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Biochemical Reaction Engineering
(CHE 505)
P. A. Ramachandran &M.P. Dudukovic
Chemical Reaction Engineering Laboratory
(CREL),
Washington University, St. Louis, MO
3-17-05 Che505_bio.doc
CHAPTER 14 ChE 505: Biochemical Reaction Engineering Application Examples Biochemical reactions are encountered in a number of environmentally important processes. Some examples are shown here
1. Secondary treatment of wastewater Wastewater undergoes first a primary treatment. Here the suspended solids are
removed. The wastewater now has colloidal matter and dissolved organic compounds. They are then treated by feeding them to microorganisms which use the dissolved organics as their food and converts the waste to CO2 and H20. The organism grow and multiply thus converting the dissolved organic to a solid waste (sludge) which can be settled and removed. The water is thus purified. Activated sludge process is a common example. Here the treatment takes place in open tanks with either surface aeration or submerged aeration to provide enough oxygen for the microorganisms. There are many other processes to do this as well, for example, the trickling filter. Nutrients such as N and P are needed for the growth of biomass and these are usually present in the wastewater. In some cases there is excess of nutrient and removal of these may be necessary (see #2 below). In other cases an external supply of nutrient has to be provided.
2. Biological nutrient removal. In some cases there is excess of nutrient in wastewater and only a part of these may
be utilized for biomass growth. The excess nutrient in discharged water has many negative environmental consequences. e.g eutrophication, (excess of nutrients leading to algae growth etc) oxygen depletion leading to fish kill etc. hence. removal of these nutrients may be necessary by additional step or as a simultaneous step during the removal of dissolved organics. The process is known as BNR (biological nutrient removal).
3. Anaerobic treatment of wastewater Under suitable conditions, wastewater can also be treated under anaerobic conditions.
In contrast, to aerobic process, the anaerobic process is an energy producing process since the byproduct is methane. The quantity of sludge produced is also smaller. Since the sludge disposal costs are often 50% of the wastewater treatment costs this represents a saving as well. Nutrient loadings are also smaller.
The treated effluent has however, a higher concentration of dissolved organic. This can be problematic and some form of post-treatment (usually a aerobic treatment) may be needed before the water can be discharged to surfaces. Thus the process is more suitable for relatively high concentration waste streams. (more than 4kg of COD per cubic meter of water).
4. Soil bioremediation Soil remediation is the removal of contaminants from soil either by convective transport techniques, such as flushing, soil vapor extraction and sparging, and by biological degradation. Bioremediation uses naturally occurring microorganisms to degrade wastes in the soil at the polluted site. Moisture and pH of the soil, availability of oxygen to the microorganisms, temperature and nutrients are important factors in treatment effectiveness. Alkanes, benzene and tetrachloroethene are examples of chemicals that can be degraded by bioremediation. COD and Yield Coefficient COD is a lumped concept useful to characterize waste water. kg /m3 is the common unit. For a solution of known composition COD can be calculated by a simple stoichiometry of combustion reaction. Organic compounds usually fit a general formula of CαHβOγNδ and the combustion reaction can be represented as
2222 224NOOHCOONOHC δβαδγβαδγβα ++→⎟
⎠⎞
⎜⎝⎛ +−++
The substrate kinetics is established by noting that as the organisms grow, the substrate is utilized. Thus the rate of substrate consumption is directly related to the microbial growth rate by a proportionality constant.
dtdSY
dtdS
UdtdX
−=−=1
Here U is called the specific substrate utilization rate. It is also the reciprocal of Y, the specific yield of organisms. Both U and Y are like stoichiometric factors familiar in reaction engineering. In many cases these can be predicted by setting up a reaction stoichiometry. A general molar representation of biomass is C5H7NO2. Some examples are shown below: Yield calculations: Example 1: Oxygen requirements and Y factor: Use of reaction stoichiometry to compute Y and COD is illustrated below. Reaction is represented by an overall stoichiometry
OHvCOvcellsnewvPOvNHvOvorganicv ismmicroorgan27265
34433221 )()( ++⎯⎯⎯⎯ →⎯+++ −
Based on this Y and O2 demand can be calculated. Assume organic is glucose and the biomass has a structure C5H7NO2. A balanced reaction is then
OHCONOHCNHOOHC 22275326126 1482283 ++→++ 3(180) 8(32) 2(17) 2(113)
Biomass produced = 226 g O2 consumed = 256 g Glucose consumed = 540 g O2 demand = 256/540 = 0.47 g / g glucose Y per glucose = 226/540 = 0.42 g cells / glucose Also U = 1/0.42 = 2.38 g glucose consumed to produce 1 g of cell. It is more convenient to write this in terms of COD (chemical oxygen demand). For this purpose COD of glucose is calculated as follows: COD of glucose:
OHCOOOHC 2226126 666 +→+ 180 192
COD = 192/180 = 1.07 g O2/g glucose Y = 0.42 g cells / g glucose x g glucose / 1.07 COD = 0.392 g cells / g COD A second approach is to write half reactions for the various oxidation and reduction processes and balance the rate of these processes. This method is described below. Consider again the substrate to be glucose. Oxidation is the electron donor half reaction, shown by equation (2), with rate Rd.
