CACHE Modules on Energy in the Curriculum
Fuel Cells
Module X (Draft 2): H2 Production by Electrolysis with a Fuel Cell
Module Author: Michael D. Gross
Module Affiliation: Bucknell University
Course:
Membrane Separations,
Text Reference:
Geankoplis 4th ed., Chapter 13
Concept Illustrated:
Problem Motivation: Fuel cells are a promising alternative energy conversion
technology. Hydrogen is currently the preferred fuel for fuel cells. The generation of pure
hydrogen is important, particularly for low temperature fuel cells such as PEMFC,
because the Pt catalyst is sensitive to impurities. One way to generate hydrogen is by
electrolysis, a process that uses electricity to decompose water into hydrogen and oxygen.
One could accomplish this with an electrolyzer, which is fuel cell operated in reverse.
The fuel cell reactions are reversed by supplying electricity from an external source.
H2O  2H+ + ½ O2 + 2e2H+ + 2e-  H2
H2O  H2 + ½ O2
The Electrolyzer reactions are: Anode:
Cathode:
Overall:
Figure 1: Electrolyzer Reactions
Electron
Flow
(Current)
e-
eH2O
O2
H2
H+
O2
H2O
H2O O2
H+
H+
Anode
Gas
Chamber
H2
Cathode
Gas
Chamber
Fuel Cell
H2
Anode
Cathode
Electrolyte
Draft 1
Cell Voltage
H2O
In
H2
H2O
H2O
Electric Load
H2
H+
O2
Figure 2: Flow Diagram for Electrolyzer
O2
Out
-1-
H2
Out
September 18, 2008
For each mole of H2O consumed, 1 mole of H2 is generated and 2 moles of electrons are
passed through the external circuit. 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 rate of hydrogen production is
directly related to the current as follows:
i  nF
dN H2
dt
dN H2
i

dt
nF
or
i  current density 
where
n
amperes coulombs

cm 2
s  cm 2
2 moles of e 1 mole of H 2
coulombs
mole of e moles of H 2

cm 2  s
F  96,485
dN H2
dt
Figure 3 shows the relationship between current density, i, and electrolyzer voltage.
There are several things to note here.
2.4
2.0
Voltage, V
1.6
Fuel Cell Mode
1.2
Electrolysis Mode
0.8
0.4
0.0
-3
-2
-1
0
1
2
3
Current Density, A•cm-2
Figure 3. A performance curve for an Electrolyzer.
Draft 1
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September 18, 2008




To be in electrolysis mode, the voltage must be greater than the voltage
corresponding to zero current. This voltage is called the open circuit voltage,
EOCV, and in Figure 3 has a value of 1.2 V.
The hydrogen reaction rate is directly proportional to the current, since for
each hydrogen molecule that reacts, two electrons are formed.
For electrolysis operating voltages (Eop) Eop > EOCV, hydrogen is produced
with an external power source. This is indicated by negative current values.
Over the entire range of current densities, there is a linear fall in voltage as the
current density increases. Assuming fast kinetics, this voltage drop is caused
by a resistance to H+ flow across the electrolyte membrane. In this case the
operating voltage can be described as Eop = EOCV – iRA, where i is current
density, R is resistance, and A is cross sectional area of the electrolyzer. In
physics and electrical engineering, this effect is referred to as Ohm’s law.
While an electrolyzer uses an electrochemical device for separation of water into
hydrogen and oxygen, it draws many parallels to typical membrane separation processes
across a dense membrane. Our primary interest is the rate of H2 production, however, H+
is the species transported across the membrane and so the flux across the membrane is
written with respect to H+. For every one mole of H2 produced, 2 moles of e- and 2 moles
of H+ are produced at the anode. The flux equations for an electrolyzer and a separation
with a dense membrane are compared below.
Draft 1
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September 18, 2008
cio
Ni
ciL
L
Figure 4. Concentration profile for a membrane process
Table 1. Flux equations for a membrane process and Electrolyzer.
Equations
Units
 mol H

2
 cm  s

Membrane
Separation
D
N i  i c iO  ciL 
L
NH 
i
nF
i
i
V
E OCV  E op  E OCV  E op 
V


RA
RA
L
E
C
C 
 cm 2 
cm 2  s
s
C
1
1
V
 V


cm 2  s
  cm cm   cm 2
 E op 
OCV
NH 



 mol H    C
mol H  mol e- 

   2 


2
C 
 cm  s   cm  s mol e
V  i  RA
Electrolyzer
  cm 2 mol H 
  

s

cm
cm 3
 
L

L  nF
E
OCV
 mol H    V
mol H  mol e- 

  



2
2
C 
 cm  s     cm mol e
 E op 
N H   flux
i  current density
E OCV  open circuit vo ltage
E op  operating voltage
RA  area specific resistance
  H  conductivi ty
L  thickness of the membrane
D  diffusivit y
c iO and c iL  concentrat ions at the membrane interfaces
Draft 1
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September 18, 2008
The only difference between the flux equations for the electrolyzer and membrane
separation is that the electrolyzer takes into account electrical units. For the electrolyzer,
voltage directly corresponds to the concentration of oxygen at each membrane interface
via the Nernst equation.
E OCV  E o 
1/ 2
RT  PH 2 PO 2
ln
nF  PH 2O




