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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
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$begingroup$
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?
electrochemistry electrolysis hydrogen electricity
New contributor
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add a comment
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$begingroup$
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?
electrochemistry electrolysis hydrogen electricity
New contributor
$endgroup$
$begingroup$
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
add a comment
|
$begingroup$
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?
electrochemistry electrolysis hydrogen electricity
New contributor
$endgroup$
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
electrochemistry electrolysis hydrogen electricity
New contributor
New contributor
edited 6 mins ago
M. Farooq
6,79710 silver badges28 bronze badges
6,79710 silver badges28 bronze badges
New contributor
asked 8 hours ago
firebanner64firebanner64
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92 bronze badges
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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
add a comment
|
$begingroup$
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
add a comment
|
1 Answer
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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 $ceO2$ 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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$begingroup$
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
$endgroup$
add a comment
|
$begingroup$
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
$endgroup$
add a comment
|
$begingroup$
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
$endgroup$
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
edited 6 hours ago
answered 7 hours ago
M. FarooqM. Farooq
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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