−+ ++→+ eHCOOHOHC 226126 41
41
241 (2)
Reduction is the electron acceptor half reaction that has two possible routes; respiration, equation (3a), and synthesis, equation ( 3b) that proceed at the rate Rc, and Rs , respectively:
A. Respiration
OHeHO 22 21
41
→++ −+ (3a)
B. Synthesis (in presence of nitrates)
OHNOHCeHCONO 227523 2811
281
2819
225
281
+→+++ −+− (3b)
In the above we have assumed that organic matter (food) is the simple sugar (glucose) and that the culture in question can utilize the nitrate ion to synthesize new biomass by equation (3b). Now, based on data, we assume that the probability of reaction (3a) occurring is fe while the probability of reaction (3b) occurring is fs. Clearly, since the electron must be used by ether route (3a) or (3b), we must have fe+fs=1. In addition, the rate of electron production must equal the rate of electron consumption so that
Rd + fs Rs + feRc = 0 (4) Now, if based on available information we conclude that fs=0.68, and hence fe=1-0.68=0.28, then substituting these values in equation (4) leads to the overall stoichiometry below. Overall stoichiometry
OHCONOHCHONOOHC 22275236126 184.014.00221.00221.0095.00221.0241
++→+++ +−
(5) Now, if we know the O2 uptake rate we can determine the nitrate or glucose consumption rate and rate of synthesis of cell mass! Unfortunately, in most real complex systems we do not know that and we need to rely on empirical yield factors defined earlier. . The factors, fe and fs can also be related to the free energy changes associated with the respiration and synthesis reaction. The free energy determines how the released energy is partitioned into respiration vs synthesis. The suggested relation for heterotropic bacteria is
R
s
e
e
s
e
GkG
ff
ff
∆∆
−=−
=1
sG∆ = energy required for cell synthesis which is calculated by Eq (1) shown later.
k = fraction of energy captured
RG∆ = energy released from the oxidation-reduction reactions (from the overall process of oxidation and respiration) The value of is free energy to convert 1-electron equivalent of carbon source to cell material. This is estimated as
sG∆
kG
GkmG
G Nc
ps
∆+∆+
∆=∆
pG∆ = free energy to convert 1- electron equivalent of carbon source to pyruvate
intermediate. m = +1 if is positive or –1 if energy is produced. pG∆
cG∆ = free energy to convert 1- electron equivalent of pyruvate to 1- electron equivalent of cells. This is normally taken as +31.41 kJ/e-eq
NG∆ = free energy per 1- electron equivalent of cells to reduce nitrogen to ammonia. This factor accounts for the fact that a nitrogen source has to be converted to ammonia if N2 is not available in the form of ammonia. The values of 17.46, 13.6, 15.81 and 0.0 are used for , , N−
3NO −2NO 2 and respectively. Tables of ∆G values are available for a
number of typical bio-chemical half reactions. The values for some reactions are attached here. An illustrative problem is solved below.
+4NH
Example: Consider acetate as a substrate and CO2 as the electron acceptor. This represents the treatment under anaerobic condition in the absence of oxygen. Find fe and fs based on free energy considerations. Take the fraction of energy used as 0.6 Solution The oxidation half reaction is represented as
−+−− +++→+ eHHCOCOOHCOOCH 3223 81
81
83
81
and has a free energy of –27.68 kJ per electron from the tabulated values. In the absence of O2 (anaerobic conditions) the CO2 acts as an electron acceptor. The respiration reaction is then represented as
OHCHeHCO 242 41
81
81
+→++ −+
and the free energy is 24.11 kJ/g. eq. of electrons Overall free energy change is therefore –27.68+24.11 = -3.57 kJ which is our ∆GR value. The free energy of pyruvate formation is now calculated. This is reaction 15 in the table.
OHCOCOOCHeHHCOCO 2332 52
101
101
51
+→+++ −−+−
∆G = 35.78 kJ. Note that the overall pyruvate formation is the sum of oxidation half reaction and the above pyruvate half reaction. Hence ∆Gp = -27.68 + 35.78 = 8.1 kJ which is positive. Energy is utilized in pyruvate formation for this substrate (acetate) and m is taken as 1. Note in a general context that puruvate formation from other substrates such as glucose releases energy and is the first step in the Kreb cycle by which we (all species) use the food energy. Assume k = 0.6.
∆Gs = 8.1 / 0.6 + 31.41 + 0 = 44.91 kJ/mole.
96.20)57.3.(6.0
91.441
=−
−=
− e
e
ff
solving gives fe = 0.9547 and fs = 0.046 g cell / g COD used. Tables:
Kinetics of Microbial Reactions The growth of the microbial concentration follows a first order dependence and can be described as:
Rx = XdtdX µ=
where X is the concentration of the micro-organisms and µ is a growth constant and Rx is the rate of growth (kinetic model) for biomass. An early model by Monod (1942) showed that the growth constant is a function of the nutrient or the substrate concentration. The relation can be described by the following equation.
SKS
s += maxµµ
where S is the concentration of the substrate and µmax is a maximum growth constant. The other constant in the model is Ks. In fact, it can be shown that if S = Ks /2 the growth rate is half the maximum value. Hence Ks is also called the half velocity constant. The Monod’s equation applies to pure cultures but is also applied to mixed cultures as a first approximation. Other models for the growth constant µ have also been proposed. Thus Haldane proposed the following model
Is K
SSK
S2max
++= µµ
where KI is often referred to as the substrate inhibition constant. In a dynamic growth situation, some organisms are born, some die and some members simply grow in mass. The death rate is added to the model by adding a decay constant and the net growth rate of microbes is represented as:
XkXSK
SXkXR ds
dx −+
=−= maxµµ
Although the model is simple (a lumped model for microbes), it has served its purpose well in many applications. However, more complete models that incorporate some of the
underlying biochemical mechanisms offer a better representation and predictability. Such models are based on metabolic pathway analysis. Such detailed models are often developed if there are certain design objectives in mind. Otherwise a simple Monod based model is often sufficient. Typical values of kinetic parameters for domestic wastewater at 20degC are in the following range: µmax 2-10 g bsCOD/g VSS. Day Ks. 10-60 mg/L bsCOD Y 0.3-0.6 mg VSs/mg bsCOD Kd 0.06-0.15 gVSS/ g VSS day Here bsCOD refers to biodegradable soluble chemical oxygen demand. VSS refers to volatile suspended solids which is commonly used as a measure of biomass concentration since this is an easily measurable quantity, Materials that can be volatilized and burned off when ignited at 500 +or- 49 deg C are classified as volatile The effect of temperature on the kinetic parameters is of some interest. The growth constant µmax increases with temperature and reaches a maximum and then decreases rapidly with temperature. This is often fitted by two Arrhenius type of parameters: The Monod constant is an inverse function of temperature and is fitted as:
⎟⎠⎞
⎜⎝⎛ −=
RTEA
K s
1exp1
The yield constant is assumed to be independent of temperature. Other effects such as product poisoning may have to be added to the model. One example is alcohol production by fermentation where higher concentrations of alcohol act as poison to the microbial population. Detailed kinetic models In this model one attempts to track the key species involved in the process rather than lumping all the species into COD.
