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Electrolysis of water 1

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markls8

Mechanical
Jul 20, 2002
16
Sorry, but it's been a while since high school chemistry -How much pressure (hydrogen and oxygen) could I expect to develop in a closed two part vessel if I applied, say 24V to water? (assume continous water supply to reach equilibrium) How much would that pressure increase if I increased V? Thanks - Brian.
 
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I think most of the pressure would be due to the heat/steam ,unless you provide efficient cooling. <nbucska@pcperipherals.com>
 
OK - then assume efficient cooling close to ambient temp.
Thanks - Brian.
 
This is a good question. I've never heard of high pressure affecting the dissociation reaction. So you are keeping voltage constant and letting the current drop as the reaction slows down? Conductivity of the water and electrode separation are constant too I assume.

Maybe the reaction would slow as the concentrations of hydrogen and oxygen dissolved in the water reached the saturation level. As the gas pressure increased it would slow the evolution of new gas from the liquid. This might inhibit bubble formation around the electrodes.

Or, the reaction might slow as the pressure of the gas above the water approached the pressure of the gas inside the newly forming bubbles so that new bubbles would not be inclined to form and rise anymore. This sounds more plausible. It is possible to calculate the pressure inside a bubble of a given diameter. It probably requires knowing the surface tension of the water/electrolyte solution and depth of the bubble. I've seen these kinds of figures long ago, I wish I could tell you where.
 
Here are some links to sites that deal with this issue. The first 2 look best. According to them, the electrolytic reaction is unaffected by pressure. The others deal with pressure & surface tension in one form or another.




 
Thanks sciguyjim - Some very helpful links. -Brian
 
Just a side note - as we deal with electrolytic treatment of effluent water - we have never found any difference at sea level, as compared to 8,000' elevation, as to the level of DO produced in the water source, utilizing the same voltage, amperage, etc. (all other factors being equal). Since I don't know the conductivity of the solution, the amperage v. voltage output, and all the compounds (metals, minerals, salts, etc.) in the solution, there is no way to properly answer the question, as pressure would effect other reactions outside that of the oxygen and hydrogen.

Hope that did'nt muddy the water too much.

Dave Orlebeke
Aquatic Technologies
 
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