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Fuel Cell Voltage Calculation

by ChaoticLlama
Tags: calculation, cell, fuel, voltage
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ChaoticLlama
#1
Apr7-09, 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 (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.
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Borek
#2
Apr8-09, 03:23 AM
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Quote Quote by ChaoticLlama View Post
simple issue; the Nernst Voltage (E) is greater than the Standard Potential (EO)
Why do you think it is an issue? Standard if for standard conditions, when you use non standard conditions potential can be either smaller or higher.
Topher925
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Apr8-09, 09:10 AM
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Quote Quote by Borek View Post
Why do you think it is an issue? Standard if for standard conditions, when you use non standard conditions potential can be either smaller or higher.
The standard voltage (E0) is the voltage at standard conditions. It's quite possible for the cell voltage to be higher than the standard voltage if the cell is operating at non-standard conditions.

I take it this is not a PEM fuel cell?

EDIT: Corrected my mistakes.

Borek
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Apr8-09, 10:07 AM
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Fuel Cell Voltage Calculation

Quote Quote by Topher925 View Post
The standard voltage (E0) is the highest possible voltage a cell can obtain.
No, it is a voltage measured in standard conditions - that is for acitivities of all reactants being 1 and at standard temperature.

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.
Topher925
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Apr8-09, 10:47 AM
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Quote Quote by Borek View Post
No, it is a voltage measured in standard conditions - that is for acitivities of all reactants being 1 and at standard temperature.
You're right, for some reason I was thinking of theoretical potential and the corrected Nernst equation.

Anyway, now that I am awake, I think you got right Borek.
ChaoticLlama
#6
Apr8-09, 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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