Mench M.M. Fuel Cell Engines

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40 Basic Electrochemical Principles

Voltmeter

2(g)

Switch

Zn anode

Cathode

compartment

Anode

compartment

Zn(s)

(aq)

2(g)

Figure 2.7 The SHE used to reference a zinc oxidation reaction. (Adapted from Ref. [5].)

In a galvanic reaction, the potential difference between the cathode and the anode is

positive, and the anode is at a lower potential compared to the cathode. As discussed, this

voltage difference is a measure of the thermodynamic potential for reaction and is related

to the chemical energy difference between products and reactants. The majority of fuel

cells operate with oxygen in air at the cathode and hydrogen fuel at the anode. Thus, at

open-circuit conditions the anode is nearly at SHE conditions and is therefore at around

0 V. The cathode potential is analogous to the top of the waterfall in Figure 2.6.

Using Table 2.1, the overall cell voltage is simply the difference in the reduction

potential between the cathode and anode:

cell

= E

cathode,red

− E

anode,red

(2.19)

There is a minus sign in Eq. (2.19) since both half-cell reactions are taken as reduction

reactions whereas Eq. (2.18) uses one reduction and one oxidation reaction. At standard

conditions (pure oxygen and hydrogen, 1 atm pressure, and 298 K), we expect an open-

circuit cell voltage of 1.23 V.

Resistance and Conductance Electrical resistance measured in ohms (), is a measure

of the potential loss associated with moving a rate of charge. Put another way, in order

to move a given rate of charge through a conductor, some voltage potential is lost. It is

important to realize the energy is not lost. Indeed, conservation of energy still applies.

Instead, the potential to do electrical work is lost and is dissipated into heat. The greater

voltage potential lost per rate of charge passed, the greater the resistance. Resistance can

be deﬁned as

R = V/A = (J/C)/(C/s) = Js/C

(2.20)

One joule-second per coulomb squared is deﬁned as one ohm, and its inverse (i.e., 1/)is

deﬁned, quite directly, as a mho. A mho is the unit of conductance, also known as a siemen

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2.3 Scientiﬁc Units, Constants, and Basic Laws 41

A = Cross-sectional area

ρ=

=Ω-m

= Linear path of ion travel

Figure 2.8 Illustration of current through a wire.

(S). The siemen is named after Werner von Siemens (1816–1892), a German electrical

engineer and businessman [7]. Electrical resistance is analogous to ﬂow potential losses in

a pipe. A small-diameter pipe has greater resistance to ﬂow than a large-diameter pipe and

suffers greater pressure loss per length for a given ﬂow rate. Similarly, a material with high

electrical resistance will have greater voltage potential loss per length for a given current.

From experience and common sense, we know that ﬂuid ﬂow pressure losses through

a pipe are a function of the geometry of the pipe, and the pressure loss is not an intrinsic

property of all pipes. Similarly, electrical resistance is not an intrinsic property of a wire.

Consider Figure 2.8, which shows a cylindrical wire axially conducting current. The resis-

tance to current (which is physically a measure of the resistance to the ﬂow of electrons) is

proportional to the cross-sectional area A and inversely proportional to the length of travel.

Obviously, the longer the wire, the greater the loss of voltage potential. An exceptionally

thin wire has more electron interaction with the wire surface, leading to increased resistance

to ﬂow compared to a larger diameter wire. Resistivity ρ (m) is an intrinsic property of

a material because it normalizes the resistance of the material for cross-sectional area (A)

and length (l), removing geometric factors:

ρ =

R = ρ

σ A

(2.21)

Some resistivities of common materials are given in Table 2.2. The inverse of resistivity,

1/ρ,istheconductivity σ [(m)

−1

]. Generally, the conductivity of a material is used if the

purpose is to conduct charged species, and its resistivity is used if it is an insulating material,

although either choice is appropriate. Metals are generally good electron conductors. Gas-

phase conductivity is negligible except at extreme temperatures where signiﬁcant gas

ionization can occur. Electrolytes are designed to be excellent ionic conductors and poor

electron conductors.

Dependence of Conductivity on Operational Parameters The conductivity of various

substances, and in particular fuel cell materials, can depend strongly on operating pa-

rameters. For example, the electron conductivity of solid materials such as graphite and

steel decreases with increasing temperature due to increased molecular collision frequency.

However, the ionic conductivity of the SOFC and PEFC electrolyte is a positively related

function of temperature. In the PEFC, water saturation in the electrolyte also plays a major

role. These functional dependencies are highly material speciﬁc and are discussed in later

chapters.