Metabolic Pathway Models As an example of metabolic pathway analysis, the growth metabolism model for E. coli was developed by Schuler and co-workers. This model shown schematically in Fig. 2 has 23 stoichiometric constants and 49 kinetic parameters. Most of the kinetic parameters could be estimated from the large number of reported previous studies on the system. Process Model for Activated Sludge Process The typical flow diagram can be represented as follows: Primary effluent Effluent Q0-Qw, Xe, S Q0, S0, X0 Q0+ QR secondary V,X,S X,S clarifier reactor QR, sludge return sludge Xu Qu,Xu underflow Sludge Qw, Xu waste Control volume/mass boundary (a)
Primary effluent Effluent Q0-Qw, Xe, S Q0, S0, X0 Reactor Q+ QR secondary Variable X,S clarifier X, S QR, sludge return sludge Xu Qu,Xu underflow Sludge waste Qw, Xu Control volume/mass boundary (b) Figure: Schematics of the two general types of activated sludge systems: (a) completely mixed; (b) plug flow Model is developed by writing mass balances for the substrate and the biomass. The model equations for a backmixed reactor are as follows: Cell Balance: Produced – Died = -In + Out
( )[ ]uwewds
XQXQQXQXVkXVSK
S+−+−=−
+ 000maxµ (7.16)
which can be rearranged as
( )d
ewuw
s
kVX
XQXQQXQSK
S+
−−+=
+000
maxµ (7.17)
The same mathematical operations may be applied to the substrate. These will not be repeated but the result will be written at once, which is
( ) XVSKY
S
s +maxµ
= ( )( ){ SQSQQSQ ww }+−+− 000 (7.18)
Eq. (7.18) may be manipulated to produce
( SSVX
YQSK
S
s
−=+ 0
0maxµ ) (7.19)
Equating the right hand sides of Eqns (7-17) and (7-19) and rearranging gives us
( ) ( ) dewuw kSS
VXYQ
VXXQXQQXQ
−−=−−+
00000 (7.20)
The ratio, cθ = ( ) 000 XQXQQXQVX
ewuw −−+, is the biomass (or mixed-liquor volatile
suspended solids, MLVSS-the numerator) in the reactor divided by the net rate of biomass wasting (the denominator). The ratio cθ represents the average time that the biomass is in the residence in the reactor. It is called by various names, such as mean cell residence time (MCRT), sludge retention time (SRT), and sludge age. The ratio
θ=0/ QV is called the nominal hydraulic retention time (NHRT). The word nominal is used here, since θ is not the actual detention time of the tank. The actual detention time is where Q( )RQQV +0/ R is the recirculated flow. Using cθ and θ in Eq. (7-20) and solving for X gives:
( )( )cd
c
kSSY
Xθθ
θ+
−=
10 (7.21)
This is known as the equation for biomass concentration in the reactor. To find S, the exit concentration of the substrate, we substitute Eq. (7.21) for X in Eq. (7.19), this leads to
c
cd
s
kSK
Sθµθ
max
1+=
+
Rearranging:
)1()1(
max cdc
cds
kkK
Sθθµ
θ+−
+= (1)
The above equation then provides the performance of the bioreactor as a function of operating conditions. The key parameters which the operator can adjust is cθ (the cell residence time). The other parameters Ks, µmax and kd are biological rate parameters which we cannot control.
An interesting effect is that the exit concentration in the bioreactor is independent of the inlet concentration. This is unlike a conventional chemical reactor.
cθ is sometimes referred to as BSRT or biomass solid retention time. The reactor performance depends on cθ (See Eq. 7.21). The concentration of cells in the bioreactor however depends on both cθ and θ. A minimum residence time is required and this is related to the inlet substrate concentration. This quantity can be obtained by setting X = 0 in Eq (7.21). This would be yield at S = S0. Now, using Eq(1) for S and solving, one obtains
⎟⎟⎠
⎞⎜⎜⎝
⎛+>
0max
11SKs
c µθ assuming kd = 0 here.
min,cθ
or 0max
0min, S
SKsc µ
θ+
=
Eq. (1) holds only if the above condition is satisfied. A plot of S vs cθ is as follows: S S0 cθ minimum Cell concentration profiles are as follows: X Low θ case High θ case min cθ
Another important parameter in the design and operation of an activated sludge plant is the recirculation ratio, . R may be obtained by performing a material balance around the secondary clarifier. Since the clarifier is not aerated, dX/dt = 0. Also at
steady state
0/ QQR R=
0=∂∂
tX . Adopting these facts and performing the material balance produces
( ) ( ) ( uwReRR XQQXQQXQQ )++−=+ 00 (7.24) Solving for the ratio gives:
( ) (( )
)XXQ
XXQXXQR
u
euweR
−−−−
==0
0
0
(7.25)
Trickling Bed System: A schematic diagram is shown in the next page. Often a first order kinetic model is used for simplicity.
a
s
QAk
SS '
ln0
=
k’ = rate constant = k Xf Xf = microbial concentration in the film As = surface area for diffusion into biomass Qa = volumetric flow rate Other detailed models based on Monod’s kinetics can also be derived.
BIOLOGICAL REACTIONS AND
BIOREACTORS: FERMENTATIONS
(CHE 471)
M.P. Dudukovic
Chemical Reaction Engineering Laboratory
(CREL),
Washington University, St. Louis, MO
1
BIOLOGICAL REACTIONS AND BIOREACTORS: FERMENTATIONS
Biological reactions are more complex than anything that we have considered so far and the reasons for this are briefly outlined below. Consider a “generic” biological reaction where some organic matter (food) is consumed by a culture of living organisms (bugs) that also consume oxygen due to their respiration (aerobic culture) and in the process generate carbon dioxide, water and more biomass (more bugs). This is represented by equation (1): Organic matter + O2 + bugs →CO2 + H2O + more bugs (1) This process involves oxidation, respiration and synthesis ( of new biomass). Oxidation is the electron donor half reaction, shown by equation (2), with rate Rd. 1
24C6H12O6 +
14
H2O →14
CO2 + H + + e−; Rd (2)
Reduction is the electron acceptor half reaction that has two possible routes: respiration, equation (3a), and synthesis, equation (3b), that proceed at the rate Rc and Rs, respectively.
