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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 (E

The only electrochemical reaction considered is H

The Nernst equation for this reaction is

E = E

'Anode In' composition is 2.78 slpm H

'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

Thanks for any help.

^{O})The only electrochemical reaction considered is H

_{2}+ 0.5O_{2}--> H_{2}OThe 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.

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