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42 Basic Electrochemical Principles

Table 2.2 Electrical and Ionic Resistivities of Selected Materials

Electron conductors Electron Resistivity at 293 K (m)

Gold 2.44 × 10

−8

Aluminum 2.28 × 10

−8

Copper 1.7 × 10

−8

Silver 1.6 × 10

−8

Stainless steel 7.2 × 10

−7

Platinum 1.1 × 10

−7

Ruthenium 7.1 × 10

−8

Palladium 10 × 10

−8

Carbon 3.5 × 10

−5a

Water (deionized) 2.5 × 10

Polytetraﬂuoroethylene (Teﬂon) 10

–10

Ionic Conductors Ionic Resistivity (m)

Naﬁon PEFC electrolyte by DuPont, fully humidiﬁed ∼10 at 353 K

SOFC electrolyte 0.1–1 at 600–1000 K

Liquid electrolytes Highly concentration, temperature,

and ion dependent

Dependent on direction and molecular structure.

2.3.2 Ohm’s Law

Ohm’s law is named after the work of Georg Simon Ohm (1789–1854), a Bavarian mathe-

matician and teacher. The basic form of Ohm’s law can be shown as

V = IR (2.22)

It is important to note that in liquid solutions Ohm’s law is true only for electrolytes without

ionic concentration gradients, which may not always be strictly accurate [8, 9], but can be

considered to be true for most fuel cell studies.

Often, students have difﬁculty following the units involved in this simple relationship

since they are often familiar with only volts, ohms, and amperes. Unit matching and

conversion are critical components of any engineering analysis, and electrochemistry is no

exception. If one breaks down the units of this relationship into units for energy, power,

and so forth, we see that Ohm’s law is dimensionally consistent:

V = IR

V = A · 

J/C = (C/s)(Js/C

) = J/COk!

Electrical Power Power is the rate of work, whether in mechanical, electrical, thermal,

or other form. Electrical power (P

) can be expressed as

= IV

(2.23)

W = (C/s)(J/C) = J/s = WOk!

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2.4 Faraday’s Laws: Consumption and Production of Species 43

If the voltage is the fuel cell voltage, the output is the electrical power from the fuel

cell. If instead the voltage in Eq. (2.23) is the voltage potential lost due to resistance, the

power generated is thermal dissipation. The delineation between electrical power and heat

dissipation rate will become clear in future chapters if it is not now. Thus, electrical power

has units of watts (J/s), as expected. Ohm’s law can also be substituted into Eq. (2.17) to

obtain alternate expressions for electrical power:

= IV = I · IR = I

(2.24)

= IV =

· V =

Since it is often the case that mechanical and electrical systems operate in conjunction

with one another, it is imperative to be able to convert among the units used for electrical

systems and thermo-mechanical systems.

Example 2.1 Simple Electrical Calculations A given circuit has a continuous 5 A DC

(direct current) and an overall resistance of 10 . Calculate

(a) the potential loss, in volts, to maintain steady state;

(b) the electrical power dissipated as heat during operation, in watts; and

SOLUTION

(a) Potential loss is found via Ohm’s law for this circuit: V = IR = 50 V.

(b) Electrical power dissipated as heat is P

= IV = 250 W.

Total heat Q =



dt = (250 W)(2 h)[1 (J/s)/W](3600 s/h) = 1800 kJ

COMMENT: The resistance shown represents the resistance in a generic electric circuit

not a fuel cell.

2.4 FARADAY’S LAWS: CONSUMPTION AND

PRODUCTION OF SPECIES

In this section we consider a fundamentally important question: How much mass of a given

reactant is required to produce a given amount of current? Conversely, how much current

is required to produce a certain amount of product? Clearly, the fundamental relationships

should be based on conservation of mass and charge. We cannot produce mass from an

electrochemical reaction, only rearrange it, and a given amount of reactant can produce a

ﬁxed number of charged species, based on the balanced chemical reaction. Consider 1 mol

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44 Basic Electrochemical Principles

of reactant in a generic electrochemical reaction:

A + ν

B → ν

C + ν

D (2.25)

where ν

is the molar (stoichiometric) coefﬁcient required to balance each species i in the

reaction equation. Consider the number of electron charges moved through the circuit to

react with 1 mol of species A in a one-electron process (e.g., for every mole of A consumed,

one electron is moved through the circuit). An example is given for the anodic portion of

the silver redox reaction:

→ Ag

+ e

−

(2.26)

We know from common experience that the overall silver redox reaction will oxidize

