CACHE Modules on Energy in the Curriculum
Fuel Cells
Module 6 (final draft): Equilibrium Coefficient and Van’t Hoff Equation for 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 conversion technology. One type of fuel
cell, a molten carbonate fuel cell (MCFC) reacts hydrogen with oxygen to produce
electricity, according to the following reactions:
Hydrogen (H2) fuel
Anode:
H2 + CO3-2  H2O + 2 e- + CO2
Cathode:
½ O2 + CO2 + 2 e-  CO3-2
Overall:
H2 + ½ O2  H2O
This fuel cell can also operate using carbon monoxide fuel, according to the following
reactions:
Carbon Monoxide (CO) fuel:
Anode:
CO + CO3-2  2 e- + 2 CO2
Cathode:
½ O2 + CO2 + 2 e-  CO3-2
Overall:
CO + ½ O2  CO2
Fundamental to the design of fuel cells is an understanding of the maximum possible
voltage that can be generated from a single cell. (This can then be used to determine the
number of cells in the fuel cell stack). After completing this module, you will be able to
evaluate the effect of different temperature on the Gibbs energy, equilibrium coefficient,
and efficiency at different temperatures using
Consider the schematic of a compressed hydrogen tank feeding a MCFC, as seen in
Figure 1 below. The electrical power generated by the MCFC can be up to 10 MW, and is
used here to power a city.
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
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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 ).
A “Green” City
e-
eH2
H2O
H2 feed line
CO3-2
O2
O2
H2
Air in
O2
O2
Anode
Gas
Chamber
O2
H2
H2O
CO3-2
H2O H2
Cathode
Gas
Chamber
O2
O2
CO3-2
O2
H2 H2O
Anode
Cathode
Electrolyte
Figure 1. Hydrogen reaction in
the MCFC
Air out
H2 and H2O out
H2 tank
Fuel Cell
Figure 2. Diagram for Fueling a City
Problem Information
Example Problem Statement:
In this example we will analyze the hydrogen reaction in the molten carbonate fuel cell.
Determine:
a) Gibbs energy G and equilibrium coefficient K at both 25 oC and 650 oC
b) The theoretical maximum voltage at standard pressure E0 at 650 oC
c) The cell efficiency at 650 oC if the cell voltage is assumed to be 700 mV. (Note
that this is a good rule of thumb for the operating voltage for a single cell).
d) Compare your results at 650 oC to the published values (see Table 7.1 of the 2nd
edition of Larminie and Dicks on page 189)
(hydrogen, G = -197 kJ/mol and Eocv = 1.02 V)
Additional information:
(Gˆ of ) H 2O ( g ) = –228.57 kJ/mol

(H of ) H 2O( g ) = -241.83 kJ/mol
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:
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
Gˆ ro 
products
i
Gˆ ofi 

i
Gˆ ofi
reac tan ts
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.
Ĝ ro  –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
Rearranging this equation, we have:
K = exp(-ΔG/RT)
Recalling that the gas constant R is 8.314
J
, we substitute for G and T to obtain
mol  K
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 3) 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 
This equation assumes that H is independent of temperature. This allows us to estimate
G without using heat capacity integrals (Note that Module 7 uses heat capacity integrals
to do this).
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
Hˆ ro 
i

Hˆ ofi 
products
i
Hˆ ofi
reac tan ts
Since H for hydrogen and oxygen are zero, we have Ĥ ro = –241830 J/mol.
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 4) 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 maximum theoretical voltage that can be obtained by converting all of the
chemical energy into electricity at standard pressure is given by:
E0 
 G
nF
where F = Faraday’s constant = 96485 C/mol e- and n is the number of moles of electrons
per mole of fuel. (This is Equation 2.2 on page 30 of the 2nd edition of Larminie and
Dicks). Recall the anode reaction is given as:
H2 + CO3-2  H2O + 2 e- + CO2
Thus 1 mole of H2 produces 2 moles of electrons (n = 2). Thus,
E0 
J
)
mol  1.04 J  1.04 V
(2)96485 C
C
 (200766
Part c. The cell efficiency is given by   E c / E 0 , where Ec is the cell voltage. Thus, at
650 oC
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
0.70 V
 0.67
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 67% of the total energy available in the fuel is
converted into electricity, with the remaining energy converted into heat. Thus, a fuel cell
could operate for a theoretical maximum of 1 hour (at 1.04 V with no heat losses) would
operate for only 40 minutes due to these irreversibilities.
Part d.. Compare your results to those published in Larminie and Dicks at 650°C, for
hydrogen (ΔG = –197 kJ/mol and E0 = 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.
Summary. In this module we have determined the Gibbs energy and equilibrium coefficient at
different temperatures. These results were used to determine the theoretical maximum cell
voltage. The cell efficiency was calculated for a specific set of operating conditions. The results
compared well with literature values.
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Home Problem Statement:
Determine, for the carbon monoxide reaction in the molten carbonate fuel cell:
a) Gibbs Energy G and equilibrium coefficient K at both 25 oC and 650 oC.
b) The theoretical maximum voltage at standard pressure E0 at 650 oC
c) The voltage efficiency at 650 oC if the cell voltage is 650 mV.
d) Compare your results at 650 oC to the published values (see Table 7.1 of the 2nd
edition of in Larminie & Dicks on page 189)
(carbon monoxide, G = -201 kJ/mol and Eocv = 1.04 V)
The following data is available at 25 oC:
(Hˆ of ) CO ( g ) = –110.5 kJ/mol
(Gˆ of ) CO ( g ) = –137.2 kJ/mol

(H of ) CO 2( g ) = -393.5 kJ/mol
(Gˆ o )
= -394.4 kJ/mol
f
CO 2 ( g )
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