# Electrolysis of water?

1. Dec 27, 2011

### cragar

When you see the bubbles coming off of the 2 electrodes in the water, why don't you see bubbles traveling in the water to the electrodes? When the water molecule splits, the oxygen and hydrogen would have to travel to their electrodes, so why don't you see them do this?
And also what causes the water molecule to split. Is the E field created from the current enough to split the polarized water molecule apart?

2. Dec 27, 2011

### Staff: Mentor

Water molecules react directly on the electrodes, not in the bulk of the solution. Reaction requires charge transfer, they are not being "torn out" by the electric field.

3. Dec 27, 2011

### cragar

When you say the reaction happens at the electrodes, So at each electrode it splits the water molecule, so why wouldn't I get both $O_2$ and $H_2$
bubbles at both ends. And when you say the reaction happens because of charge transfer, is the water molecule transferring an electron to one of the electrodes and then it becomes unstable and separates into Hydrogen and Oxygen.
Thanks for your response by the way.

4. Dec 27, 2011

### epenguin

All chemical reactions are some kind of electron transfer. The most subtle ones are some kind of rearrangement of where the electrons in a molecule are. But electrolysis is a particularly obvious kind of electron transfer.

H2O dissociates into H+ and OH- (a small fraction of these is present in any aqueous solution). So to which electrode can you imagine either of these being attracted and what could it do in the way of giving up/acquiring an electron in what chemical reaction?

Last edited: Dec 28, 2011
5. Dec 28, 2011

### Staff: Mentor

If there is a potential difference between electrodes it means that they are charged differently, doesn't it? In other words - there is an excess of electrons on one, and a lack on another. Both gases are produced in different reactions - one consumes electrons, other gives them away. The reaction that consumes electrons goes only on the electrode with excess electrons, while the one that produces them goes on the other electrode.

(This is not contradicting what epeneguin wrote, this is just another part of the full explanation).

6. Dec 28, 2011

### cragar

ok thanks for your posts, I think I am starting to understand a little better.

7. Dec 28, 2011

### Antiphon

So far we have an image of H20 turning into H (which bubbles up at one electrode) then an OH migrating through the liquid to the other electrode where (?) it gets the remaining H torn off (but this one doesn't bubble up) and the O goes up as a bubble here.

Not a very nice picture yet.

8. Dec 28, 2011

### Staff: Mentor

Perhaps the best (even if not necessarily correct) way of dealing with the problem is to write two reaction equations:

2H+ + 2e- -> H2(g)

4OH- -> 2H2O + 1O2(g) + 4e-

As you see it is not necessary that hydrogen "gets torn off but doesn't bubble up".

That being said I am not sure what the exact mechanism of the electrolysis is, could be it changes depending on the pH. While it is easy to write reactions involving H+ and OH-, and while it is obvious these ions are attracted to cathode and anode, in low or high pH solutions their concentrations are very low, but we don't observe any slowing of electrolysis. Still it is solution conductivity that is the bottleneck. As concentration of these ions changes by 14 orders of magnitude, could be their presence doesn't matter much, and it is just water molecules that react. It can also be that water autodissociation is so fast, even these 14 orders of magnitude don't matter.