(tarnish) silver over time. In this oxidation process, 1 mol of electrons and 1 mol of

silver cation is produced per mole of silver reacted. However, not every reaction is a 1

mol–one electron process. Faraday’s constant F represents the charge per mole of equivalent

electrons, that is,

F =

6.023 × 10

electrons/mole equivalent

6.242 × 10

electrons/C

= 96,485 C/eq (2.27)

Faraday’s constant can also be written in terms of ampere-hour:

F = 96,485 C/eq

3600 C/Ah

= 26.8Ah/eq (2.28)

The equivalent electrons (eq) is very important but is commonly omitted. Many electro-

chemical reactions do not exchange 1 mol of electrons for 1 mol of reactant, as in Eq.

(2.26). For example, consider the ORR that occurs at the cathode of many fuel cells:

−

+ 4H

+ O

→ 2H

O (2.29)

Here, 4 mol of electrons react per mole of oxygen. In this case, the charge carried per mole

of oxygen reacted would be 4F. The scaling factor n is deﬁned as the number of electrons

transferred per mole of species of interest.

n =

number of electrons

mole of species of interest

= eq/mol (2.30)

Thus, the combination of nF is the charge passed per mole of species of interest, and the

units of nF are coulombs per mole. Note that the unit “mole” is speciﬁc to the chosen

species. That is, it is really a mole of the species of interest. The choice of the species of

interest makes a difference, and n simply permits determination of the relationship between

charge passed and reactant consumption (or product generation) of any species chosen. For

example, in Eq. (2.29), n is 4 eq electrons/mol O

for oxygen consumption. Alternatively,

considering water produced as the species of interest, the value of n is 2, and there are 2F

coulombs passed per mole of H

O produced.

Now, consider the HOR that occurs at the anode of many fuel cells:

→ 2e

−

+ 2H

(2.31)

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2.4 Faraday’s Laws: Consumption and Production of Species 45

Here, there are 2F coulombs passed per mole of H

(n = 2 eq electrons/mol H

). Note

that it is possible that some purely chemical reactions can also occur in parallel with

electrochemical reactions. In this case, the purely chemical component of the reaction must

also be known to determine the relationship between charge passed and consumption or

generation in the overall multi-reaction process.

Faraday’s Laws In the early 1830s, Michael Faraday reported that the quantity of species

electrolytically separated was proportional to the total charge passed, establishing a link

between the ﬂow of charge and mass [9]. This became the basis for Faraday’s ﬁrst law of

electrolysis:

1. For a speciﬁc charge passed, the mass of the products formed are proportional to

the electrochemical equivalent weight of the products.

Although Faraday’s results were purely experimental, the important implication from this

proportional relationship was that electrical current was a result of discrete particles we

now understand to be ions and electrons that are conserved as part of an overall balanced

reaction equation. This understanding enabled quantiﬁcation of many electrical phenomena

and helped reveal the nature of protons and electrons.

In his work, Faraday also discovered that the mass of the product species was directly

proportional to the charge passed. This law provides a connection between the charge passed

and the mass generated or consumed in a reaction through the balanced electrochemical

reaction equations. Faraday’s second law of electrolysis is stated as follows:

2. The amount of product formed or reactant consumed is directly proportional to the

charge passed.

The second law is the most important for fuel cell study. This result is what we could

expect from modern common sense and conservation of mass; the current generated is

proportional to the mass reacted or produced:

m ∝ I (2.32)

Considering a purely electrochemical reaction, with an understanding of the discrete nature

of charged particles that carry current, it seems obvious that there is a proportional rela-

tionship between charge passed and consumption or production of the involved species.

The proportionality is based on the balanced electrochemical reaction equation and can be

written as

(2.33)

Let us examine the parameters and units involved:

= rate of molar consumption or production of species x (mol of x/s)

I = current (A)

A = superﬁcial electrode area (cm

)

i = current density, = I /A (A/cm

)

n = equivalent electrons per mole of reactant x (eq/mol)

F = charge carried on one equivalent mole (C/eq)

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46 Basic Electrochemical Principles

The superﬁcial (or geometric) electrode area A is the active area of the electrode as viewed

from above. For a given electrochemical reaction, Eq. [2.33] can be used to determine the

amount of reactant consumed to produce a given current and the amount of product formed

for a known current. The actual amount of reactant delivered to an electrode can be greater

than determined by Faraday’s law but can never be less. The actual amount of product

produced in a given reaction will never be greater than predicted.

