. What s in a slutin? Hw far des a reactin g? 2. What factrs influence hw far a reactin ges and hw fast it gets there? 3. Hw d atmic and mlecular structure influence bserved prperties f substances? A Macrscpic Cmparisn f Gases, Liquids, and Slids (Review frm Ch. ) State Shape and Vlume Cmpressibility Ability t Flw Gas Cnfrms t shape and vlume f cntainer High High Liquid Cnfrms t shape f cntainer; Very lw Mderate vlume limited by surface Slid Maintains its wn shape and vlume Almst nne Almst nne Energies f Phase Changes (Sme review frm Chapter 6, Sme New) heat f fusin ( H FUS ): enthalpy change fr the melting f mle f a substance e.g., H2O(s) H2O(l) H = HFUS = +6.02 kj heat f vaprizatin ( H VAP ): enthalpy change fr the vaprizatin f mle f a substance e.g., H2O(l) H2O(g) H = HVAP = +44.0 kj At a phase change, all added/remved energy ges int the phase change and nne ges int changing the temperature. Thus, any measurement f the heat invlved when passing thrugh a phase change has t be brken int steps. (Als recall frm Chapter 6: Hess s Law, Specific Heat, q=mcδt) e.g., Cnsider cnverting.5 ml f ice at 25 C t liquid water at 35 C. What quantity f heat is required by this prcess in J? The mlar heat capacity f ice is 37.6 J ml C. The mlar heat capacity f liquid water is 75.4 J ml C. The heat f fusin f water is 6.02 kj ml. 37.6 J ml C q = (.50 ml) 0 C ( 25 C) 6.02 kj 000 J + (.50 ml) ml kj 75.4 J + (.50 ml) (35 C C) ml C Heat f fusin and heat f vaprizatin are bth ALWAYS endthermic. Fr a substance, the heat f vaprizatin is always larger than the heat f fusin. Big Questin 2 350:53-002/80
Equilibrium Nature f Phase Changes Figure 2.5 Liquid-vapr equilibrium Effects f Temperature and Intermlecular Frces n Vapr Pressure Figure 2.6: Effect f temperature n mlecular speeds Figure 2.7 Vapr pressure vs. temperature and intermlecular frces Phase Diagrams (belw: ~Figures 2.0, 2.) Big Questin 2 2 350:53-002/80
Intermlecular Frces (Big Questin #3 Alert!) Big Questin 2 3 350:53-002/80
Figure 2.5 Biling pints f Perid 2, 3, 4, and 5 hydrides Figure 5.34 H-bnds in DNA Table 2.7 Mlar Mass & Biling Pint Figure 2.8 Dispersin frces & mlecular shape Big Questin 2 4 350:53-002/80
Slutin Prcess Slutin frmatin generally depends n relative strengths f intermlecular frces. A slutin frms if the driving frces can cmbine t give a favrable energy change. We can mdel the slutin prcess as a 3-step prcess and use Hess s Law t determine the verall enthalpy f the slutin prcess ( HSOLN):. Separatin f slute particles ( Hslute endthermic) 2. Separatin f slvent particles ( Hslvent endthermic) 3. Mixture f slute and slvent particles ( Hmix exthermic) H SOLN = H slute + H slvent + H mix Figure 3.4 Case #. Slutin frmatin & the rle f enthalpy (especially imprtant fr inic slids in water) Figure 3.5 Big Questin 2 5 350:53-002/80
