Thermodynamics and Electrochemistry Of Low Temperature ... · Friedrich Wilhelm Ostwald –visionary ideas: “I do not know whether all of us realise fully what an imperfect thing

Thermodynamics and Electrochemistry

Of Low Temperature Fuel Cells

Frano BarbirFrano Barbir

Professor, FESB, Split, Croatia

Joint European Summer School for Fuel Cell and Hydrogen Technology

energy partners

Prof. Dr. Frano Barbir

heat

What is fuel cell?What is fuel cell?

Fuel cell is an electrochemical energy converter.

It converts chemical energy of fuel (H2) directly into electricity.

Fuel cell is like a battery but with constant fuel and oxidant supply.

hydrogen

DC electricity

_

+

BATTERYoxygen

water

FUEL CELL

Invention of fuel cell

Fuel cell historyFuel cell historyChristian Schönbein

(1799-1868):

Prof. of Phys. and Chem.

in Basel, first

observation of fuel cell

effect

(Dec. 1838, Phil. Mag.)

William Robert Grove

1839

(1811 –1896):

Welsh lawyer turned

scientist, demonstration

of fuel cell

(Jan. 1839, Phil. Mag.)


Scientific curiosity

Friedrich Wilhelm Ostwald

(1853-1932, Nobel prize 1909):

founder of “Physical Chemistry”,

provided much of the theoretical

Fuel cell historyFuel cell history

1839

provided much of the theoretical

understanding of how fuel cells operate

Friedrich Wilhelm Ostwald – visionary ideas:

“I do not know whether all of us realise fully what an

imperfect thing is the most essential source of power

which we are using in our highly developed engineering

– the steam engine”

Energy conversion in combustion engines

limited by Carnot efficiency limited by Carnot efficiency

unacceptable levels of atmospheric pollution

vs.

Energy conversion in fuel cells

direct generation of electricity

highly efficient, silent, no pollution

Transition would be technical revolution

Prediction: practical realization could take a long time!

Source: Z. Elektrochemie, Vol. 1, p. 122-125, 1894.


Reinvention



1839 1939

Industrial curiosity


Reinvention



1960’s1839 1939

Space program



Reinvention



Space program

1960’s1839 1939

PEM Fuel Cells used in Gemini program

Apollo program used alkaline fuel cells

Space Shuttle uses alkaline fuel cellsSpace Shuttle uses alkaline fuel cells

Renewed interest in PEM fuel cells

Courtesy of UTC Fuel Cells

First PEM Fuel Cell

Grubb and Niedrach (GE)Gemini Space Program

1960s

Space program



Reinvention



Space program

1990’s

Entrepreneural phase

1960’s1839 1939

Space program



Reinvention



Space program

1990’s

Entrepreneural phase

1960’s1839 1939

Birth of new industry

Automobiles

Buses

Scooters

Bicycles

Golf-carts

Distributed power generation

Cogeneration

Back-up power

Fuel Cell Applications Fuel Cell Applications -- DemonstrationsDemonstrations

Back-up power

Portable power

Space

Airplanes

Locomotives

Boats

Underwater vehicles

Why fuel cells

Promise of high efficiency

Promise of low or zero emissions

Run on hydrogen/fuel may be produced from

indigenous sources/issue of national security

Simple/promise of low cost

No moving parts/promise of long life

Modular

Quiet

electrodehydrogen




Fuel cell is like a battery but with constant fuel and oxidant supply.

electrolyte

electrode

hydrogen

oxygen

water

FUEL CELL

electrodehydrogen




electrolyte

electrode

hydrogen

oxygen

water

Types of fuel cellsTypes of fuel cells(based on electrolyte)

Alkaline

Acid polymer membrane

or Polymer Electrolyte Membrane

or Proton Exchange Membrane

PEM

Phosphoric acid

Molten carbonate

Solid oxide

- Zinc/Air fuel cell is not a fuel cell

- Direct methanol fuel cell is PEM fuel cell

Reactions in fuel cells

H2 O2H+

H2 O2

H2OOH-

H2O

depleted oxidant and

product gases out

depleted fuel and

product gases out

load

e-

AFC

PEMFC

65-220 °C

60-80 °CH2 O2

H2OH+

H2O2

H2O

CO3=

H2 O2

H2OO=

CO2

CO2

oxidant infuel in

anode cathodeelectrolyte

PEMFC

PAFC

MCFC

SOFC

(CO)

(CO)

(CH4)

60-80 °C205 °C

650 °C

600-1000 °C

Types of fuel cellsTypes of fuel cells(based on electrolyte)

Alkaline

Acid polymer membrane

or Polymer Electrolyte Membrane

or Proton Exchange Membrane

PEM

Phosphoric acid

Molten carbonate

Solid oxide

Why PEM Fuel Cells?Why PEM Fuel Cells?