Example 2.2 Faraday’s Law Calculations Consider a single hydrogen fuel cell at 4 A

current output:

Anode oxidation:

→ 2e

−

+ 2H

Cathode reduction:

−

+ 4H

+ O

→ 2H

Global reaction:

→ H

(a) What is the molar rate of H

consumed for the electrochemical reaction?

(b) What is the molar rate of O

consumed for the electrochemical reaction?

reaction? Assume air is a mixture of 21% oxygen and 79% nitrogen by volume.

(d) What is the maximum molar ﬂow rate of air delivered for the electrochemical

reaction?

(e) What is the rate of water generation at the cathode in grams per hour? The molecular

weight of water is 18 g/mol.

(f) Can the generation rate of water be greater or less than the value predicted in part

(e)?

SOLUTION (a) Consider the hydrogen as the “reactant of interest,” x in Faraday’s law.

Also recall that 1 A = 1 C/s:

→ 2H

+ 2e

−

n = 2 electron eq/mol H

4C/s

(2 e

−

eq/mol H

)(96,485 C/eq)

= 2.072 × 10

−5

mol H

(b) Here, we are looking for the molar oxygen ﬂow rate of oxygen consumed, which

is determined from the oxidation–reduction reaction:

+ 2e

−

→ H

n = 4 electron eq/mol O

4C/s

(4 e

−

eq/mol O

)(96,485 C/eq)

= 1.036 × 10

−5

mol O

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2.4 Faraday’s Laws: Consumption and Production of Species 47

rate of air, which contains 79% nitrogen. The only required reactant here is oxygen, and

the molar ﬂow rate of oxygen required was solved in part (b). To achieve the same ﬂow

rate of oxygen in air, we simply must divide by the mole fraction of reactant (oxygen) in

the air mixture. Note that nitrogen ﬂow is inert here and simply “goes along for the ride”:

0.21

air

4C/s

(0.21 × 4e

−

eq/mol O

)(96,485 C/eq

= 4.94 × 10

−5

mol O

(d) There is no maximum of reactant supplied. The amount of reactant supplied is

governed by, for example, the pumps and blowers that make up the system. Faraday’s law

only imposes a minimum amount of mass required to generate a given current.

(e) Here, we are to determine the rate of water generation in grams per hours, which

will require some minor unit conversion. The ﬁrst step is to solve for the molar rate of water

generation, which occurs at the cathode:

+ 2e

−

→ H

n = 2 electron eq/mol H

4C/s

(2 e

−

eq/mol H

(96,485 C/eq) = 2.072 × 10

−5

mol H

O/s

Now simply convert to the appropriate units using the molecular weight:

× MW =

4C/s

(2 e

−

eq/mol H

O)(96,485 C/eq)

(18 g/mol)(3600 s/h)

= 1.34 g

(f) The value for water produced cannot be greater for a given current, since this would

violate conservation of mass. Normally, the products cannot be less than that predicted

either, unless another electrochemical reaction (called a side reaction) takes place that

consumes some of the charge passed.

COMMENTS: Notice the small numbers when units of moles per second are used. Be

sure to keep enough signiﬁcant digits. Also note from part (e) that a hydrogen fuel cell is

also a water generator. The water produced must be removed from the fuel cell to allow

reactant to reach the catalyst layer. In high-temperature systems, the water product is vapor,

while lower temperature fuel cell systems can produce liquid efﬂuent, which complicates

design.

Example 2.3 Fuel Cell Stack Calculations Consider a 20-cell stack operating steadily

in series with 100 cm

active area per electrode, with a current density of 0.8 A/cm

.The

fuel cell nominal voltage is 0.6 V per plate.

(a) Determine the water production in grams per hour for this stack.

(b) Determine the stack voltage and electrical power output.

SOLUTION (a) This example is quite similar to the previous one, but with an additional

complication that this is now a stack of fuel cells. Since they are in series, each fuel cell

c02 JWPR067-Mench December 19, 2007 17:26 Char Count=

48 Basic Electrochemical Principles

produces the same total current (0.8 A/cm

× 100 cm

= 80 A), and thus the total water

produced is simply 20 times the rate of water generation of an individual fuel cell in the

stack. Consider the ORR at the cathode:

−

+ 4H

+ O

→ 2H

For water production,

n = 2 electron eq/mol H

(0.8C/s · cm

100 cm

)

(2 e

−

eq/mol H

O)(96,485 C/eq)

= 4.15 × 10

−4

mol H

O/s per fuel cell

Now simply scale for the entire stack and to the appropriate units using the molecular

weight and number of plates in series:

O,stack

= 20

· MW

= (20 cells)(4.15 ×10

−4

mol H

O/s per fuel cell)(18 g/mol)(3600 s/h)

= 538 g H

O/ h per stack

(b) Since the plates are in series, the stack voltage is

20 cells × 0.6 V per fuel cell = 12 V per stack

The cells are all in series and therefore all carry the same current of 80 A. The stack electric

power is simply P

= IV, or 960 W.

