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# THE NERNST EQUATION by 938285

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```									                    CACHE Modules on Energy in the Curriculum

Fuel Cells
Module Title: Using Plug Flow Reactor Equations for Fuel Cell Voltages
Module Author: Jason Keith
Author Affiliation: Michigan Technological University

Course: Kinetics and Reactor Design

Text Reference: Fogler (4th edition), section 1.4 (plug flow reactor), section 3.2 and
appendix C (Nernst equation); Davis and Davis (1st edition), section 1.1 and appendix A
(Nernst equation) and section 3.4 (plug flow reactor)

Concepts: Plug flow reactor equation, Gibbs free energy change, Nernst equation

Problem Motivation:
Fuel cells are a promising alternative energy technology. One type of fuel cell, a proton
exchange membrane fuel cell reacts hydrogen and oxygen together to produce electricity.
Fundamental to the design of fuel cells is an understanding of gas pressures on the
electrical energy production. This is reflected through the Nernst equation.

Consider the schematic of a compressed hydrogen tank feeding a proton exchange
membrane fuel cell, as seen in the figure below. The electricity generated by the fuel cell
is used here to power a laptop computer. We are interested in analyzing the kinetics
within the fuel cell, and how the effect of gas pressure impacts the open circuit voltage in
the fuel cell.

Computer

H2 feed line
Air in

Anode           Cathode
Gas              Gas
Chamber          Chamber

Air / H2O out
H2 out
H2 tank                       Fuel Cell
Problem Information
Example Problem Statement:
The effect of the gas pressure on the potential of a proton exchange membrane fuel cell at
100 oC (with water vapor output) is given by the Nernst equation:

RT  PH 2 PO 22 
1/
E  Eo       ln          
2 F  PH 2O 
         

Consider the flow of fuel in a bipolar plate of a proton exchange membrane fuel cell as
shown in the schematic below:

H2 in
O2 / H2O in

The flow of all gases is parallel and at 100 oC. The channel length 0 < x < 1.

Assume the feed partial pressures are PH2,in = 1 bar, PO2,in = 0.29 bar, and PH2O,in = 0.1
bar.

Assume that the oxygen and water flows are in such excess that the partial pressure of the
oxygen and water are constant along the length of the channel. However, hydrogen is fed
in more limited quantities such that the partial pressure of hydrogen gas is given as:

PH 2  PH 2,in exp(  Dx )

In this equation, D is a parameter that relates reaction time and residence time within the
channel.

a) For parameter values of D = 1 and 10, what is the cell voltage E at the channel
entrance and at the channel exit?

b) For D = 1 plot the following:
 on one graph plot the hydrogen partial pressure as a function of x
 on another graph show the voltage E as a function of x

Example Problem Solution:
Part a)
Step 1) Derivation of this equation is provided in section 2.5 of Larminie and Dicks, and
is reproduced here.

We know that for the reaction H2 + ½ O2 → H2O at low pressures, the Gibbs energy is
given as:
 P P1 / 2 
G  Go  RT ln  H 2 O 2 
 P        
 H 2O 

Recognizing that G  2FE we can obtain

RT  PH 2 PO 22 
1/
E  Eo       ln          
2 F  PH 2O 
         

where Eo is the open circuit voltage at a particular temperature. These are tabulated
below:

Form of water product            Temperature, oC            Eo (V)
Liquid                          25                    1.23
Liquid                          80                    1.18
Gas                          100                    1.17
Gas                          200                    1.14
Gas                          400                    1.09
Gas                          600                    1.04
Gas                          800                    0.98
Gas                         1000                    0.92

Step 2) We can look up the value of Eo at the standard pressures. This is given as 1.17 V
from the data table above.

Step 3) Determine entrance and exit hydrogen concentrations for parameter D.
 When D = 1 and D = 10, the entrance hydrogen partial pressure is given as 1 bar.
 For D = 1, the exit hydrogen partial pressure is given as
PH 2  PH 2,in exp(  Dx )  1  exp( 1)  0.368 bar
   For D = 10, the exit hydrogen partial pressure is given as
PH 2  PH 2,in exp(Dx)  1 exp(10)  4.5 105 bar

Step 4) Determine Nernst voltage at each condition:
 When D = 1 and D = 10, the entrance voltage is given as:
8.314 J/mol  K    373 K     1 V  1(0.29)1 / 2 
E  1.17                                    ln             1.197 V
2        96485C/mol 1 J/C  0.1 
          
