 #1
HelloCthulhu
 150
 3
 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/horizonminipemelectrolyzerinstructionsfcsu010.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.
https://www.fuelcellstore.com/manuals/horizonminipemelectrolyzerinstructionsfcsu010.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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