Okay. I'm very, very confused.
The initial question is: predict the products of the electrolysis of a 1 mol/L solution sodium chloride.
This is my work:
The four different possible equations are:
2H2o + 2e- -------> H2 + 2OH- (V= -0.83)
O2 + 4H+ + 4e- -------> 2H2O (V= 1.23)
Na+ + e- ---------> Na (V= -2.71)
Cl2 + 2e- -----> 2Cl- (V= 1.36)
So the reaction of O2 and the reaction of Cl2 are the reduction reactions, and the reaction of 2H20 and Na are the oxidation reactions.
[ could someone tell me if I'm using the right 2 equations for water? ]
Then I combine the pairs of half reactions:
Cl2 reaction with the 2H2o reaction: (2.19)
Cl2 reaction with the Na reaction: (4.07)
O2 reaction with the 2H2o reaction: (2.06)
O2 reaction with the Na reaction: (3.94)
[i used the equation E cell= Ecathode-Eanode. However, I'm confused because its an electrolysis reaction, and I'm supposed to get negative numbers.]
And then I know that the reaction of 02 with 2H2O requires the least amount of energy, so...
O2 + 4H+ + 4e- ---------> 2H2O (v=1.23)
H2 + 2OH- -----> 2H2O + 2e- (flipped) (V= +0.83)
2H2 + o2 + 4OH- ----------> 6H2O (+0.40)
ANNNDDD that what I did. And I don't get the right answer, as my products are supposed to be hydrogen and oxygen. Could someone PLEASE correct my work?
Water is reduced in preference to Na^+ at the cathode. Water is also prreferentially oxidized relative to the chloride at the anode.
2H2O + 2e ==> H2 + 2OH^-
2H2O ==> O2 + 4H^+ + 4e
Multiply equation 1 by 2 and add.
4H2O + 4e ==> 2H2 + 4 OH^- (cathode)
2H2O ==> O2 + 4H^+ + 4e (anode)
6H2O ==>2H2 + 4 OH^- + 4H^+ + O2
6H2O ==> 2H2 + 4H2O + O2
net equation: 2H2O ==> 2H2 + O2
I just reread your answer and although you did a lot of unnecessary work I think your error was that you didn't multiply the equations involving water (the last set you wrote) to make the electons equal. Therefore, the final equation wasn't balanced and that's why the H^+ and OH^- didn't combine and add out to give 2H2O on the left.