COMMENTS: This works out to be about 538 cm

of water per hour. Larger fuel cell

engines have a signiﬁcant amount of water production to manage. Consider that an auto-

motive fuel cell engine should be on the order of 100 kW, and about 932 kg (around 2 lb)

of water per minute would be produced! Much of this leaves the system as vapor, however.

2.5 MEASURES OF REACTANT UTILIZATION

EFFICIENCY

There are various metrics utilized to quantify the efﬁciency of different aspects of an

electrochemical reaction. One type of efﬁciency for a purely electrochemical reaction is

based on species consumption. For a galvanic process, there will be a minimum amount

of reactant required for a given reaction, as calculated by Faraday’s law, Eq. [2.33]. In

practice, we are not constrained to provide exactly the minimum amount of reactant. For a

given current, there is a calculated minimum amount of reactant, but there is no maximum.

The actual ﬂow rate of reactants is a function of the pumps and blowers that are used

for reactant delivery. Obviously, the more ﬂow delivered, the higher the parasitic power

required, so we generally seek to deliver something close to the minimum requirement. The

Faradic efﬁciency is a measure of the percent utilization of reactant in a galvanic process:

theoretical required rate of reactant supplied

actual rate of reactant supplied

(2.34)

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2.5 Measures of Reactant Utilization Efﬁciency 49

Faradic efﬁciency is often called the fuel utilization efﬁciency (µ

) when applied to the fuel

in a galvanic redox reaction:

theoretical required rate of fuel supplied

actual rate of fuel supplied

(2.35)

For an electrolytic process, some side reactions and inefﬁciencies may occur and result in

less than complete conversion. The current efﬁciency is deﬁned as:

actual rate of species reacted or produced

theoretical rate of species reacted or produced

(2.36)

Stoichiometric Ratio In fuel cell parlance, the term stoichiometry is deﬁned as the in-

verse of the Faradic efﬁciency. Students may be confused with this terminology, since the

stoichiometric condition typically describes a balanced chemical reaction equation with no

excess oxidizer. Here, the term stoichiometry is used slightly differently, and its meaning is

similar to the deﬁnition of equivalence ratio used in combustion. Unlike chemical reactions,

the reduction and oxidation reactions are separated by electrolyte, so each electrode can

have a discrete stoichiometry:

The cathodic stoichiometry is deﬁned as:

f,c

actual rate of oxidizer delivered to cathode

theoretical rate of oxidizer required

(2.37)

The anodic stoichiometry is deﬁned as:

f,a

actual rate of fuel delivered to anode

theoretical rate of fuel required

(2.38)

To avoid confusion the reader should be aware that other symbols for stoichiometry, besides

λ, are commonly used in the literature, including ζ and ξ . The theoretical rate of reactant

required is calculated by Faraday’s law, and the actual rate of reactant delivered is a function

of the fuel or oxidizer delivery system. One important point is worth mentioning: Fuel cells

must always have an anode and cathode stoichiometry greater than 1. For a value less than

unity, the current speciﬁed could not be produced. For reasons explained in Chapter 4, a

stoichiometry of exactly 1 is not possible either, so that a Faradic efﬁciency of 100% is not

possible on the anode or cathode for a single pass of reactant.

Example 2.4 Stoichiometry and Utilization Consider a portable 20 cm

active area fuel

cell operating steadily at 0.75 V, 0.6 A/cm

. The fuel utilization efﬁciency is 50%, and the

cathode stoichiometry is 2.3. The fuel cell is expected to run for three days before being

recharged. The cathode operates on ambient air, and the anode runs off of compressed

hydrogen gas.

(a) Determine the volume of the hydrogen fuel tank required if it is stored as a

compressed gas at 200 atm (20.26 MPa), 298 K.

(b) How large would a pure oxygen container be if it was used to provide the oxi-

dizer? Consider 200 atm (20.26 MPa) storage pressure and 298 K average ambient

temperature.

Fuel recirculators can be used to increase the effective faradic efﬁciency to 100%, but we are talking about a

single pass of reactant through the fuel cell here.