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Amperage for the electrolysis of water?


What are possible side reactions for the generation of hydrogen via electrolysis?Calculate how much hydrogen or oxygen will be produced in the electrolysis of wateraluminum hydroxide as an electrolyteElectrolysis of Water: Overvoltage?High voltage AC water electrolysis not workingAlkaline Water Electrolysis with RbOHVoltage needed for electrolysis of waterBest Settings for Electrolysis of water






.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty margin-bottom:0;








1












$begingroup$


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?










share|improve this question









New contributor



firebanner64 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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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
    7 hours ago

















1












$begingroup$


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?










share|improve this question









New contributor



firebanner64 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.






$endgroup$













  • $begingroup$
    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
    7 hours ago













1












1








1


1



$begingroup$


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?










share|improve this question









New contributor



firebanner64 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.






$endgroup$




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?







electrochemistry electrolysis hydrogen electricity






share|improve this question









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share|improve this question









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share|improve this question




share|improve this question








edited 6 mins ago









M. Farooq

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asked 8 hours ago









firebanner64firebanner64

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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
    7 hours ago
















  • $begingroup$
    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
    7 hours ago















$begingroup$
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
7 hours ago




$begingroup$
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
7 hours ago










1 Answer
1






active

oldest

votes


















2














$begingroup$

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 $ceO2$ 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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    1 Answer
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    active

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    active

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    2














    $begingroup$

    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 $ceO2$ 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






    share|improve this answer











    $endgroup$



















      2














      $begingroup$

      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 $ceO2$ 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






      share|improve this answer











      $endgroup$

















        2














        2










        2







        $begingroup$

        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 $ceO2$ 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






        share|improve this answer











        $endgroup$



        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 $ceO2$ 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







        share|improve this answer














        share|improve this answer



        share|improve this answer








        edited 6 hours ago

























        answered 7 hours ago









        M. FarooqM. Farooq

        6,79710 silver badges28 bronze badges




        6,79710 silver badges28 bronze badges
























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