The maximum electrical energy output, and the potential difference between the cathode and anode is achieved when the fuel cell is operated under the thermodynamically reversible condition. This maximum pos- sible cell potential is the reversible cell potential. The net output volt- age of a fuel cell at a certain current density is the reversible cell potential minus the irreversible potential which is discussed in this sec- tion, and can be written as [1]:
(5-3) where Vrev Eris the maximum (reversible) voltage of the fuel cell, and
Virrevis the irreversible voltage loss (overpotential) occurring at the cell.
The maximum electrical work (Welec) a system can perform at a con-
stant temperature and pressure process is given by the negative change in Gibbs free energy change (G) for the process. This equation in molar quantities is:
(5-4) The Gibbs free energy represents the net energy cost for a system cre- ated at a constant temperature with a negligible volume, minus the energy from the environment due to heat transfer. This equation is
Welec 5 2G
Vsid 5 Vrev2 Virrev LHV5 2hrx5 2,017,020 J mol C3H8 HHV5 2hrx5 2,193,020 J mol C3H8 5 22,017,020 J molC3H8 5 22,193,020 J mol C3H8
valid at any constant temperature and pressure for most fuel cell sys- tems. From the second law of thermodynamics, the maximum useful work (change in free energy) can be obtained when a “perfect” fuel cell operating irreversibly is dependent upon temperature. Thus, Welec, the
electrical power output is:
(5-5) where G is the Gibbs free energy, H is the heat content (enthalpy of formation), T is the absolute temperature, and S is entropy. Both reac- tion enthalpy and entropy are also dependent upon the temperature. The absolute enthalpy can be determined by the system temperature and pressure and is usually defined as combining both chemical and thermal bond energy. The change in the enthalpy of formation for the chemical process can be expressed from the heat and mass balance:
(5-6) where imihithe summation of the mass times the enthalpy of each
substance leaving the system, and jmjhjis the summation of the mass
times the enthalpy of each substance entering the system. A simple dia- gram of the heat and mass balance is shown in Figure 5-1. Detailed dis- cussions of mass and heat (energy) balances are described in Chapters 16 and 9 respectively.
The potential of a system to perform electrical work by a charge, Q (coulombs) through an electrical potential difference, E in volts is [2]:
Welec EQ (5-7)
If the charge is assumed to be carried out by electrons:
(5-8) Q5 nF H 5
i mihi2
j mjhj Welec 5 G 5 H – T S Products mi (H2O) Reactants mj (Hydrogen and Oxygen) LOAD Q (Heat) Figure 5-1 Fuel cell heat and mass balance.
where n is the number of moles of electrons transferred and F is the Faraday’s constant (96,485 coulombs per mole of electrons). Combining the last three equations, the maximum reversible voltage provided by the cell can be calculated:
(5-9) where n is the number of moles of electrons transferred per mol of fuel consumed, F is Faraday’s constant, and Er is the standard reversible
potential.
The relationship between voltage and temperature is derived by taking the free energy, linearizing about the standard conditions of 25C, and assuming that the change in enthalpy (H) does not change with temperature:
(5-10)
where Eris the standard-state reversible voltage, and Grxnis the stan-
dard free-energy change for the reaction. The change in entropy is neg- ative; therefore, the open circuit voltage output decreases with increasing temperature. The fuel cell is theoretically more efficient at low temperatures. However, mass transport and ionic conduction is faster at higher temperatures and this more than offsets the drop in open-circuit voltage.
In the case of a hydrogen–oxygen fuel cell under standard-state conditions:
H2 (g) 1/
2 O2 (g) l H2O (l)
(H –285.8 KJ/mol; G –237.3 KJ/mol)
At standard temperature and pressure, this is the highest voltage obtainable from a hydrogen–oxygen fuel cell. Most fuel cell reactions have theoretical voltages in the 0.8 to 1.5 V range. To obtain higher voltages, several cells have to be connected together in series.
