Amount of water electrolyzed during PEM electrolysis

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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+(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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Answers and Replies

  • #2
Borek
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Q = 54,000C
Looks off at first sight.

Besides, most of your numbers have no units, so it is very difficult to make something out of them.
 
  • #3
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Courtesy of Wikipedia :

at STP,
Hydrogen: 0.08928 g/L
Oxygen: 1.429g/L
 
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  • #4
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Looks off at first sight.

Besides, most of your numbers have no units, so it is very difficult to make something out of them.
Apologies. I added the units to the numbers.
 
  • #5
Borek
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Relevant Equations:: I * t * mw / n * F
This is not an equation, this is an expression and it is only a part of the Faraday's law of electrolysis.

In Faraday's law mw is not a molecular weight, but molar mass.

mw = 45033g
Molar mass is a property of a molecule. All you need to calculate molar mass is a molecular formula. mw for hydrogen is 2 g/mol and for oxygen is 32 g/mol. No idea what you did :(
 
  • #6
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This is not an equation, this is an expression and it is only a part of the Faraday's law of electrolysis.

In Faraday's law mw is not a molecular weight, but molar mass.



Molar mass is a property of a molecule. All you need to calculate molar mass is a molecular formula. mw for hydrogen is 2 g/mol and for oxygen is 32 g/mol. No idea what you did :(
I used these values when I first started learning about Faraday's Laws of Electrolysis:

(1 * 60 * 4) / (4*96,500) = 0.0006ml

(1 * 60 * 32) / (4*96,500) = 0.0049ml

If this is correct, I thought I could algebraically substitute the values for H2 and O2 with 7ml and 3.5ml respectively and solve for molar mass. But I thought I did something wrong, because I wasn't producing twice the amount of H2. I'm very lost.
 
  • #7
Borek
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Again, you wrote these things without units, so there is no way to find out why you think you got 0.0006 mL (which is a bubble barely visible to a naked eye). Using molar mass in the equation you should get mass of the product, not its volume.

This is a simple plug and chug exercise, but you need to start with a correct equation and to be consistent about units. Try to google the first Faraday's law of electrolysis and copy the whole equation. Post it here.
 
  • #8
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https://www.ausetute.com.au/faradayl.html

First Law of Electrolysis : The mass of a substance produced by electrolysis is proportional to the quantity of electricity used.

Q/m

Q = quantity of electricity in coulombs
m = mass produced
 
  • #9
Borek
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You do understand the difference between equation and expression?

You need Faraday's law in the form

amount of substance produced = some function of the charge
 
  • #11
Borek
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Yes, that's much better. Actually the most convenient form of the equation is in wikipedia - it doesn't use some magic "proportionality constant", but calculates it from the first principles (that's more or less what you have in the first post, it was just incomplete there).

See if you can take the wikipedia equation and plug your numbers there.
 
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  • #12
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Ok, let's see if I got it right this time.

2 moles of H2O undergoes electrolysis at 1 amp for 1 min. What is the mass of H2 and O2 gas produced?

Cathode: 4 H+(aq) + 4e− → 2H2(g)

Anode: 2 H2O(l) → O2(g) + 4 H+(aq) + 4e−

m = ZQ = mass produced

Q = quantity of electricity in coulombs; I*t

Z = molar mass/valence electrons transferred; g/C

m=(t*I*Z)/F)

F = 96,5000

t = 60

I = 1A


Z = 4g/4e- = 1g/C H2
Z = 32g/4e- = 8g/C O2


(60s*1A*1g)/(96,500C) = 0.000622ml H2
6.22e-7L/22.4L = 2.77678571e-8 mol*2g = 5.55357142e-8g H2

(60s*1A*8g)/(96,500C) = 0.004974ml O2
4.974e-6L/22.4L= 2.22053571e-7mol*32g = 7.10571427e-6g O2
 
  • #13
chemisttree
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Wasn’t your question how much water was being electrolyzed when 7 mL H2 and 3.5 mL O2 are being produced?
If so, you are off the road and upside down in the ditch!

mL O2—> mol O2 (@STP) —> mol H20 —> g H2O

Forget Faraday!
 
