**2. The thermodynamics of a fuel cell**

A fuel cell, also known as a galvanic or voltaic cell, is a well-known example of a device that works by changing chemical energy into electrical energy, which is exhibited in terms of cell potential and electrical current output. The maximum possible electrical energy output and the corresponding electrical potential difference between the cathode and anode are achieved when the fuel cell is operated under thermodynamically reversible conditions, as shown in **Figure 1**, a fuel cell system to which fuel and oxidant streams enter and product stream exits. Unfortunately, it is inevitable that some of the energy will be dissipated as heat.

The overall electrochemical reactions occurring inside the fuel cell system boundary are described as follows:

$$\text{H}\_2\text{ (fuel)} + \text{O}\_2\text{ (oxidant)} \rightarrow \text{W} + \text{Q} + \text{H}\_2\text{O}\text{ (product)}\tag{1}$$

where W is the rate of electrical work done by the system and Q is the rate of heat transferred into the system from the surroundings at constant pressure and temperature.

Electrical work is, in general, described by the relation:

$$\mathbf{W} = \mathbf{E} \mathbf{I} \mathbf{A} \mathbf{t} \tag{2}$$

where E° is also called the reversible voltage, because it is the maximum possible voltage without any irreversible losses. This is the maximum possible voltage of an electrochemical cell, since it is attained assuming a reversible process. If we are looking at the redox reaction on a per-mole-of-fuel basis, the absolute Gibbs function is equivalent to the molar specific value. All fuel cell losses are associated with deviation from this maximum. Since F and n are constants for a particular global redox reaction, the functional dependence of the maximum possible voltage of an electrochemical cell is related strictly to the dependencies of the Gibbs free energy, namely, temperature and pressure of the reactants and products. If all the potential chemical energy for a reaction went into electrical work and there was no heat transfer, there would be no entropy change; dG = dH. In this case, we can show that:

Erev ¼ � <sup>Δ</sup><sup>H</sup>

<sup>f</sup> <sup>þ</sup> RT ln aCc

where a's are the thermodynamic activity coefficients for the reacting species. To

RT nF

where I is the standard voltage evaluated at 1 atm pressure for all components and II accounts for the thermodynamic activity dependence on the Nernst voltage.

i. For an ideal gas, a = Pi/P°, where Pi is the partial pressure of the species of

nF ln

where the partial pressures are evaluated at the particular electrode where the reaction involving the species occurs. Using this expression, we can solve for the expected maximum (Nernst) voltage for a given fuel cell reaction. Two important

PC Po � �*<sup>c</sup>* PD Po � �*<sup>d</sup>*

" #

PA Po � �*<sup>a</sup>* PB Po � �*<sup>b</sup>*

ii. For water vapour, the partial pressure of the vapour cannot exceed the saturation pressure, Psat, which is a function of temperature. Thus, the reference pressure is set to Psat, and a = Pv/Psat, which is the relative humidity, RH. This can normally be considered to be 1.0 in the immediate molecular region of the water-generating electrode. This is a reasonable assumption because water generation is always at the catalyst surface and the activity of water here is 1.0. Also, the reaction itself is not limited by the product water

For a general reaction or process of A and B giving products C and D:

Eo |{z} <sup>I</sup> �

<sup>Δ</sup>Gf <sup>¼</sup> <sup>Δ</sup>G<sup>o</sup>

E T, P ð Þ¼

interest and P° is the reference pressure, 1 atm.

E T, P ð Þ¼ <sup>E</sup><sup>o</sup> � RT

For a generic reaction:

*Fuel Cell Thermodynamics*

*DOI: http://dx.doi.org/10.5772/intechopen.90141*

convert to voltage, we can divide it by nF:

concentration at this surface:

points are as follows:

**5**

nF (7)

� � (9)

(10)

(11)

aA þ bB ! cC þ dD (8)

aDd

aDd

aA<sup>a</sup> aB<sup>b</sup>

ln aC<sup>c</sup>


aAa aBb � �

where E is the cell voltage and I is the current. In a fuel cell reaction, electrons are transferred from the anode to the cathode, generating a current.

The amount of electricity (IΔt) transferred when the reaction occurs is given by nF, where n is the number of electrons transferred and F is Faraday's constant = 96,493 coulombs. The electrical work can hence be calculated as:

$$\mathbf{W} = -\mathbf{n} \mathbf{F} \mathbf{E}\_{\text{cell}} \tag{3}$$

The Gibbs free energy is the maximum amount of work done on the system:

$$\mathbf{W}\_{\rm el} = \Delta \mathbf{G} \tag{4}$$

$$
\Delta \mathbf{G} = -\mathbf{n} \mathbf{F} \mathbf{E}\_{\text{cell}} \tag{5}
$$

Hence the maximum cell potential or the reversible cell potential becomes:

**Figure 1.** *Simple H2/O2 fuel cell diagram.*

*Fuel Cell Thermodynamics DOI: http://dx.doi.org/10.5772/intechopen.90141*

energy output which is the electrical energy produced, and the energy input is the

A fuel cell, also known as a galvanic or voltaic cell, is a well-known example of a

device that works by changing chemical energy into electrical energy, which is exhibited in terms of cell potential and electrical current output. The maximum possible electrical energy output and the corresponding electrical potential difference between the cathode and anode are achieved when the fuel cell is operated under thermodynamically reversible conditions, as shown in **Figure 1**, a fuel cell system to which fuel and oxidant streams enter and product stream exits. Unfortu-

The overall electrochemical reactions occurring inside the fuel cell system

where W is the rate of electrical work done by the system and Q is the rate of heat transferred into the system from the surroundings at constant pressure and

where E is the cell voltage and I is the current. In a fuel cell reaction, electrons

The amount of electricity (IΔt) transferred when the reaction occurs is given by

The Gibbs free energy is the maximum amount of work done on the system:

Hence the maximum cell potential or the reversible cell potential becomes:

Erev ¼ � <sup>Δ</sup><sup>G</sup>

H2 ð Þþ fuel O2 ð Þ! oxidant W þ Q þ H2O product ð Þ (1)

W ¼ EIΔt (2)

W ¼ �nFEcell (3)

Wel ¼ ΔG (4) ΔG ¼ �nFEcell (5)

nF (6)

nately, it is inevitable that some of the energy will be dissipated as heat.

Electrical work is, in general, described by the relation:

are transferred from the anode to the cathode, generating a current.

nF, where n is the number of electrons transferred and F is Faraday's constant = 96,493 coulombs. The electrical work can hence be calculated as:

enthalpy of hydrogen.

**2. The thermodynamics of a fuel cell**

*Thermodynamics and Energy Engineering*

boundary are described as follows:

temperature.

