_4^(-) +5RCH_2OH+12H^(+) rarr4Mn^(2+) +5RCO_2H+11H_2O#

Is charge balanced as well as mass? It is your problem?

What would we observe in this reaction? The strongly (and beautiful) deep purple colour of permanganate would dissipate to give the ALMOST colourless #Mn^(2+)# ion.