Thanks for the answers!
I wonder if NaOH would be better than NaCl in understanding this context.
So with a basic solution there's gonna be a LOT more OH
- than there would otherwise be with pure water.
Looking up the half reactions in a basic solution I see:
[itex]\ce{2H2O(l) + 2e- -> H2(g) + 2OH- (aq)}[/itex]
[itex]\ce{4OH- (aq) -> O2(g) + 2H2O(l) + 4e-}[/itex]
So the high amount of OH
- will allow this to happen way more often than pure water with barely any H
+. I see that. And I can see how HCl would do something similar. But NaCl doesn't change the pH like these strong base/acids would. I'm still not sure how IT would help as far as providing more H
+/OH
-.
why does it need to conduct electricity? Isn't the circuit completed by the ionic current of the H+ ions?
It is. But there are so few of them the resistance is very high (ultrapure water has a specific resistivity of about 18 MΩm). High resistance means low currents and pretty slow electrolysis.
Right. Maybe I'm misunderstanding exactly why this helps, though. By reducing the resistance in the water, what I imagine happening is that electrons can flow
through the water from the cathode to the anode. But if these electrons are just going through the water, they're not actually contributing to the process of electrolysis. That is, every electron leaving the cathode on its way to the anode is one electron that would have otherwise reduced a hydrogen to H
2. And every electron that makes its way to the anode is one fewer electron that oxygen would have given up to oxidize to O
2.
To put it even further, if 100. A of current is being supplied by my power source and 99 A goes
through the salt water, that only leaves 1 A of current actually contributing to the electrolysis.
I'm guessing my mental model of electrons going
through the water is incorrect.