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1、2-1 CHAPTER 2:FUNDAMENTAL RELATIONSHIPS 1.Physical units 2.Faradays law 3.Efficiencies 4.Ionic and electronic conduction 5.Ionic conductance,mobility,Kohlrauschs laws 6.Grotthuss transport 7.Strong&weak electrolytes(Arrhenius&Ostwald laws)8.Some additional fundamentals:Field,Potential,Electrical for
2、ce,Poissons&Laplaces equations 9.Movement of an ion in an electrical field 10.Waldens Laws 11.Debye-Huckel-Onsager theory for strong electrolytes 12.Non-aqueous electrolytes.13.Molten salts 14.Heterogeneous media:Maxwells and Bruggemanns models.Emulsions,gas evolving electrodes.15.Conductivity measu
3、rements 2-2 CHAPTER 2:FUNDAMENTAL RELATIONSHIPS Review of Physical Quantities MECHANICAL Force:f=m x a f=Dyne or Nt m=gr or Kg a=acceleration,cm/s2 or m/s2 Work(or energy):W=F x S W=erg;Joule;Watt s;KWH;Cal;BTU Power:;WJPWatt KW HPts 746 W=1 HP ELECTRICAL Charge:q cb F 1 Faraday=96,480 cb=105 cb Cur
4、rent:Current densitya Power=P=V x I Watt=Ampere x Volt Work or Energy=W=P t=V I T THERMAL Thermal energy:Q cal Heat flux a current density units:qCbIAmpts2222;IAmAAAiorororASFAcmcmdmftqCalqts222222100110.1 1;1000mAmAAcmAAASFcmcmmAdmdmft2-3 Faradays Laws:The quantitative treatment of electrochemical
5、processes was initiated by Faraday.In 1813,Michael Faraday,a 22 year old bookbinder apprentice joined Davy in his laboratory at the Royal Institute as an assistant.On the basis of experimental observations he formulated in 1834 two laws:The amount of primary product formed at an electrode by electro
6、lysis is directly proportional to the quantity of electricity passed.)(Itqw 2-1 The amounts of various primary products formed at an electrode by the same quantity of electricity are proportional to their relative molecular,or relative atomic masses divided by the change in their charge numbers for
7、the electrode process.wMqn 2-2 Faradays laws can combined as M I twF n 2-3 defining the proportionality constant F=Faradays number.Its numerical value is defined by F=L.e L is the Avogadro-Lochschmidt number and e the electron charge,i.e.1 F=1 Faraday=L x e=6.02 x 1023x 4.8 x 10-10(esu)/3 x 109(esu/
8、cb)=96,500 cb/gr equiv.=26.8 Ahr/gr equiv.w =gr I=Amp i=A/cm2 q=cb t=sec M=At.Wt.=gr/gr.Atom n=number of e transferred in electrode rxn Equiv/Mole 2-4 Faradays Laws Faradays Laws provides a relationship between the amount of charge passed to the amount of material reacted in the electrode reactionw
9、q=ItwMqn12Combine:MwItn1 MwItF nDefining Faradays constant:w-weight M Formula weightq chargen number of electrons I currenti current densityt timeh deposit thicknessF=96,480 cb=26.8 AHr(=L*e)L Avogadro No.(6.02 x1023atoms/mole)e electron charge(1.6x10-19cb/electron)1 MhitF nDeposit thickness:iNnFIon
10、ic Flux:M/sec-cm2 orgr-ion/sec-cm2 2-5 Example 1:Faradays Law:An electrolytic cell for depositing copper from copper sulfate solution is placed in series with another for deposition silver from silver nitrate solution.The cathode in this latter solution is initially 0.1 gr lighter than that in the c
11、opper sulfate solution.If electrolysis is carried out at a constant current of 0.1 ampere,when will the two cathodes be of the same weight?(Assume 100%efficiency.)2000001221:0.12:00.11211163.5460.19650021Half ReactionsCueCuAgeAgInitial WeightCathodegCathodegEquationgweight platedweight platedweight
12、platedmol emol Cug CuwAtCmol emol Cuwe 000211107.8680.196500111268sec21minight platedmol emol Cug CuwAtCmol emol Cut 2-6 2-7 2-8 EFFICIENCIES IN ELECTROCHEMICAL SYSTEMS Fraction of current or voltage which is useful Efficiency 1 or 100%Electrolyzer,Plating CellBattery Faradaic Coulombic Efficiency C
