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HelloCthulhu

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- Homework Statement
- If 7ml of hydrogen and 3.5 ml of oxygen are produced at 1A per minute during electrolysis, how much water is being electrolyzed per minute?

- Relevant Equations
- I * t * mw / n * F

I = current

t = time in seconds (s)

mw = molecular weight

n = moles of electrons

F = Faraday's Constant = 96,500C/mol

I've been doing a little independent research into PEM fuel cells and came across this Horizon mim fuel cell. I read the following specs, but didn't see anything about the volume of water being electrolyzed.

https://www.fuelcellstore.com/manuals/horizon-mini-pem-electrolyzer-instructions-fcsu-010.pdf

- Input Voltage: 1.8V ~ 3V (D. C.)

- Input Current: 0.7A

- Hydrogen production rate: 7ml per minute at 1A

- Oxygen production rate: 3.5ml per minute at 1A

I tried using Faraday's Law of Electrolysis to solve the initial volume of water undergoing electrolysis, but my answer is much larger than I expected. Here's my attempt:

If 7ml of hydrogen and 3.5 ml of oxygen are produced at 1A per minute during electrolysis, how much water is being electrolyzed per minute?

Cathode: 4 H+(

Anode: 2 H2O(

I * t * mw / n * F

I = current

t = time in seconds (s)

mw = molecular weight

n = moles of electrons

F = Faraday's Constant = 96,500C/mol

hydrogen = (1A*60s*mw)/(4*96,5000C/mol) = 7ml

mw = 45033g

oxygen = (1A*60x*mw) / (4*96,500C/mol) = 3.5ml

mw = 22516.6g

67549.6g of water

I'm pretty sure I'm using the wrong equation and I'm feeling really lost. Any help is greatly appreciated.

https://www.fuelcellstore.com/manuals/horizon-mini-pem-electrolyzer-instructions-fcsu-010.pdf

- Input Voltage: 1.8V ~ 3V (D. C.)

- Input Current: 0.7A

- Hydrogen production rate: 7ml per minute at 1A

- Oxygen production rate: 3.5ml per minute at 1A

I tried using Faraday's Law of Electrolysis to solve the initial volume of water undergoing electrolysis, but my answer is much larger than I expected. Here's my attempt:

If 7ml of hydrogen and 3.5 ml of oxygen are produced at 1A per minute during electrolysis, how much water is being electrolyzed per minute?

Cathode: 4 H+(

*aq*) + 4e− → 2H2(*g*)Anode: 2 H2O(

*l*) → O2(*g*) + 4 H+(*aq*) + 4e−I * t * mw / n * F

I = current

t = time in seconds (s)

mw = molecular weight

n = moles of electrons

F = Faraday's Constant = 96,500C/mol

hydrogen = (1A*60s*mw)/(4*96,5000C/mol) = 7ml

mw = 45033g

oxygen = (1A*60x*mw) / (4*96,500C/mol) = 3.5ml

mw = 22516.6g

67549.6g of water

I'm pretty sure I'm using the wrong equation and I'm feeling really lost. Any help is greatly appreciated.

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