Heterogeneous Reaction Engineering: Theory and Case

Studies

Module 8Module 8Electrochemical Processes and

Reactor Design

P.A. Ramachandranrama@wustl.edu

Chemical Reaction Engineering Laboratory

OUTLINE

Cell thermodynamicsTechnological examplesEnvironmental aspectsKinetics of electrode reactionsTransport effects

Half-Reactions and electrodesRedox reactions: Reactions in which there is a transfer of electrons

from one substance to another.Half-reactions: Redox reactions are expressed as the sum of two

half-reactions.Electrode: Metallic conductorAnode: Electrode where oxidation occursCathode: Electrode where reduction occurs

Example: 2H2(g)+O2(g)→2H2O (aq)

Anode (Oxidation) reaction: Cathode (Reduction) reaction:2H2 → 4H+ + 4e- O2+4H++4e-→2H2O

+

Electrons move through external circuit

H2 H+ H+

Anode Porous Cathodeseparator

Current flow from cathode to anode

Schematic of a fuel cell

O2O2

H2O

2H2 → 4H+ + 4e- O2+4H++4e-→2H2O

-

Details of a fuel cell

Schematic of an electrolyte cell

GnFEr Δ=0

Cell Thermodynamics

where n = number of electrons transferredF = Faraday’s constant = 96500 C/g-mole

It is common practice to write all half-reactions as two oxidation reactions and the overall reaction is the difference of the two:

2H2 → 4H+ + 4e- Ea = 0 (by convention)2H2O → O2+4H++4e- Ec = 1.23V

E0r = Ea-Ec = -1.23V (standard conditions)

If the reaction potential is negative, the reaction is spontaneous → FUEL CELL

Cell Thermodynamics cont’d

The free energy change of reaction is

where ap is the product of activities of all products.

⎟⎟⎠

⎞⎜⎜⎝

⎛+Δ=Δ

r

p

aa

RTGG ln0

⎟⎟⎠

⎞⎜⎜⎝

⎛+=

r

prr a

anFRTEE ln0

( )TETTETE ref ∂∂

−+=0

0)(

nFS

TE 00 Δ

=∂∂

Nernst Equation

The effect of temperature in the reaction equilibrium can be calculated in a similar manner. A linear equation is often used:

From thermodynamics, it can be shown thatwhere ΔS0 is the entropy change for the reaction.

Electrolysis of HClConsider the oxidative electrolysis of HCl. The overall reaction is

2HCl+1/2O2 = Cl2+H2O Cell potential = 0.13 V

Direct electrolysis2HCl →H2+Cl2 Cell potential = 1.36V

Comparing the numbers, it is seen that the oxidative electrolysis is preferable if the cathodic process of oxidation of H2O can be carried out a reasonable rate.

The issue becomes now how to promote the reaction and also find cathodes which are corrosion resistant.

This is an example of process intensification. (coupling of exo- and endothermic reactions)

Voltage Balance

Similar to the heat balance in chemical reactors

( ) )()(0 metaliRsolutioniREV CAT ++−++= ηη

where VT = overall voltageE0 = decomposition voltage predicted from thermodynamics = ErηA = anode overpotentialηC = cathode overpotentialiR = voltage drop due to a resistance in the solution and metal

Lower the current density, lower the voltage drop, but the reaction rate is also correspondingly lower.

Schematic of voltage balance

IR metal

IR solutionEcηc

Ea

ηa

Cell Voltage VT

Voltage balance exampleThe electrolysis of NaCl is as follows:

NaCl+H2O → NaOH + 1/2Cl2 + 1/2H2

The free energy change for this process from standard thermodynamic data is 50 kcal/mole which corresponds to E0 of 2.17V.

Voltage balance for a diaphragm cell for electrolysis of NaCl:Decomposition voltage 2.17VAnode overvoltage 0.03VCathode overvoltage 0.30VSolution IR drop 0.35VDiaphragm drop 0.60VMetal hardware 0.20V

Total: 3.65VCathode reaction is slow requiring larger overpotential. The drop in the diaphragm is also large due to transport limitations of ions across the membrane.

2NaCl+2H2O → 2NaOH + Cl2 + H2Economics of equal masses of NaOH and Cl2 being produced as well as reintroduction of soda ash process for NaOH

Older process:Anode reaction:Cl-(aq)→1/2Cl2+e-

at a graphite anode or Ti/IrO2 (newer cells)

Cathode reaction:old style mercury cellNa+(aq)+e-→Na/HgNa/Hg+H2O→NaOH+1/2H2(separate reaction)

Electrochemical processes: Chloralkali process

Advantage: H2 and Cl2 formed in separate stepsDisadvantage: cost and toxicity of Hg

Newer technology: Diaphragm or membrane separated cells to separate H2(g) and Cl2(g) streams with Nafion as a cation exchange membraneMERCURY free

Anode reaction:Cl-(aq)→1/2Cl2+e-

Cathode reaction:H+(aq)+e-→1/2H2

Chloralkali process cont’d

Electrochemical processes: Aluminum Production (Electrowinning)