The concentration of oxygen is typically expressed as a partial pressure of oxygen, P O2.
For more information regarding the Nernst equation, refer to Module 9, under
Thermodynamics.
Draft 1
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September 18, 2008
Example Problem
A company is developing a new car powered by a fuel cell system that runs on H2. You
have been asked to consider generating the H2 by electrolysis with a fuel cell. The H2
tank to be used is 10 liters in volume and a fill-up requires a pressure of 500 psi.
a. Calculate the current required to operate at a voltage of 1.8V.
b. Calculate the rate of hydrogen production per membrane area and the total
membrane area required to fill the tank in 2 minutes.
Consider the following specifications of the system.
60% conversion of H2O
Eo = 1.172 V
The cathode pressure is maintained at 1 atm
The anode pressure is maintained at 1 atm
Membrane thickness = 100 μm
Membrane conductivity (σ) = 0.1 S/cm (S = 1/Ω)
Electrolysis T = 373 K (assume water is in the gas phase)
H2 storage tank T = 298 K
Example Problem Solution
Part a
Step 1. Calculate the partial pressure of H2O and O2 in the anode for 60% conversion of
H2O.
PH2O = 1 atm·(1-conversion fraction) = 1 atm·(1-0.6) = 0.4 atm
PO2 = 1 atm·conversion fraction·
1 mol O 2 produced
= 1 atm·0.6·0.5 = 0.3 atm
2 mol H 2 O reacted
Step 2. A constant anode pressure of 1 atm specified in the problem requires
normalization of the partial pressures to 1 atm.
Normalized PH2O =
Normalized PO2 =
Draft 1
0.4 atm H 2 O
 0.57 atm H 2 O
0.4 atm H 2 O  0.3 atm O 2
0.3 atm O 2
 0.43 atm  O 2
0.3 atm O 2  0.4 atm H 2 O
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September 18, 2008
Step 3. Calculate the open circuit voltage for the given conditions using the Nernst
equation.




J
8.314
 373K
 1  0.431/2 
mol  K

  1.174 V
 1.172 V 
ln
mol e C
 0.57 
2
 96485
mol H 2 O
mol e -
E OCV  E o 
1/ 2
RT  PH 2 PO 2
ln
nF  PH 2O
Step 4. Calculate the resistance of the membrane.
RA 
l


0.01cm
 0.1  cm 2
S
0.1
cm
Step 5. Generate the linear equation of the V-i curve as shown in Figure 3 for electrolysis
mode.
Eop = EOCV - iRA = 1.174 - 0.1i
Step 6. Calculate the current required to achieve a voltage of 1.8 V given the V-i curve
relationship.
1.8V  1.174V  0.1  cm 2  i
i
1.8V  1.174V
 6.26 A  cm 2
 0.1
It is standard notation to express current as a negative value if current is being supplied to
the electrolyzer. For the purposes of calculating a rate of H2 generation, as in Part b, a
positive current value will be used.
Part b.
Step 1. Calculate the rate of H2 generation with the current calculated in part a.
dN H2
i


dt
nF
mol H 2
6.26 A  cm 2
 3.24 x10 5
mol e
C
cm 2  s
2
 96485
mol H 2
mol e -
Note: 1 A = 1 C/s
Draft 1
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September 18, 2008
Step 2. Convert the tank pressure from units of psi to atm.
 1 atm 
  34atm
500 psi
 14.696 psi 
Step 3. Calculate the total amount of H2 required to fill the tank. The temperature of the
H2 storage tank was given as 298 K.
Assuming an ideal gas.
n
PV

RT
34 atm  10 L
 13.9 mol H 2
L  atm
0.08206
 298 K
mol  K
Step 4. Calculate the total membrane area required to generate 13.9 mol of H2 in 2
minutes.
Area 
mol H 2 dN H2
13.9 mol H 2

 3590 cm 2  0.358 m 2
dN H2
mol H 2
t
120 s  3.24x10 -5
dt
cm 2  s
Summary
For an electrolyzer unit consisting of 10cm x 10cm electrolyzer cells arranged in a stack,
a minimum of 36 cells would be required to fulfill the above requirements.
Draft 1
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September 18, 2008
Home Problem
A company is developing a new truck powered by a fuel cell system that runs on H2, and
you have been asked to consider generating the H2 by electrolysis with a fuel cell. The H2
tank to be used is 20 liters in volume and a fill-up requires a pressure of 750 psi.
a. Calculate the rate of H2 production as a function of H2O conversion. Consider
H2O conversions of 0.1%, 20%, 40%, 60%, 80%, 99.9%.
b. Calculate the membrane area required to fill up the tank in 10 minutes as a
function of H2O conversion.
Consider the following specifications of the system.
60% conversion of H2O
Eo = 1.172 V
The anode pressure is maintained at 2 atm
The cathode pressure is maintained at 2 atm
Membrane thickness = 100 μm
Membrane conductivity (σ) = 0.05 S/cm (S = 1/Ω)
Electrolysis T = 373 K (assume water is in the gas phase)
H2 storage tank T = 298 K
Draft 1
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September 18, 2008