13、urrent Actual Product Wt charge:Actual Wt Theoretical Wt discharge:Theoretical Wt reactedTheoretical Product Wt Actual Wt reacted Voltaic Efficiency E0 charge:E0 Actual Voltage discharge:Terminal Voltage Applied Voltage E0 Current Efficiency 1:Competing reactions(in electrolyte&cell hardware)Product
14、 decomposition Product recombination Voltaic Efficiency 1:Ohmic overpotential(in cell,electrodes,and external circuitry Activation overpotential Mass transport limitation 2-9 Example:Efficiencies in Electrochemical cells:Assuming that an empty aluminum beer can weighs 20 g,calculate how many hours y
15、ou could operate a 100-watt light bulb at 110V for the same amount of electricity that is consumed in producing the aluminum for the beer can.The aluminum reduction proceeds according to:2Al2O3+3C 4Al+3CO2 The Faradaic efficiency is 90%.The theoretical Standard Potential(EMF)for the reaction is 1.8V
16、.However,a voltage of 4.64 is actually applied across the cell.(For simplicity,you may assume that the light bulb operates on D.C.)Determine:a)the voltage efficiency of the process b)the power efficiency of the process c)where you think the voltage is dissipated and what you could do about it d)how
17、many hours can you operate the light bulb?e)how expensive it is to refine the aluminum(assuming a low cost of electricity,5/kW hr).(The purpose of this problem is to realize the importance of irreversibilities in electrochemical systems,the physical implication of the magnitude of the Faraday number
18、,and the cost of electrochemical processes.)2-10 Solution to Example 1:0.901.84.641.8)38.8%4.64)0.900.38834.9%)sourcesof dissipation:ohmic,activation,mass transfer overpotentials 1ptsolutionFaradaicActualVoltageActualPowerFaradaicVoltageEMFVVVEMFVaVVbc30s:bring electrodes closer to minimize IR losse
19、s,try different electrodes(geometry,etc)different electrolyte,mixing,.):3132026.98d two possible solutionsi coulombic basedAleAlmol Almg Alg Al96500214603111000.90911021460323608765.60.909)21460321460314.643070.903600100bulbFaradaicol excbcbmol Almol eWbulb currentAAtotal chargecbtshrcurrentAii powe
20、r basedcbcbWshrVP tVW hrcbVs%3073100)$0.0565.6100$0.3281000)$0.053071000efficient light bulbW hrhrWei coulombic basedhrWW hrii power basedW hrW hr 2-11 Example:Faradays Law and Faradaic Efficiencies:-Hydrogen fuel cells have been suggested for automotive application.My current gasoline tank holds ab
21、out 18Gal giving me a range of about 250 miles.My car is about 200 HP,but I am willing to drop to 100 HP for the benefit of going electric.Please calculate the volume of a hydrogen tank that will give you a similar range.Assume two cases:a.the hydrogen is at 500 atm.b.the hydrogen is liquefied.(what
22、 is the pressure for this case?)Assume that the fuel cell is air breathing(i.e.,no need to carry oxygen).Assume that the fuel cell operates at 0.8 V(instead of the theoretical x Vwhat is x?)and 10%of the hydrogen crosses over and reacts on the oxygen cathode.You may need to calculate the energy cont
23、ent of gasoline assume that it is isopentane.Make any reasonable assumptions that you need.(The purpose of this problem is to use simple engineering estimations to critically analyze technical proposals).2-12 5122225122221.230.80.650.90.585/18250150856179.30393.5actualVFMkJmolFkJmolFkJmolFFEMFVVVmec
24、hanical electrical interfaceEnergy Estimategal of isopentanemilesHPC HOCOH OHof C HHof OHof COHof H 0241.835393.56241.83179.33239.183239.18586.520%379956250kJmolkJkJkJkJmolmolmolmolFkJmolOHUseful EnergyWith Gas EfficiencymolkJ useful energy goesmiles 22220.65 0.9 0.5850.2937995613101930.294965001131