Heroult-Hall process:Al2O3 is purified and then electrolyzed using cryolite Na3AlF6 as the supporting electrolyte 15 wt%, 1000C

Overall chemistry:2Al2O3 + 3C → 4Al + 3CO2(exact Al species are not known)

Disadvantage:pollution problems related with HF and polycyclic aromatic hydrocarbons

Electrowinning of Al cont’dEnergy aspects and costs• High electricity requirement ~ 15,000 kWh/t Al• Thermodynamic cell potential is -1.18V

(it would be -2.2V if O2 was produced at the anode instead of CO2)Contributions to Ecell 1.8VAnode/cathode overvoltages 0.5ViR drops (anode, cathode) 1.1ViR drop (electrolyte) 1.5VTotal 4.3V

• CostsPurified bauxite 30 % Labor 16%Electricity 23 % Carbon anodes 7%Capital 17% Other materials 7%

Electrochemical processes: Monsanto’s adiponitrile process

Adiponitrile is an intermediate in the production of nylon[6.6]2CH2=CH-C≡N+2H++2e-→N≡C-(CH2)4-C≡N

Exact mechanism is not known.

Older process (Monsanto, 1965): Pb cathode, PbO2/AgO anode, Et4N+EtOSO3

- supporting electrolyte

Newer technology:Undivided cell, carbon steel anodes and corrosion inhibitors, Cd-

plated carbon steel cathodes, 15% Na2HPO4 supporting electrolyte

Monsanto’s adiponitrile process cont’d

Energy aspects and costsOld style cell New style cell

Voltages (V)reversible cell potential -2.50 -2.50overpotentials -1.22 -1.87electrolyte iR -6.24 -0.47membrane iR -1.69 --

Total -11.65 -4.84

Energy (kWht-1) 6700 2500

Paired ElectrosynthesisThe synchronous utilization of anodic and cathodic reactions is named paired electrosynthesis.

Example:*

N-aryl triazoles and tetrazoles can be synthesized in one-step by the reaction of cathodically generated heterocyclic anions with the anodically generated aromatic cation-radicals.

*: http://www.electrosynthesis.com/news/w6content.html

Electrochemical approaches to Pollution Control

Electrochemical techniques offer many advantages such as*:• Environmental compatibility• Versatility• Energy efficiency• Safety• Selectivity• Cost effectiveness• Amenability to automation

*: Environmental Electrochemistry, Fundamentals and Applications in Pollution Abatement, K. Rajeshwar, J. Ibanez,Academic Press, 1997

Electrochemical approaches to Organic Pollutants

• Phenols• Aromatic amines

Example: Degradation of Aniline with a Nafion cationexchange membraneC6H5NH2+2H2O→C6H4O2+3H++4e-+NH4

+

C6H4O2+6H2O →C4H4O4+12H++12e-+2CO2C4H4O4+4H2O →12H++12e-+4CO2

• Halogenated and nitro derivatives• Waste biomass• Carboxylic acid anions• Tributylphosphate• Chlorinated organics

Electrochemical approaches to Inorganic Pollutants

• Cyanide• Thiocyanate• Oxynitrogen ions

Example: NO3-+H2O+2e-=NO2

-+2OH-

• Oxychloride species

Equilibrium Potential (Half Reaction)R→ O+n+ne-

Driving force for the oxidation reaction is ( )eoR nμμμ +−

RgRR CTR ln0 += μμ

sOgOO nFCTR φμμ ++= ln0

mee Fφμμ −= 0

Combining all the equations we get:

⎟⎟⎠

⎞⎜⎜⎝

⎛+=

R

Ogeq C

CnF

TREE ln0

Rg = gas constantCi = concentration of species iμi

0 = chemical potential of i in the standard stateφs = electric potential of the solutionφm = electric potential of the metalEeq = φm - φsE0= standard potential for half reaction

OverPotential and Rate of Reaction

eqm EE −=ηSurface overpotential = Actual applied potential – Equilibrium potential

If η < 0, reduction is favoredη > 0, oxidation is favored

( )βηnfCkr Rff exp= TRFfg

=

( )[ ]ηβ−−= 1exp nfCkr Obb

( ) [ ]ηββη )1(expexp −−−= nfCknfCkrateNet ObRf

Kinetic Model:

β: symmetry factor between 0 and 1

Standard Rate Constant

( )0exp nfEkk fb −=

( )0exp Enfkk fo β−=

( )( ) ( )( )( )( )00 1expexp EEfnCEEnfCkr ORo −−−−−= ββ

0* EE −=η

( ) ( )( )( )** 1expexp ηββη −−−= fnCnfCkr ORo

Thermodynamic consistency

Current and Exchange Currenti = nFr

( ) ( )[ ][ ]ηββη −−−= 1expexp0 nfnfii

β = 0.5 gives the simplified form:

( )2/sinh2 0 ηnfii =

( )βηnfii exp0=

If η is large and reverse reaction is negligible

)log(iba +=η Tafel Equation, plot n vs log(i)

Butler-Volmer

Butler-Volmer empricial form( ) ( )[ ][ ]ηββη −−−= 1expexp0 nfnfii Butler-Volmer

nA βα =

nC )1( βα −=

( ) [ ][ ]ηαηα ffii CA −−= expexp0

i0, αa and αc are fitted parameters.