Case #2. Slutin frmatin and the rle f entrpy (imprtant fr mixtures f cvalent cpds.) Figure 3.6B Many mixtures f like nnplar cmpunds frm despite HSOLN being nearly zer. Entrpy drives the frmatin f such slutins. Case #3. N slutin frmatin Figure 3.9 slubility f inic cpds. vs. temp. Inic cmpunds d nt disslve in nnplar slvents. Sme inic cmpunds are insluble in water. Bth situatins result frm Hslute being much, much larger than Hmix such that HSOLN is far t endthermic t allw the prcess t ccur. Figure 3.6A The mral f the stry a rule f thumb: LIKE disslves LIKE Factrs that Affect Slubility. Structure Effects r Vitamin C Vitamin A Big Questin 2 6 350:53-002/80
2. Pressure Effects Sgas = kh Pgas Henry s Law gases are mre sluble in liquids at higher pressure 3. Temperature Effects Generally, increasing temperature increases the rate f disslving, but has varius effects n the amunt f slute that will disslve. Gases: mre sluble in liquids at lwer temperature Slids: depends n HSOLN Nn-Liquid Slutins Gas-gas e.g., air all gases are infinitely sluble in ther gases. Slid-slid e.g., allys, waxes substitutinal ally vs. interstitial ally Beynd Slutins: Cllids & Suspensins Suspensin Cllid Slutin hetergeneus mixture Dispersed substance larger than hmgeneus mixture settles ut eventually simple mlecules, but small enugh t nt settle ut (sizes 000 nm, des nt separate suspended particles includes the range f visible visually unifrm visible t naked eye wavelengths) cllids scatter light beams Big Questin 2 7 350:53-002/80
Chapter 6 Chemical Kinetics the study f rates f chemical reactins. Hw fast is a reactin? The kinetics f a reactin must be determined experimentally. in general, reactin rate = change in cncentratin change in time rate = Δ Δt = d d t The differential frm f the rate shws that the rate = the slpe f the tangent line t the curve in a graph f cncentratin vs. time. (Rates are always described as psitive numbers.) Δ(H2O) S, fr 2H2O2 2H2O + O2, rate f appearance f H2O = Δt Δ(O 2) Δ(H2O 2) rate f appearance f O2 = rate f disappearance f H2O2 = Δt Δt Big Questin 2 8 350:53-002/80
The rate f reactin changes during the curse f the reactin. Lking at the H2O2 data: Δ(H2O 2) rate = Δt Δ(H2O 2) rate2 = Δt Δ(H2O 2) rate3 = Δt Δ(H2O 2) rate4 = Δt In general, fr aa + bb cc + dd, Δ a Δt = Δ(B) b Δt 0.500 ml L = 4 2.60 s 0.250 ml L = 4 2.60 s 0.25 ml L = 4 2.60 s 0.0625 ml L = = Δ(C) c Δt 4 2.60 s = Δ(D) d Δt = 2.3 0 5 ml L s =.6 0 5 ml L s = 0.58 0 5 ml L s = 0.29 0 5 ml L s relatinships between/amng rates in terms f different reactants and prducts Δ(H2O 2) e.g., hydrgen perxide disappears at twice the rate at which xygen appears: Δt Δ(O ) 2 Δt = 2 Differential Rate Laws A general relatinship between species cncentratin and reactin rate is called a rate law. Assumptins:. nly frward reactins (i.e., write rate laws in terms f reactants) 2. equilibrium avided We bserved that the rate f reactin depended n (H2O2), but hw exactly? rate (H2O2) n n = rder f reactin rate = k (H2O2) n k = rate cnstant Δ(H2O 2) rate = = k (H2O2) n Δt a differential rate law: rate as a functin f cncentratin CAUTIONS!. n nt necessarily equal t stichimetric cefficient. 2. n and k are experimentally determined. 3. rates vary fr each prduct/reactant. Must specify. e.g., Cnsider this reactin: 2ClO2(aq) + 2OH (aq) ClO3 (aq) + ClO2 (aq) + H2O(l) and the fllwing initial rate data: (ClO 2 ) /ml L (OH ) /ml L initial rate/ml L s 0.0500 0.00 5.77 0 2 0.00 0.00 2.32 0 0.00 0.050.5 0 Determine the rate law fr the reactin and the value f k (with units). Big Questin 2 9 350:53-002/80
