Voltage breakdown of distilled water

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The discussion focuses on the DC breakdown voltage of distilled water, noting that it varies based on electrode surface finish and the presence of contaminants. Distilled water can differ significantly from ultra-pure water due to variations in purity and exposure to air, which introduces carbon dioxide and alters pH. The high specific resistance of ultra-pure water necessitates higher voltages to achieve current flow. The conversation also touches on the importance of electrode material and activation energy in determining breakdown voltage. Finally, the Nernst equation is suggested as a method to derive potential limits, emphasizing the need to account for pH and ohmic drop in calculations.
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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
 
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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
 
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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.
 
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?
 
Borek said:
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.
 
axawire said:
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.



 
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