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I am electrolysing water using several different potential catalysts for my eight grade science fair project. The electrolysis will be running at 2 to 2.5 volts over nickel electrodes, as I do not need a long term solution. How much amperage should I put on my DC power supply? Thanks and I would appreciate simplified answer compared to what you would normally give. Yes, I do understand ohms law.

Edit: I did some calculation and if I assume the resistance of the electrolyte is lets say 5 ohms, then we can assume a current of 0.5A at 2.5 volts. Is this correct?

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    $\begingroup$ Not exactly. To use Ohms law on an electrolysis cell, you have to subtract the cell voltage. Only when you increase the voltage above that level, a current starts to flow. $\endgroup$
    – Karl
    Commented Sep 15, 2019 at 17:36

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In electrochemistry, the rate of electrolysis depends on rate of charge entering the cell. Second most important point is that one can either control potential or current but not both during electrolysis i.e. you cannot have both values set. If you are fixing potential at 2.5 V, the value of current is not in your hand. I don't think Ohm's law remains valid during electrolysis in general.

So the main question one should ask is: What rate of electrolysis do I need? For example, you may want 200 mL of $\ce{O2}$ in 60 min. From here you calculate the charge Q needed to collect this volume of oxygen. Current is charge per unit time (3600 s) and this is your required current above the decomposition voltage of current. See second figure: http://www1.lsbu.ac.uk/water/electrolysis.html

See the sample calculation here: https://www.quora.com/What-amount-of-current-is-required-to-produce-0-224-cc-STP-of-O_2-at-anode-during-the-electrolysis-of-water

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