Integrated Rate Laws Differential rate laws give infrmatin abut rate as a functin f cncentratin, but what abut cncentratins as a functin f time? We need t integrate the differential rate law. The calculus invlved fr rders 0,, and 2 is belw it s nly necessary t use the final results and nt wrry abut the derivatin. Zer-Order d = k dt Rate = t = k0 = k d r Rate = = k dt d = k dt First-Order Rate = t = k r d Rate = = k dt d = k dt Secnd-Order Rate = t = k2 r d Rate = = k 2 dt d = k dt 2 d = k dt 2 d = k 2 t d t 0 d = k t d t = k(t 0) = kt 0 Half-Life let = 2 at t = t ½ 2 = kt ½ + 2 = kt ½ 2k = t ½ = kt + an integrated rate law: cncentratin as a functin f time d d = k dt = k t d t ln ln = k(t 0) ln ln = kt ln = kt + ln + = kt + 0 = k(t 0) = kt = kt Half-Life let = 2 at t = t ½ ln 2 = kt ½ + ln ln ln 2 = kt ½ +ln Half-Life let = 2 at t = t ½ = kt ½ + 2 2 = kt ½ + ln 2 = kt ½ ln 2 = kt ½ ln 2 k = t ½ Big Questin 2 0 350:53-002/80 = kt ½ = t ½ k
Summary f Rate Laws Order Differential Rate Law Integrated Rate Law (straight-line frm) 0 d Rate = = k dt = kt + d Rate = = k dt ln = kt + ln 2 d Rate = = k 2 dt = kt + Using the Integrated Rate Laws Example: What are the rder and rate cnstant at 25 C fr the decmpsitin f N2O5? 2N2O5(g) 4NO2(g) + O2(g) Experimental methd: place 0.000 ml f N2O5 in a.000 L flask and measure (N2O5) as a functin f time. Experimental data: time/s (N 2 O 5 )/ml L 0 0.000 50 0.0707 00 0.0500 200 0.0250 300 0.025 400 0.00625 Apprach: Make a guess at the reactin rder, n. Plt the experimental data using the crrespnding integrated rate law. The integrated rate law that gives a straight line will give the values f n and k. Integrated rate law plts and reactin rders. Half-Life, t ½ the amunt f time required fr the reactant cncentratin t decrease t half its initial value i.e., the amunt f time t reach = 2 Yu can derive the frmulas fr half-life frm the integrated rate laws; see abve. Only the first-rder half-life frmula is independent f the initial cncentratin. Big Questin 2 350:53-002/80
Cllisin Mdel f Chemical Kinetics Experimentally, reactin rate is affected by:. Cncentratin 2. Temperature 3. Orientatin Figure 6.6 Figure 6.8 Activated Cmplex r Transitin State : unstable species present as reactant bnds break and prduct bnds frm. can t generally be islated. (Belw: Fig. 6.20) activatin energy, Ea: energy required t vercme the barrier t reactin. Big Questin 2 2 350:53-002/80
Evaluating E a a a 889, Svante Arrhenius k = Zρ e RT = A e RT E RT ln k = ln Ae a E a E RT a = ln A + ln e = ln A RT Ea ln k = R T + ln A E E Z: frequency factr ρ: steric factr (rati between the experimental value f the rate cnstant and the ne predicted by cllisin thery) Determine k at varius temperatures; plt ln k vs. T. Or measure k at tw temperatures (T and T2): k ln k 2 Ea = R T T2 CAUTIONS! need R = 8.345 J ml K T must be in K Ea will be calculated in J, nt kj Catalysts increase rate f reactin are nt cnsumed verall by reactin (disappear, but later reappear) prvide alternative reactin pathways with lwer activatin energies hmgeneus catalyst: in same phase as reactants (e.g., all aqueus r all gaseus) hetergeneus catalyst: in different phase frm reactants (e.g., slid surface fr gases t react n) Big Questin 2 3 350:53-002/80