A. Respiration 14
O2 + H + + e− →12
H2O ; Rc (3a)
B. Synthesis (in presence of nitrates)
128
NO3− +
522
CO2 +1928
H + + e− →128
C5H7NO2 +1128
H2O ; Rs (3b)
In the above we have assumed that organic matter (food) is the simple sugar (glucose) and that the culture in question can utilize the nitrate ion to synthesize new biomass by equation (3b). Now, based on data, we assume that the probability of reaction (3a) occurring is fc while the probability of reaction (3b) occurring is fs. Clearly, since the electron must be used by either route (3a) or (3b), we must have fc + fs = 1. In addition, the rate of electron production must equal the rate of electron consumption so that Rd + fsRs + fcRc = 0 (4) Now, if based on available information we conclude that fs = 0.62, and hence fc = 1-062=0.38, then substituting these values in equation (4) leads to the overall stoichiometry below.
2
Overall stoichiometry
124
C6H12O6 + 0.0221 NO3− + 0.095 O2 + 0.0221H +
→ 0.0221C5H7NO2 + 0.14CO2 + 0.184 H2O (5)
Now, if we know the O2 uptake rate we can determine the nitrate or glucose consumption rate and rate of synthesis of cell mass! Unfortunately, in most real complex systems we do not know that and we need to rely on empirical yield factors which will be defined below. We need now to describe the rate of biomass (bugs) growth, substrate (food, e.g. sugar) consumption rate and oxygen consumption rate. Biomass Growth Rate Typically, Monod’s model is used where the rate of formation of the biomass, Rx, is proportional to the biomass concentration present in the system, X, as shown by equation (6) Rx = μX (6) The specific growth constant is given by Michaelis Menten kinetic form and depends on the limiting substrate concentration, S, as shown below.
μ =μmS
Ks + S (7)
The above ignores the death term in the cell rate which has to be included for mature cultures, so that a more complete description of the net biomass formation rate is:
Rx =μmSXKs + S
− kd X (8)
Substrate Consumption Rate The substrate consumption rate is customarily given in terms of the biomass production rate and the appropriate yield coefficient. Please, note that the yield coefficient may change with conditions but is most frequently used as constant:
−Rs =
1Yx / s
Rx (9)
3
Yx / s = yield coefficient =
(mg / L )biomass produced(mg / L) substrate consumed
(10)
Oxygen Consumption Rate This rate must also be expressed in terms of an empirical yield coefficient as shown below:
−Ro =
1Yx / o
Rx (11)
Yx / o =
(mg / L) biomass produced(mg / L) oxygen consumed
(12)
Balance on well mixed compartment (CSTR at transient conditions) Well mixed CSTR at unsteady state can now be described by writing the mass balance on biomass, substrate and oxygen as follows:
V dX
dt= Q(Xo − X ) + RxV (13)
t = 0 X = Xi
V dS
dt= Q(So − S) − (−Rs )V (14)
t = 0 S = Si
V dCo
dt= Q(Cos
−Co) + ksaV (Co* − Co ) − (−Ro )V (15)
oio CCt == 0 Let dilution rate
D =QV
(16)
Then, rewriting equation (13) to (15) above yields:
4
dXdt
= D(Xo − X ) +μmS
Ks + SX (17)
dSdt
= D(So − S) −1
Yx / s
μmSKs + S
X (18)
dCo
dt= D(Cos
− Co ) −1
Yx / o
μmSKs + S
X + kLa(Co* − Co ) (19)
In a batch D = 0.
In steady state CSTR ddt
= 0. Let us look at a steady state CSTR.
In steady state CSTR with no fresh culture in the feed stream ,X0 = 0, so that
D = μ (20)
D =μmS
Ks + S→ Solve for S (21)
S =KsD
μm − Dfor D < Dcr (22)
In case that the dilution rate is larger than the critical dilution rate given below, washout occurs and only the steady state at zero biomass concentration is possible.
D > Dcr =μmSo
Ks + So
⇒ X = 0 (23)
For admissible values of the dilution rate D the biomass steady state concentration in the effluent (and hence in the reactor due to CSTR assumptions) is given by
X = Xss = Yx / s(So − S) = Yx / s So −
KsDμm − D
⎡
⎣ ⎢
⎤
⎦ ⎥ (24)
The production rate of the biomass per unit reactor volume is the product DX To maximize it we need
ddD
(DX ) = 0 which yields :
5
Dopt = μm 1−
Ks
Ks + So
⎛
⎝ ⎜
⎞
⎠ ⎟
1/ 2⎡
⎣ ⎢ ⎢
⎤
⎦ ⎥ ⎥ (25)
Xopt = Yx / s So + Ks 1−Ks
Ks + So
⎛
⎝ ⎜
⎞
⎠ ⎟
−1/ 2⎛
⎝ ⎜ ⎜
⎞
⎠ ⎟ ⎟
⎡
⎣
⎢ ⎢
⎤
⎦
⎥ ⎥ (26)
See diagram below:
S
X
S
DX
D Dopt Dcr
DX
Electrochemical Processes
(CHE 505)
M.P. Dudukovic
Chemical Reaction Engineering Laboratory
(CREL),
Washington University, St. Louis, MO
Electro_chapter16.doc 3-31-05
Chapter 16 Electrochemical Processes This chapter considers some applications of electrochemical transport and reactions in environmental engineering. First we review some application examples. Then we provide a review of electrochemical thermodynamics and kinetics of electrode processes. These concepts are useful for design of electrochemical systems for waste treatment. The transport in electrolytic solutions is then discussed followed by application to membrane processes assisted by an electric field (electrodialysis). The educational outcome of this chapter is as follows:
• To indicate application areas where the use of an electric potential is beneficiary • To understand the minimum voltage (Faraday’s law, Nernst equation) and the
concept of voltage balance • To understand the kinetics of electrode processes in terms of current-voltage
relations (Butler-Volmer equations) • To understand transport in charged membranes. • To do preliminary design of electrochemical reactors.
