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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 Sep 15 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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If you're choosing the voltage, then you won't get to choose the current. This is because the resistance is already chosen for you, as it is a physical property of your electrochemical cell. Therefore, two of three variables are known in Ohm's Law, which means the third, in this case current, is fixed by the other two, voltage and resistance. For example, using your DC power supply set to constant voltage mode, slowly turn up the voltage from zero. Now watch the current reading as your continue to increase the voltage. At some voltage, the current will spike, as an electrochemical reaction is initiated. In your case, this will be water splitting, evident from gas generation at the electrodes.

The resistance of your electrochemical cell will be dependent upon a variety of factors. However, you don't need to theoretically determine this. The DC power supply is providing all the data you need to determine the cell's resistance! It outputs V and I, so all you need to do is use Ohm's Law to get the resistance (R=V/I). Now you can use this resistance value to calculate the current at different voltages, say 2.5 V. Note that the cell's resistance is dependent on factors like electrode spacing, temperature, and water composition. Hence, the resistance value you derive will only be valid under those conditions.

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  • $\begingroup$ Resistance of an electrolyte is never quoted with a DC current. This defies the fundamental principles of conductometry because you never wish to have electrolysis during current measurement. $\endgroup$ – M. Farooq Sep 30 at 4:40
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    $\begingroup$ @M.Farooq Ah, I slipped up. You are correct $\endgroup$ – Tunk Sep 30 at 4:44

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