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I want to electrolyze water as a personal project, but I have found conflicting power recommendations on google. Some sources recommended a higher voltage, claiming that higher amps will just destroy the electrodes, while others insist a higher amperage is required, and extra voltage will be wasted heating up the water. Unable to find one absolute result, I'm turning here- assuming not a crazy amount of amps or volts, what is a good ratio to get the best results? (an explanation of the answer would be appreciated.)

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Ideally, it takes 1.23 volts to split water. Less voltage than this will give essentially no current, no electrolysis. If you want more H2 + O2, you raise the potential. (Of course, you use an electrolyte like H2SO4 or NaOH with electrodes that will not corrode. Platinum is OK, also gold. Oh, yeah, also stainless steel for ordinary people.) If the surface becomes covered with bubbles of gas, the effective area of the electrode is reduced, so the current will drop. So you raise the voltage, and bubbles come pouring off. Then the electrolysis uses more energy (volts x amps). 1 amp for 96,500 seconds (27 hours) will give you 1 gram of hydrogen and 8 grams of oxygen. The product is dependent on the current, and the current is dependent on how high you raise the voltage.

It's not as slow as watching grass grow, but even at furious bubbling, it's surprising how slowly the volumes build up.

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  • $\begingroup$ Thanks for your answer! I could use some clarification, though. What actually causes the bubbles, amperage or voltage? $\endgroup$ – Rafael Jan 30 '18 at 5:08
  • $\begingroup$ When an electron flows to/from an ion, it is a current. The voltage is how hard you have to push it. So the quantity of product (H2, O2) is proportional to the electron flow, which is current in amperes. 1 ampere for 96,500 seconds is one Faraday = 1 mole of electrons. $\endgroup$ – James Gaidis Jan 31 '18 at 4:41

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