For nonstandard conditions, the reversible voltage of the fuel cell may be calculated from the energy balance between the reactants and
EH2/O25 2 2237.3 KJ/mol 2 mol * 96,485 C/mol 5 1.229V Er5 a dE dTbsT 2 25d 5 S nFsT 2 25d Er5 2Grxn nF 5 2 H 2 TS nF G 5 2 nFEr
the products [7]. The theoretical potential Et for an electrochemical
reaction is expressed by the Nernst equation:
(5-11) where R is the ideal gas constant, T is the temperature, aiis the activ-
ity of species i, viis the stoichiometric coefficient of species i, and Eris
the standard-state reversible voltage, which is a function of tempera- ture and pressure.
The hydrogen–oxygen fuel cell reaction is written as follows using the Nernst equation:
(5-12) where E is the actual cell voltage, Eris the standard-state reversible volt-
age, R is the universal gas constant, T is the absolute temperature, N is the number of electrons consumed in the reaction, and F is Faraday’s con- stant. If the fuel cell is operating under 100C, the activity of water can be set to 1 because liquid water is assumed. At a pressure of 1.00 atmos- pheres absolute (as it is at sea level on a normal day), and if the acid elec- trolyte has an effective concentration of 1.00 moles of Hper liter, the ratio of 1.001/2/1.00 1, and ln 1 0. Therefore, E Er. The standard
electrode potential is that which is realized when the products and reac- tants are in their standard states.
At standard temperature and pressure, the theoretical potential of a hydrogen–air fuel cell can be calculated as follows:
The potential between the oxygen cathode where the reduction occurs and the hydrogen anode at which the oxidation occurs will be 1.229 volts at standard conditions with no current flowing. When a load connects the two electrodes, the current will flow as long as there is hydrogen and oxygen gas to react. If the current is small, the efficiency of the cell (measured in voltages) could be greater than 0.9 V, with an efficiency greater than 90 percent. This efficiency is much higher than the most complex heat engines such as steam engines or internal combustion engines, which can only reach a maximum 60 percent thermal efficiency. By assuming the gases are ideal (the activities of the gases are equal to their partial pressures, and the activity of the water phase is equal to unity), equation 5-11 can be written as:
(5-13) Et5 Er2 RT nF lnc qi a pi p0b vi d E5 1.229 28.314sJ/smol * Kdd * 298.15 2 * 9,6485sC/mold ln 1 1* 0.211/25 1.219V E5 Er2 RT 2F ln aH2O aH2a1/2 O2 Et5 Er2 RT nFlnc qi avi i d
where piis the partial pressure of species i, and p0is the reference pres-
sure. For ideal gases or an estimate for a nonideal gas, partial pressure of species A, pA∗ can be expressed as a product of total pressure PAand
molar fraction Aof the species:
pA∗ xAPA (5-14)
If the molar fraction for the fuel is unknown, it can be estimated by taking the average of the inlet and outlet conditions [11]:
(5-15)
where A is the stoichiometric flow rate, xAnode is the molar ratio of
species 2 to 1 in dry gas, and xC,Anodeis:
(5-16)
The molar fractions are simply ratios of the saturation pressure (Psat)
at a certain fuel cell temperature to the anode and cathode pressures. The water saturation temperature is a function of cell operating tem- perature. For a PEM hydrogen–oxygen fuel cell, Psat can be estimated
using [10]:
log10Psat –2.1794 0.02953∗T 9.1837
10–5∗T2
1.4454 10–7
∗T3 (5-17)
where T is the cell operating temperature in C.
If the current is large, the cell voltage falls fairly rapidly due to var- ious nonequilibrium effects. The simplest of these effects is the voltage drop due to the internal resistance of the cell itself. According to Ohm’s law, the voltage drop is equal to the resistance times the current flow-
ing. At maximum current density of 1 amp/cm2, the cell can drop
0.5 volts.