   For D = 1, the exit voltage is given as:
8.314 J/mol  K      373 K      1 V  0.368(0.29)1 / 2 
E  1.17                                     ln                   1.181V
2         96485C/mol 1 J/C        0.1         

   For D = 10, the exit voltage is given as:
8.314 J/mol  K     373 K      1 V  0.000045(0.29)1 / 2 
E  1.17                                    ln                      1.036 V
2         96485C/mol 1 J/C        0.1           

Part b)
We can use the equation PH 2  PH 2,in exp(  Dx ) as a parameter in the equation:
RT  PH 2 PO 22 
1/
E  Eo       ln           . The plots of PH2 and E are shown in the figures below.
2 F  PH 2O 
         
Home Problem Statement:
The effect of the gas pressure on the potential of a proton exchange membrane fuel cell at
100 oC (with water vapor output) is given by the Nernst equation:

RT  PH 2 PO 22 
1/
E  Eo     ln          
2 F  PH 2O 
         

Consider the flow of fuel in a bipolar plate of a proton exchange membrane fuel cell as
shown in the schematic below:

H2 in
O2 / H2O in

The flow of all gases is parallel and at 100 oC. The channel length 0 < x < 1.

Assume the feed partial pressures are PH2,in = 1 bar, PO2,in = 0.29 bar, and PH2O,in = 0.1
bar.

Assume that the oxygen and water flows are in such excess that the partial pressure of the
oxygen and water are constant along the length of the channel. However, hydrogen is fed
in more limited quantities such that the partial pressure of hydrogen gas is given as:

PH 2  PH 2,in exp(  Dx )

In this equation, D is a parameter that relates reaction time and residence time within the
channel.

a) For parameter values of D = 2 and 5, what is the cell voltage E at the channel entrance
and at the channel exit?

b) For D = 2 and PO2,in = 0.58 bar (double the oxygen partial pressure), what is the cell
voltage E at the channel entrance and channel exit?
Example Problem Solution:
Part a)
Step 1) Derivation of this equation is provided in section 2.5 of Larminie and Dicks, and
is reproduced here.

We know that for the reaction H2 + ½ O2 → H2O at low pressures, the Gibbs energy is
given as:

 PH 2 PO 22 
1/

G  Go  RT ln             
  PH 2O    

Recognizing that G  2FE we can obtain

RT  PH 2 PO 22 
1/
E  Eo     ln          
2 F  PH 2O 
         

where Eo is the open circuit voltage at a particular temperature. These are tabulated
below:

Form of water product            Temperature, oC            Eo (V)
Liquid                          25                    1.23
Liquid                          80                    1.18
Gas                          100                    1.17
Gas                          200                    1.14
Gas                          400                    1.09
Gas                          600                    1.04
Gas                          800                    0.98
Gas                         1000                    0.92

Step 2) We can look up the value of Eo at the standard pressures. This is given as 1.17 V
from the data table above.

Step 3) Determine entrance and exit hydrogen concentrations for parameter D.
 When D = 2 and D = 5, the entrance hydrogen partial pressure is given as 1 bar.
 For D = 2, the exit hydrogen partial pressure is given as
PH 2  PH 2,in exp(  Dx )  1  exp( 1)  0.135 bar
   For D = 5, the exit hydrogen partial pressure is given as
PH 2  PH 2,in exp(  Dx )  1  exp( 5)  0.0067 bar

Step 4) Determine Nernst voltage at each condition:
 When D = 2 and D = 5, the entrance voltage is given as:
8.314 J/mol  K    373 K     1 V  1(0.29)1 / 2 
E  1.17                                    ln             1.197 V
2        96485C/mol 1 J/C  0.1 
          
   For D = 2, the exit voltage is given as:
8.314 J/mol  K      373 K     1 V  0.135(0.29)1 / 2 
E  1.17                                    ln                   1.165 V
2         96485C/mol 1 J/C       0.1         

   For D = 5, the exit voltage is given as:
8.314 J/mol  K      373 K      1 V  0.0067(0.29)1 / 2 
E  1.17                                     ln                    1.117 V
2         96485C/mol 1 J/C         0.1         


Part b)
We follow the calculation for D = 2 in step 4, but with PO2 = 0.58 bar to give:

8.314 J/mol  K    373 K    1 V  0.135(0.58)1 / 2 
E  1.17                                  ln                  1.171V
2        96485C/mol 1 J/C     0.1         


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