Studies Module 8 Module 8 [email protected]

Reactor Design P.A. Ramachandran

Heterogeneous Reaction Electrochemical Processes and

Engineering: Theory and Case Chemical Reaction Engineering Laboratory OUTLINE

¾ Cell thermodynamics ¾ Technological examples ¾ Environmental aspects ¾ Kinetics of electrode reactions ¾ Transport effects Half-Reactions and electrodes

Redox 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 conductor Anode: Electrode where oxidation occurs Cathode: Electrode where reduction occurs

Example: 2H2(g)+O2(g)→2H2O (aq)

Anode (Oxidation) reaction: Cathode (Reduction) reaction: + - + - 2H2 → 4H + 4e O2+4H +4e →2H2O Schematic of a fuel cell

Current flow from cathode to anode

Electrons move through external circuit

+ + H2 H H O2 - + O2

H2O Anode Porous Cathode separator + - + - 2H2 → 4H + 4e O2+4H +4e →2H2O Details of a fuel cell Schematic of an electrolyte cell Cell Thermodynamics

0 nFEr = ΔG where n = number of electrons transferred F = 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

0 E r = Ea-Ec = -1.23V (standard conditions) If the reaction potential is negative, the reaction is spontaneous → FUEL CELL Cell Thermodynamics cont’d

0 ⎛ a p ⎞ The free energy change of reaction is ΔG = ΔG + RT ln⎜ ⎟ ⎝ ar ⎠ where ap is the product of activities of all products. RT ⎛ a ⎞ 0 ⎜ p ⎟ Er = Er + ln⎜ ⎟ Nernst Equation nF ⎝ ar ⎠ The effect of temperature in the reaction equilibrium can be calculated in a similar manner. A linear equation is often used: ∂E 0 E(T ) = E 0 + ()T − T ref ∂T

From thermodynamics, it can be shown that ∂E 0 ΔS 0 = where ΔS0 is the entropy change for the reaction. ∂T nF Electrolysis of HCl Consider the oxidative electrolysis of HCl. The overall reaction is

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

Direct electrolysis

2HCl →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

VT = E0 +η A + (−ηC )+ iR (solution) +iR(metal) where VT = overall voltage E0 = decomposition voltage predicted from thermodynamics = Er ηA = anode overpotential ηC = cathode overpotential iR = 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

ηa

Ea IR solution Cell Voltage VT

Ec ηc IR metal Voltage balance example The 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 2.17V Anode overvoltage 0.03V Cathode overvoltage 0.30V Solution IR drop 0.35V Diaphragm drop 0.60V Metal hardware 0.20V Total: 3.65V Cathode reaction is slow requiring larger overpotential. The drop in the diaphragm is also large due to transport limitations of ions across the membrane. Electrochemical processes: Chloralkali process

2NaCl+2H2O → 2NaOH + Cl2 + H2 Economics 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 cell Na+(aq)+e-→Na/Hg

Na/Hg+H2O→NaOH+1/2H2 (separate reaction)

Advantage: H2 and Cl2 formed in separate steps Disadvantage: cost and toxicity of Hg Chloralkali process cont’d

Newer technology:

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

Anode reaction: - - Cl (aq)→1/2Cl2+e Cathode reaction: + - H (aq)+e →1/2H2 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’d Energy 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 1.8V Anode/cathode overvoltages 0.5V iR drops (anode, cathode) 1.1V iR drop (electrolyte) 1.5V Total 4.3V •Costs Purified 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 costs Old style cell New style cell Voltages (V) reversible cell potential -2.50 -2.50 overpotentials -1.22 -1.87 electrolyte iR -6.24 -0.47 membrane iR -1.69 --

Total -11.65 -4.84

Energy (kWht-1) 6700 2500 Paired Electrosynthesis

The 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 cation exchange membrane + - + C6H5NH2+2H2O→C6H4O2+3H +4e +NH4 + - C6H4O2+6H2O →C4H4O4+12H +12e +2CO2 + - C4H4O4+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 μR − (μo + nμe ) μμ μ 0 R = μR + RgT lnCR Rg = gas constant 0 C = concentration of species i = + R T lnC + nFφ i O O g O s 0 μ μi = chemical potential of i in μ 0 the standard state e = e − Fφm φs = electric potential of the solution Combining all the equations we get: φm = electric potential of the metal

RgT ⎛ C ⎞ E = φ - φ E = E 0 + ln⎜ O ⎟ eq m s eq ⎜ ⎟ E0= standard potential for half nF ⎝ CR ⎠ reaction OverPotential and Rate of Reaction Surface overpotential = Actual applied potential – Equilibrium potential

