Electrolysis using Displacement current?

Normally, electrolysis occurs at the anode and cathode, not in the water in the middle. Your diagram shows an insulation layer that prevents contact of the water from the anode and cathode.

What kind of electrolysis reaction are you expecting? The reduction reaction (see below) require electrons, but the number of free electrons in pure water due to displacement current will be exceedingly small. You should be able to calculate the energy needed for an oxidation reaction, and compare that with energy available from the displacement current in the immediate vicinity of an H2O molecule.

It sounds difficult to make that work. But you did not specify the voltage level, and if we keep increasing the voltage pulse magnitude eventually something(?) should happen. Neither did you specify how much electrolysis is needed to declare success (a single molecule or more?).

https://en.wikipedia.org/wiki/Electrolysis_of_water

 

Pretty sure the short answer is no...

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Edit: Sorry about the formatting of this post. For some reason the editor is preventing me fix it.

 

http://physics.stackexchange.com/qu...ter-not-used-as-a-dielectric-in-the-condenser

It would take a very very strong field.

 

I'm not sure if there is a distinction between "pure water" and "distilled water" , if yes it might make a big difference. You should research it more for pure water.

 

Is the "strong electric field" across the water or across the insulators?

 

For that to be true the water would need to have a higher impedance than the insulators.

 

In general - dielectric breakdown yield a plasma, and a plasma underwater would at least be a steam explosion and considering you are separating into Hydrogen and Oxygen - and then you have a plasma (ignition source), really does not seem practical to do this. Part of the benefit of traditional electrolysis is the Oxygen and Hydrogen are generated at different locations. Distilled or De-ionized Water - ether may "work" if you measure the conductivity first.

And then - As CW points out you still have the electrodes insulation - so as the water approaches a breakdown (effectively lowering the resistance of the water) then the the voltage applied will be across the electrodes' insulation.

 

You cannot electrolyse water with just voltage or just current. It takes power/energy.