An exercise I have tried to solve is this
"Show electrode semiractions that take place during the electrolysis of a solution 1,0 M of KOH." My proceeding to solve this problem
At the cathode, the negative pole, occurs the reduction.
Species wich can undergo reduction are K+ and H2O.
It is necessary however to see what semielement algebraically has the HIGHEST potential of reduction.
1) K+ + e- --> K
2) 2 H2O + 2 e- --> 2 OH- + H2 .
Since [K+] e [OH-] are 1 M, respective potentials are:
1) E(K+/K) = E°(K+/K) = -2,93
2) E(H2O/H2) = E°(H2O/H2) = -0,83 .
Therefore at the cathode the reduction of water to hydrogen takes place.
This is right.
We now consider the anode, the positive pole, which the oxidation takes place at.
Here the oxidizable species are H2O and OH-.
It is necessary to see however what semielement algebraically has the LOWEST potential of reduction.
3) 2 H2O --> 4 H+ + O2 + 4 e-
4) 4 OH- --> O2 + 2 H2O + 4 e- .
Semielement potentials in this case:
3) E(O2/H2O) = E°(O2/H2O) + 0,0591/n log [H+]^4 =
= 1,23 + 0,0591/4 * (-56) = 0,40 .
(note that [H+]=10^-14 M, since [OH-] is 1 M; therefore [H+]^4 is 10^-56)
4) E(H2O/OH-) = E°(H2O/OH-) = 0,40
(since [OH-] is 1 M). My question
Why my book is sure that the only possible reaction at the anode is the 4), if he could according to the calculations happen in the same way also the 3) (water oxidation)?
The oxygen overvoltage does not enter there because oxygen forms in both cases. Where is the explanation
Thank you for your answer.
Be clear and precise please!