CACHE Modules on Energy in the Curriculum
Fuel Cells
Module Title: Hydrogen Adsorption and Catalyst Surface Coverage
Module Author: Jason Keith
Author Affiliation: Michigan Technological University
Course: Kinetics and Reaction Engineering
Text Reference: Fogler (4th edition), Section 10.2.3
Concepts: Given the mechanism for hydrogen adsorption determine the surface coverage
on a platinum catalyst.
Problem Motivation:
Fuel cells are a promising alternative energy conversion technology. One type of fuel
cell, a proton exchange membrane fuel cell (PEMFC) reacts hydrogen with oxygen to
produce electricity (Figure 1). Fundamental to the design of fuel cells is an understanding
of the effect of kinetics on the fuel cell performance.
Consider the schematic of a compressed hydrogen tank (2000 psi, regulated to 10 psi)
feeding a proton exchange membrane fuel cell, as seen in Figure 2 below. We will now
focus on the voltage / current relationship of the fuel cell.
-
-
e
e
H2
H2O
O2
H+
H2
H2O
H2
O2
O2
H+
H2
Computer
(Electric
Load)
Pressure
regulator
H2 feed line
Air in
H2
H2
H2
H2
H2
H2O
H2O
H+
H+
O2
Anode
Cathode
Electrolyte
Figure 1. Reactions in the PEMFC
H2 out
H2 tank
Fuel Cell
Air / H2O out
Figure 2. Diagram for fueling a laptop.
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J.M. Keith
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October 7, 2008
The PEMFC reactions are:
Anode:
Cathode:
Overall:
H2
→ 2H+ + 2e½ O2 + 2H+ + 2e- → H2O
H2 + ½ O2
→ H2O
For each mole of hydrogen consumed, two moles of electrons are passed through the
electric load. To convert electron flow (moles of electrons/s) to electrical current
(coulombs/s or amps), one would use Faraday’s constant: F  96,485 coulombs / mole of
electrons. The primary objective of a fuel cell is to deliver energy to the electric load. To
calculate the energy delivery rate (also know as power) one would multiply the current
times the cell voltage: Power = Current · Voltage. (Recall the unit conversions:
Coulomb  Volt  Joule and Joule / s  Watt ).
Background
Figure 3 shows a “polarization plot” which is the relationship between current density i
(fuel cell current per unit area of the electrode, in units of milliamps per square
centimeter) and cell voltage Vc (in units of volts). There are several things to note here:
Theoretical
voltage of 1.2 V
Overvoltage
Rapid drop
(kinetic losses)
Linear drop
(ohmic losses)
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J.M. Keith
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Rapid drop at
higher currents
(mass transfer
losses)
October 7, 2008






The theoretical maximum voltage of this fuel cell is 1.2 V. This is called the
theoretical, or “open circuit voltage” VOCV .
The hydrogen reaction rate (in moles per second, for example) is directly
proportional to the current (in coulombs per second, or amperes), since for each
hydrogen molecule that reacts, two electrons are formed.
Any drop from this maximum value is termed “overvoltage.” It is desired to
minimize the overvoltage so that the fuel cell can operate as efficiently as
possible.
There is a critical current density called the “exchange current density” with
symbol io. For current densities i < io, the cell voltage is equal to the theoretical
value. For current densities i > io, there is a rapid fall in cell voltage, due to a slow
reaction rate constant (kinetics). It is desired to have as high a value of io as
possible, and as rapid kinetics as possible.
At current densities between 100 mA/cm2 and about 1000 mA/cm2, there is a
linear fall in voltage as the current density increases. This effect is due to the fact
that there is a resistance to current and ion flow within the fuel cell. As the current
increases, the voltage drop will increase. In physics and electrical engineering,
this effect is referred to as Ohm’s law. It is desired to have as small a resistance as
possible.
At very high current densities (greater than 1000 mA/cm2), the hydrogen reaction
rate is high. However, the hydrogen cannot diffuse to the electrode fast enough to
react. Thus, mass transfer is limiting and the voltage rapidly drops to zero.
Consider the adsorption and surface reaction of hydrogen as described below and as
shown in Figure 4.
H
H
H
Figure 4a. Hydrogen
molecule in presences of
vacant sites.
H
Figure 4b. Adsorbed Hydrogen.
+

