CACHE Modules on Energy in the Curriculum
Fuel Cells
Module Title: Thermodynamics: Gibbs Energy, Equilibrium Coefficient, and Fuel Cell
Efficiency
Module Author: Jason Keith
Author Affiliation: Michigan Technological University
Course: Thermodynamics
Text Reference: Sandler (2nd ed.), section 9.1
Smith, Van Ness, and Abbott (7th ed.), section 13.3 and 13.4
Concepts: Equilibrium coefficient, Van’t Hoff’s 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 nonideal equations of state,
as hydrogen fuel is often stored at high pressure. After completing this module, you will
be able to size compressed fuel tanks for fuel cell applications. In this module you will
calculate Gibbs energy, equilibrium coefficient, and efficiency at different temperatures.
Consider the schematic of a compressed hydrogen tank feeding a proton exchange
membrane fuel cell, as seen in figure 1 below. The electricity generated by the fuel cell is
used here to power a laptop computer. We are interested in determining the effect of
inefficiencies on the amount of time the laptop can be operated from the compressed
hydrogen tank.
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Computer
(Electric Load)
H2 feed line
Air in
Anode
Gas
Chamber
Cathode
Gas
Chamber
Air / H2O out
H2 out
H2 tank
Fuel Cell
Problem Information
Example Problem Statement:
In this example we will analyze the hydrogen reaction in the molten carbonate fuel cell.
The reaction is given by:
Hydrogen (H2) fuel
Anode:
2 H2 + 2 CO3-2  2 H2O + 4 e- + 2 CO2
Cathode:
O2 + 2 CO2 + 4 e-  2 CO3-2
Overall:
2 H2 + O2  2 H2O
Determine, for the hydrogen reaction:
a) Gibbs energy G and equilibrium coefficient K at both 25 oC and 650 oC
b) The open circuit voltage E at 650 oC
c) The voltage efficiency at 650 oC if the cell voltage is assumed to be 750 mV. Note
that a good rule of thumb for the operating voltage for a single cell is close to 700
mV. What impact does this have on the fuel cell performance?
d) Compare your results to the published values in Larminie & Dicks
(hydrogen, G = -197 kJ/mol and Eocv = 1.02 V)
Example Problem Solution:
Part a.
Step 1) Calculating the Gibbs energy per mole of reactant (in this case, hydrogen) for the
above overall reaction at 25°C (298 K) requires the use of the given data according to:
ΔGformation = ΣΔGreact - ΣΔGprod
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= ΔGH20(g) – ΔGH2 –
1
ΔGO2
2
(1)
Remembering that the Gibbs energy of any elemental substance (hydrogen and oxygen
are elemental substances) is zero, we obtain the Gibbs energy of formation of vapor
water.
ΔGformation = -228.57 kJ/mol at 298 K
Step 2) We can then determine the equilibrium coefficient, K, which is defined according
to the relationship (This is equation 13.11 on page 490 of the 7th edition of Smith, Van
Ness, and Abbott):
ΔG = -RT ln K
(2)
Rearranging equation 2, we have:
K = exp(-ΔG/RT)
Step 3) Recalling that the gas constant R is 8.314
(3)
J
, we substitute for G and T to
mol  K
obtain for the formation of water vapor at 298 K:
J


 228572


mol

K = exp 
J


298K 
 8.314
mol

K


K = 1.16 x1040 at 298 K (such that ln K = 92.256)
This large value suggests the reaction goes almost to completion.
Step 4) To calculate the Gibbs Energy and K at 650°C (923 K), first the value of K must
be calculated at the different temperature. This will be performed using Van’t Hoff’s
equation (This is equation 13.15 on page 492 of the 7th edition of Smith, Van Ness, and
Abbott):
 K   H  1 1 
  
ln   
K
R
 1
 T T1 
(4)
This equation assumes that H is independent of temperature. Although not true it does
allow us to estimate G without using heat capacity integrals.
Using the H for combustion of hydrogen into water vapor yields H = -241830 J/mol.
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Substitution for T, H, R, and K gives:
ln K  92.256 
241830 J/mol  1
1 



8.314 J/mol  K  923 K 298 K 
Thus, ln K = 26.16
K = 2.3 x 1011 at 923 K
Step 5) To calculate the Gibbs Energy, we will use the equation directly.
ΔG = –RT ln K
=  (8.314
J
)(923K ) ln( 2.3  1011 )
mol  K
ΔG = - 200.8 kJ/mol at 923 K
Part b. The Open Circuit Voltage (OCV) is the maximum theoretical voltage that can be
obtained by converting all of the chemical energy into electricity. The equation used to
calculate this value (This is equation 2.2 on page 30 of the 2nd edition of Larminie and
Dicks) is:
EOCV 
 G
nF
where F = Faraday’s constant = 96485 C/mol e- and n is the number of moles of electrons
per mole of fuel. Recall the anode reaction is given as:
2 H2 + 2 CO3-2  2 H2O + 4 e- + 2 CO2
Thus 2 moles of H2 produce 4 moles of electrons. Thus, on a per mole of hydrogen basis,
n = 2. Thus,
E ocv 
J
)
mol  1.04V
(2)96485 C
 (200766
E ocv  1.04
J
 1.04V
C
Part c. The voltage efficiency is given by   Ec / Eocv , where Ec is the cell voltage.
Thus, at 650 oC
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
0.75 V
 0.72
1.04 V
We note that the efficiency is a ratio of the actual cell voltage divided by the maximum
theoretical voltage. This means that 72% of the total energy available in the fuel is
converted into electricity, with the remaining energy converted into heat. Thus, a fuel cell
that had enough fuel thermal energy input (due to H) to operate for a theoretical
maximum of 1 hour would operate for only 43 minutes due to these irreversibilities.
Note that we will calculate heat losses in a future module on transport phenomena.
Part d.. Compare your results to those published in Larminie and Dicks at 650°C, for
hydrogen (ΔG = –197 kJ/mol and Eocv = 1.02 V.)
The published results compare well with the calculated values for the hydrogen reaction
of –200.8 kJ/mol and 1.04 V. Thus, the assumption of constant H used in calculating
the K values does not provide a large amount of error.
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