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2 Thermodynamics and Fuel Cells

2 Thermodynamics and Fuel Cells

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Fuel Cells

Ecell = Ecathode − Eanode = 1.23 − 0.00 = 1.23 V


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.


Chemistry of Sustainable Energy  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


  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


  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:


Thermodynamic Data for Hydrogen, Oxygen, and Water




H2O (l)

H2O (g)

ΔH (J/mol), 298 K, 1 atm





ΔS (J/mol ⋅ K), 298 K, 1 atm





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.


Fuel Cells

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


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


Calculation of ΔG:

   ∆G = −241, 845 J/mol − 298 K (−44.42 J/mol ⋅ K ) = −228, 608 J/mol


Calculation of Ecell:

Ecell =

−228, 608 J/mol

= 1.185 V

(2 mol e − /mol H 2 )(96, 487 C/mol e − )


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.  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).


Chemistry of Sustainable Energy

Reversible cell potential (V)


298 K






500 K









Pressure (atm)




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).


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


Fuel Cells

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




= max =

= rev × 100% (6.15)

Heating value of fuel used

− ∆H




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).



6.4.1 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


Current (amps) =

Potential (volts)

Resistance (ohms)

Power (watts ≡ J/s) = Current (amps) × Potential (volts)



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,

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