Thread: power draw for electrolysis of water

Can anyone tell me if there is an optimal voltage required for electrolyzing water? I might have seen somewhere that it requires between 1.6 and 2 volts, but i can't confirm this. Or does the voltage required depend on the volume of water? I have my self believing that the volts required would remain constant, but the amps would fluctuate depending on the volume of water. I'd appreciate it if someone set me straight on this one.

Essentially I am trying to determine the voltage and amperage required for electrolyzing 1 liter of water. All advice is greatly welcomed. Thanks

well you don't want the current to get to high otherwise your electrodes will heat up and boil the water so that could set a maximum limit

Get a 12v car battery charger and play around with it.

I don't believe the volume of water matters - it is only between the electrodes that anything happens. You want these to be close for faster reaction, I think preferably just two large plates facing each other. The current is mainly "line of sight" so most bubbles appear on facing surfaces.

Amps increase as you shorten the distance between electrodes.

The space between anode & cathode may be screened with something porous (fine fabric) to keep the bubbles from mixing. This won't add resistance.

An electrolyte helps. Salt yields a number of nasty outcomes. About a gram of washing soda or baking soda per liter works nicely.

Stainless steel will produce a toxic gas, don't use it. Plain steel is safe, effective, and readily available in useful shapes. Carbon or platinum are good in theory. Where to get plates of those?

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There is no optimal voltage for the electrolysis of water, just as there is no optimal voltage for producing incandescant light. In general, the more electrical power you provide, the greater the output.

Ohm's law still applies. It is the current that does all the work. Since you normally want higher current, you must supply higher voltage.

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Thanks for the replies, that pretty much answers my questions and more. I appreciate the help.