**Figure 1.**

**4**

*Simple H2/O2 fuel cell diagram.*

where E° is also called the reversible voltage, because it is the maximum possible voltage without any irreversible losses. This is the maximum possible voltage of an electrochemical cell, since it is attained assuming a reversible process. If we are looking at the redox reaction on a per-mole-of-fuel basis, the absolute Gibbs function is equivalent to the molar specific value. All fuel cell losses are associated with deviation from this maximum. Since F and n are constants for a particular global redox reaction, the functional dependence of the maximum possible voltage of an electrochemical cell is related strictly to the dependencies of the Gibbs free energy, namely, temperature and pressure of the reactants and products. If all the potential chemical energy for a reaction went into electrical work and there was no heat transfer, there would be no entropy change; dG = dH. In this case, we can show that:

$$\mathbf{E\_{rev}} = -\frac{\Delta \mathbf{H}}{\mathbf{n} \mathbf{F}} \tag{7}$$

For a generic reaction:

For a general reaction or process of A and B giving products C and D:

$$\mathbf{a}\mathbf{A} + \mathbf{b}\mathbf{B} \to \mathbf{c}\mathbf{C} + \mathbf{d}\mathbf{D} \tag{8}$$

$$
\Delta \mathbf{G}\_{\mathbf{f}} = \Delta \mathbf{G}^{\mathbf{o}}{}\_{\mathbf{f}} \; + \text{RT } \ln \left[ \frac{\mathbf{a} \mathbf{C}^{\mathbf{c}} \mathbf{a} \mathbf{D}^{\mathbf{d}}}{\mathbf{a} \mathbf{A}^{\mathbf{a}} \mathbf{a} \mathbf{B}^{\mathbf{b}}} \right] \tag{9}
$$

where a's are the thermodynamic activity coefficients for the reacting species. To convert to voltage, we can divide it by nF:

$$\mathbf{E}\left(\mathbf{T},\mathbf{P}\right) = \underbrace{\mathbf{E}^{\circ}}\_{\mathbf{I}} - \underbrace{\frac{\mathbf{RT}}{\mathbf{nF}} \quad \ln\left[\frac{\mathbf{a}\mathbf{C}^{\circ}\mathbf{aD}^{\mathrm{d}}}{\mathbf{a}\mathbf{A}^{\circ}\mathbf{a}\mathbf{B}^{\mathrm{b}}}\right]}\_{\mathbf{II}}\tag{10}$$

where I is the standard voltage evaluated at 1 atm pressure for all components and II accounts for the thermodynamic activity dependence on the Nernst voltage.


$$\mathbf{E}\left(\mathbf{T},\mathbf{P}\right) = \mathbf{E}^{\circ} - \frac{\mathbf{RT}}{\mathbf{nF}} \ln \left[ \frac{\left(\frac{\mathbf{p}\varepsilon}{\mathbf{p}^{\circ}}\right)^{\varepsilon} \left(\frac{\mathbf{p}\_{\mathrm{D}}}{\mathbf{p}^{\circ}}\right)^{d}}{\left(\frac{\mathbf{p}\_{\mathrm{A}}}{\mathbf{P}^{\circ}}\right)^{d} \left(\frac{\mathbf{p}\_{\mathrm{B}}}{\mathbf{P}^{\circ}}\right)^{b}} \right] \tag{11}$$

where the partial pressures are evaluated at the particular electrode where the reaction involving the species occurs. Using this expression, we can solve for the expected maximum (Nernst) voltage for a given fuel cell reaction. Two important points are as follows:

i. The Nernst equation is a result of the equilibrium established at the electrode surfaces. A particular gradient can exist between the concentration of a species in the channel of a fuel cell and the electrode, especially under highcurrent-density conditions, which cannot be considered a true thermodynamic equilibrium situation anyway.

Usually, only gas species produce an appreciable volume change. Assuming that

¼ � <sup>Δ</sup>ngRT

nFp (18)

ln P (19)

H ¼ U þ PV (20)

H ¼ Qp (21)

(22)

∂E ∂P 

Erev <sup>¼</sup> Eo

system is equal to the heat gained or lost in the system:

T

where Δng represents the change in the total number of moles of gas upon reaction. Pressure, like temperature, turns out to have a minimal effect on revers-

> rev � ð Þ <sup>Δ</sup><sup>n</sup> RT nF

Enthalpy is the thermodynamic quantity that states the total heat content of the system, which is the sum of all internal process in a closed system [2]. For homogeneous systems, enthalpy is solely based on the size of the system as it is an extensive property. Enthalpy (H) is the sum of the internal energy of the system (U) and the product of pressure (P) and volume (V) of the system. The change in enthalpy in a

The enthalpy change (ΔH) for a reaction in a fuel cell indicates the full amount of heat released by the reaction at a constant pressure; hence, enthalpy is simply

At constant pressure and volume, the thermodynamic properties of the cell are related to the behaviour of its potential and are defined from the Gibbs-Helmholtz

> ∂T

p

H2 þ ½ O2 ! H2O ΔH < 0 (24)

<sup>p</sup> � Ecell (23)

f) and absolute entropies of forma-

<sup>Δ</sup><sup>H</sup> <sup>¼</sup> <sup>Δ</sup><sup>G</sup> � <sup>T</sup> <sup>∂</sup>Δ<sup>G</sup>

In accordance with Eq. (5), in terms of electrochemical processes, change in

T

The overall reaction in Eq. (1) is the same as the reaction of hydrogen combus-

f) obtained from the basic thermodynamic data (see **Table 1**), the heat of

tion. Combustion is an exothermic process, which means that there is energy

formation of both liquid and vapour water can be calculated using the equation

<sup>Δ</sup><sup>H</sup> <sup>¼</sup> nF T *<sup>∂</sup>*Ecell

the ideal gas law applies, we can write Eq. (17) as:

*DOI: http://dx.doi.org/10.5772/intechopen.90141*

**2.1 Enthalpy of reaction for a fuel cell**

equal to the heat released:

enthalpy can be written as:

released in the process [4, 5]:

From the table of enthalpies of formation (h<sup>o</sup>

equation [3]:

tion (so

**7**

above to form:

ible voltage:

*Fuel Cell Thermodynamics*

ii. Only species directly involved in the electrochemical reaction of Eq. (8) are represented directly in the activity terms of Eq. (10). Species not participating in the electrochemical charge transfer reaction only indirectly alter the voltage through the species mole fractions of the participating species.

For the H2/O2 fuel cell potential, the open-circuit voltage is the maximum operating voltage (when no current is flowing) and is determined by the chemical thermodynamics of the overall cell reaction. The Nernst equation provides a relationship between the standard potential (E<sup>o</sup> ) for the cell reaction and the opencircuit voltage, where it can be determined at the partial pressures of reactants and products at temperature (T):

$$\mathbf{E}\left(\mathbf{T},\mathbf{P}\right) = \mathbf{E}^{\diamond} - \frac{\mathbf{RT}}{2\mathbf{F}} \quad \ln\left[\frac{\left(\frac{\text{yH}\_2\text{OPcathode}}{\text{Psat}}\right)}{\left(\frac{\text{yH}\_2\text{Pamode}}{\text{P}^\diamond}\right)\left(\frac{\text{yO}\_2\text{Pcathode}}{\text{P}^\diamond}\right)^{1/2}}\right] \tag{12}$$