Simple

Quick start-up

Fast response

High efficiency

High power density (kW/kg and kW/l)

Zero emissions

PEM Fuel Cell Basic ComponentsPEM Fuel Cell Basic Components

proton

exchange

membrane

porous

electrode

porous

electrode

catalyst

layer

Porous electrode structure

Gas diffusion layer surface

(carbon fiber paper)

80 µµµµm

proton

exchange

membrane

porous

electrode

PEM Fuel Cell Basic ComponentsPEM Fuel Cell Basic Components

porous

electrode

catalyst

layer

MEA – Membrane Electrode Assembly

cell frame / bipolar plate

Pt-catalyst

current collector /

gas diffusion layer

MEA Cross-Section

current collector /

gas diffusion layer

cell frame / bipolar plate

Pt-catalyst

polymer membrane

PEM Fuel Cell: How does it work?PEM Fuel Cell: How does it work?

membraneelectrode electrode

MembranePorous

electrode

structure

Carbonsupport

Platinum

2H+

collector

plate

collector

plate

2e-

O2

oxygen feed

How does it work?How does it work?

membraneelectrode electrode

external load

2H+

1/2O2 + 2H+ + 2e- → H2O

H2OH2

hydrogen feed

H2 → 2H+ + 2e-

MembranePorous

electrode

structure

Carbonsupport

Platinum

electrode membrane electrode

oxygen feed

collector

plate

O2

external load

2e-


oxygen feed

O2

PEM Fuel Cell - Basic Principle

2H+

H2 → 2H+ + 2e–

H2

2e-

H2O

hydrogen feed

bi-polar

collector

plate

2H+

H2 H2O

hydrogen feed

2e-

½O2 + 2H+ + 2e– → H2O

end plate

oxidant in

hydrogen out

bus plate

bi-polar collector plates

anode

coolant in

FUEL CELL STACKFUEL CELL STACK

oxidant + water out

hydrogen in

tie rod

membrane

anode

cathode

coolant out

Actual fuel cell stacksActual fuel cell stacks

BallardTeledyne

UTC Fuel Cells

Honda

Hydrogenics

exhaust

pressure

regulator

compressor

load

DC/

DC

hydrogen tankgorivni članak

Fuel cell needs supporting equipment/balanceFuel cell needs supporting equipment/balance--ofof--plantplant

pressure

regulator

compressor

heat

exchanger

water

pump

MM

M

fan

heat exchanger/

humidifier

battery

©2002 by Frano Barbir.

compressor

battery

heat

exchangers

hydrogen tank

pressure

regulator DC/DC inverter

Actual fuel cell systemActual fuel cell system

fuel cell stack

battery

humidifierwater tank

water pump

el. motor


Understand the difference between:Understand the difference between:

Single cell

StackStack

Systemexhaust

pressure

regulator

pressure

regulator

compressor

heat

exchanger

water

pump

load

MM

M

DC/

DC

hydrogen tank

fan

gorivni članak

heat exchanger/

humidifier

battery

Fuel Cell Theory Fuel Cell Theory -- ThermodynamicsThermodynamics

Hydrogen side: anode – hydrogen oxidation

Oxygen side: cathode – oxygen reduction

Mnemonic: reduction of oxygen on cathode: red-cat

Basic reactions

Anode reaction: H2 →→→→ 2e– + 2H+ oxidation

Cathode reaction: ½O2 + 2e– + 2H+ →→→→ H2O reduction

Overall reaction: H2 + ½O2 →→→→ H2O

Same as combustion of hydrogen

Combustion of hydrogen is an exothermic reaction – heat is generated

How much heat?

H2 + ½O2 →→→→ H2O + heat

From the tables:

Heat of formation: ∆∆∆∆H = (h ) – (h ) – ½(h ) = – 286 – 0 – 0 = – 286 kJ/mol

Heat of Formation

∆∆∆∆ means difference between products and reactants

Heat of formation: ∆∆∆∆H = (hf)H2O – (hf)H2 – ½(hf)O2 = – 286 – 0 – 0 = – 286 kJ/mol

mol is a measure of quantity

1 mol of hydrogen contains 2 grams of hydrogen (2.0159 to be more precise)

1 mol of oxygen contains 32 grams of oxygen

1 mol contains 6.022 x 1023 molecules (Avogadro’s number)

Chemistry can be confusing!!!

Don’t forget – fuel cell is an electrochemical device

Enthalpy = heat of formation

Negative heat means heat is released →→→→ exothermic reaction

Forget about plus and minus →→→→ use common sense

Of course that combustion of hydrogen and oxygen produces heat.

How much heat? →→→→ 286 kJ/mol

This enthalpy of hydrogen combustion reaction is also called:

Hydrogen Heating Value (or to be more precise Higher Heating Value)

obviously – how much heat we can get from hydrogen

Important note: all chemical tables refer to 25ºC.

That means that both reactants and products are at 25 ºC.

Heat of combustion – heating value

1 mol of H2 + ½ mol of O2

2 g of H2 + 16 g of O2

T = 25 °C (298.15 K)

PV = GRT or PV = NRT

1 mol of H2O (l)

18 g of H2O (l)

T = 25 °C (298.15 K)

Heat = 286 kJ


1 mol H2 + x mol O2

2 g of H2 + 32x g of O2

T = 25 °C (298.15 K)

PV = GRT or PV = NRT

1 mol of H2O (g) + (x- ½) mol O2

18 g of H2O (g)

T = 25 °C (298.15 K)

Heat = 241 kJ


Saturation pressure at 25°C:

Psat = 3.169 kPa

Maximum water content in air(oxygen)

- at saturation:

1 mol of H2O (g) + (x- ½) mol O2

18 g of H2O (g)

T = 25 °C (298.15 K)

Heat = 241 kJ

mol H2O/mol O2 = PH2O/PO2

= 3.169/(101.3 – 3.169)

= 0,0323

x – ½ >>>>= 1/0.0323 = 31

X >>>> = 31.5

1 mol H2 + 32 mol O2

Liquid water produced: 286 kJ/mol

Water vapor produced: 241 kJ/mol

Higher heating value

Lower heating value

The difference is: 45 kJ/mol Heat of evaporation

45 kJ/mol

18 g/mol= 2.5 kJ/g = 2500 kJ/kg

Higher and lower heating value

18 g/mol= 2.5 kJ/g = 2500 kJ/kg

Why do we care about heating value?