Application Areas
1. Metal salts recovery from waste water and other process streams Metal salts are frequently found in wastewater. These can be electrochemically processed to recover the metals. Thus we not only treat the waste streams but also generate value from waste. Potential for metal recovery and some statistical data is shown in the book by Allen and Rosselot. From their data it is seen that the metals in many waste streams are significantly underutilized. Thus there is considerable scope of process development and optimum design in this area.
2. Treatment of organic pollutants
Some pollutants such as phenol, aromatic amines, halogenated nitroderivatives can be oxidized to CO2 and water using electrochemical routes
3. Electric field assisted membrane separations (Electrodialysis)
In this process one uses semipermeable ion selective membranes to separate ionic components of the solution. An electric field is applied across the system to facilitate the transport of the ions across the selective membranes. Details of the process and design aspects are considered in a later section.
Additional examples are discussed in the ‘Environmental Electochemistry’ book by K. Rajeshwar.
Thermodynamics Consider a reaction A +B C+D. From thermodynamics if the free energy change is negative the reaction is spontaneous and energy can be recovered from the reaction. The energy is usually released as heat, for example, combustion reaction of fuels such as hydrogen but can also be released as electrical energy if the reaction is carried out electrochemically as pair of oxidation and reduction reactions at the anode and cathode respectively. The cell potential under idealized condition is related to the standard free energy change of the reaction as:
00 GnFEcell ∆−= n is the number of electrons transferred and F is the Faradays’ constant = 96500 C/ g-eq and represents the charge to be supplied for 1 g- eq of electron of transported in the external circuit. If the reaction has a positive change in free energy then the reaction is not spontaneous and needs energy input to carry out the reaction. Again if the reaction is carried out as an electrochemical reaction, the energy can be supplied in form of electrical potential. The minimum energy needed to be supplied is related to the free energy change in the reaction. This value defined here as the decomposition potential or the reaction potential given as:
000 GnFEE dr ∆== where ∆G0 is the free energy change on the overall reaction under standard conditions. Note that 00
cellr EE −= Thus if is negative, the reaction will yield energy (a galvanic cell ot fuel cell) and vice versa (an electrolyzer or electrochemical reactor).
0rE
The decomposition potential is more usually calculated by writing the overall reaction as a sum of oxidation and reduction reaction separately. These are called as half reactions. For example consider the electrolysis of silver nitrate which can be represented by the following reaction:
2323 2/12222 ONOHAgOHAgNO +++→+ −+ or more simple since the species are ionized as:
2 Ag+ + H20 2 Ag + 2 H++ ½ O2 This can be represented as two half reactions. Oxygen production at the anode (Oxidation reaction) H20 2 H+ + 2e- + ½O2 (i) Silver formation at the cathode (reduction reaction) Ag+ + e- Ag (ii) The potential values for a number of common reactions are tabulated in electrochemistry books. Tables 1 and 2 reproduce some of these values. Note that these values have the same sign as the overall free energy change of the reaction and hence represent the reaction potentials and not the cell potential. Thus if the value is negative it means that the corresponding half reaction has a negative free energy change and is therefore spontaneous. The standard conditions are defined as 1M concentrations for liquid phase components and 1 atm pressure for gas phase species. The overall potential is then calculated as:
cathodeanode EEE −= where both the anode and cathode reaction are written as oxidation reactions. The overall reaction is therefore the difference between the anode reaction and cathode reaction. The actual cathode reaction is the reduction and proceeds in the opposite direction to those shown in the tables. However, both reactions are written as oxidation for the purpose of using these tables and the difference is the net reaction.
Example 1: Find the minimum potential needs to decompose AgNO3 form a 1M solution The oxidation reaction is decomposition of water at the anode. From table 1, we find that this reaction has potential change of 1.229V associated with it. (Reaction 15) The oxidation of silver has a potential of 0.7991V. (Reaction 15 in Table 2) The overall potential change is:
cathodeanoder EEE −= = 1.229 -0.799 = 0.43V which is the minimum potential that needs to be applied to carry out the reaction. Example 2: Consider the hydrogen-oxygen fuel cell. We can represent the two half reactions (both as oxidation) in the following manner. Anode: Oxidation of hydrogen (Reaction 4 of Table 1) H2 2 H+ + 2e- (E = 0 by convention) Cathode Oxidation of water (reaction i above) E= 1.229V
cathodeanoder EEE −= = 0-1.229= -1.229V
Ecell (= - Er) therefore +1.229V a positive value here showing that reaction is spontaneous. The reaction generates energy of 1.229V as a fuel cell under these conditions An illustration of a PEM (proton exchange membrane as the porous separator) based fuel cell is schematically shown in Figure 1.
Schematic of a fuel cell
Electrons move through external circuit
Current flow from cathode to anode
H2 H+ H+ O2
Anode Porous Cathode separator
+-
O2+4H++4e-→2H2O 2H2 → 4H+ + 4e-
H2O
O2
If the reactants are not at the standard conditions, a correction for activity is applied as shown below. The free energy change of reaction is
⎟⎟⎠
⎞⎜⎜⎝
⎛+∆=∆
r
p
aa
RTGG ln0
where ap is the product of activities of all products raised to its stoichiometric number in accordance with law of mass action. Similarly ar is the activity of the reactants. Dividing by nF and converting in terms of reaction potential we obtain
⎟⎟⎠
⎞⎜⎜⎝
⎛+=
r
prr a
anFRTEE ln0
This equation is referred to as the Nernst equation. The effect of temperature in the reaction equilibrium can be calculated in a similar manner. A linear equation is often used:
( )TETTETE ref ∂∂
−+=0
0)(
From thermodynamics, it can be shown that
nFS
TE 00 ∆
=∂∂
where is the entropy change for the reaction. 0S∆ Example 3: A hydrogen fuel cell is operated at 80C. Find the maximum potential generated in this cell. Solution: For the H2+O2 reaction, the entropy change was found to be =0.1638 J/K mole. 0S∆Hence /nF = 8.49 x 100S∆ -4 V/K E(80C) = -1.23+ 8.49x10-4 (80-25) = -1.18 V Example 4: Conditions with cobalt deposition. Cobalt is to be recovered from a solution of cobalt sulfate at a concentration of 0.005M. Find the reaction potential and also find the potential for the competing reaction of hydrogen evolution at the cathode if the pH of the solution is 1. The standard potential for oxidation of cobalt is 0.277 V. Solution: The Nernst equation is applied to the Co oxidation. n=2 here.