To understand how the reversible voltage varies with temperature and pressure, respectively, we have previously shown that the Gibbs free energy is related to the reversible cell voltage by Eq. (5):

$$
\Delta \mathbf{G} = -\mathbf{n} \mathbf{F} \mathbf{E}\_{\text{cell}}
$$

At constant pressure, the above relationship produces a Maxwell relation that links the change in open cell voltage with temperature T (a measurable quantity) to the change in entropy S [1]:

$$
\left(\frac{\partial \mathbf{E}}{\partial \mathbf{T}}\right)\_{\mathbf{Q}} = \left(-\frac{\partial \mathbf{S}}{\partial \mathbf{Q}}\right)\_{\mathbf{T}}\tag{13}
$$

$$\left(\frac{\partial \mathbf{E}}{\partial \mathbf{T}}\right)\_{\mathbf{P}} = \frac{\Delta \mathbf{S}}{\mathbf{n} \mathbf{F}}\tag{14}$$

$$
\Delta \mathbf{S} = \mathbf{n} \mathbf{F} \left( \frac{\partial \mathbf{E}}{\partial \mathbf{T}} \right)\_{\mathbf{p}} \tag{15}
$$

$$\mathbf{E\_{rev}} = \mathbf{E^{o}}\_{\text{rev}} + \frac{\Delta \mathbf{S}}{\mathbf{nF}} \left(\mathbf{T} - \mathbf{T}^{o}\right) \tag{16}$$

At constant temperature, Eq. (5) produces an equation that links voltage with pressure, p, to the change in volume:

$$
\left(\frac{\partial \mathbf{E}}{\partial \mathbf{P}}\right)\_{\mathbf{T}} = -\frac{\partial \mathbf{v}}{\mathbf{n} \mathbf{F}}\tag{17}
$$

If the volume change of the reaction is negative (if fewer moles of gas are generated by the reaction than consumed, for instance), then the cell voltage will increase with increasing pressure.

i. The Nernst equation is a result of the equilibrium established at the electrode surfaces. A particular gradient can exist between the concentration of a species in the channel of a fuel cell and the electrode, especially under high-

ii. Only species directly involved in the electrochemical reaction of Eq. (8) are represented directly in the activity terms of Eq. (10). Species not participating in the electrochemical charge transfer reaction only indirectly alter the voltage through the species mole fractions of the participating species.

For the H2/O2 fuel cell potential, the open-circuit voltage is the maximum operating voltage (when no current is flowing) and is determined by the chemical thermodynamics of the overall cell reaction. The Nernst equation provides a rela-

circuit voltage, where it can be determined at the partial pressures of reactants and

2 6 4

yH2Panode Po

To understand how the reversible voltage varies with temperature and pressure, respectively, we have previously shown that the Gibbs free energy is related to the

ΔG ¼ �nFEcell

At constant pressure, the above relationship produces a Maxwell relation that links the change in open cell voltage with temperature T (a measurable quantity) to

> ¼ � <sup>∂</sup><sup>S</sup> ∂Q � �

> > <sup>¼</sup> <sup>Δ</sup><sup>S</sup>

∂T � �

¼ � <sup>∂</sup><sup>v</sup>

ΔS nF P

) for the cell reaction and the open-

3 7

<sup>5</sup> (12)

(13)

(15)

yH2OPcathode Psat � �

> Po � �<sup>1</sup>*=*<sup>2</sup>

� � yO2Pcathode

T

nF (14)

<sup>T</sup> � <sup>T</sup><sup>o</sup> ð Þ (16)

nF (17)

current-density conditions, which cannot be considered a true

thermodynamic equilibrium situation anyway.

tionship between the standard potential (E<sup>o</sup>

*Thermodynamics and Energy Engineering*

E T, P ð Þ¼ Eo � RT

2F

∂E ∂T � �

Erev <sup>¼</sup> Eo

Q

∂E ∂T � �

P

<sup>Δ</sup><sup>S</sup> <sup>¼</sup> nF <sup>∂</sup><sup>E</sup>

rev þ

∂E ∂P � �

At constant temperature, Eq. (5) produces an equation that links voltage with

T

If the volume change of the reaction is negative (if fewer moles of gas are generated by the reaction than consumed, for instance), then the cell voltage will

ln

products at temperature (T):

reversible cell voltage by Eq. (5):

the change in entropy S [1]:

pressure, p, to the change in volume:

increase with increasing pressure.

**6**

Usually, only gas species produce an appreciable volume change. Assuming that the ideal gas law applies, we can write Eq. (17) as:

$$
\left(\frac{\partial \mathbf{E}}{\partial \mathbf{P}}\right)\_{\mathbf{T}} = -\frac{\Delta \mathbf{n}\_{\mathbf{g}} \mathbf{R} \mathbf{T}}{\mathbf{n} \mathbf{F} \mathbf{p}}\tag{18}
$$

where Δng represents the change in the total number of moles of gas upon reaction. Pressure, like temperature, turns out to have a minimal effect on reversible voltage:

$$\mathbf{E\_{rev}} = \mathbf{E^{o}\_{rev}} - \frac{(\Delta \mathbf{n})\mathbf{R}\mathbf{T}}{\mathbf{n}\mathbf{F}} \text{ ln } \mathbf{P} \tag{19}$$

#### **2.1 Enthalpy of reaction for a fuel cell**

Enthalpy is the thermodynamic quantity that states the total heat content of the system, which is the sum of all internal process in a closed system [2]. For homogeneous systems, enthalpy is solely based on the size of the system as it is an extensive property. Enthalpy (H) is the sum of the internal energy of the system (U) and the product of pressure (P) and volume (V) of the system. The change in enthalpy in a system is equal to the heat gained or lost in the system:

$$\mathbf{H} = \mathbf{U} + \mathbf{P}\mathbf{V} \tag{20}$$

The enthalpy change (ΔH) for a reaction in a fuel cell indicates the full amount of heat released by the reaction at a constant pressure; hence, enthalpy is simply equal to the heat released:

$$\mathbf{H} = \mathbf{Q}\_{\mathbf{p}} \tag{21}$$

At constant pressure and volume, the thermodynamic properties of the cell are related to the behaviour of its potential and are defined from the Gibbs-Helmholtz equation [3]:

$$
\Delta \mathbf{H} = \Delta \mathbf{G} - \mathbf{T} \left( \frac{\partial \Delta \mathbf{G}}{\partial \mathbf{T}} \right)\_{\mathbf{p}} \tag{22}
$$

In accordance with Eq. (5), in terms of electrochemical processes, change in enthalpy can be written as:

$$
\Delta \mathbf{H} = \mathbf{n} \mathbf{F} \left[ \mathbf{T} \left( \frac{\partial \mathbf{E} \text{cell}}{\mathbf{T}} \right) \mathbf{p} - \mathbf{E}\_{\text{cell}} \right] \tag{23}
$$

The overall reaction in Eq. (1) is the same as the reaction of hydrogen combustion. Combustion is an exothermic process, which means that there is energy released in the process [4, 5]:

$$\text{H}\_2 + \text{Me} \bullet \text{O}\_2 \rightarrow \text{H}\_2\text{O} \qquad \Delta \text{H} < \text{O} \tag{24}$$

From the table of enthalpies of formation (h<sup>o</sup> f) and absolute entropies of formation (so f) obtained from the basic thermodynamic data (see **Table 1**), the heat of formation of both liquid and vapour water can be calculated using the equation above to form:

The heat (or enthalpy) of a chemical reaction is the difference between the heat of formation of products and reactants. This means [6]:

$$
\Delta \mathbf{H}\_{\mathbf{f}} = \sum \mathbf{products} - \sum \mathbf{reactants}
$$

$$
= (\mathbf{h\_{f}})\_{\mathbf{H\_{2}O}} - \left[ (\mathbf{h\_{f}})\_{\mathbf{H\_{2}}} + \mathbf{1/2} \left( \mathbf{h\_{f}} \right)\_{\mathbf{O\_{2}}} \right] \tag{25}
$$

measure of the unavailable energy in a closed thermodynamic system that is usually considered to be a measure of the system's disorder, known as the second law of thermodynamics. The total entropy of a system increases over time, as the molecu-

between the initial state and the final state, the system is going through a reversible change. Since entropy represents the unavailable energy used in the system, a system of zero entropy optimises the work output of the system. The fuel cell generates the amount of electricity and rejects an amount of thermal energy Q to its environment. As there is heat transfer, and it is a real system, there must be an increase in entropy. The amount of heat rejected and the maximum amount of electrical power that a fuel cell will generate can be determined by formulating the

The entropy of H2 and O2 will disappear, but the new entropy of H2O and heat formation will appear. As long as the process is reversible, which is the assumption for the fuel cell, the entropy appearing in the rejected heat can be written as [8]:

<sup>Δ</sup><sup>S</sup> <sup>¼</sup> *<sup>Δ</sup>*Qrev

The equation for the change of entropy (ΔS) is equal to the change in enthalpy

As there is no heat transfer in electrical work, the entropy is zero. The entropy in

mol*:*<sup>k</sup> � 1 mol H2 <sup>þ</sup> <sup>0</sup>*:*<sup>205</sup> kJ

¼ 0*:*070 kJ*=*mol*:*K

<sup>Q</sup> min <sup>¼</sup> <sup>T</sup> <sup>X</sup>Sreactants �XSproducts � � <sup>¼</sup> 298 k 0ð Þ *:*<sup>0134</sup> � <sup>0</sup>*:*<sup>070</sup>

an H2/O2 fuel cell can be calculated using the absolute entropy values given in **Table 1**. The second law of thermodynamics requires that in a fuel cell, there will be a net increase in entropy. Therefore, the entropy that shows up in the rejected heat and the product water (liquid water) must be greater than the entropy contained in

H2 þ O2 ! H2O þ Q (26)

<sup>T</sup> <sup>þ</sup>Xproducts <sup>≥</sup> <sup>X</sup>reactants (28)

Q ≥T products ½ � � reactants (29)

mol*:*<sup>k</sup> � 1 mol H2O lð Þ

<sup>T</sup> (27)

mol*:*<sup>k</sup> � ½ mol O2

lar disorder increases. Therefore, if the system is in equilibrium, the change

entropy changes occurring in the cell:

*DOI: http://dx.doi.org/10.5772/intechopen.90141*

*Fuel Cell Thermodynamics*

the reactants (H2 and O2) [9, 10]: Entropy gain ≥ entropy loss

(ΔQ) divided by the temperature (T) of the system.

Q

<sup>X</sup>Sreactants <sup>¼</sup> <sup>0</sup>*:*<sup>131</sup> kJ

To calculate the amount of heat rejected per mole of H2:

¼ 0*:*2335 kJ*=*mol*:*K

The minimum of heat released during the reaction:

¼ 48*:*72 kJ per mole of H2

**9**

<sup>X</sup>Sproducts <sup>¼</sup> <sup>0</sup>*:*<sup>070</sup> kJ

ΔHf, H2O(l) = HHV = �285.8 kJ/mol and ΔHf, H2O (g) = LHV = �241.8 kJ/mol.

The enthalpy of the hydrogen combustion reaction (Eq. (25)) is also called hydrogen's heating value. The 285.83 kJ/mol is known as hydrogen's higher heating value (HHV), which means that 1 mol of hydrogen is fully combusted with ½ mol of oxygen and cooled down to 25°C. If hydrogen is combusted with sufficient excess oxygen and cooled down to 25°C, the value will become 241.82 kJ/mol, which is known as hydrogen's lower heating value (LHV) [7]. The difference between the LHV and HHV of 44.01 kJ/mol is equal to the molar latent heat of water vaporisation at 25°C.

In the heating value for reactions involving water as a product, there is a choice in the calculation of thermodynamic voltages between a high heating value (HHV) and a low heating value (LHV), defined as follows for a given reaction:


Note that all calculations are based on HHV or LHV and do not necessarily correspond to the actual physical state of the product water at the fuel cell electrode. The terms HHV and LHV are used in combustion calculations as well, where the product water is nearly always in the gas phase. The difference between the two values is proportional to the latent heat of vaporisation of the liquid. The use of the LHV (gas-phase vapour product) will result in a lower calculated thermal voltage, since some energy is used for the latent heat of vaporisation of the liquid. In practice, the LHV is completely appropriate for high-temperature fuel cells, but the HHV is also commonly used. An important point regarding low-temperature fuel cells that is often confusing is that the choice of HHV or LHV is arbitrary and 100°C is not a point of demarcation between the two. Often 100°C is thought of as a natural boundary between the HHV and LHV because it is the phase change temperature of water at 1 atm pressure. The delineation between liquid and gas, however, is more complex and is related to the local vapour pressure and total pressure.

#### **2.2 Entropy (S)**

The concept of entropy is one of the thermodynamic parameters that are important to the science of fuel cells to understand. Entropy is defined as the


#### **Table 1.**

*Enthalpies of formation and absolute entropies of formation of fuel cell reactants and products (at 25°C and 1 atm).*

The heat (or enthalpy) of a chemical reaction is the difference between the heat

<sup>Δ</sup>Hf <sup>¼</sup> <sup>X</sup>products �Xreactants

ΔHf, H2O(l) = HHV = �285.8 kJ/mol and ΔHf, H2O (g) = LHV = �241.8 kJ/mol. The enthalpy of the hydrogen combustion reaction (Eq. (25)) is also called hydrogen's heating value. The 285.83 kJ/mol is known as hydrogen's higher heating value (HHV), which means that 1 mol of hydrogen is fully combusted with ½ mol of oxygen and cooled down to 25°C. If hydrogen is combusted with sufficient excess oxygen and cooled down to 25°C, the value will become 241.82 kJ/mol, which is known as hydrogen's lower heating value (LHV) [7]. The difference between the LHV and HHV of

In the heating value for reactions involving water as a product, there is a choice in the calculation of thermodynamic voltages between a high heating value (HHV)

• High heating value: It is assumed all the product water is in the liquid phase.

• Low heating value: It is assumed all the product water is in the gas phase.

Note that all calculations are based on HHV or LHV and do not necessarily correspond to the actual physical state of the product water at the fuel cell electrode. The terms HHV and LHV are used in combustion calculations as well, where the product water is nearly always in the gas phase. The difference between the two values is proportional to the latent heat of vaporisation of the liquid. The use of the LHV (gas-phase vapour product) will result in a lower calculated thermal voltage, since some energy is used for the latent heat of vaporisation of the liquid. In practice, the LHV is completely appropriate for high-temperature fuel cells, but the HHV is also commonly used. An important point regarding low-temperature fuel cells that is often confusing is that the choice of HHV or LHV is arbitrary and 100°C is not a point of demarcation between the two. Often 100°C is thought of as a natural boundary between the HHV and LHV because it is the phase change temperature of water at 1 atm pressure. The delineation between liquid and gas, however, is more complex and is related to the local vapour pressure and total pressure.