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  • #14
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Apologies. I've felt very very lost about how to approach this solution. Let's see if I get this correct:

7ml H2 =0.007L/22.4L = 0.00031M(STP) H2

3.5ml O2 = 0.0035L/22.4L = 0.00016M(STP) O2

0.00047M H20 * 18g = 0.00846g H20
 
  • #15
chemisttree
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Apologies. I've felt very very lost about how to approach this solution. Let's see if I get this correct:

7ml H2 =0.007L/22.4L = 0.00031M(STP) H2

3.5ml O2 = 0.0035L/22.4L = 0.00016M(STP) O2

0.00047M H20 * 18g = 0.00846g H20
Ohhh! So close! (Did you really add those mol H2 and mol O2 together?)
Write out the electrolysis equation first and tell me how many moles of water you expect.
 
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  • #16
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Thank for your correction! I think I may understand how to do this now...

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?

7ml H2 =0.007L/22.4L = 0.0003125 H2(STP)
3.5ml O2 = 0.0035L/22.4L = 0.00015625 O2(STP)

Cathode: 0.000625 H+(aq) + 0.000625e−→0.0003125 H2(g)
Anode: 0.0003125 H2O(l) → 0.00015625O2(g) + 0.000625 H+(aq) + 0.000625e−

0.0003125M H2O *18g = 0.005625g H2O
 
  • #17
chemisttree
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Not correct.

Still waiting for the balanced equation....
 
  • #18
epenguin
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Did you reproduce the question verbatim? I do not like the word order produced at 1A per minute during electrolysis. 1A per minute? Produced per minute during electrolysis by 1A surely, though I guess what was meant was understandable.

As to the question how much water is being a electrolysed per minute, it is reasonable to give the answer in moles of H2O per minute. I think people have tried to tell you that there is superfluity in the question. You don't need to know anything about the current to know how any moles of H2 there are in 7mL gas. Then how many moles of H2O those moles came from is not difficult either.

After that you should sure check to see whether this corresponds to the coulombs that have passed. There should even be some curiosity about it – what would a disagreement mean?
 
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  • #19
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Did you reproduce the question verbatim? I do not like the word order produced at 1A per minute during electrolysis. 1A per minute? Produced per minute during electrolysis by 1A surely, though I guess what was meant was understandable.

As to the question how much water is being a electrolysed per minute, it is reasonable to give the answer in moles of H2O per minute. I think people have tried to tell you that there is superfluity in the question. You don't need to know anything about the current to know how any moles of H2 there are in 7mL gas. Then how many moles of H2O those moles came from is not difficult either.

After that you should sure check to see whether this corresponds to the coulombs that have passed. There should even be some curiosity about it – what would a disagreement mean?
Did I calculate the moles of H2 and O2 STP correctly?

7ml H2 =0.007L/22.4L = 0.0003125 H2(STP)
 
  • #20
epenguin
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Did I calculate the moles of H2 and O2 STP correctly?

7ml H2 =0.007L/22.4L = 0.0003125 H2(STP)
That's right for H2. You don't even need the O2, though it would be alarming if it were not near to half that.
 
  • #21
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Thank you for that confirmation! Now I just need to figure out how to balance the equation.

Cathode: 0.000625 H+(aq) + 0.000625e−→0.0003125 H2(g)
Anode: 0.0003125 H2O(l) → 0.00015625O2(g) + 0.000625 H+(aq) + 0.000625e−

0.0003125M H2O *18g = 0.005625g H2O

Did I get any of this correct?
 
  • #22
epenguin
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Well as I said, it is reasonable to state the number of moles.
And to get the number of grams of water from that, sure you multiply by the molecular mass, 18. I might almost ask though, who would care? I can imagine in a class experiment someone measuring the gas volumes, and the current and time, but quite a lot more trouble to do it so carefully that they are going to verify by weighing that 5.6 mg of water have disappeared.
So more scientifically significant would be to check the coulombs as I mentioned before.
Whether those other equations are the ones wanted I leave to the others, but I think they were looking for ordinary and clarifying chemical equations, without these numbers in them.
Quite soon I think you are going to find it more convenient to use the ×10n number system – all these zeros make your eyes go funny and you have to tediously currently count and recount them, like bank Iban numbers!
 
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  • #23
chemisttree
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That's right for H2. You don't even need the O2, though it would be alarming if it were not near to half that.
I would use O2... you don’t even need H2.
 
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  • #24
epenguin
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I would use O2... you don’t even need H2.
I did think of that, however the moles of H2 evolved equals the moles of H2O electrolysed whereas for O2 you have the additional step of multiplying by 2, indeed of deciding whether to multiply or divide, so I thought it was best to keep it simple.
 
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  • #25
chemisttree
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Yeah, I know but I had already given the process based on O2 in #13. Didn’t want to confuse the OP by switching to H2.
 
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