There is no combustion in fuel cell!

We use heating value as a measure of energy going into the fuel cell:

mol/s x kJ/mol = kJ/s = kW Beauty of SI system!

In a fuel cell both heat and electricity are generated.

But how much electricity?

Can all heat be converted into useful work (electricity)?

Remember 2nd law of thermodynamics?

Remember entropy?

∆∆For mechanical processes: W = Q - T∆∆∆∆S

For chemical reactions: ∆∆∆∆G = ∆∆∆∆H - T ∆∆∆∆S

∆∆∆∆G = Gibbs free energy


∆∆∆∆G = 237.2 kJ/mol (at 25ºC)

Electrical work = charge x voltage

Mechanical work = distance x force

Charge =

Electrical power = current x voltage

Mechanical power = flow rate x pressure

electrons/molecule of H2 molecules/mol Coulombs/electron

2 6.022x1023 1.602x10-19X X

Faraday’s constant

Theoretical cell voltage

∆∆∆∆G = – 2FE

E = voltage

Faraday’s constant

96,485 C/electron mol

Electrical work =

= – ∆∆∆∆G

2F=

237,200

2x96.485= 1.23

J/mol Ws/mol

C/mol As/mol

Coulomb = Amper.second

= = V

V

Beauty of SI system!

(at 25ºC)

∆∆∆∆G = ∆∆∆∆H - T ∆∆∆∆S


hf (J/mol) s (J/mol,K)

H2 0 130.59

O2 0 205.14

H2O (l) –285,838 70.05

H2O (g) –241,827 188.83

More thermodynamics – effect of temperature

∆∆∆∆h = f(T)

∆∆∆∆s = f(T)

∆∆∆∆H = (hf)H2O – (hf)H2 – ½(hf)O2 = – 286 – 0 – 0 = – 286 kJ/mol

∆∆∆∆S = (s)H2O – (s)H2 – ½(s)O2 = 70.05 – 130.59 – ½x205.14 = – 163.11 J/mol,K

Cp = f(T)

∫∫∫∫++++====T

15298

p15298T dTchh.

. ∫∫∫∫++++====T

15298

p15298T dTcT

1ss

.

.

(at 25ºC)

∆∆∆∆G = f(T)

31.00

33.00

35.00

37.00

39.00

41.00

43.00S

pe

cif

ic h

ea

t (J

/mo

l,K

)

Cp = f(T)

H2O(g)

O2

H2

25.00

27.00

29.00

273 373 473 573 673 773 873 973 1073

Temperature (K)

Sp

ec

ific

he

at

(J/m

ol,

K)

H2

Dashed lines: equations from Larminie&Dicks pp.392-393

Solid lines: equations from Fuel Cell Handbook

∆∆∆∆H and ∆∆∆∆G as a function of temperature

200

250

300

350d

H o

r d

G (

kJ/m

ol)

G (

kJ/m

ol)

0

50

100

150

0 200 400 600 800 1000 1200

temperature (C)

dH

or

dG

(kJ/m

ol)

H (l)

G (l)

H (g)

G (g)

∆∆ ∆∆H

or ∆∆ ∆∆

G (

kJ/m

ol)

Theoretical cell voltage – function of temperature

1.05

1.1

1.15

1.2

1.25

1.3C

ell

Po

ten

tial

(V)

liquid water

water vapor

0.8

0.85

0.9

0.95

1

0 200 400 600 800 1000 1200

temperature (C)

Cell

Po

ten

tial

(V)

Efficiency of any process

Fuel cell (ideal) efficiency

Theoretical Fuel Cell Efficiency

Ein Eout

Edeg

Eout < Einηηηη =

Eout

Ein

∆∆∆∆H ∆∆∆∆G∆∆∆∆G 237.2

∆∆∆∆H ∆∆∆∆G

Q = T ∆∆∆∆S∆∆∆∆G < ∆∆∆∆H

ηηηηfc = ∆∆∆∆G

∆∆∆∆H

237.2

286.0= = 0.83

Remember:∆∆∆∆G

2F= 1.23 V

∆∆∆∆H

2F= 1.48 VBy analogy: thermoneutral voltage

Fuel cell (ideal) efficiency ηηηηfc = = 0.831.23 V

1.48 V

Sometimes the lower heating value is used to express the efficiency

In internal combustion engines the products leave at a temperature

above 100ºC – product water is steam indeed.

Therefore, the heat of condensation cannot be utilized, and because

of that the lower heating value is used to calculate the engine efficiency.

Hydrogen higher heating value: 286 kJ/mol

Hydrogen lower heating value: 241 kJ/mol

2237G .============

∆ηEfficiency based on higher heating value: 830

0286

2237

H

G.