The concept of entropy is one of the thermodynamic parameters that are important to the science of fuel cells to understand. Entropy is defined as the

Hydrogen, H2 0 0.131 0 Oxygen, O2 0 0.205 0 Water (liquid), H2O (l) �285.8 0.070 �237.2 Water (vapour), H2O (g) �241.8 0.189 �228.6

**<sup>f</sup> (kJ/mol) so**

*Enthalpies of formation and absolute entropies of formation of fuel cell reactants and products (at 25°C and*

**<sup>f</sup> (kJ/mol.K) ΔG<sup>o</sup>**

**<sup>f</sup> (kJ/mol)**

**ho**

44.01 kJ/mol is equal to the molar latent heat of water vaporisation at 25°C.

and a low heating value (LHV), defined as follows for a given reaction:

**2.2 Entropy (S)**

**Table 1.**

*1 atm).*

**8**

¼ ð Þ hf H2O � ð Þ hf H2 þ 1*=*2 hð Þ<sup>f</sup> O2

h i

(25)

of formation of products and reactants. This means [6]:

*Thermodynamics and Energy Engineering*

measure of the unavailable energy in a closed thermodynamic system that is usually considered to be a measure of the system's disorder, known as the second law of thermodynamics. The total entropy of a system increases over time, as the molecular disorder increases. Therefore, if the system is in equilibrium, the change between the initial state and the final state, the system is going through a reversible change. Since entropy represents the unavailable energy used in the system, a system of zero entropy optimises the work output of the system. The fuel cell generates the amount of electricity and rejects an amount of thermal energy Q to its environment. As there is heat transfer, and it is a real system, there must be an increase in entropy. The amount of heat rejected and the maximum amount of electrical power that a fuel cell will generate can be determined by formulating the entropy changes occurring in the cell:

$$\rm H\_2 + O\_2 \rightarrow H\_2O + Q \tag{26}$$

The entropy of H2 and O2 will disappear, but the new entropy of H2O and heat formation will appear. As long as the process is reversible, which is the assumption for the fuel cell, the entropy appearing in the rejected heat can be written as [8]:

$$
\Delta \mathbf{S} = \begin{array}{c} \Delta \mathbf{Q} \mathbf{rev} \\ \mathbf{T} \end{array} \tag{27}
$$

The equation for the change of entropy (ΔS) is equal to the change in enthalpy (ΔQ) divided by the temperature (T) of the system.

As there is no heat transfer in electrical work, the entropy is zero. The entropy in an H2/O2 fuel cell can be calculated using the absolute entropy values given in **Table 1**. The second law of thermodynamics requires that in a fuel cell, there will be a net increase in entropy. Therefore, the entropy that shows up in the rejected heat and the product water (liquid water) must be greater than the entropy contained in the reactants (H2 and O2) [9, 10]:

Entropy gain ≥ entropy loss

$$\frac{\text{Q}}{\text{T}} + \sum \text{products} \ge \sum \text{reactants} \tag{28}$$

$$\mathbf{Q} \ge \mathbf{T} \text{ [products -- reactants]} \tag{29}$$

To calculate the amount of heat rejected per mole of H2:

$$
\sum \mathbf{S}\_{\text{reactants}} = 0.131 \,\frac{\text{kJ}}{\text{mol} \,\text{k}} \times 1 \,\text{mol} \,\text{H}\_2 + 0.205 \,\frac{\text{kJ}}{\text{mol} \,\text{k}} \times 14 \,\text{mol }\,\text{O}\_2
$$

$$
= 0.2335 \,\text{kJ/mol} \,\text{K}
$$

$$
\sum \mathbf{S}\_{\text{products}} = \mathbf{0}.070 \,\frac{\text{kJ}}{\text{mol} \,\text{k}} \times 1 \,\text{mol }\,\text{H}\_2\mathbf{O} \,\text{(l)}
$$

$$
= \mathbf{0}.070 \,\text{kJ/mol} \,\text{K}
$$

The minimum of heat released during the reaction:

$$\begin{aligned} \text{Q}\_{\text{min}} &= \text{T} \left( \sum \text{S}\_{\text{reactants}} - \sum \text{S}\_{\text{products}} \right) = 298 \text{ k} \left( 0.0134 - 0.070 \right) \\ &= 48.72 \text{ kJ per mole of } \text{H}\_2 \end{aligned}$$

As heat capacity effects are generally minor, ΔH and ΔS values are usually assumed to be independent of temperature. A simplified entropy calculation can be with entropy values obtained from **Table 1**:

$$
\Delta \mathbf{S}\_{\mathbf{f}} = (\mathbf{s}^{\diamond})\_{\mathrm{H}\_{2}\mathrm{O}} - \left[ (\mathbf{s}^{\diamond})\_{\mathrm{H}\_{2}} + \mathbf{1}/2 \, (\mathbf{s}^{\diamond})\_{\mathrm{O}\_{2}} \right] \tag{30}
$$

where n is the amount of exchanged electrons and F is Faraday's constant. For the hydrogen oxidation or water formation, n = 2. The free enthalpies ΔG of water

ΔGf, H2O lð Þ ¼ �237*:*13 kJ*=*mol or ΔGf,H2O gð Þ ¼ �228*:*57 kJ*=*mol

Another property that is important in thermodynamics and the study of fuel cells is the specific heat. The specific heat of a solid or liquid is usually defined as the heat required to raise unit mass through 1 degree temperature rise. For a gas there are an infinite number of ways in which heat may be added between any two temperatures, and hence a gas could have an infinite number of specific heats. However, only two specific heats for gases are defined, as specific heat at constant

For a reversible ideal gas, a non-flow process at constant pressure and at con-

For a substance that is modelled as incompressible, the specific heats are

Cp <sup>¼</sup> Cv <sup>¼</sup> dQ

For an ideal gas in which a fuel cell is assumed to be, the specific enthalpy

The specific heat can also be related to specific entropy at temperature T:

H2 + 1/2O2 ! H2O (l) �285.8 �237.1 �0.163 1.23 H2 + 1/2O2 ! H2O (g) �241.8 �228.6 �0.045 1.18

*Enthalpies, entropies and Gibbs free energy of H2/O2 fuel cell reaction in (kJ/mol.K) and the resulting*

ðT

298*:*15

hT ¼ h298*:*<sup>15</sup> þ

where h298.15 is the enthalpy at a reference temperature.

<sup>o</sup> <sup>¼</sup> �ΔGfH2O gð Þ

dQ ¼ Cp dT (34)

dQ ¼ Cv dT (35)

hf ¼ dQ ¼ CpdT (37)

**ΔH (kJ/mol) ΔG (kJ/mol) ΔS (kJ/mol.K) E (V)**

dT (36)

CPdT (38)

2F ¼ �1*:*18 kJ*=*mol

The corresponding potential is therefore (**Table 2**):

2F ¼ �1*:*23 V and Eg

volume, cv, and specific heat at constant pressure, cp.

stant volume is given as, respectively:

assumed to be equal, Cp = Cv [15]:

depends only on temperature:

formation are either:

*Fuel Cell Thermodynamics*

*DOI: http://dx.doi.org/10.5772/intechopen.90141*

<sup>o</sup> <sup>¼</sup> �ΔGfH2O lð Þ

**2.4 Specific heat capacity**

El

and

**Table 2.**

**11**

*theoretical cell potential at 25°C.*

ΔSf, H2O (l) = HHV = �0.163 kJ/mol.K and ΔSf, H2O (g) = LHV = �0.045 kJ/mol.K.