25、01931.23210005326)50053260.082TVFMFuel Cell Energy neededkJkJ neededMoles H neededmol eCkJkJVX mol Hmol Hmol eJXmol HaatmmolnRTVP 296298260.568.8500)0.0710215326149.939.60.07101000L atmmol KgmLKLgalatmb liquid HgmLLmolLgalmolgmL 2-13 Conduction Metals:lRresistivitycmA Typical metallic resistivity at
26、 20C:Ag 1.6 x 10-6.cm Cu 1.7 Al 2.8 Ni 7.8 Fe 10 Hg 96 Carbon(fiber)4-5000 Graphite 1360 Electrolytes:=1/=-1cm-1 =S/cm =f(concentration,Z,vi,composition,T,.)0.1 1 S/cm (106 more resistive than metals!)Z C S cm2 eq.-1()jjZzzzvalence Implication of limited conductivity:Power dissipation(narrow gaps,lo
27、w current density)Non-uniform current distribution Structure of water C C 2-14 Ionic and Electronic Conduction:Current passage,whether it takes place in a metallic conductor or in electrolytic solution,involves the transport of charged species through a medium.The movement of these species is impede
28、d by collisions with the medium and is therefore associated with a certain amount of energy dissipation,expressed as a voltage drop.a)The electrochemical interface:electronic conducting phase-ionic conducting phase a.the electrode phase(electronic conducting phase):metal(10-6 cm),semiconductor b.the
29、 electrolyte phase(ionic conducting phase):ionic solution(10-6 cm),molten salt(1 cm),ionic conducting solid(5 cm,300C),a plasma b)Metals:In metallic conductors the only charge carriers are electrons in the conduction band and they interact through(a)lattice collisions and(b)collisions with other ele
30、ctrons not in the conduction band.The voltage drop is related to the current through Ohms law:VI or V=R I volts=Ohms*Amperes The resistance,R,of any electrical conductor,is proportional to its length,L,and inversely proportional to its cross sectional area,A:It-is therefore possible to write;is an i
31、ntrinsic property of the conductor and called the specific resistance or resistivity(cm).LRALRA2-15 Typical metallic resistivities at 20C:10-4 10-6 ohm cm Metal Resistivity(cm)Ag 1.6 Cu 1.7 Al 2.8 Ni 7.8 Fe 10 Hg 96 Carbon(Filament)4000 5000 Graphite 1360 Temperature coefficient 0C-1,is positive for
32、 metals(0.5%/0C)and negative for carbon(-0.05%/0C),and semiconductors.c)Electrolytes:conduction is by movement of positive and negative ions in the applied field.Because of the very much smaller number of charge carriers,as well as the large resistance offered to the motion of ions(reasonably closel
33、y approximated by Stokes resistance for the motion of spheres),the conductivity of electrolytes dissolved in water is very much lower than those of metals.Moderately concentrated solutions,in the range of 1-10 moles/liter have resistivities in the order of 1 l0cm at room temperature.The temperature
34、coefficient of resistivity of electrolytes is negative,and approximately equal and opposite in magnitude of the temperature coefficient of viscosity:2-3%/C.1T2-16 Typical values:1 N KC1:10 cm 1 N CuSO4:40 cm DI Water:106-108 cm(Megaohm water)In electrochemistry we are more concerned with conductance
35、 of a solution rather than its resistivity.So we define a new term Conductivity,such that:Thus higher the conductivity,lower is the resistivity and hence the material(or solution)is a better conductor.It should be noted that typically the conductivity of metals is over 106(a million)times larger tha
36、n the conductivity of electrolytes.That is why we can often(but not always)neglect the potential drop across metal wires in our calculations involving electrochemical systems.Importance of Conductivity:a.Power dissipation in electrochemical cells P=I2R=I2L/A Implications:1.Narrow gap between the ele
