Voltage breakdown of distilled water

[axawire]

I can't seem to find this value...

Does anyone know what the DC breakdown voltage of distilled water is at standard temp and pressure.

Thanks

 

[mgb_phys]

It's not very well defined as it depends critically on the surface finish of the electrodes since any irregularities cause cavitation which start a breakdown.

There is some data here http://www.waterfuelconverters.com/SandiaNationalLabsData.html

 

[Borek]

Don't forget "distilled water" is quite often not what you think it is. In fact it can be distilled, or RO or DI water. Then, it can be ultra pure water. Each of these has slightly different combination of contaminants (similar, but not repetable between samples from different sources).

Next problem: such water if allowed to contact with air, quite fast gets saturated with carbon dioxide. That lowers pH to around 5.5, changing half reaction potentials. In theory these changes should cancel out (same change on both electrodes), but you have to know that you are no longer working with pure water.

Next problem: really pure water has very high specific resistance (ultra pure is sometimes listed as 18MΩ water, not without a reason). While it doesn't change half reaction potentials, it forces you to use high voltage just to force any current flowing through the system.

And finally - as mgb_phys pointed out - a lot depends on the electrode material and finish. As far as I am aware it hasn't anything to do with cavitation, rather with activation energy, google electrochemical overvoltage.

 

[axawire]

Thank you for your replies, do any of you know off hand if there is a way to mathematically derive a lower limit on the strength of the E-Field or Potential based on the assumption you only have H2O and electrode effects do not exist?

 

[mgb_phys]

As far as I am aware it hasn't anything to do with cavitation, rather with activation energy, google electrochemical overvoltage.

Interesting - I had assumed it was microbubbles.

 

[Borek]

Thank you for your replies, do any of you know off hand if there is a way to mathematically derive a lower limit on the strength of the E-Field or Potential based on the assumption you only have H2O and electrode effects do not exist?

Check Nernst equation. You have two half reactions - oxidation and reduction going on on two electrodes. Each half reaction has its own potential - these are given in standard half reaction potential tables. You have to account for pH, as standard potentials are given for standard state, which means pH=0. Then there is ohmic drop. Potential needed for both simultaneous reactions is in the range of volt or two.