#### **2.3 Gibbs free energy**

From the second law of thermodynamics, the change in free energy, or maximum useful work, can be obtained when a 'perfect' fuel cell operating irreversibly is dependent upon temperature. Therefore, Wel, the electrical power output, is [11]:

$$
\Delta \mathbf{G} = \Delta \mathbf{H} - \mathbf{T} \,\Delta \mathbf{S} \tag{31}
$$

where H is the total energy of the system, S is the 'unavailable' energy and G is the 'free' energy, or the energy available to do useful work.

The change in Gibbs free energy varies with both temperature and pressure. It can be shown that for a H2/O2 fuel cell:

$$
\Delta \mathbf{G}\_{\mathbf{f}} = \Delta \mathbf{G}^{\diamond}{}\_{\mathbf{f}} + \text{RT } \ln \left[ \frac{\mathbf{p} \mathbf{H}\_2 \mathbf{O}}{\mathbf{p} \mathbf{H}\_2 p \sqrt{\mathbf{O}\_2}} \right] \tag{32}
$$

where ΔG<sup>o</sup> f is the change in Gibbs free energy at standard pressure, which varies with the temperature T of the fuel cell, in Kelvin; pH2 , pO2 and PH2O are the partial pressure of the hydrogen, oxygen and vapour, respectively; and R is the universal gas constant (8.314 J/(kg.K)). The fact that the value of ΔG<sup>o</sup> <sup>f</sup> is negative means that the energy is released from the reaction [12–14].

For the H2/O2 fuel cell reaction, the change in Gibbs free energy is:

$$
\Delta \mathbf{G}\_{\rm f} = \mathbf{G}\_{\rm f, products} - \mathbf{G}\_{\rm f, reactants} = \mathbf{G}\_{\rm f,H\_2O} - \left[ \mathbf{G}\_{\rm f,H\_2} + \mathbf{G}\_{\rm f,O\_2} \right] \tag{33}
$$

The maximum possible electrical energy output and the corresponding electrical potential difference between the cathode and anode are achieved when the fuel cell is operated under the thermodynamically reversible condition. This maximum possible cell potential is called 'reversible cell potential', one of the significantly important parameters for FC.

From Eq. (23) we calculated ΔH to be 285.830 kJ/mol for hydrogen's HHV and 241.98 kJ/mol for hydrogen's LHV, while for ΔS in Eq. (28), the entropy of reaction is 0.163 kJ/mol for HHV and for LHV 0.044 kJ/mol.

To calculate Gibbs free energy for H2/O2 fuel cell reaction:

$$\Delta\text{G}\_{\text{f}}\text{ (l)} = \text{HHV} = -285.8 \text{ kJ/mol} - (273.15 \text{ K}) \ (-0.1633 \text{ kJ/mol}) = -237.1 \text{ kJ/mol}$$

$$\Delta\text{G}\_{\text{f}}\text{ (g)} = \text{LHV} = -241.8 \text{ kJ/mol} - (273.15 \text{ K}) \ (-0.045 \text{ kJ/mol}) = -228.6 \text{ kJ/mol}$$

The fact that the value of ΔG<sup>o</sup> <sup>f</sup> is negative means that the energy is released from the reaction. From Eq. (6), the potential or reversible open cell voltage Eo of any electrochemical device is defined as:

$$\mathbf{E}^{\bullet} = -\frac{\Delta \mathbf{G}}{\mathbf{n} \mathbf{F}}$$

*Fuel Cell Thermodynamics DOI: http://dx.doi.org/10.5772/intechopen.90141*

where n is the amount of exchanged electrons and F is Faraday's constant. For the hydrogen oxidation or water formation, n = 2. The free enthalpies ΔG of water formation are either:

$$
\Delta \mathbf{G}\_{\text{f}, \text{H}\_2\text{O} \text{ (l)}} = -237.13 \text{ kJ/mol} \text{ or } \Delta \mathbf{G}\_{\text{f}, \text{H}\_2\text{O} \text{ (g)}} = -228.57 \text{ kJ/mol}
$$

The corresponding potential is therefore (**Table 2**):

$$\mathrm{E\_l^{\circ}} = \frac{-\Delta \mathrm{G\_f} \mathrm{H\_2O(l)}}{2\mathrm{F}} = -1.23 \text{ V and } \mathrm{E\_g}^{\circ} = \frac{-\Delta \mathrm{G\_f} \mathrm{H\_2O(g)}}{2\mathrm{F}} = -1.18 \text{ kJ/mol}$$

#### **2.4 Specific heat capacity**

Another property that is important in thermodynamics and the study of fuel cells is the specific heat. The specific heat of a solid or liquid is usually defined as the heat required to raise unit mass through 1 degree temperature rise. For a gas there are an infinite number of ways in which heat may be added between any two temperatures, and hence a gas could have an infinite number of specific heats. However, only two specific heats for gases are defined, as specific heat at constant volume, cv, and specific heat at constant pressure, cp.

For a reversible ideal gas, a non-flow process at constant pressure and at constant volume is given as, respectively:

$$\mathbf{dQ} = \mathbf{C} \mathbf{p} \,\mathrm{d}\mathbf{T} \tag{34}$$

and

As heat capacity effects are generally minor, ΔH and ΔS values are usually assumed to be independent of temperature. A simplified entropy calculation can be

ΔSf, H2O (l) = HHV = �0.163 kJ/mol.K and ΔSf, H2O (g) = LHV = �0.045 kJ/mol.K.

From the second law of thermodynamics, the change in free energy, or maximum useful work, can be obtained when a 'perfect' fuel cell operating irreversibly is dependent upon temperature. Therefore, Wel, the electrical power output, is [11]:

where H is the total energy of the system, S is the 'unavailable' energy and G is

The change in Gibbs free energy varies with both temperature and pressure. It

pressure of the hydrogen, oxygen and vapour, respectively; and R is the universal

ΔGf ¼ Gf,products � Gf,reactants ¼ Gf,H2O � Gf,H2 þ Gf,O2

The maximum possible electrical energy output and the corresponding electrical potential difference between the cathode and anode are achieved when the fuel cell is operated under the thermodynamically reversible condition. This maximum possible cell potential is called 'reversible cell potential', one of the significantly impor-

From Eq. (23) we calculated ΔH to be 285.830 kJ/mol for hydrogen's HHV and 241.98 kJ/mol for hydrogen's LHV, while for ΔS in Eq. (28), the entropy of reaction

ΔGf ðÞ¼ l HHV ¼ �285*:*8 kJ*=*mol � ð Þ� 273*:*15 K ð 0*:*1633 kJ*=*mol޼�237*:*1 kJ*=*mol ΔGf ð Þ¼ g LHV ¼ �241*:*8 kJ*=*mol � ð Þ� 273*:*15 K ð 0*:*045 kJ*=*mol޼�228*:*6 kJ*=*mol

the reaction. From Eq. (6), the potential or reversible open cell voltage Eo of any

<sup>E</sup><sup>o</sup> ¼ � <sup>Δ</sup><sup>G</sup> nF

<sup>f</sup> is negative means that the energy is released from

For the H2/O2 fuel cell reaction, the change in Gibbs free energy is:

<sup>f</sup> <sup>þ</sup> RT ln pH2O

pH2*p*√O2

f is the change in Gibbs free energy at standard pressure, which varies

" #

<sup>o</sup> ð ÞH2 <sup>þ</sup> <sup>1</sup>*=*2 s<sup>o</sup> ð ÞO2 h i

ΔG ¼ Δ H � T ΔS (31)

(30)

(32)

, pO2 and PH2O are the partial

� � (33)

<sup>f</sup> is negative means that

<sup>o</sup> ð ÞH2O � <sup>s</sup>

with entropy values obtained from **Table 1**:

*Thermodynamics and Energy Engineering*

**2.3 Gibbs free energy**

where ΔG<sup>o</sup>

tant parameters for FC.