.

.============

∆∆

η

Efficiency based on lower heating value: 9800241

2237

H

G.

.

.============

∆∆

η

Remember:

∆∆∆∆G/2F = 1.23 V

∆∆∆∆HHHV /2F = 1.482 V

∆∆∆∆HLHV /2F = 1.254 V

Efficiency is also:

(case of condensing boilers)

4821

Vcell

.====η

2541

Vcell

.====η

HHV

LHV

Maximum efficiency for mechanical systems

Qin

Wout

Qdeg

Wout < Qin

TH

TC

Carnot efficiency

or

Efficiency of Carnot engine

ηηηη = Wout

Qin

= 1 –TC

THQin TH

Carnot efficiency is the maximum efficiency any process/engine

taking place between the two temperatures can have.

However, from any practical point of view the Carnot efficiency

is meaningless!

An engine operating with Carnot efficiency would produce zero power!

0.5

0.6

0.7

0.8

0.9

eff

icie

ncy

Max efficiency = 1 – Tc/Th

Efficiency at

max. power = 1 – Tc/Th

Carnot efficiency

0

0.1

0.2

0.3

0.4

0.5

0 20 40 60 80 100 120

power

eff

icie

ncy

An engine operating with Carnot efficiency would produce zero power!

A fuel cell operating with ideal efficiency would produce zero power!

E = – ∆∆∆∆G

2Fis maximum cell voltage

Similarity between Carnot efficiency and fuel cell ideal efficiency

E = 2F

is maximum cell voltage

Maximum voltage is achieved when there is no current.

Power = current x voltage

0.3

0.4

0.5

0.6

0.7

0.8

0.9

eff

icie

ncy l

imit

Difference between Carnot efficiency and fuel cell ideal efficiency

fuel cell

Carnot

0

0.1

0.2

0.3

0 200 400 600 800 1000 1200

temperature (C)

eff

icie

ncy l

imit

Ideal cell Voltage at different temperature and pressure

∆∆∆∆G = ∆∆∆∆H - T ∆∆∆∆S

nF

ST

nF

H

nF

G ∆∆∆−−−−==== At 25ºC V231

nF

G.====

∆V4821

nF

H.====

∆V2520

nF

ST.====

∆

ET = 1.482 – 0.000845T Note: ∆∆∆∆H and ∆∆∆∆S are f(T)

For T<373K ET ≈≈≈≈ 1.482 – 0.000845TFor T<373K ET ≈≈≈≈ 1.482 – 0.000845T

All of the above considerations were at ambient pressure

P = 1 atm, P = 101.3 kPa; P = 1.013 bar, P = 14.698 psi; P = 0 psig

Does the ideal voltage change with pressure?

Unfortunately – yes!

A little bit of basic thermodynamics

– dG = VmdP

For an ideal gas:

PVm = RT

– dG = RTdP

P

– G = – G0 + RT ln P

P0

Fuel cell reaction: H2 + ½ O2 →→→→ H2O

⋅⋅⋅⋅++++====

OH

50OH

0P

2

22

P

PP

nF

RTEE

.

ln

Nernst Equation

⋅+

∆−=

∆−

OH

50

2O2H0

2P

PP

nF

RT

nF

G

nF

G .

ln

Summary

E0 = E1atm,298.15K = 1.23 V

Ideal fuel cell voltage

⋅⋅⋅⋅++++−−−−====

OH

50OH

TP

2

22

P

PP

F2

RTT00084504821E

.

, ln..

⋅⋅⋅⋅

++++−−−−====50

OH 22PP

T00004310T00084504821E

.

ln...

This is voltage at 0 current

So called open circuit voltage (OCV)

⋅⋅⋅⋅++++−−−−====

OH

OH

TP

2

22

P

PPT00004310T00084504821E , ln...

Caution: valid only for low temperatures

Ideal Cell voltage: 1.23 V

Function of temperature, pressure and concentration of reactants:

At 80ºC: 1.186 V

Air instead of O2: 1.174 V

Actual voltage is always lower!

0.97

(H2/O2 at 25ºC, 1 atm)

0.97

Ideal Cell voltage: 1.23 V

Function of temperature, pressure and concentration of reactants:

At 80ºC: 1.186 V

Air instead of O2: 1.174 V

Actual voltage is always lower!

(H2/O2 at 25ºC, 1 atm)

0.97 1.1740.950.900.800.700.600.500.97

current

vo

ltag

e

0.97

1.174

0

0.950.900.800.700.600.50

Relationship between current and voltage

Butler-Volmer equation

( ) ( ) ( )

−α−−

−α−=

RT

EEF1

RT

EEFii rr0 expexp

i0 = exchange current density

αααα = charge transfer coefficientαααα = charge transfer coefficient

F = Faraday’s constant

R = gas constant

T = temperature

E = potential

Er = reversible potential

αααα ≅≅≅≅ 1

Note: in some literature nαααα is used instead of ααααn = 2 for the anode and n = 4 for the cathode