Reactin Mechanisms reactin mechanism: sequence f individual steps by which a reactin prceeds. The slwest step gverns the verall rate. Mechanistic steps MUST add t verall reactin. be physically reasnable. (Fr example, termlecular steps are unreasnable.) crrelate with bserved rate law. e.g., NO2 + CO CO2 + NO bserved rate law: rate = k(no2) 2 e.g., 2 NO + H2 N2O + H2O bserved rate law: rate = k(no) 2 (H2) Prpsed mechanism: 2 NO N2O2 fast N2O2 + H2 N2O + H2O slw The slw secnd step causes sme N2O2 t build up waiting t react. The first step reaches equilibrium. (N2O 2) K =, s (N2O2) = K(NO) 2 2 (NO) The slw step gverns the verall rate f the reactin: rate = k(n2o2)(h2) But N2O2 can nt be included in the rate law since it is nt a reactant.. Big Questin 2 4 350:53-002/80
350:54 Experiment 4 (Chemical Kinetics) 6 I (aq) + BrO3 (aq) + 6 H3O + (aq) 3 I2 (aq) + Br (aq) + 9 H2O(l) [BrO rate = rate f lss f BrO3 3 ] = t rate = 3 (rate f frmatin f I2) = + Δ[I 2] 3 Δt rate = k [I ] m [BrO3 ] n [H3O + ] p The three pairings can be accmplished in a ttal f fur experiments at rm temperature, as cvered in the Prcedure. Experiments 3 and 4 in Table 2 in the Prcedure fulfill the requirements f case a: [I ]3 [I ]4 but [BrO3 ]3 = [BrO3 ]4 and [H3O + ]3 = [H3O + ]4 Given these cncentratin values, we can use equatin 4 t write the initial rates f reactin in the tw experiments: rate3 = k ([I ]3) m ([BrO3 ]4) n ([H3O + ]4) p rate4 = k ([I ]4) m ([BrO3 ]4) n ([H3O + ]4) p The rati f the rates is m [I ] 3 3 3 m 4 [I ] 4 4 rate [I ] rate [I ] m In equatin 5, rate3, rate4, [I ]3 and [I ]4 are numbers that yu measure experimentally. Only the value f the expnent m is unknwn. Slve fr m using the laws f lgarithms: rate 3 [I ] 3 ln mln rate 4 [I ] 4 ln rate3 rate4 m ln [I ] [I ] 3 4 Table 2. Cmpnents f Slutins A and B fr Rm Temperature Experiments Exp. Cmpnents f Slutin A (ml) Cmpnents f Slutin B (ml) H2O Na2S2O3 KI KBrO3 HCl 0 2.00 2.00 4.00 2.00 2 0 2.00 2.00 2.00 4.00 3 0 2.00 4.00 2.00 2.00 4 2.00 2.00 2.00 2.00 2.00 3H 0 2.00 4.00 2.00 2.00 3C 0 2.00 4.00 2.00 2.00 Big Questin 2 5 350:53-002/80
First Law f Thermdynamics Energy can be cnverted frm ne frm t anther, but can nt be created r destryed. ΔEUNIV = ΔESYS + ΔESURR = 0 Reactin Spntaneity spntaneus : ccurring withut utside interventin Entrpy, S Entrpy is a measure f the tendency f energy t spread ut, t diffuse, t becme less cncentrated a measure f the number f ways energy can be distributed amng the mtins f particles. a measure f a driving frce. nt necessarily cnserved. Entrpy is NOT a driving frce itself disrder r a measure f disrder applicable t macrscpic bjects Ludwig Bltzmann (late 800s): S = k ln W Secnd Law f Thermdynamics In any spntaneus prcess, the entrpy f the universe increases. qsys qsurr Experimentally, fr a reversible prcess, ΔSSYS = and ΔSSURR = at cnstant T T T Example: An irn skillet at 500 K is allwed t cl in the kitchen (300 K). The skillet transfers 500 J f heat t the kitchen. Why des this ccur spntaneusly? Big Questin 2 6 350:53-002/80