**10**

ΔSf ¼ s

the 'free' energy, or the energy available to do useful work.

<sup>Δ</sup>Gf <sup>¼</sup> <sup>Δ</sup>G<sup>o</sup>

gas constant (8.314 J/(kg.K)). The fact that the value of ΔG<sup>o</sup>

with the temperature T of the fuel cell, in Kelvin; pH2

the energy is released from the reaction [12–14].

is 0.163 kJ/mol for HHV and for LHV 0.044 kJ/mol.

The fact that the value of ΔG<sup>o</sup>

electrochemical device is defined as:

To calculate Gibbs free energy for H2/O2 fuel cell reaction:

can be shown that for a H2/O2 fuel cell:

$$\mathbf{dQ} = \mathbf{C} \mathbf{v} \,\mathrm{d}\mathbf{T} \tag{35}$$

For a substance that is modelled as incompressible, the specific heats are assumed to be equal, Cp = Cv [15]:

$$\mathbf{C}\mathbf{p} = \mathbf{C}\mathbf{v} = \frac{\mathbf{d}\mathbf{Q}}{\mathbf{d}\mathbf{T}}\tag{36}$$

For an ideal gas in which a fuel cell is assumed to be, the specific enthalpy depends only on temperature:

$$\mathbf{h}\_{\mathbf{f}} = \mathbf{d} \mathbf{Q} = \mathbf{C} \mathbf{p} \mathbf{d} \mathbf{T} \tag{37}$$

$$\mathbf{h\_{T}} = \mathbf{h\_{298.15}} + \int\_{298.15}^{\mathbf{T}} \mathbf{C\_{P}d} \mathbf{T} \tag{38}$$

where h298.15 is the enthalpy at a reference temperature. The specific heat can also be related to specific entropy at temperature T:


**Table 2.**

*Enthalpies, entropies and Gibbs free energy of H2/O2 fuel cell reaction in (kJ/mol.K) and the resulting theoretical cell potential at 25°C.*

*Thermodynamics and Energy Engineering*

$$\text{d}\mathbf{S} = \frac{\text{d}\mathbf{Q}}{\text{T}} = \text{Cp}\,\frac{\text{dT}}{\text{T}}\tag{39}$$

(41)

The heat capacity values for hydrogen, oxygen and water as a function of temperature are shown in **Figure 2**. The data were obtained from Kabza [4].

In the process of energy conversion in a fuel cell, the initial chemical energy between the enthalpy of the products and reactants is converted into electrical energy and thermal energy, as stated in the first law of thermodynamics. The efficiency of any energy conversion device is defined as the ratio between useful

> <sup>η</sup> <sup>¼</sup> actual electrical work maximum available work

> > <sup>Δ</sup><sup>H</sup> <sup>¼</sup> <sup>Δ</sup><sup>H</sup> � <sup>T</sup>Δ<sup>S</sup>

Themaximum possible thermodynamic efficiency of a fuel cell can be written as [9]:

In the case of a fuel cell, the useful energy output is the electrical energy produced, and the energy input is the enthalpy of hydrogen, that is, hydrogen's HHV. Assuming that all of the Gibbs free energy can be converted into electrical energy (the reaction is reversible), the maximum theoretical efficiency of a fuel cell is [23, 24] (**Figure 3**):

The LHV has higher efficiency compared to HHV, because the reversible effi-

The expected fuel cell efficiency is not always achieved due to thermodynamic

Other than calculating energy quantities during the conversion of chemical energy to electrical energy, there is also the matter of electron flow through

<sup>286</sup>*:*<sup>02</sup> x 100% <sup>¼</sup> 83%

� 100% ¼ 94*:*5%

<sup>η</sup> ¼¼ <sup>1</sup> � <sup>T</sup>Δ<sup>S</sup>

<sup>Δ</sup><sup>H</sup> (43)

<sup>Δ</sup><sup>H</sup> (44)

<sup>η</sup> <sup>¼</sup> <sup>Δ</sup><sup>G</sup>

<sup>η</sup> <sup>¼</sup> <sup>Δ</sup><sup>G</sup>

<sup>η</sup> <sup>¼</sup> <sup>Δ</sup><sup>G</sup>

*Energy inputs and outputs for a fuel cell as an energy conversion device [25].*

and electrochemical irreversible losses [28].

**4. Irreversible losses**

**Figure 3.**

**13**

<sup>Δ</sup><sup>H</sup> <sup>¼</sup> <sup>237</sup>*:*<sup>34</sup>

<sup>Δ</sup><sup>H</sup> <sup>¼</sup> <sup>228</sup>*:*<sup>74</sup> 241*:*98

ciency of the fuel cell decreases as the operating temperature increases [27].

For hydrogen's LHV, the fuel cell efficiency would be [26]:

**3. Fuel cell thermodynamic efficiency**

*DOI: http://dx.doi.org/10.5772/intechopen.90141*

*Fuel Cell Thermodynamics*

energy output and energy input [19–22]:

$$\mathbf{s}\_{\rm T} = \mathbf{s}\_{298.15} + \int\_{298.15}^{\rm T} \frac{\mathbf{1}}{\mathbf{T}} \mathbf{C}\_{\rm P} \mathbf{d} \mathbf{T} \tag{40}$$

The specific enthalpy and entropy for H2/O2 fuel cell are given by [16]:

$$\begin{array}{l} \Delta \mathbf{H} = \left[ \mathbf{h}^{\bullet}\_{\mathbf{f}, \mathrm{H}\_{2} \mathrm{O}} + \mathbf{C} \mathbf{p}, \, \mathrm{H}\_{2} \mathrm{O} \left( \mathrm{T} - 2 \mathbf{9} \mathbf{8.15} \right) \right]\_{\mathrm{H}\_{2} \mathrm{O}} - \mathsf{M} \left[ \mathbf{h}^{\bullet}\_{\mathbf{f}, \mathrm{O}\_{2}} + \mathbf{C} \mathbf{p}, \, \mathrm{O}\_{2} \left( \mathrm{T} - 2 \mathbf{98.15} \right) \right]\_{\mathrm{O}\_{2}} \\\ - \left[ \mathbf{h}^{\bullet}\_{\mathrm{f}, \mathrm{H}\_{2}} + \mathbf{C} \mathbf{p}, \, \mathrm{H}\_{2} \left( \mathrm{T} - 2 \mathbf{98.15} \right) \right]\_{\mathrm{H}\_{2}} \end{array}$$

$$\begin{aligned} \Delta \mathbf{S} &= \left[ \mathbf{h}^{\bullet}\_{\mathrm{f,H\_2O}} + \mathbf{C} \mathbf{p}, \, \_{\mathrm{H\_2O}} \ln \frac{\mathbf{T}}{298.15} \right]\_{\mathrm{H\_2O}} - \mathbb{M} \left[ \mathbf{h}^{\bullet}\_{\mathrm{f,O\_2}} + \mathbf{C} \mathbf{p}, \, \_{\mathrm{O\_2}} \ln \frac{\mathbf{T}}{298.15} \right]\_{\mathrm{O\_2}} \\ &- \left[ \mathbf{h}^{\bullet}\_{\mathrm{f,H\_2}} + \mathbf{C} \mathbf{p}, \, \_{\mathrm{H\_2}} \ln \frac{\mathbf{T}}{298.15} \right]\_{\mathrm{H\_2}} \end{aligned} \tag{42}$$