CoCF
RTEE ln2
0 += =0.2089V
The oxidation of water is the anodic reaction. E = 1.229V Hence, the voltage required for Co deposition is 1.02V. For hydrogen oxidation, the standard reaction potential can be corrected for pH as follows:
( )2ln2
02
++= HF
RTEH where H+=10-pH
For a pH of 1, the value is –0.0592.
Hence, ( )(pH )F
RTEH 303.22= =0.052 x 2.303 = 1.1698 V
The reaction potential for hydrogen evolution is 1.17 V which is comparable to that for Co deposition. Hence both reactions are likely to occur at this pH. From an application point of view, a relatively high pH is needed for the Co deposition to become the favored reaction. Thus if the wastewater is acidic then Co recovery is difficult by electrochemical process. Voltage Balance This is similar to the heat balance in chemical reactors. The overall voltage needed can be expressed as ( ) ( ) )()( metalIRmembraneIRsolutionIREV CArT +++−++= ηη where = decomposition voltage predicted from thermodynamics rE Aη = anode overpotential = Ea in figure Cη =cathode overpotential = Ec in figure IR=voltage drop due to a resistance in the solution and metal. IR drop in the metal is low in general but can not be ignored of cells that are connected in series. The components of the above voltage balance are all functions of the current density except of course the term which is determined by thermodynamics. Lower the current density is lower the voltage drop, but the reaction rate is also correspondingly lower. To quantify the overpotential-current relations, the reaction kinetics is needed as discussed in the next section.
rE
Kinetics of electrode processes The rate of electrochemical reactions are often expressed in terms of current density rather than in terms of the moles produced per unit time per unit area of the electrode surface. We define the current density as i which is A/m2 s. The total current at the cathode is the same as the current at the anode. The current density is related to the rate of reaction by the following equation which is a consequence of Faraday’s law. i = n F r where r is the rate of reaction in moles produced/m2s To provide an expression for the rate, r or the equivalent current density consider an electrode reaction represented as a reversible reaction of the type:
−+ +⇔ neOR n where R and O represent the reduced and oxidized species and n is the number of electrons transferred. The rate is affected by the potential difference between the metal and the adjacent solution. Let E represent this value. smE φφ −= . Then the rate of forward reaction (oxidation) is increased by increasing the value of E, i.e. by having a higher potential at the metal compared to the solution). Similarly, the rate of backward reaction is improved by decreasing E. The effect of potential difference on the rate is thus equivalent to the effect of temperature and the rate is found to depend exponentially on the value of E. The rate is also dependent on the concentration of the species. For as simple elementary reaction the dependency is linear. Combining the effect of both concentration and temperature we can write the rate as: The rate of forward reaction is therefore proportional to ( )EnfCk Rf βexp while the rate of backward reaction is proportional to ( )( )EnfCk Ob β−− 1exp where β is a factor between 0 and 1 that represents the efficiency of activation due to electric potential. The factor f represents F/R T which is a factor similar to that in the Arrhenius equation for the effect of rate on temperature. Net rate is therefore equal to ( )EnfCk Rf βexp - ( )( )EnfCk Ob β−− 1exp . The rate constants kf and kb are, however, not independent and must meet the thermodynamic consistency. At equilibrium the rate is zero. Also CR and CO are related by Nernst equation which is expressed now in the following rearranged form
( )[ ]0exp rrR
o EEnfCC
−=
Setting the net rate as zero and also using the CR and C0 values from the above version of the Nernst equation, we can drive the following thermodynamic relation between the forward and backward rate constants.
( )0exp rfb nfEkk = where is the equilibrium potential for the reaction at the standard conditions. (The students may wish to verify the algebra for clarity).
0rE
The rate constant kf is often expressed in terms of a rate constant k0 defined below which is also called as a standard rate constant.
( )00 exp rf Enfkk β−=
Similarly,
( ) ( )( )00
0 1expexp rrfb EnfknfEkk β−== Substituting and rearranging, the rate can be expressed as ( ) ( )[ ]( ) ( )( )[ ]0
00
0 1expexp EEnfCEEnfCkr R −−−−−=− ββ (A) Expressing in terms of current, we have
( )[ ]( ) ( )( )[ ]00
00 1expexp EEnfCEEnfCnFki R −−−−−= ββ (B)
Note that if k0 and β are the two kinetic parameters in this ‘rate’ form. This provides a working model to represent the kinetics of electrode processes. However, it is more common in the field to express the rate in terms of an overpotential at the electrode surface. That is referred to as the difference between the actual applied potential and the equilibrium potential as η, the surface overpotential
req EEEE −=−=η (C) If η is greater than zero, the driving force term is positive and hence oxidation is favored. If η is less than zero, the driving force term is negative and hence the reverse reaction (reduction) is favored. If η is zero the system is at equilibrium and the net rate is zero.