Larminie&Dicks suggest ααααa = 0.5 and ααααc = 0.1 to 0.5

Newman specifies αααα in a range between 0.2 and 2.0

Factors affecting the exchange current densityCatalyst

Roughness/actual surface area

Temperature

Reactants concentration

Pressure

−−

=

γ

Crefr

ccref

T

T

RT

E

P

PLaii 1

00 exp

ref

refr

ccTRTP00

i0ref = exchange current density at reference conditions (T,P) in A/cm2 Pt surface

ac = catalyst specific area (for Pt theoretically: 2,400 cm2/mg

practically: 600-1,000 cm2/mg

in electrode: 420-700 cm2/mg

Lc = catalyst loading (today 0.3 to 0.5 mg/cm2)

Ec = activation energy (66 kJ/mol)

R = universal gas constant (8.314 J/molK)

γγγγ = pressure coefficient (0.5 to 1.0)

( ) ( ) ( )

−α−−

−α−=

RT

EEF1

RT

EEFii rr0 expexp

( ) ( ) ( )

−α−−

−α−=

RT

EEF1

RT

EEFii

araaaraaa0a

,,, expexp

(E – Er) = overpotential

On the fuel cell anode:

Er,a = 0 definition of zero potential – hydrogen electrode

(Ea – Er,a) > 0

( ) ( )

−α−−=

RT

EEF1ii

araaa0a

,, exp

negative sign means current is leaving the electrode

– oxidation reaction

( ) ( ) ( )

−α−−

−α−=

RT

EEF1

RT

EEFii

crcccrccc0c

,,, expexp

On the fuel cell cathode:

Er,c = 1.23 V

(E – E ) < 0(Ec – Er,c) < 0

( )

−α−=

RT

EEFii

crccc0c

,, exp

Types of losses in fuel cell:

Activation losses (activation polarization)

Resistive (Ohmic) losses

Mass transfer losses (concentration polarization)

Internal current and fuel crossover

Activation polarization losses

Tafel equation: ∆∆∆∆Vact = a + b . log(i) (empirical)

α=∆

0i

i

F

RTV ln

∆α=

RT

VFii 0 exp

i = current density (mA/cm2)

i0 = exchange current density (mA/cm2)

R = universal gas constant = 8.314 J/mol

T = temperature of the process (K)T = temperature of the process (K)

α = charge transfer coefficient

n = number of electrons involved (2)

F = Faraday’s constant = 96,485 C

( ) ( )iF

RTi

F

RTV 0 lnln

α+

α−=∆

( ) ( )00 iF

RT32i

F

RTa log.ln

α−=

α−=

F

RT32bα

= . Tafel slope

Activation polarization losses

α=∆

0i

i

F

RTV ln

∆∆∆∆Vact = a + b . log(i)

0

0.2

0.4

0.6

0.8

1

1.2

po

ten

tia

l lo

ss

(V

)

i0a

slope = b

log scale

For αααα = 1, T = 333 K ⇒⇒⇒⇒ b = 0.066 V/dec

0

0.2

0.4

0.6

0.8

1

1.2

0 500 1000 1500 2000

current density (mA/cm²)

po

ten

tial

loss (

V)

0

0.0001 0.001 0.01 0.1 1 10 100 1000 10000


Resistive losses

Ohmic law: ∆∆∆∆V = I.R = i.Ri

I = current (Amps)

R = resistance (Ohms)

i = current density (A/cm2)

Ri = aerial resistance (Ohm-cm2)

A

dR ρ==== ρ = resistivity (Ohm-cm)

d = distance, length (cm)

A = cross sectional area (cm2)

Ri = R.A

0

0.2

0.4

0.6

0.8

1

1.2

0 500 1000 1500 2000


po

ten

tial

loss (

V)

Resistance through:

Membrane (ionic)

Electrodes

Bi-polar plates

Interfacial contacts


oxygen feed

collector

plate

O2

external load


O2

2e-

V

bus

plate

Resistive losses

2H+

H2 H2O

hydrogen feed

bi-polar

collector

plate

2H+

H2 H2O

hydrogen feed

2e-

½O2 + 2H+ + 2e– → H2O

2e-

H2 → 2H+ + 2e–

Concentration polarization or mass transport losses

Po

iLi

P ii

PPP

L

00 −−−−====

L0 i

i1

P

P−−−−====

−−−−====

====ii

i

F2

RT

P

P

F2

RTV

L

L0 lnln∆

−−−−−−−−====

Li

i1

F2

RTV ln∆or

0

0.2

0.4

0.6

0.8

1

1.2

0 500 1000 1500 2000


po

ten

tia

l lo

ss (

V) theoretical

more realistic

⋅=∆d

icVconc exp

Fick’s First Law of Diffusion:(((( ))))

ACCD

N SB

δ−−−−⋅⋅⋅⋅

====

N = flow rate, mol/s

D = diffusion coefficient, cm2/s

CB = bulk concentration, mol/cm3

CS = concentration at

catalyst surface, mol/cm3

δ= diffusion path, thickness

of diffusion layer, cm

A = active area, cm2

nF

IN ====

====

S

B

C

C

nF

RTV ln∆Voltage gain/loss

(Nernst equation)

Alternative way to express concentration polarization

A = active area, cm2

(((( ))))δ

SB CCDnFi

−−−−⋅⋅⋅⋅⋅⋅⋅⋅====

δB

L

CDnFi

⋅⋅⋅⋅⋅⋅⋅⋅====

LB

S

i

i1

C

C−−−−====

−=∆

L

L

ii

i

nF

RTV ln

Voltage loss due to concentration polarization:

Limiting current CS = 0

0.6

0.8

1

1.2p

ote

nti

al

loss (

V)

Voltage losses compared

0

0.2

0.4

0.6

0 500 1000 1500 2000


po

ten

tial

loss (

V)

activation

resistive

concentration

Actual fuel cell voltage

concresactPTcell VVVEV ∆∆∆ −−−−−−−−−−−−==== ,

(((( )))) (((( )))) (((( )))) (((( ))))caconcanconcrescaactanactPTcell VVVVVEV ∆∆∆∆∆ −−−−−−−−−−−−−−−−−−−−==== ,

Losses occur on both anode and cathode:

−+⋅−

α−=

Li

0PTcell

i

i1

nF

RTRi

i

i

F

RTEV lnln,

0.8

0.2

0.4

0.6

Cathode solid phase potential

Anode overpotential

Ecell

E

DL DLCL CL

membrane1 A/cm2

po

ten

tial (V

)

0

0.2

-0.4

-0.2

Anode solid phase potential

Anode overpotential

Cathode overpotential

electrolyte phase potential

Eloss

Eeq

po

ten

tial (V

)

0 50 150100nominal cell thickness (µµµµm)

0.6

0.8

1

1.2cell

po

ten

tial

(V)

Fuel cell polarization curve

activation losses

resistance losses

theoretical (ideal) voltage

0

0.2

0.4

0.6

0 500 1000 1500 2000


cell

po

ten

tial

(V)

mass transport

lossesresulting V vs. i curve


0.6

0.7

0.8

0.9

1p

ow

er

den

sit

y (

W/c

m²)

Power density vs. current density

Power density = V.i (W/cm2 or mW/cm2)

0

0.1

0.2

0.3

0.4

0.5

0.6

0 500 1000 1500 2000


po

wer

den

sit

y (

W/c

m²)


0.5

0.6

0.7

0.8

0.9

1

cell

po

ten

tial

(V)

Cell potential vs. power density

0

0.1

0.2

0.3

0.4

0.5

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7

power density (W/cm²)

cell

po

ten

tial

(V)


2e-

O2

e- loss

Internal currents and fuel crossover

2H+

1/2O2 + 2H+ + 2e- → H2O

H2OH2

H2 → 2H+ + 2e-

H2 loss


Hydrogen consumption in fuel cell

Faraday’s law:

I = nFN

(Amps = Amp-second/mol x mol/s) nF

IN ====

Hydrogen consumption in fuel cell: NH2 = I/(2F)

mol/s

Oxygen consumption in fuel cell: NO2 = ½ NH2 = I/(4F)

Water generation in fuel cell: NH2O = I/(2F)


Both internal currents and hydrogen crossover mean loss of current:

Iloss = 2FNloss

This means that even at 0 external current there is hydrogen consumption.

+

α−=

0

lossPTcell

i

ii

F

RTEV ln,

Internal currents and fuel crossover

0.900

0.920

0.940

0.960

0.980

1.000

1.020

0 2 4 6 8 10 12

current density (mA/cm2)

ce

ll p

ote

nti

al

(V)

iloss = 1 mA/cm2

iloss = 2 mA/cm2

iloss = 4 mA/cm2

iloss = 8 mA/cm2


Fuel cell efficiency4821

Vcell

.====η HHV

Efficiency = Electricity out

Hydrogen in=

I.V

N.HHHV

=I.V

F2

IN ====

I.HHHV/(2F)

Vcell

1.482=

Cell potential vs. power densityefficiency

0.5

0.6

0.7

0.8

0.9

1

cell

po

ten

tial

(V)

cell

eff

icie

ncy (

HH

V)

0.7

0.6

0.5

0.4

0

0.1

0.2

0.3

0.4

0.5

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7


cell

po

ten

tial

(V)

cell

eff

icie

ncy (

HH

V)

0.3

0.2

Fuel cell efficiency4821

Vcell

.====η HHV

Efficiency = Electricity out

Hydrogen in=

I.V

N.HHHV

=I.V

F2

IN ====

I.HHHV/(2F)

Vcell

1.482=

Hydrogen in = hydrogen converted into current + hydrogen loss

II++++ iVIV

F2

IIN loss++++====

loss

cell

loss

cell

ii

i

4821

V

II

I

4821

V

++++====

++++====

..η⇒⇒⇒⇒

electrolyte

electrode

electrode

hydrogen in

oxygen in

hydrogen out

oxygen outelectrolyte

electrode

electrode

hydrogen in

oxygen in

hydrogen out

oxygen out

In some cases excess hydrogen is fed into the fuel cell – some H2 is wasted

fucellV

η=η4821.

hydrogen consumed

hydrogen suppliedηηηηfu =

Fuel cell efficiency vs. power density

0.5

0.6

0.7

0.8

0.9

1eff

icie

ncy (

HH

V)

0

0.1

0.2

0.3

0.4

0.5

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7


eff

icie

ncy (

HH

V)

Additional slides

0.6

0.8

1

1.2c

ell

po

ten

tia

l (V

)Tafel = 77 mV/dec

Tafel = 66 mV/dec

Tafel = 55 mV/dec

−+⋅−

+

α−=

Li

0

lossPTcell

i

i1

nF

RTRi

i

ii

F

RTEV lnln,

0

0.2

0.4

0.6

0 500 1000 1500 2000


ce

ll p

ote

nti

al

(V)