The values of molar entropy and enthalpy of formation at 298.15 K are given in **Table 1** [17]. Eqs. (38) and (40) can be used to determine the specific heat capacity, Cp, at constant pressure. Specific heat relationships are generally modelled with a high-order polynomial, such as those listed below for hydrogen fuel cell gases, valid in the range of 300–1000 K [18]. It is reported that over a range of temperatures, Cp is not constant, while over the range of 300–350 K, the obtained Cp values are 0.6% accurate [4].

Hydrogen, H2:

$$\begin{aligned} \text{Cp (T)} &= 3.057 + 2.677 \times 10^{-3} \text{ (T)} - 5.810 \times 10^{-6} \text{ (T)}^2 + 5.521 \times 10^{-9} \text{ (T)}^3 \\ &- 1.812 \times 10^{-12} \text{ (T)}^4 \end{aligned}$$

Oxygen, O2:

$$\begin{split} \text{Cp} \left( \text{T} \right) &= 3.626 - 1.878 \times 10^{-3} \left( \text{T} \right) + 7.055 \times 10^{-6} \left( \text{T} \right)^{2} - 6.764 \times 10^{-9} \left( \text{T} \right)^{3} \\ &+ 2.156 \times 10^{-12} \left( \text{T} \right)^{4} \end{split}$$

For H2o:

$$\begin{array}{l} \text{Cp (T)} = \text{4.070} - \text{1.808} \times \text{10}^{-3} \text{ (T)} + \text{4.152} \times \text{10}^{-6} \text{ (T)}^2 - \text{2.964} \times \text{10}^{-9} \text{ (T)}^3\\ \text{ + 0.807} \times \text{10}^{-12} \text{ (T)}^4 \end{array}$$

**Figure 2.** *Specific heat values for hydrogen, oxygen and water as a function of temperature.*

dS <sup>¼</sup> dQ

The specific enthalpy and entropy for H2/O2 fuel cell are given by [16]:

H2O

H2

H2

sT ¼ s298*:*<sup>15</sup> þ

298*:*15

298*:*15

f,H2O <sup>þ</sup> Cp, H2O ð Þ <sup>T</sup> � <sup>298</sup>*:*<sup>15</sup> � �

f,H2O <sup>þ</sup> Cp, H2O ln <sup>T</sup>

f,H2 <sup>þ</sup> Cp, H2 ln <sup>T</sup>

� <sup>1</sup>*:*<sup>812</sup> � <sup>10</sup>�<sup>12</sup> ð Þ <sup>T</sup> <sup>4</sup>

<sup>þ</sup> <sup>2</sup>*:*<sup>156</sup> � <sup>10</sup>�<sup>12</sup> ð Þ <sup>T</sup> <sup>4</sup>

<sup>þ</sup> <sup>0</sup>*:*<sup>807</sup> � <sup>10</sup>�<sup>12</sup> ð Þ <sup>T</sup> <sup>4</sup>

*Specific heat values for hydrogen, oxygen and water as a function of temperature.*

� �

� �

f,H2 <sup>þ</sup> Cp, H2 ð Þ <sup>T</sup> � <sup>298</sup>*:*<sup>15</sup> � �

<sup>Δ</sup><sup>H</sup> <sup>¼</sup> <sup>h</sup><sup>o</sup>

<sup>Δ</sup><sup>S</sup> <sup>¼</sup> <sup>h</sup><sup>o</sup>

accurate [4].

Hydrogen, H2:

Oxygen, O2:

For H2o:

**Figure 2.**

**12**

� <sup>h</sup><sup>o</sup>

*Thermodynamics and Energy Engineering*

� ho

<sup>T</sup> <sup>¼</sup> Cp dT

298*:*15

H2O � ½ h<sup>o</sup>

� ½ h<sup>o</sup>

The values of molar entropy and enthalpy of formation at 298.15 K are given in **Table 1** [17]. Eqs. (38) and (40) can be used to determine the specific heat capacity, Cp, at constant pressure. Specific heat relationships are generally modelled with a high-order polynomial, such as those listed below for hydrogen fuel cell gases, valid in the range of 300–1000 K [18]. It is reported that over a range of temperatures, Cp is not constant, while over the range of 300–350 K, the obtained Cp values are 0.6%

Cp Tð Þ¼ <sup>3</sup>*:*<sup>057</sup> <sup>þ</sup> <sup>2</sup>*:*<sup>677</sup> � <sup>10</sup>�<sup>3</sup> ð Þ� <sup>T</sup> <sup>5</sup>*:*<sup>810</sup> � <sup>10</sup>�<sup>6</sup> ð Þ <sup>T</sup> <sup>2</sup> <sup>þ</sup> <sup>5</sup>*:*<sup>521</sup> � <sup>10</sup>�<sup>9</sup> ð Þ <sup>T</sup> <sup>3</sup>

Cp Tð Þ¼ <sup>3</sup>*:*<sup>626</sup> � <sup>1</sup>*:*<sup>878</sup> � <sup>10</sup>�<sup>3</sup> ð Þþ <sup>T</sup> <sup>7</sup>*:*<sup>055</sup> � <sup>10</sup>�<sup>6</sup> ð Þ <sup>T</sup> <sup>2</sup> � <sup>6</sup>*:*<sup>764</sup> � <sup>10</sup>�<sup>9</sup> ð Þ <sup>T</sup> <sup>3</sup>

Cp Tð Þ¼ <sup>4</sup>*:*<sup>070</sup> � <sup>1</sup>*:*<sup>808</sup> � <sup>10</sup>�<sup>3</sup> ð Þþ <sup>T</sup> <sup>4</sup>*:*<sup>152</sup> � <sup>10</sup>�<sup>6</sup> ð Þ <sup>T</sup> <sup>2</sup> � <sup>2</sup>*:*<sup>964</sup> � <sup>10</sup>�<sup>9</sup> ð Þ <sup>T</sup> <sup>3</sup>

1

ðT

<sup>T</sup> (39)

<sup>T</sup>CPdT (40)

f,O2 <sup>þ</sup> Cp, O2 ð Þ <sup>T</sup> � <sup>298</sup>*:*<sup>15</sup> � �

298*:*15

O2

f,O2 <sup>þ</sup> Cp, O2 ln <sup>T</sup>

� �

O2

(41)

(42)

The heat capacity values for hydrogen, oxygen and water as a function of temperature are shown in **Figure 2**. The data were obtained from Kabza [4].
