2 Thermodynamics and Fuel Cells
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Ecell = Ecathode − Eanode = 1.23 − 0.00 = 1.23 V
(6.4)
This value represents the maximum possible amount of energy that can be generated
by this fuel cell under standard conditions, provided that the cell is operating in a
thermodynamically reversible fashion.
N.B. Things can get quite confusing when talking about cell potential. There is Ecell,
Ecell(reversible), and ETN, the thermoneutral cell potential. E°cell is the cell potential at standard conditions. E°cell(reversible) just spells out that this is the cell potential when the cell is
operating under reversible conditions. It represents the thermodynamic maximum, but it
is not the 100% theoretically possible maximum voltage. The theoretical maximum is
called the “thermoneutral” cell potential (ETN, also referred to as the “potential voltage”
to make things really confusing), which does not take into account the energy losses that
must result by the second law of thermodynamics, vide infra. The thermoneutral cell
potential is the voltage a fuel cell would exhibit if all the chemical energy of the fuel
and oxidant is converted perfectly into electric energy—a thermodynamic impossibility.
6.2.2 Cell Potential and Gibbs Free Energy
For a given redox couple, the Gibbs free energy, ΔG, corresponds to the net energy
that this pair can produce. The relation between Gibbs free energy, enthalpy, and
entropy is given by
ΔG = ΔH – TΔS(6.5)
where ΔH amounts to the thermal energy that is generated and ΔS the entropic cost.
These concepts are related in the equation*
ΔG = −nFEcell(6.6)
where F is the Faraday constant of 9.65 × 104 C/mol and n is the number of mols of
electrons transferred in the balanced electrochemical equation. Note that this equation illustrates the connection of sign to the spontaneity of the reaction: a negative
ΔG relates to a positive Ecell; therefore, the cell potential must be positive in order for
the electrochemical reaction to proceed as written. Since an electrochemical reaction
is ultimately about transferring some quantity of electrons, ΔG is also related to the
amount of charge, q. Since q = nF
ΔG = qEcell = −nFEcell(6.7)*
In the end analysis, ΔG is simply the cell potential recast in joules to encompass the
molar amount of charge.
*
You may be wondering “where did that negative sign come from?!” It all goes back to thermodynamics and the convention of sign, as discussed in Section 3.2. Free-energy change (ΔG) represents the
maximum possible work that can be done. Because the work is done on the surroundings, the negative
sign must appear.
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6.2.2.1 State of Water
There is an important caveat in calculating the cell potential for the hydrogen fuel
cell: the water produced in the cell may be a liquid or a gas depending upon the
conditions. This has a profound effect on the thermodynamic values and, therefore,
E°cell because of the heat of vaporization of water (recall the discussion of lower heating value (LHV) and higher heating value (HHV) in Section 1.3). When calculating
E°cell, one must make sure to use the correct thermodynamic data (see Table 6.3). To
illustrate, we will first confirm the value of E°cell for H2O (l) by using thermodynamic
data (instead of the half-cell reduction potentials).
1.Calculation of ΔG° using ΔG = ΔH – TΔS. Referring to Table 6.3, at 1 atm
and 25°C, the enthalpies for H2 and O2 are both zero by definition and the
enthalpy of formation for H2O (l) is −285,830 J/mol. Therefore, ΔH is calculated as
∆H ° = H prod − Hreact = −258, 830 J/mol − 0 J/mol = −258, 830 J/mol
(6.8)
The change in entropy is calculated in a similar fashion, taking into
account the stoichiometry of the reaction:
∆S ° = S prod − Sreact = 69.92 J/mol ⋅ K − (130.68 J/mol ⋅ K + 0.5 mol O2 /mol H 2
× 205.14 J/mol ⋅ K ) = −163.2 J/mol ⋅ K
(6.9)
Substituting in our calculated values for ΔH° and ΔS°, we find ΔG°:
∆G ° = −285, 830 J/mol − 298 K (−163.2 J/mol ⋅ K ) = −237,196 J/mol (6.10)
2.Calculation of E°cell from ΔG°. Since 2 mol of electrons are transferred for
each mol of hydrogen fuel (recall Equations 6.1 and 6.2), calculation of the
standard, reversible cell potential E°cell is straightforward by rearrangement of Equation 6.6:
TABLE 6.3
Thermodynamic Data for Hydrogen, Oxygen, and Water
Substance
H2
O2
H2O (l)
H2O (g)
ΔH (J/mol), 298 K, 1 atm
0
0
−285,830
−241,845
ΔS (J/mol ⋅ K), 298 K, 1 atm
130.595
205.14
69.92
188.83
Source: Adapted from Li, X. 2006. Principles of Fuel Cells. 1st ed. New York: Taylor & Francis.
Note: Enthalpy of vaporization for water = 44,010 J/mol.
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Ece° ll =
−237,196 J/mol
= 1.2292 J/C
C = 1.23 V
(2 mol e − /mol H 2 )(96, 487 C/mol e − )
As expected, the value matches that obtained from the measured half-cell
potentials.
If the water produced is not in the liquid state but is instead a gas, a different value
for Ecell will result given the different values for enthalpy and entropy in the formation of the product, in water. Referring again to the thermodynamic values provided
in Table 6.3 and running through the calculations as described above, we find that
the maximum energy produced is less when water is produced in the gaseous state
(1.18 V < 1.23 V) (Equations 6.11 through 6.14). This is because energy is lost in the
vaporization of water from liquid to gas.