0.8

1

1.2c

ell

po

ten

tia

l (V

)

io = 0.0003 mA/cm²

io = 0.003 mA/cm²

io = 0.03 mA/cm²

−+⋅−

+

α−=

Li

0

lossPTcell

i

i1

nF

RTRi

i

ii

F

RTEV lnln,

0

0.2

0.4

0.6

0 500 1000 1500 2000


ce

ll p

ote

nti

al

(V)

0.2

0.4

0.6

0.8

1

1.2

ce

ll p

ote

nti

al

(V)

ix = 0 mA/cm²

ix = 2 mA/cm²

ix = 20 mA/cm²

1.2ix = 0 mA/cm²

−+⋅−

+

α−=

Li

0

lossPTcell

i

i1

nF

RTRi

i

ii

F

RTEV lnln,

0

0.2

0 500 1000 1500 2000


0.6

0.7

0.8

0.9

1

1.1

0 20 40 60 80 100


ce

ll p

ote

nti

al

(V)

ix = 2 mA/cm²

ix = 20 mA/cm²

0.8

1

1.2c

ell

po

ten

tia

l (V

)

R = 0.1 Ohm-cm²

R = 0.15 Ohm-cm²

R = 0.2 Ohm-cm²

−+⋅−

+

α−=

Li

0

lossPTcell

i

i1

nF

RTRi

i

ii

F

RTEV lnln,

0

0.2

0.4

0.6

0 500 1000 1500 2000


ce

ll p

ote

nti

al

(V)

0.8

1

1.2c

ell

po

ten

tia

l (V

)

iL = 1200 mA/cm²

iL = 1600 mA/cm²

iL = 2000 mA/cm²

−−−−++++⋅⋅⋅⋅−−−−

++++−−−−====

Li

0

lossPTcell

i

i1

nF

RTRi

i

ii

nF

RTEV lnln, α

0

0.2

0.4

0.6

0 500 1000 1500 2000


ce

ll p

ote

nti

al

(V)

0.8

1

1.2c

ell

po

ten

tia

l (V

)

P = 300 kPa

P = 200 kPa

P = 101.3 kPa

−+⋅−

+

α−=

Li

0

lossPTcell

i

i1

nF

RTRi

i

ii

F

RTEV lnln,

Operating pressure

0

0.2

0.4

0.6

0 500 1000 1500 2000


ce

ll p

ote

nti

al

(V)

++++−−−−====

============

3P0

loss3PT3Pcell

i

ii

nF

RTEV

,

,, lnα

(((( )))) (((( )))) (((( ))))loss3P03T3Pcell iiF250

RTi

F250

RTP

F4

RTEV ++++

⋅⋅⋅⋅−−−−

⋅⋅⋅⋅++++++++==== ======== ln

.ln

.ln ,,

(((( )))) (((( )))) (((( ))))loss1P01T1Pcell iiRT

iRT

PRT

EV ++++−−−−++++++++==== ======== lnlnln ,, (((( )))) (((( )))) (((( ))))loss1P01T1Pcell iiF250

iF250

PF4

EV ++++⋅⋅⋅⋅

−−−−⋅⋅⋅⋅

++++++++==== ======== ln.

ln.

ln ,,

++++

====−−−− ========1

3

F

RT

1

3

F4

RTVV 1Pcell3Pcell lnln,,

Vcell,P=3 – Vcell,P=1 = 0.039 V (at 60º)

0.6

0.8

1

1.2c

ell

po

ten

tia

l (V

)air

oxygen

−+⋅−

+

α−=

Li

0

lossPTcell

i

i1

nF

RTRi

i

ii

F

RTEV lnln,

Air vs. oxygen

0

0.2

0.4

0.6

0 500 1000 1500 2000


ce

ll p

ote

nti

al

(V)

V0560210

1

F

RT

210

1

F4

RTVV 1Pcell3Pcell .

.ln

.ln,, ====

++++

====−−−− ========

Evidence of Mass Transport Losses

Source: T.P. Ralph and M.P. Hogarth, Platinum Metals Review, Vol. 46, No. 1, pp. 3-14, 2002

Constant current

Constant voltage

Constant resistance

Constant power

Variable power

0.6

0.7

0.8

0.9

1

ce

ll p

ote

nti

al

(Vo

lts

)V OP1

OP2

R=2ΩΩΩΩcm2 1 0.5

w=1W/cm2

0

0.1

0.2

0.3

0.4

0.5

0.6

0 200 400 600 800 1000 1200 1400 1600 1800 2000

current density (mA/cm2)

ce

ll p

ote

nti

al

(Vo

lts

)V OP1

0.2

w=1W/cm

0.8

0.6

0.4

0.2

Operational conditions.

Operating pressure

Flow rates of reactants

Operating temperature

Humidity of reactants

What else can affect the polarization curve?

Heat transfer

Fluid mechanics

Fluid mechanics/chemistry

PsychrometryHumidity of reactants Psychrometry

Thermodynamics of moist gases

We must know some basics!

We must know fuel cell inner working!

What else can affect the polarization curve?

