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Author Topic: How does atmospheric pressure affect electrolysis?  (Read 6394 times)

Niggle

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How does atmospheric pressure affect electrolysis?
« on: 11/04/2015 10:29:53 »
Does the efficiency of electrolysis of water change when atmospheric pressure is changed? Thanks.
« Last Edit: 17/04/2015 16:53:17 by chris »

evan_au

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Re: Electrolysis question.
« Reply #1 on: 11/04/2015 12:11:10 »
For the extreme case where atmospheric pressure is reduced to 0 (a vacuum), all the water will boil away, and electrolysis will stop.

For the other extreme case, where atmospheric pressure is increased to above 1 GPa, the water will turn to ice, and electrolysis will stop.

For more moderate, day-to-day changes in atmospheric pressure, I don't know (but I am guessing that any effect would be slight).

See: http://commons.wikimedia.org/wiki/File:Phase_diagram_of_water.svg

chiralSPO

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Re: Electrolysis question.
« Reply #2 on: 11/04/2015 13:07:50 »
Atmospheric pressure will have some effects on the efficiency, even at pressures more moderate than the extreme cases Evan mentioned.

Two effects come to mind immediately (and there may be more):

The first is a thermodynamic effect--the atmosphere contains oxygen, which is a product of water electrolysis. By increasing the atmospheric pressure, you also increase the amount of oxygen that remains dissolved in the electrolyte, and this makes the production of oxygen slightly less favorable (a higher voltage is required). The voltage required to drive the oxygen production reaction (as with all electrochemical half-reactions) is related to the concentrations of product and reactant (see: http://en.wikipedia.org/wiki/Nernst_equation)

For the reaction 2 H2O → O2 + 4 H+ + 4 e

E = E0 – (R*T*ln([H2O]2*[O2]–1*[H+]–4))/F

where R is the universal gas constant, T is absolute temperature, F is Faraday's number, and [something] denotes the concentration of something.

The second effect is more of an engineering consideration--the higher the pressure is, the smaller the volume of the bubbles produced (inverse proportionality). This means that for a given rate of hydrogen and oxygen production, less of each electrode will be covered by bubbles (which are effective insulators, and block the reaction). Or, turned around, at higher pressures you can form more product per unit surface area than at low surface area.

The overall change in efficiency, given these competing effects, would be hard to quantify, as it is a function of the temperature, pressure, pH, electrode area, applied voltage, conductivity of the solution and probably a few more parameters I can't think of at the moment.

At reasonable pressures, both effects are minimal. If you want to increase the efficiency of your electrolysis substantially, the easiest variables to play with are composition of the electrolyte, electrode material, and cell geometry.

Niggle

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Re: Electrolysis question.
« Reply #3 on: 11/04/2015 22:06:58 »

chiralSPO

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Re: Electrolysis question.
« Reply #4 on: 12/04/2015 01:15:56 »
Are you actually doing electrolysis, or just thinking about it? I would be interested to hear how you're going about it if you are actually electrolyzing...

Niggle

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Re: Electrolysis question.
« Reply #5 on: 13/04/2015 22:00:12 »
I am only thinking about it at this stage as I am doing a lot of research into making a water powered car.

chiralSPO

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Re: Electrolysis question.
« Reply #6 on: 14/04/2015 21:07:34 »
you cannot power a car with water. It takes energy to split water into hydrogen and oxygen, and that energy has to come from somewhere (and it necessarily takes more energy to split the water than you would get by burning said hydrogen to make water again). There might be benefits to harvesting "waste" energy from your hydrocarbon-fueled car and using it to generate hydrogen, which can be pumped into the engine, but I am not convinced that there could be any dramatic improvement of mileage.

The Naked Scientists Forum

Re: Electrolysis question.
« Reply #6 on: 14/04/2015 21:07:34 »