Although the net rate is zero at equilibrium, we may consider the rate to be balance of the forward and backward reaction. bf rrr −= and at equilibrium bf rr = . From Eq(A), we can write
( )[ ]0*0
** (exp EEnfCkrr rRbf −== β The corresponding current is referred to as the exchange current.
**0 bf nFrnFri ==
We use this expression in (B) and also express E in terms of the overall potential. Hence the current can be expressed as a function of overpotential as:
( ) ( )( )[ ]ηββη −−−= 1expexp0 nfnfii This equation is known as the Butler-Volmer equation. In many cases the parameter nβ is grouped into a parameter αa. Similarly, the parameter n(1-β) is grouped into a parameter αc. and the Butler-Volmer equation is written as:
( ) (( ))ηαηα ffii ca −−= expexp0 (B) and all the three parameters i0, αa and αc are then treated as fitted constants. The physical significance of these parameters are as follows: io represents the exchange current and is similar to the rate constant. Its value can vary over a wide range. 0.001 to 10 A/m2 range for example. This parameter is sensitive to the electrode conditions as well as to surface contamination and other factors which affect the electron transfer process at the electrode. The parameter αa is a measure of how an applied potential favors the anode (oxidation) reaction. Similarly, αc is a measure of cathodic process. The values for these parameters are in the range of 0.2 to 2. In some cases, they are both taken as 0.5 which says that the potential affects each reaction in the same manner. If more than one reaction occurs then both reactions must be described by these equations and the parameters i0 and β fitted for each reaction. The overpotential for each reaction must also be calculated or measured with respect to its own equilibrium potential. If β = 0.5, the Butler-Volmer equation simplifies to a hyperbolic form:
( )2/sinh2 0 ηnfii = If the overpotential η is sufficiently large, then the reverse reaction is negligible and Eqn (B) can be expressed as:
( )βηnfii exp0=
which can be rearranged to a linear form
)ln(iba +=ηwhere a and b are system specific constants which are fitted from the experimental data. The above equation is known as the Tafel equation. If η is small, the Butler-Volmer equation reduces to a linear form:
ηnFii 0= Example : Kinetic model fitting A sample of plot of overpotential vs current density is shown in Fig 5.10 for Cu deposition/dissolution in acidic solution. The data can be fitted as
( ) ([ ])ηη 4.19exp1.58exp01.0 −−=i where i is in A/m2. Note that mA/cm2 is used in this figure.
Design Example: Voltage balance revisited We now show how the Butler-Volmer equation can be used to calculate the voltage needed to deposit a metal at a certain rate. Copper is to be deposited at a rate of 0.005 mole/sec from a solution of CuSO4. Find (a) the current, (b) voltage required if the operating current density is 965 A/m2 and (c) the electrode surface area. Neglect hydrogen evolution and use the Butler-Volmer equation given earlier for Cu deposition.
For oxygen evolution at the cathode use the following Tafel type of model: ( )ia 10log05.08.0 +=η where i is in A/m2.
Solution: From Faraday’s law, the current needed is nF times the rate of deposition of Cu = 2 x 96500 x 0.005 = 965 A. For a current density 965 A/m2 use then an electrode surface area of 1 m^2. Using the Butler-Volmer equation for cathode and solving for ηc we have a cathode overpotential of 0.1982V. The anode overpotential is calculated using i of 0.1 in the units of A/cm2 as 0.75V. The thermodynamic decomposition potential (assuming a standard concentration) is obtained as 1.229V-0.337V = 0.9V. Note that if the concentration of Cu++ is not 1M, then the Nernst correction is required. The total voltage needed is (ignoring the solution and membrane drop) 0.9+0.75+0.1982=2.44V If the solution conductivity is known, the solution IR drop can be added to the voltage drop. If we assume a drop of 0.2V then the voltage needed is 2.66V. The power needed is I.V = 965 x 2.66 or 2560 W for this process. In the above example, we assumed no mass transfer resistance. Additional voltage is needed to facilitate mass transfer of ions if there is significant resistance to mass transfer. Transport effects The reaction rate is often affected by the transport of ions near the electrode. The film model is often convenient to represent the transport. The rate of transport to the surface is represented as: ( )sbL CCk − where is mass transfer coefficient to the surface. Traditionally mass transfer data are avialble for non-charged species. A correction to this is needed for charged species due to the contribution by migration due to electron field. In this discussion, we simply assume that
is the corrected value.
Lk
Lk
Lk The transfer rate can be expressed in terms of the current as:
( )sbL CCnFki −= The reaction rate (current density) is given for a simple case where the reverse reaction is ignored. as:
( ηαfCC
iib
S exp0= ) (A)
where i0 is now corrected by the ratio of surface to bulk concentration.
The maximum rate of transport occurs when the surface concentration is near zero and can be expressed as: bLCk The corresponding current is called the limiting current and is the maximum value of current possible at an electrode surface: bLL CnFki = The mass transfer rate to the electrode can then be represented as a current in terms of the limiting current as:
⎟⎟⎠
⎞⎜⎜⎝
⎛−=
b
sL C
Cii 1 (B)
The current given by Eq (A) and (B) are the same at steady state. Hence b
s
CC
can be
eliminated.
( )ηαfiii
CC
L
L
b
s
exp0+=
The current with mass transfer effects included in the analysis is then given from Eq (A) as
)exp(
)exp(
0
0
ηαηαfii
fiii
L
L
+= ©
For large overpotential, we can then show that the current i→iL and hence the current vs overpotential curve reaches an asymptote of iL. Example: Calculate the current for different values of overpotential for Cu deposition. Use the value for limiting current of 1000 A/m2 which is a representative value for a well stirred solution. Solution: We use the Butler-Volmer model for the kinetics with i0 = 0.01 A/m2 and αf = 58.1 as before. Then using Eq (C) we can find i for various values of η and the results are tabulated below. η,V 0.2 0.5 0.6 i, A/m2 526.74 1000 1000
b
s
CC
is calculated as 0.4733 at η = 0.2 indicating significant transfer resistance.
Electrodialysis The method of electrodialysis can be understood by the simple arrangement shown in Fig A. Here cation and anion selective membranes are placed alternatively with two electrodes at the end. In the diagram only two pairs are shown but for larger scale applications one usually stacks more pairs with the feed entering at alternative pairs. The electrodes cause an electric current to pass through the system and cause a migration of cations towards the negative electrode and the anions to the anode. Because of the alternative spacing of cation and anion permeable membranes, cells of concentrated and dilute salts are formed.
