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Fuel Cell Voltage Calculation 
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#1
Apr709, 06:47 PM

P: 62

Hello, I'm doing work in Fuel Cells and am having difficulty with a simple issue; the Nernst Voltage (E) is greater than the Standard Potential (E^{O})
The only electrochemical reaction considered is H_{2} + 0.5O_{2} > H_{2}O The Nernst equation for this reaction is E = E^{O}  (RT / 2F) ln (PH2O / (PH2 * SQRT(PO2)) 'Anode In' composition is 2.78 slpm H_{2} and 0.049 sccm H_{2}O; at about 1 atm 'Cathode In' is air supplied at 8.34 slpm, at about 1 atm When I calculate the pressures, I get P H2O = 0.00619 kPa = 6.19 Pa P H2 = 101.319 kPa = 101319 Pa P O2 = 21.28 kPa = 21280 Pa which gives me roughly ln(10^7), thus a positive loss of roughly 0.65 V (while running at approximately 1028 K). Thanks for any help. 


#2
Apr809, 03:23 AM

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#3
Apr809, 09:10 AM

P: 1,672

I take it this is not a PEM fuel cell? EDIT: Corrected my mistakes. 


#4
Apr809, 10:07 AM

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Fuel Cell Voltage Calculation
Besides, if you take a look at the Nernst equation: [tex]E = E_0 + \frac {RT} {nF} ln Q[/tex] and then at the conditions mentioned by ChaoticLlama (T=1028K) it is is obvious that if the reaction quotient is higher than 1 (not that difficult  just activity of the oxidized form must be higher than activity of the reduced form), observed potential skyrockets. I can be missing something, but that's what I see here. 


#5
Apr809, 10:47 AM

P: 1,672

Anyway, now that I am awake, I think you got right Borek. 


#6
Apr809, 12:09 PM

P: 62

thanks for your replies.
I did some research on this topic myself and found out where some of my confusion lies. on the site Fuel Cell Knowledge, it does clearly say in the Gibbs and Nernst Potential document that the Nernst voltage can be either higher or lower than the standard potential; whereas I thought that Nernst must be lower. Secondly, I mixed up some values in the spreadsheet I created which created even more confusion. The reactants for the reaction quotient were correctly identified (PH2 & PO2). However, instead of using product water for the quotient, I used water from the Anode In stream. *facepalm* I'm modifying my approach slightly, to instead consider the partial pressure of oxygen on the cathode side versus the anode side. And you are right, Topher, I'm working with SOFCs (hence the high operating temperature). Thanks! 


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