Understanding Nernst Voltage in Fuel Cells at High Temperatures

In summary, the Nernst Voltage (E) is greater than the Standard Potential (EO) and this causes a loss of potential.
  • #1
ChaoticLlama
59
0
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.
 
Last edited:
Chemistry news on Phys.org
  • #2
ChaoticLlama said:
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.
 
  • #3
Borek said:
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.
 
Last edited:
  • #4
Topher925 said:
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.
 
  • #5
Borek said:
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.
 
  • #6
thanks for your replies.

I did some research on this topic myself and found out where some of my confusion lies.

on the site http://www.fuelcellknowledge.org/research_and_analysis/cell_performance/index.html", 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!
 
Last edited by a moderator:

1. What is a fuel cell and how does it work?

A fuel cell is an electrochemical device that converts the chemical energy from a fuel, such as hydrogen, into electrical energy. It does this through a process called oxidation, where the fuel is split into positively charged ions and negatively charged electrons. The electrons are then directed through an external circuit, creating an electric current, while the positively charged ions combine with oxygen to form water. This process continues as long as there is a supply of fuel and oxygen.

2. What is the voltage of a fuel cell?

The voltage of a fuel cell is the measure of its electrical potential difference between the anode (negative) and cathode (positive) electrodes. This potential difference is created by the chemical reactions happening within the cell and can be calculated using the Nernst equation or measured directly using a voltmeter.

3. How do you calculate the voltage of a fuel cell?

The voltage of a fuel cell can be calculated using the Nernst equation, which takes into account the electrode potentials, the concentrations of the reactants and products, and the temperature. Alternatively, the voltage can be measured directly using a voltmeter, which gives a more accurate reading of the actual voltage output of the cell.

4. What factors affect the voltage of a fuel cell?

The voltage of a fuel cell can be affected by several factors, including the type of fuel being used, the concentration of reactants and products, the temperature, the pressure, and the design of the cell. Higher concentrations of reactants and products and higher temperatures typically result in a higher voltage output, while using different types of fuels can also impact the voltage.

5. How is fuel cell voltage used in practical applications?

Fuel cell voltage is a crucial factor in determining the efficiency and performance of a fuel cell in practical applications. It is used to calculate the power output of the cell and to compare the performance of different types of fuel cells. It is also used in the design and optimization of fuel cell systems for various applications, such as powering vehicles, homes, and electronic devices.

Similar threads

  • Advanced Physics Homework Help
Replies
1
Views
651
Replies
1
Views
2K
Replies
4
Views
7K
  • Engineering and Comp Sci Homework Help
Replies
1
Views
2K
Back
Top