Cathode rinse
Cathode rinse
Na+Na+
Cl-
E(-ve) C A Cathode Feed S C = cation permeable membrane A = anion permeable membrane Figure A. Schematic of an electrodialysis membran The ion permeable membranes are essentially sheeexchange resins have a fixed group attached to a pgroup Na+ ion for example. The fixed group renegative charge and only positive ions can beLikewise the anion exchange resin has affixed
Concentrated solution
Anode rinse
Anode rinse
Na+
Cl- Cl-
Desalted solution
C A E(-ve) Anode
olution
e arrangement
ts of ion-exchange resins. Thus cation olymer denoted as RSO3
- and a labile pels the negative ions since it has a transported across this membrane.
group of the type NR3+ (quarternary
ammonium groups) and are selective to only negative ions because the fixed group NR3+
will repel positive ions. The electrodes are neutral and do not participate in the reaction. However reactions are needed for a current to flow. The electrolysis of water is the main reaction which takes place in this system and can be represented as: Anode reaction: 2OH- ½ O2 + H2O + 2e-
Cathode reaction: 2H+ + 2e- H2 If the solution has chloride ion then chlorine formation is another competing reaction which can take place at the anode. 2Cl- Cl2 + 2 e-
Standard potential for reaction set 1 is 0.401 V while for reaction set 2 is 1.3595V using the values in Table 1. It may be noted that the extent of these reactions are small and only a small fraction of the water gets electrolyzed. Pretreatment of the solution is needed in some cases. For example suspended solids larger than 10nm in diameter need to be removed or else they will plug the membrane. Calcium salt precipitation over the electrode could also be a problem and these are prevented by keeping the solution slightly acidic. As a rule of thumb standard ion exchange is preferred if the salt concentration is in the range of 500ppm. For larger concentrations in the range of 500 to 5000ppm electodialysis is more economical. For even more concentrated solution, the reverse osmosis is the process of choice. Some advantages of the electrodialysis process are as follows: Some ionic dissolved substances which cannot be separated by conventional methods can be removed by this process. Unlike ion exchange those do not require a periodic regeneration step. Some disadvantages of the process are noted below. Membranes can be fouled leading to poor separation, e.g. due to organic contaminants. Multipass operation is usually needed to achieve a high removal of ionic contents. Electrical energy costs and hence the operating costs are high. Design of electrodialysis unit Current required for electrodialysis can be calculated using the Faraday’s law of electrolysis. If ∆N is the g-eq of ions migrated from one electrode to another the current is given as F times ∆N. If the concentration change is ∆C then the g-eq transferred is z times ∆C where z is the valence of the ions. Hence the current needed is given by:
I = z F Q ∆C / η where η is the current efficiency. The corresponding equation for the current infor a stack of cells is given as: I = = z F Q ∆C / η n where n is the number of cell pairs. Note that one cell pair represents one cation plus one anion exchange membranes. Current density is another important parameter which is needed for the design. It is defined as the current per unit area of membrane in the direction perpendicular to the direction of current flow. Current density to solution (CD/N) normality ratio is an important a parameter. Too high value a value for (CD/N) indicates that there is insufficient charge to carry the current. Too low indicates a poor rate of membrane transport. A value of 50 to 500 A/m2 is used in design and once this is set the membrane area is fixed. The phenomena of concentration polarization becomes important at large current densities. Membrane area = current / current density The resistance of the electrodialysis unit to treat a particular type of waste water is usally determined experimentally since it depends on the composition of the wastewater and the type of membrane used. Once the resistance R and the current is known the power consumption can be calculated as Voltage applied V = current times resistance Hence Power P = V x I = R x I2
The voltage is actually determined by the various potential drops in the system similar to that for an electrolyzer. These terms are listed below
1. Equilibrium potenital for the cathode and anode reactions. 2. overpotential at the cathode and anode 3. voltage needed to overcome the ohmic resistance of the electrolytes in each
compartment. 4. voltage needed to overcome the resistance of the membrane. 5. voltage needed to overcome the concentration polarization caused by the mass
transfer resistance at each membrane surface. For a given current all these terms can, in principle, be calculated and the overall potential drop in the system can be calculated. The contribution of the first two terms above is small and usually the last three terms are mainly responsible for the voltagae
drop. However, for a quick design, the measured value of resistance is used in order to estimate the power needed. Often no detailed voltage balance is performed. These simple calculations are useful for a first level analysis and the procedure is illustrated by the following example. Example 4: 4000 m3 per day of water containing 2500mg/L of NaCl is to be treated in an electrodialysis unit consisting of 240 cell pairs to reduce the salt concentration by 50%. The current efficiency is taken as 90%. Assume a voltage drop of 1V per stack and a design current density of 400A/m2. Determine the power needed and the membrane surface area to be provided. Solution: Q = 4000/24/3600 = 0.0463 m3 /s Cin= 2500/58.5 mg- mol/L = 42.735 g-mol/m3
From mass balance we first determine the g-eq of charge transferred. Z= 1 for NaCl. ; ∆C = Cin * 0.5 for 50% conversion = 21.368 g-mol/m3
N = number of cell pairs = 240. The current needed is then calculated as: I = z F Q ∆C / η n = 442 A From the recommended current density value of 400, the needed membrane surface area is: Area = 442./400 = 1.105m2 area needed. Total voltage = 1V x 240 = 240V The power needed is therefore 240 x 442 W = 106.08 kW It is also instructive to calculate the fraction of the water electrolyzed. Using Faraday’s law one mole of H2O is equal to 2g-eq of H+ and needed charge of 96520 * 2 = 193040.C/g-mol of water. Hence for a current of 442A or 442C/s we will electrolyze 442/193040 = 0.0023 g mol/s of water. This is equal to 3.5609kg/day of water compared to feed of 4E+06 kg/day. Hence the quantity of water electrolyzed is very small.
References 1 K. Rajeshwar.Environmental Electochemistry’ 2. Goodridge and Scott. Electrochemical process engineering. 3. Hine. Electrode processes and Electrochemical engineering. 4. Seader and Henley. Separation process principles. 5. Newman. J and Thomas_Alyea K. E. Electrochemical systems 6. Allen, D, T. and Rosselot, K. S., Pollution prevention for chemical processes.