37、ctrodes problems:electrode shorting,mechanical strength,chemicals supply.2.Low current density(i=I/A),often below 100 mA/cm2 in industrial processes.3.Cooling may be required Illustration:Electrochemical machining is a typically high current operation,Assume that the cathode tool and the part are 1
38、cm apart and a current density of 100 Amp/cm2 is applied.The electrolyte is 1 M KC1(10 cm).Calculate the electrolyte flow rate required.P=12R=(100)210=105 Watt=2.5x104 cal/sec 1 Watt 1/4 cal/sec Rate of heating of electrolyte=25,000 C/sec 111/cmS cm2-17 High flow rate is required to remove this heat
39、.Assume inlet electrolyte temp.at 30C and outlet at 90C:T=90C 30 C=60C Current Distribution:This is very rarely a problem in metals in which the current is assumed to fill the entire conductor.Due to the much higher electrolyte resistance,the current will be distributed in electrochemical systems in
40、 such a manner that the voltage balance is maintained.The ohmic drop in the electrolyte is just one component of the many voltage drops associated with electrochemical cells,however often it is very important and most often leads to extremely non-uniform current distribution(the so-called primary di
41、stribution).We will later spend time in learning how to evaluate this current distribution in a variety of geometries.Here it will suffice to say that the problem of non-uniform current distribution is an important one and may lead to inefficient use of electrochemical cells and to serious damages(b
42、urned membrane or deposits).Determination of Conductivity:Early D.C.measurements of electrolyte conductivities indicated that Ohms law is not obeyed.This,of course,was an artifact due to non-reversible polarization at the electrodes.Nowadays,AC Wheatston bridges in the audio range(1-4 K Cycle)with c
43、apacitance compensation are used for simple and accurate conductivity determination.Conductivity is effected by:Concentration C Number of charges per ion z Ionic velocity v Temperature T Pressure,Magnetic and electric fields(to a much smaller extent)25,000400/0.4/24/min1 1 60PQGcc sL sLCTx x2-18 2-1
44、9 2-20 2-21 2-22 Law of Independent Ionic Mobilities Kohlrausch observed that the equivalent conductivity data at infinite dilution,0,exhibited some unusual properties.The difference between equivalent conductivities of different salts with common cation or anion was constant:salt 0 difference salt-
45、.0 differenceKCl 149.86 23.41KC1 149.86 4 9NaC1 126.45 XNO3144.96 KNO3 144.96 23.41 NaCl 126.45 4.9 NaNO3 121.55 NaNO3121.55 KJ 150.32 LiC1115.03 23.41 4.9 NaJ 126.91 LiNO3110.1 23.41KClNaClKNa 334.9KClNaNOClNO This led Kohlrausch to the conclusion that the individual contribution of each ion to the
46、 total conductivity of the solution(at infinite dilution)is specific to the ion,is additive,and is independent of the other ions in the solution.Therefore,we can write:000 where 00,are the equivalent ionic conductivities or limiting ionic conductivities.Using this technique,tables of limiting ionic
47、conductivities have been prepared,from which the conductivities of various salts can be obtained by simple addition.It turns out,and we will see later why,that there is a correlation between the solvated ionic radii and the limiting ionic conductivities.2-23 Table 2-1:Equivalent ionic conductivities
48、 and solvated ionic radii It should be recognized that these values are valid only for infinitely dilute solutions,and appropriate corrections should be introduced for concentrated solutions.For C 0(repulsion)If qi and qj different sign,f MHz)Zero current measurement(Wheatstone Bridge):At all times:332244xxI RI RI RI R At Balance(Galv=I=0)33244xRRLRRL Rx R2 L3(R3)L4(R4)Power Supply