Figure 4c. Desorbed
products.
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J.M. Keith
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October 7, 2008
The following mechanism has been shown to occur at the anode in a fuel cell catalyst
(typically platinum), where S denotes a catalyst site and HS denotes adsorbed hydrogen
atoms.
H2 + 2 S ↔ 2 HS (Tafel reaction)
(1)
HS ↔ H+ + S + e- (Volmer Reaction)
(2)
Reaction 1 can be thought of as equilibrium between figures 4a and 4b. Reaction 2 can be
thought of as equilibrium between figures 4b and 4c. The protons (H+) in Equation 2
flow through the membrane, and the electrons (e-) in Equation 2 flow through the load
In this module we will use reaction rate expressions to determine the surface coverage of
hydrogen in the regime where the kinetic losses are dominant (i < 200 mA/cm2).
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J.M. Keith
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October 7, 2008
Example Problem Statement: The anode (fuel electrode) in a proton exchange
membrane fuel cell is made of a 0.26 mg platinum catalyst. A surface area analysis of this
catalyst yielded 250 m2/g. The cross-sectional area of the fuel cell is 8.5” x 11” and the
exchange current density for a feed of pure hydrogen at 1 atm pressure has been
determined to be io = 4 mA/cm2.
a) Determine the maximum number of active sites for the catalyst C t , max
b) Assume that the metal dispersion is 40% (this is the percent of platinum atoms
that are surface atoms, which are available for hydrogen to adsorb). Thus,
determine the number of active sites Ct.
c) At the exchange current density the forward and reverse reaction rates are
equal. Thus, determine the forward hydrogen reaction rate (-rH2,forward) from
the relationship io  nF (rH 2, forward ) / A . In this expression, n is the number of
moles of electrons formed from one mole of hydrogen, F is Faraday’s
constant, 96485 C/mol e-, and A is the cross-sectional area of the cell.
d) The rate determined in part c) is equal to the forward and reverse reaction rate
for hydrogen adsorption using the Tafel reaction mechanism. The ratio of the
forward to reverse reaction rates k1/k-1 is equal to 2.7 x 105 atm-1. Determine
the forward and reverse reaction rate constants and the fraction of the surface
that has adsorbed hydrogen.
Example Problem Solution:
Part a)
The maximum number of active sites can be determined using the catalyst mass, the
molecular weight of the catalyst, and Avogadro’s number to give:
Ct ,max 
0.26 mg
g
mol 6.02  10 23 atoms
 8.0  1017 atoms  8.0  1017 sites
1000 mg 195 g
mol
(3)
Part b)
The dispersion can be used to calculate the total number of active sites.
Ct  8.0  1017 sites (0.40)  3.2  1017 sites
(4)
Part c)
Given the current density we can solve for the forward (or reverse) reaction rate as:
 rH 2, forward 
Ai o 8.5  11in 2 2.54 2 cm 2 4 mA 1C/s mol e 
mol H 2

 1.25  10 5

2
2
nF
s
mol e
in
cm 1000 mA 96485 C
2
mol H 2
(5)
Part d)
Step 1) At the exchange current density, the forward and reverse rates are equal to each
other. Thus, the overall rate of hydrogen reaction is equal to zero:
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J.M. Keith
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October 7, 2008
 rH 2  rH 2, forward  (rH 2,reverse )  0
(6)
Step 2) Writing out the elementary steps in the Tafel reaction into Equation 6 above
gives:
 rH 2  0  k1 PH 2 C v2  k 1C H2  S
(7)
where PH2 is the partial pressure of hydrogen (1 atm), Cv is the concentration of vacant
catalyst sites S, and C H  S is the concentration of catalyst sites with adsorbed hydrogen
H S .
Step 3) Solving Equation 7 for C H  S gives:
C H S  Cv
k1 PH 2
k 1
(8)
Step 4) Performing a total site balance gives:
Ct  C v  C H S
(9)
Step 5) Combining Equations 8 and 9 gives:
Cv 
Ct
(10)
kP
1 1 H2
k 1
Step 6) Combining the forward reaction in Equation 7 with Equation 10 gives:
 rH 2, forward 
k1 PH 2 Ct2

k1 PH 2
1 
k 1




(11)
2
Step 7) We can rearrange Equation 11 to solve for the forward rate constant k1.

kP
(rH 2, forward ) 1  1 H 2
k 1

k1 
2
PH 2 Ct



2
(12)
Step 8) Substituting, we find that:
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J.M. Keith
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October 7, 2008
k1 

(1.25  10 5 mol H 2 /s ) 1  2.7  10 5 atm 1 (1atm )
1atm (3.2  10 sites )
17
Alternatively, k1Ct2  3.39

2
2
 3.30  10 35
mol H 2
(13)
s  atm  site 2
mol H 2
s  atm
Step 9) The reverse reaction rate constant is given as:
mol H 2
s  atm  site 2  1.22  10 40 mol H 2
1
s  site 2
2.7  10 5
atm
3.3  10 35
k 1 
(14)
Step 10) Combining Equations 9 and 10 give the fractional surface coverage as:
C H S
C
 1 v 
Ct
Ct
k1 PH 2
k 1
(15)
kP
1 1 H2
k 1
Step 11) Substituting into Equation 15 gives:
2.7  10 5 atm 1 (1atm )
C H S

 0.9981
Ct
1  2.7  10 5 atm 1 (1atm )
(16)
Analysis: The surface is nearly fully covered with adsorbed hydrogen. In the homework
problem we will investigate the effect of changing the overall reaction rate on the surface
coverage.
2nd Draft
J.M. Keith
Page 7
October 7, 2008
Home Problem Statement: The anode (fuel electrode) in a proton exchange membrane
fuel cell contains a 0.26 mg platinum catalyst. A surface area analysis of this catalyst
yielded 250 m2/g. The cross-sectional area of the fuel cell is 8.5” x 11” and the exchange
current density for a feed of pure hydrogen at 1 atm pressure has been determined to be io
= 4 mA/cm2.
Use the forward and reverse reaction rate constants from the example problem.
a) For a reaction rate –rH2 = 6.25 x 10-4 mol/s, determine the current density in
mA/cm2.
b) For the conditions from part a) determine the surface coverage of adsorbed
hydrogen.
c) Discuss the validity of the following statement: “Under the conditions of this
home problem kinetics are controlling the hydrogen reaction.”
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J.M. Keith
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October 7, 2008