Calculation of ΔH:
∆H = H prod − Hreact = −241, 845 J/mol − 0 J/mol = −241, 845 J/mol (6.11)
Calculation of ΔS:
∆S = S prod − Sreact = 188.83 J/mol ⋅ K − (130.68 J/mol ⋅ K + 0.5 mol O2 /mol H 2
× 205.14 J/mol ⋅ K ) = −44.42 J/mol ⋅ K
(6.12)
Calculation of ΔG:
∆G = −241, 845 J/mol − 298 K (−44.42 J/mol ⋅ K ) = −228, 608 J/mol
(6.13)
Calculation of Ecell:
Ecell =
−228, 608 J/mol
= 1.185 V
(2 mol e − /mol H 2 )(96, 487 C/mol e − )
(6.14)
The take-home message in these calculations is that it is important to pay close
attention to the physical states and thermodynamic values of the compounds
involved.
6.2.2.2 Effect of Temperature and Pressure
Because Ecell is ultimately a reflection of Gibbs free energy, it stands to reason that
temperature will influence the operating potential of a fuel cell. The TΔS term in the
Gibbs free energy relationship (Equation 6.5) reveals that the influence of temperature is tied to entropic changes. Almost all fuel cell reactions have a negative change
in entropy, so an increase in temperature means that Ecell decreases with temperature.
This is true, for example, for the hydrogen fuel cell, where a decrease of about 0.3 V
over a temperature range spanning 300–1300 K is observed (Li 2006).
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Reversible cell potential (V)
1.22
298 K
1.21
1.20
1.19
1.18
1.17
500 K
1.16
1.15
1.14
0
10
20
30
40
Pressure (atm)
50
60
70
FIGURE 6.4 Impact of temperature and pressure on cell potential. (Reprinted with permission from Li, X. 2006. Principles of Fuel Cells. New York: Taylor & Francis.)
An aside regarding temperature: the operating temperature of a fuel cell impacts considerably more than just the cell potential. Of course, there are advantages to operating
at a higher temperature: the rate of the redox reaction increases leading to an increase
in the current density and the power of the cell. Poisoning of noble metal catalysts
decreases at higher temperatures as well. But with higher temperatures can come
increased degradation of the fuel cell materials and additional mechanical problems,
particularly with polymer components.
The effect of pressure on Ecell is significant for those fuel cell systems that have
gaseous reactants and/or products. In this case, the Ecell is dependent upon the change
in the number of moles of gas species in the electrochemical reaction. Again, most
fuel cell reactions trend in the same direction—most see a decrease in the number of
moles of gas species formed. This results in an increase in Ecell but at an even lower
level than the impact of temperature. Figure 6.4 graphically illustrates the impact of
temperature and pressure for the hydrogen/oxygen fuel cell (Li 2006).
6.3 EFFICIENCY AND FUEL CELLS
Our treatment of thermodynamics in Chapter 2 is revisited here as we consider
the efficiency of fuel cells: even though fuel cells are more efficient than, say, a
mechanical heat engine, they are still limited by the restrictions posed by the second law of thermodynamics. Some energy will be lost as waste heat even under
ideal operating conditions, even for a fuel cell. In the following discussion of fuel
cell efficiency, it should be noted that the focus is strictly on the electrical efficiency
of the fuel cell: efficiency in terms of the energy generated by the electrochemical reaction only. Because fuel cells can generate high-quality heat that may be
captured and used as part of the process, a separate value for efficiency—the CHP
efficiency (for combined heat and power, also known as cogeneration power)—is
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sometimes reported. The potential for >80% CHP efficiency exists for fuel cells
(Vogel et al. 2009).
The efficiency of a fuel cell operating under thermodynamically reversible conditions is
hrev =
Electricity produced
W
∆G
E
= max =
= rev × 100% (6.15)
Heating value of fuel used
− ∆H
∆H
H
Etn
Note that the efficiency is a ratio of free energy to enthalpy or of the reversible cell
potential as it compares to the thermoneutral cell potential. The thermoneutral cell
potential for the hydrogen/oxygen redox pair is 1.48 V; comparing that to the E°cell
of 1.23 V (liquid water) gives a reversible efficiency of 83% for the hydrogen/oxygen
fuel cell. (A word of warning: this efficiency calculation is based on the HHV for
hydrogen gas.)
It cannot be stressed enough that this “reversible efficiency” is an ideal, whereas
practical, everyday fuel cells invariably operate under irreversible conditions.
Incomplete reaction, by-product formation, voltage losses due to resistance, and
other miscellaneous issues are reality. For comparison, the practical efficiency of
fuel cells typically ranges from about 50% to 65% (based on the LHV of the fuel
used), while steam and gas turbines have practical efficiencies in the realm of 40%
(also based on LHV, Li 2006).
6.4 CELL PERFORMANCE: WHERE DO INEFFICIENCIES
COME FROM?
6.4.1 Voltage, Current, and Power
Given that even the ideal, reversible fuel cell cannot operate with 100% efficiency,
where do these losses in voltage come from? Before we can answer this question, we need to review the relationships between voltage, current, resistance, and
power:
Current (amps) =
Potential (volts)
Resistance (ohms)
Power (watts ≡ J/s) = Current (amps) × Potential (volts)
(6.16)
(6.17)
These relationships allow us to understand the two extreme conditions in electrical circuits: an open circuit (zero current, maximum voltage) and a short circuit
(maximum current, zero voltage). When the circuit is open, the anode and cathode
are not connected and no electricity can flow, hence the voltage is at its maximum.
This is the open-circuit voltage (VOC). The open circuit voltage is not a theoretical
quantity: it is the actual measurement of voltage when the circuit is open. Hence,
in a perfect fuel cell at standard temperature and pressure, VOC could, in theory,