H2 and fuel cells

“Why is the H2 myth so persistent? I believe it fulfills a deep psychological need.”

Dr. Reuel Shinnar, CEP Magazine, ‘Demystifying the Hydrogen Myth’, November 2004

“H2 and fuel cells may have many applications, but they will not solve large-scale energy or pollution problems.”

Voltage Balance ExampleAn electrolytic reaction is used to generate chlorine: At the anode: Cl-→1/2Cl2+e-

At the cathode: H2O+e-→OH-+1/2H2

Show how the reactor voltage changes with current density.Temperature is 250C and Cl2 and H2 are at standard conditions.Cathodic reaction is ηc = -(0.17+0.06log(i))Anodic reaction is ηa = 0.0527+0.0277log(i)where i is in mA/cm2

The conductivities are as follows:Anolyte conductivity = ka = 18.0 mho m-1

Separator conductivity = ka/3Catholyte conductivity = 40.0 mho m-1

The anode and cathode interelectrode gaps are 3 mm and separator thickness is 1 mm.

Voltage Balance Solution

0rE

= Ea-EcFrom electrochemical tablesEa=1.36VEc=-0.828V

0rE

=2.188V

This is positive and hence energy has to be supplied (Electrolyzer)Choose a value for current density i = 1 mA/cm2

Anodic overvoltage = 0.0527 VCathodic overvoltage = (-ηc)=0.17 V

Solution Cont’d

R = 4x10-4 ohm m2 = (V /A) m2

Voltage drop across solution = (V/A)i

i = 10 A/m2

ΔVsolution = 4x10-3 V

Total voltage for a current density of 1 mA/cm2 is 2.188+0.0527+0.17+0.004=2.4147V

The cathodic overpotential is the main factor here and the cathode reaction is therefore rate limiting.

c

c

m

m

a

a

kkkRδδδ

++=1

Transport EffectsCAb

CAs

Rate to surface = kL*(CAb-CAs)

kL* = mass transfer coefficient enhanced by migration due

to field = fkL where f is the order of 2.

Expressing in terms of current ⎟⎟⎠

⎞⎜⎜⎝

⎛−=

Ab

AsAbL C

CCnFki 1*

The rate of reaction expressed as current is ( )βηnfCCii

Ab

As exp0=neglecting the reverse reaction.

Eliminating the surface concentration we fixedRL iii111

+=

where AbLL Cnfki *= (limiting current)

( )βηnfiiR exp0=

Resistance of solution = R where R isc

c

m

m

a

a

kkkRδδδ

++=1

Transport Effects: Migration Terms

dydF

RTzCD

uCdy

dCDN jjj

jj

jjφ

−+−=

The last term is the migration term and

is the potential gradient.dydφ

Voltage Balance for H2 fuel cell

( )IB

B

LC

C

CLA

A

A

RiLii

ii

i

FRT

iii

i

FRTVV −⎟⎟

⎠

⎞⎜⎜⎝

⎛−

⎪⎭

⎪⎬

⎫

⎪⎩

⎪⎨

⎧

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛

−−

⎪⎭

⎪⎬

⎫

⎪⎩

⎪⎨

⎧

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛

−−= −−

σαα,

0,1

,

0,10

121sinh

121sinh

Power density vs Voltage

( )IB

B

LC

C

CLA

A

A

RiLii

ii

i

FiRT

iii

i

FiRTiVP 22

,

0,1

,

0,10

121sinh

121sinh −⎟⎟

⎠

⎞⎜⎜⎝

⎛−

⎪⎭

⎪⎬

⎫

⎪⎩

⎪⎨

⎧

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛

−−

⎪⎭

⎪⎬

⎫

⎪⎩

⎪⎨

⎧

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛

−−= −−

σαα

Homework 1. Methane is being considered for use as the fuel in a fuel cell. The

overall reaction is CH4 + 2O2 → CO2 + 2 H2OIn acid electrolyte, oxygen is reduced at the cathode.a) Write the electrolyte reactions. Note that a balanced reaction can

be written using methane and oxygen as anodic reactants and oxygen as the cathodic reactant. Try it, why is this scheme unreasonable?

b) Estimate the standard cell potential.

2. Overpotential measurements at 25C for copper dissolution in well-stirred electrolyte yielded the following results:i (mA/cm2) 1.2, 2.4, 4.8, 9.7, 20, 40, 60, 200, 2000ηs (mV) 1.5, 3.0, 6.0, 9.0, 18, 30, 36, 60, 104

Determine i0 and αa.

Homework 3. A reaction follows Butler-Volmer kinetics with αa = αc = 0.5 and

i0 = 1 ma/cm2.• Determine the increase in reaction rate when the overpotential is

increased from 0.1 V to 1.1 V.• For a chemical reaction following Arrhenius behavior, determine

the temperature required to achieve the same increase. Assume that the rate is initially measured at 25C and that the activation energy is 100 kJ.