Unsteady conditions (flooding/drying)

Contaminants

Crossover leaks

Aging (loss of catalyst surface, crossover leaks,

change in membrane properties)

Summary

Voltage losses:

Activation polarization

Ohmic (resistive losses)

Concentration polarization or mass transport losses

Internal currents and reactants crossover

++++ iRTiiRT

Resulting polarization curve:

4821

Vcell

.====ηFuel cell efficiency:

−−−−++++⋅⋅⋅⋅−−−−

++++−−−−====

L

i

0

lossPTcell

i

i1

nF

RTRi

i

ii

nF

RTEV lnln, α

Effect of operating pressure

Effect of oxygen vs. air

PEM Specific Issues – Water transport, gas diffusion

PEM Fuel Cell:

Membrane/Electrode Processes

e-e-

e-

e-

e-e-

Hydrogen

Oxygen

e-

H2 → 2H+ + 2 e- 2H+ + 1/2O2 + 2 e- → H2O

e-

+

+

+ +

+

+

e-e-

e-

Platinum

Carbon

Functional Group

Polymer Electrolyte

e- Electron

Materials Science and Technology Division

Water content and conductivity

Source: T.A. Zawodzinski, et al., J. Electrochem. Soc., Vol. 140, No. 7, pp. A1981-A1985, 1993

λλλλ = water molecules per sulfonate groupλλλλ = water molecules per sulfonate group

Water content and conductivity

Source: T.A. Zawodzinski, et al., J. Electrochem. Soc., Vol. 140, No. 7, pp. A1981-A1985, 1993

Nafion conductivity: ~0.1 S/cm

1/conductivity = resistivity = 10 Ohm-cm

Nafion 115 thickness: 5 mils = 0.0127 cm

Nafion conductivity

Resistance = 10 Ohm-cm x 0.0127 cm = 0.127 Ohm-cm2

At 1 A/cm2 it would cause 0.127 V loss

Nafion conductivity is a function of water content and temperature!

Nafion resistivity – function of membrane thickness?

• unisotropic structure

• resistance from the electrodes

• thickness of the membrane in the cell different from measured before assembly

Source: F.N. Buchi and G.G. Scherer, J. Electrochem. Soc., Vol. 148, No. 3, pp. A181-A188, 2001

• Electro-osmotic

drag removes

water from the

anode side.

With thicker 0.6

0.8

1.0

HF R

ES

ISTA

NC

E (ohm

cm

2)

or C

ELL V

OLTA

GE

(V

)

Cell Resistance and Performance:

PEM Thickness Effects

Increasing

Membrane

Thickness

With thicker

membranes,

back diffusion

of water is

difficult - the

anode side loses

water content.

0.0

0.2

0.4

0.0 0.5 1.0 1.5 2.0 2.5 3.0

HF R

ES

ISTA

NC

E (ohm

cm

or C

ELL V

OLTA

GE

(V

)

CURRENT DENSITY (A/cm2)

Fg 3, TF69,109,74,117 Ox's

(Neat

Oxygen

Cathodes)

Materials Science and Technology Division

Membrane resistance – function of current density

Source: F.N. Buchi and G.G. Scherer, J. Electrochem. Soc., Vol. 148, No. 3, pp. A181-A188, 2001

Permeation of gases through the membrane

Permeability is a product of diffusivity and solubility:

Pm = D x S

cm2/s x mol/cm3atm =mol

cm.s.atm

For ideal gas: Pvm = RT; at standard conditions vm = 0.024 m3/mol

mol.cm

cm2.s.atmor

For ideal gas: Pvm = RT; at standard conditions vm = 0.024 m3/mol

vm = 24,060 cm3/mol

cm3.cm

cm2.s.atm

cm3.cm

cm2.s.cmHg1 atm = 75 cmHg = 1010 barrers

Example:

DH2 = 6.65x10-7 cm2/s

SH2 = 2.2x10-5mol/cm3atm

Pm = 1.5x10-11mol.cm-1s-1atm-1

1.5x10-11mol.cm-1s-1atm-1 x 24,060 cm3/mol =

= 3.6x10-7 cm3.cm.cm-2.s-1atm-1=

= 3.6x10-7 cm3.cm.cm-2.s-1atm-1/75 cmHg/atm =

= 4.8x10-9 cm3.cm.cm-2.s-1cm-1Hg = 48 Barrers

Permeation rate, N (mol/s)

N = Pm.A.∆P/d

N = I/nF

i = I/A = nF.Pm.∆P/d

Example:

Pm = 1.5x10-11mol.cm-1s-1atm-1

A = 100 cm2

d = 51 µm = 0.0051 cm

N = 2.94x10-7mol/s

I = nFN = 0.057 A

NV = N.24060 (cm3/mol) = 0.125×I cm3/s

NV = 7.5 cm3/min/Amp

d = 51 µm = 0.0051 cm∆P = 1 atm (partial pressure)

I = nFN = 0.057 A

i = I/A = 0.00057 A/cm2 = 0.57 mA/cm2

Additional variable (complication)

DH2 = f(T) = 0.0041e-2602/T

0.00E+00

1.00E-06

2.00E-06

3.00E-06

4.00E-06

5.00E-06

0 20 40 60 80 100 120

temperature (°C)

dif

fusiv

ity (

cm

²/s)

Permeation of gases

through Nafion

Source: T. Sakai, et al., J. Electrochem. Soc., Vol. 133, No. 1, pp. 88-92, 1986

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