Electrolysis using Displacement current ?

[Proud Teslian]

The Figure shows two insulated metal electrodes kept in pure water with a small gap between them.

 

When sufficiently large Voltage pulses applied between electrodes, there is a flow of displacement current through the insulators and water, but will this split the water molecules?

 

All your answers are welcomed !

[studiot]

The first thing you need to do is show a proper circuit diagram with all the circuit elements in place.

 

Then intelligent comment can be made.

Edited May 21, 2016 by studiot

[Proud Teslian]

what do you think is missing ? i have animated the idea for easier understanding ? if you tell me what's missing i'll explain it to you!

[studiot]

I didn't ask for an explanation I asked for a proper circuit diagram.

 

Your pretty picture does not conform to your description/explanation above.

 

In particular the equipment and all connections to create this are not shown.

 

 

When sufficiently large Voltage pulses applied between electrodes

 

Displacement current does not flow through the insulators, that is a convenient fiction to simplify the maths.

[Proud Teslian]

OK ...To say simple above idea tries to achieve dielectric Breakdown of water using strong E field.

 

H2O has 65KV / mm strength, ......but heat and input frequency will lower the Breakdown voltage and make the idea engineerable.

[John Cuthber]

 

H2O has 65KV / mm strength,

Water conducts (unless you freeze it)

[Sensei]

kept in pure water with a small gap between them.

By "pure water" what do you mean?

 

Distilled water?

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

 

Deionized water?

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

 

I have tried to do electrolysis of distilled water (distilled by me, a few minutes prior electrolysis), with 230 Volts, and after hours (the whole night >12 h) there was not a single bubble of Hydrogen nor Oxygen in graduated cylinders (to catch gas) above electrodes.. I used 24 carat Gold (14 nor 18 can't be used, because soon you will have some Ag⁺ and/or Cu²⁺ ions floating and destroying experiment) on either positive and negative electrode. Ammeter showed 0 A, and watt meter showed 0.00 kWh energy used.

 

Then I dropped (literally) single drop of tap water to container (approximately 2 L of distilled water), and electrolysis started, ammeter started showing I>0A.

Maybe 0.01 A, I don't remember exactly, it was long time ago, but the smallest current it can show.

Edited May 22, 2016 by Sensei

[John Cuthber]

The conductivity of pure water is quite well documented.

 

For example

https://us.mt.com/us/en/home/supportive_content/know_how/Paper-THOR-Fundamentals-Cond-Res-08-2004/jcr:content/download/file/file.res/Fundamental_Conductivity_and_Resistivity_of_Water_Oct09.pdf

 

Pure water contains H⁺ and OH ^- ions- each at a concentration of about 10^-7 molar.

These give rise to a resistivity of about 18 Mohm cm near room temperature.

[studiot]

Please note that conductivity has nothing whatsoever to do with displacement current, which the OP was asking about.

That is due to the capacitive effect between the insulated (not 24 carat gold) electrodes.

 

However since the OP will not tell us more about the complete circuit, including earthing, upon which the displacement current depends, nothing more can be deduced.

[Enthalpy]

With pure water, as we used in semiconductor manufacturing, the conduction current is minute, so the current pulse not too long results essentially from water's polarization. Whether the electrodes are insulated will change little.

 

The polarization is mainly the orientation of the molecules in the case of water - or rather, of large groups rather than individual molecules, which explains water's big permittivity. Other compounds have their molecules oriented almost independently, in which case the ambient heat defeats the torque created by the external electric field, or their molecules may be nonpolar, and then the deformation of the molecules (atoms' position, orbitals' shape) is the only response to the external field.

 

And, no, the orientation of the water molecules resulting from the voltage pulse will not split them. For electrolysis, you need preexisting ions that join the electrodes and get neutralized there.

OK ...To say simple above idea tries to achieve dielectric Breakdown of water using strong E field.

H2O has 65KV / mm strength, ......but heat and input frequency will lower the Breakdown voltage and make the idea engineerable.

 

First, don't trust dielectric strengths. They are not reproducible, and they decrease (though not quickly) with a bigger distance. They decrease also with the insulator's volume and with time, suggesting that some random process triggers the discharge.

 

I'd have to double-check if heat eases the dielectric breakdown of water. For sure, short pulses don't ease the breakdown, quite the opposite: they help insulators survive the field, and significantly.

 

You don't need short pulses to make sparks in water. Put a needle in deionized water, insulate the electrode everywhere else, put a high DC voltage, zap.

 

I just wonder why one should want a high voltage for electrolysis. The charge makes the produced amount, and to save energy, you want to provide it at a voltage as low as possible. This also implies concentrated solutions, not pure water. And it demands many wide electrodes close to an other, as is done industrially.

[studiot]

 

enthalpy

And, no, the orientation of the water molecules resulting from the voltage pulse will not split them. For electrolysis, you need preexisting ions that join the electrodes and get neutralized there.

 

 

A good deal of valuable input, particularly this +1

 

 

 

enthalpy

Put a needle in deionized water, insulate the electrode everywhere else

But please note the OP drawing has the electrodes entirely encased in insulation.

[aristyllus]

Could be you'll have better luck in the chemistry section. Electrolysis is a big part of electrochemistry!