Redox reactions, cellular respiration - glucose?

Question

So I want to give the aerobic harvesting of energy from glucose in cellular resp as an example of a redox reaction. I have to show the increase/decrease in oxidation number to demonstrate the oxidation/reduction reactions, however, the number of e- doesn't cancel, i.e isn't equivalent...

C6H12O6 + 6O2 --> 6CO2 + 6H2O + (energy – ATP + heat)
Oxidation half-reaction
C6H12O6 --> CO2
a + b + c = 0 a + b = 0
a + (+12) + (-12) = 0 a + (-4) = 0
a = 0 a = +4
ON of C = 0 ON of C = +4
increase of 4

Reduction half-reaction
O2 --> H2O
ON of O = 0 ON of O = -2
decrease of two

Shouldn't the number of 'freed' electrons be equivalent???
thanks!

Answer 1

Well done.

You have calculated the oxidation numbers correctly.

and you are right that the total number of electrons lost by oxidation should be equal to the number of electrons gained during the reduction. BUT be careful: the total number of electrons must be balanced and not the changes in the oxidation numbers of the reactants.

Here for C you go from 0 to +4 so you have a change of 4 per atom of C. But you have 6 C atoms participating in the reaction thus in total 4*6=24 e lost (oxidation)

For O you go from 0 to -2 so you have a change of 2. But you have 6*2=12 O being reduced in the reaction thus you have 12*2=24 e gained (note that O in C6H12O6 is already at -2 so its oxidation number is not changed).

So you see that atlhough the changes in oxidation number are not the same (4 vs 2) the total number of e is balanced (24e) because of the different stoichiometry of the reactants.