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Topic: Hydrogen-oxygen fuel cell volumes  (Read 6218 times)

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Offline cabaal

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Hydrogen-oxygen fuel cell volumes
« on: April 16, 2010, 05:32:54 PM »
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The hydrogen-oxygen fuel cell is described in Section 19.6. (A) What volume of H2(g), stored at 25°C at a pressure of 155 atm, would be needed to run an electric motor drawing a current of 8.5 A for 3.0 h? (B) What volume (liters) of air at 25°C and 1.00 atm will have to pass into the cell per minute to run the motor? Assume that air is 20% O2 by volume and that all the O2 is consumed in the cell. The other components of air do not affect the fuel-cell reactions. Assume ideal gas behaviors.

2H2(g) + O2(g)  :rarrow: 2H2O(l)
cell = 1.23 V

I have no idea where to start. The book doesn't mention anything about how to do volumes of gasses or what to do with amps and hours. Could someone help me? :(
« Last Edit: April 16, 2010, 06:23:47 PM by cabaal »

Offline Borek

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Re: Hydrogen-oxygen fuel cell volumes
« Reply #1 on: April 16, 2010, 06:05:04 PM »
Current times time gives charge. Charge divided by faraday's constant gives equivalents. Equivalents give moles. Moles of gas give volume.
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Offline cabaal

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Re: Hydrogen-oxygen fuel cell volumes
« Reply #2 on: April 16, 2010, 06:33:14 PM »
8.5 Amps * 3 hours * (3600 sec/1 hour) * (1 C/1 Amp*sec) = 91800 C * (1 mol e-/96500 C) = 0.951 mol e-
0.951 mol e- * (2 mol H2/4 mol e-) = 0.476 mol H2
V = (nRT)/P = (0.476 mol H2*0.0821*298 K)/1500 atm = 0.00776 L H2

Does that look correct?

Offline cabaal

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Re: Hydrogen-oxygen fuel cell volumes
« Reply #3 on: April 16, 2010, 06:40:03 PM »
For part (B):

0.951 mol e- * (1 mol O2/4 mol e-) = 0.238 mol O2
V=(0.238 mol O2 * 0.0821 * 298K)/1 atm = 5.82 L / 20% = 29.11 L

But I don't think that's right, because it gives me liters and not liters/second like the problem asks? Where did I go wrong?

edit: Actually my answer doesn't make sense at all. I have no idea how to do part B because it's asking for liters of air and I found liters of oxygen.

Offline Borek

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Re: Hydrogen-oxygen fuel cell volumes
« Reply #4 on: April 17, 2010, 03:56:05 AM »
8.5 Amps * 3 hours * (3600 sec/1 hour) * (1 C/1 Amp*sec) = 91800 C * (1 mol e-/96500 C) = 0.951 mol e-
0.951 mol e- * (2 mol H2/4 mol e-) = 0.476 mol H2
V = (nRT)/P = (0.476 mol H2*0.0821*298 K)/1500 atm = 0.00776 L H2

Does that look correct?

Not bad, just why 1500 atm?
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Offline Borek

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Re: Hydrogen-oxygen fuel cell volumes
« Reply #5 on: April 17, 2010, 03:58:46 AM »
But I don't think that's right, because it gives me liters and not liters/second like the problem asks? Where did I go wrong?

You have calculated amount of oxygen spent during 3 hours. How many per sec then?

Quote
edit: Actually my answer doesn't make sense at all. I have no idea how to do part B because it's asking for liters of air and I found liters of oxygen.

Once you know volume of oxygen needed you have to calculate volume of air that contains this volume of oxygen. You are told oxygen makes up 1/5 of air.
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Offline cabaal

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Re: Hydrogen-oxygen fuel cell volumes
« Reply #6 on: April 17, 2010, 10:06:24 AM »
8.5 Amps * 3 hours * (3600 sec/1 hour) * (1 C/1 Amp*sec) = 91800 C * (1 mol e-/96500 C) = 0.951 mol e-
0.951 mol e- * (2 mol H2/4 mol e-) = 0.476 mol H2
V = (nRT)/P = (0.476 mol H2*0.0821*298 K)/1500 atm = 0.00776 L H2

Does that look correct?

Not bad, just why 1500 atm?

Oops. For some reason I confused 155 atm with 1500 atm.  :o

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