η = Em − Eeq If η < 0, reduction is favored η > 0, oxidation is favored

Kinetic Model: F f = rf = k f CR exp()nfβη RgT

β: symmetry factor between 0 and 1 rb = kbCO exp[]− nf (1− β )η

Net rate = k f CR exp(nfβη )− kbCO exp[− nf (1− β )η] Standard Rate Constant

0 kb = k f exp(− nfE ) Thermodynamic consistency β 0 ko = k f exp(− nfβE )

0 0 r = ko (CR exp(nf (E − E ))− CO exp(n − f (1− β )(E − E )))

η * = E − E 0 βη

* β * r = ko (CR exp(nf )− CO exp(n − f (1− )η )) Current and Exchange Current

i = nFr

[]i = i0 exp()nfβη − exp[− nf (1− β )η] Butler-Volmer

β = 0.5 gives the simplified form:

i = 2i0 sinh()nfη / 2 If η is large and reverse reaction is negligible

i = i0 exp()nfβη

η = a + blog(i) Tafel Equation, plot n vs log(i) Butler-Volmer empricial form

[]i = i0 exp()nfβη − exp[− nf (1− β )η] Butler-Volmer

α A = βn

αC = (1− β )n

[]i = i0 exp()α A fη − exp[−αC fη]

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 Example An 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. 0 Temperature is 25 C 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: -1 Anolyte conductivity = ka = 18.0 mho m Separator conductivity = ka/3 Catholyte conductivity = 40.0 mho m-1 The anode and cathode interelectrode gaps are 3 mm and separator thickness is 1 mm. Voltage Balance Solution

0 Er = Ea-Ec From electrochemical tables

Ea=1.36V Ec=-0.828V

0 Er =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 V

Cathodic overvoltage = (-ηc)=0.17 V Solution Cont’d 1 δ δ δ = a + m + c R ka km kc

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

Voltage drop across solution = (V/A)i i = 10 A/m2 -3 ΔVsolution = 4x10 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. Transport Effects * CAb Rate to surface = kL (CAb-CAs) * kL = mass transfer coefficient enhanced by migration due CAs to field = fkL where f is the order of 2. ⎛ C ⎞ * ⎜ As ⎟ Expressing in terms of current i = nFkLCAb ⎜1− ⎟ ⎝ CAb ⎠ The rate of reaction expressed as current is C As i = i0 exp()nfβη neglecting the reverse reaction. C Ab 1 1 1 Eliminating the surface concentration we fixed = + i iL iR * (limiting current) where iL = nfk LC Ab

iR = i0 exp(nfβη)

1 δ δ δ Resistance of solution = R where R is = a + m + c R ka km kc Transport Effects: Migration Terms

dC D C z dφ N = −D j + C u − j j j F j j dy j RT dy The last term is the migration term and dφ is the potential gradient. dy α

Voltage Balanceα for H2 fuel cell ⎧ ⎛ i ⎞⎫ ⎧ ⎛ i ⎞⎫ RT ⎪1 ⎜ i ⎟⎪ RT ⎪1 ⎜ i ⎟⎪ ⎛ L ⎞ −1 A,0 −1 C,0 ⎜ B ⎟ V = V0 − sinh ⎨ ⎜ ⎟⎬ − sinh ⎨ ⎜ ⎟⎬ − i⎜ ⎟ − i()RI AF 2 1− i C F 2 1− i σ ⎪ ⎜ i ⎟⎪ ⎪ ⎜ i ⎟⎪ ⎝ B ⎠ ⎩ ⎝ A,L ⎠⎭ ⎩ ⎝ C,L ⎠⎭ α

Power densityα vs Voltage

⎧ ⎛ i ⎞⎫ ⎧ ⎛ i ⎞⎫ iRT ⎪1 ⎜ i ⎟⎪ iRT ⎪1 ⎜ i ⎟⎪ ⎛ L ⎞ −1 A,0 −1 C,0 2 ⎜ B ⎟ 2 P = iV0 − sinh ⎨ ⎜ ⎟⎬ − sinh ⎨ ⎜ ⎟⎬ − i ⎜ ⎟ − i ()RI A 2 1− i C 2 1− i σ F ⎪ ⎜ i ⎟⎪ F ⎪ ⎜ i ⎟⎪ ⎝ B ⎠ ⎩ ⎝ A,L ⎠⎭ ⎩ ⎝ C,L ⎠⎭ Homework

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

overall reaction is CH4 + 2O2 → CO2 + 2 H2O In 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, 2000

ηs (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.