If the total energy based upon the higher heating value could be con- verted into electrical energy, then a theoretical potential of 1.48 V per cell could be obtained. The theoretical potential based upon the lower heating value is also shown. Because of the TS limitation, the max- imum theoretical potential of the cell is 1.229 V. This is the voltage that could be obtained if the free energy could be converted entirely to electrical energy without any losses. The actual work in the fuel cell
xC,Anode5 Psat PA xA5 1 2 xC,Anode 1 1 axAnode 2 b a1 1 a zA szA21dbb
is less than the maximum useful work because of other irreversibili- ties in the process. These irreversibilities (irreversible voltage losses) are the activation over potential (vact), ohmic overpotential (vohmic), and
concentration overpotential (vconc). This is shown by the following
equation:
(5-18) Virrevin equation 5-18 is substituted back into equation 5-3 to account
for the irreversible voltage losses to obtain an accurate fuel cell net output voltage. The variables in equation 5-18 will be discussed in more detail in Chapters 6–8. Chapter 6 covers fuel cell electrochemistry and discusses activation potential, Chapter 7 covers fuel cell charge trans- port and discusses ohmic overpotential, and Chapter 8 covers fuel cell mass transport and concentration overpotential. Figure 5-2 illustrates the fuel cell voltage losses that need to be considered when designing fuel cells.
Example 5-4 Determine the reversible cell potential as a function of tem- perature at 650C and 1 atm. Assume that Tref 25C. The reaction is
H21/2O2: H2O
Virrev5 yact1 yohmic 1 yconc
Figure 5-2 Hydrogen–oxygen fuel cell performance curve at equilibrium. 1.48 1.30 1.25 1.10 1.00 0.90 0.80 0.70
Single Cell Voltage
T
∆
S
Higher Heating Value (HHV)
Lower Heating Value (LHV) Gibbs Free Energy
Activation Polarization
Ohmic Polarization
Concentration Polarization
Current Density, A/ft2
125 100 75 50 25 0
The entropy for the fuel cell reaction at the standard reference tempera- ture and pressure, and the reversible cell potential is:
For gaseous water vapor at T 25C and 1 atm:
Therefore, the desired expression is:
For every degree of temperature increase, reversible cell potential is reduced by 0.2302 mV. At a temperature of 650C:
The reversible potential is reduced from 1.185 to 1.041 V when the tem- perature increases from 25C to 650C.
Example 5-5 Determine the reversible cell potential for the following reaction:
H2(g) + 1/2O2(g) + H2O (l)
The molar fraction of H2in the fuel stream is 0.5 and the molar fraction of O2in the oxidant stream is 0.21. The remaining species are chemically inert.
Ers650,1d 5 1.185 V 2 0.2302 3 1023 V K * s650 2 25dK 5 1.041V ErsT, Pd 5 1.185 V 2 0.2302 3 1023 V K * sT 2 Trefd ssTref, Pd nF 2 244.42 J/smol2fuel*Kd
2mol2e2/mol2fuel396,487C/mol2e2520.2302 310
23V/K Er5 G nF 5 2228,588.84 J molK a2mol2e2 mol2fuel * 96,487 C mol2e2b 5 1.185 J C 5 1.185V
G 5 2241,826 molKJ 2 s298Kd * 44.42 molKJ 5 2228,588.84 molKJ
2 a130.68 J mol2H2O1 1 2 * 205.14 J mol2O2bb G 5 a2241,826 molJ 2H2O2 a0 1 1 2 * 0bb2T *a188.83 J mol2H2O G 5 ahH2Osgd2 ahH21 1 2 hO2bb 2 T *asH2Osgd2 asH21 1 2sO2bb G 5 H – T S
Since the fuel cell operates at the standard temperature and pressure, T 25C and P is 1 atm. Since the reactant streams are not pure, the reversible cell potential for the reaction is:
The molar fraction of the reactants are 0.5 and 0.21, and K can be calculated as follows:
The reversible cell potential at standard pressure and temperature when pure H2and O2are used as reactants is 1.229 V.
The cell potential is decreased as a result of dilute reactant products, but not as drastically decreased as one would expect.