Question regarding electrolysis setups?

Question

Heres the info: Two containers filled with water and both containing an added sodium electrolyte. A pipe connects the two containers which allows free passage of the water and therefore the electrolyte. A negative electrode is applied to the left container and a positive electrode applied to the right container. Why does electrolysis not take place in this setup?

Answer 1

You don't mention anything that completes the external circuit. Is there a battery connecting the two electrodes? Also, the "pipe" is supposed to be what is called a salt bridge. This means that there should be a salt solution in it. If the pipe contains pure water, the water will act as an insulator and nothing can happen.

Answer 2

You need to use platinum foil electrodes. If you use another metal as the anode, there is a chance that the oxygen will react with the metal instead of being released as a gas. For instance if you used and iron based electrode you might get iron oxide forming at the anode.

Answer 3

Lancenigo di Villorba (TV), Italy

I read answers which preceed present mine.
I lack information on the external energy supply...what are its technical notes, e.g. supplied tension?
I lack information on the electrolyte's nature, you wrote it is a sodium's salt but it is not a satisfacent communication...e.g. iodide's sodium is a sodium's salt but its employment as electrolyte lead to iodide's electrolysis, instead water's electrolysis!
I lack information on electrode's nature...who preceed me thought iron or mild carbon steels are corrodible matters, hence they can react instead evolved anodic oxygen's gas bubbles.
It may be true, but you could you ever considerate supplied tension at electrodes (e.g. anode who may evolves oxygen bubbles) and current flow in electrical circuit.
DON'T WORRY! You are not constricted to buy platinum's rods (not cheaps), you can substitue them with two graphite's rods (e.g. pencil's tips, C as carbon).

I hope you choose this answer.
Have nice experiments.