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ChaoticLlama
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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 (EO)
The only electrochemical reaction considered is H2 + 0.5O2 --> H2O
The Nernst equation for this reaction is
E = EO - (RT / 2F) ln (PH2O / (PH2 * SQRT(PO2))
'Anode In' composition is 2.78 slpm H2 and 0.049 sccm H2O; 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.
The only electrochemical reaction considered is H2 + 0.5O2 --> H2O
The Nernst equation for this reaction is
E = EO - (RT / 2F) ln (PH2O / (PH2 * SQRT(PO2))
'Anode In' composition is 2.78 slpm H2 and 0.049 sccm H2O; 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.
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