Hi
I understand that in the electrolysis of water, there are these half equations
1. Cathode (reduction): 2 H2O(l) + 2e− → H2(g) + 2 OH−(aq) E=-0.83
2. Anode (oxidation) 2 H2O(l) → O2(g) + 4 H+(aq) + 4e− E=1.23
3. Cathode(reduction) 2H+(aq) + 2e− → H2(g) E=0
4. Anode(oxidation) 4 OH- (aq) → O2(g) + 2H2O(l) + 4e- E=0.4
And from what I understand, all four occur. But two will be the main ones that occur.
In Neutral conditions, we'd get (1,2) as the main ones
In Acidic conditions we'd get (3,2) as the main ones
In Basic conditions we'd get (1,4) as the main ones.
And from what I understand, the E values there are SEP - standard electrode potentials, particualrly, E(red), so, for reduction. And they need to be converted into "Actual Potentials" in order to be of any use, is that right?
'cos from what I understand, in order to predict the products that will be the main products, so the half equations that are the main ones. One would look at the half equations for the Cathode, and pick the one with the highest reduction potential. And since E(ox) = E(red)*-1 The half equation with the lowest E(red) value is the one with the highest E(ox), and thus the main one at the Anode. So for the Anode I should pick the lowest E value. But the problem I am having here, is that the E values don't reflect the fact that in neutral conditions I should get (1,2) . In acidic (3,2). And in Basic (1,4).
e.g. I get
3. Cathode(reduction) 2H+(aq) + 2e− → H2(g) E=0 <--- Highest E for cathode as 0> -0.83
4. Anode(oxidation) 4 OH- (aq) → O2(g) + 2H2O(l) + 4e- E=0.4 <--- Lowest E for Anode as 0.4 < 1.23
But i'm not sure that any conditions are meant to produce main ones of (3,4).
So, i'm thinking the E values would be different for e.g. pH 3 (for acidic), or pH(10) for basic.
Let's assume room temperature and atmospheric pressure.
I'm wondering what the actual potentials are, and whether that rule about highest E for cathode, and lowest E for Anode, will then apply?
Thanks