Alternate frm f Secnd Law: Energy always flws as heat frm ht bjects t cld nes. The Secnd Law allws predictins abut what shuld ccur, but says nthing abut when. Spntaneus prcesses can be very, very slw. Smething hlds back sme spntaneus prcesses; therwise all spntaneus prcesses wuld be instant. Obstructins t the Secnd Law make life pssible! ΔS and Spntaneity Recap ΔSUNIV + : spntaneus prcess ΔSUNIV : nnspntaneus prcess; reverse prcess spntaneus ΔSUNIV = 0 : prcess at equilibrium (n net tendency in either directin) Gibbs Free Energy, ΔG We wuld like t be able t predict spntaneity based n a system variable rather than a universe ne. ΔSUNIV = ΔSSYS + ΔSSURR (cnstant T) and ΔH = qsys (cnstant P) Heat is exchanged between the system and surrundings: qsurr And ΔSSURR = T Substituting: ΔSUNIV = ΔSSYS Multiplying by T: ΔHSYS = qsys = qsurr = ΔHSURR ΔH T = SURR ΔH T SYS ΔHSYS = T TΔSUNIV = TΔSSYS + ΔHSYS = ΔHSYS TΔSSYS ΔGSYS = ΔHSYS TΔSSYS ΔG = ΔH TΔS ΔSSURR in terms f system ΔSUNIV in terms f system TΔS: energy units ΔG Gibbs Free Energy ΔG and Spntaneity ΔSUNIV = ΔG T SYS ΔG : spntaneus prcess ΔG + : nnspntaneus prcess; reverse prcess spntaneus ΔG = 0 : prcess at equilibrium (n net tendency in either directin) Gibbs Free Energy is the amunt f energy that is free t d useful wrk in a spntaneus prcess. ΔG = wmax (in thery yu never get that much wrk in real prcesses) If ΔG, wrk that can be dne n surrundings by system If ΔG +, wrk that must be dne n system by surrundings t frce the nnspntaneus prcess t ccur. Big Questin 2 7 350:53-002/80
Many bdily prcesses are kept frm reaching equilibrium by bichemical cupling. A system at equilibrium can nt d any useful wrk. Equilibrium: pint f minimum free energy and maximum entrpy Third Law f Thermdynamics The entrpy f a perfect, crystalline substance at abslute zer is 0. Standard State cnditins cmpunds gas: pressure f bar (00 kpa, ~ atm) pure liquid/slid slutin: M cncentratin elements frm in which element exists at bar and T specified ΔHF and ΔGF f an element in its standard state 0/ ΔH, ΔS, T, and Spntaneity. ΔH, ΔS +: always spntaneus (ΔG always ) because bth ΔH and ΔS are favrable 2. ΔH +, ΔS : always nnspntaneus (ΔG always +) because bth ΔH and ΔS are unfavrable 3. ΔH, ΔS : sign f ΔG depends n T and magnitudes f ΔH and ΔS 4. ΔH +, ΔS +: sign f ΔG depends n T and magnitudes f ΔH and ΔS Big Questin 2 8 350:53-002/80
Fr reactins that change in spntaneity depending n T, we can calculate the crssver temperature : Example: Hg(l) Hg(g) ΔH = 60.83 kj ΔS = 97.49 J K ΔG, ΔG, and K ΔG = free energy change when reactants in standard states are cnverted t prducts in standard states Why bther with ΔG?. Can cmpare relative tendencies f reactins t ccur 2. Need it t calculate ΔG under nnstandard cnditins ΔG = ΔG + RT ln Q 3. Prvides infrmatin abut equilibrium psitin At equilibrium, ΔG = 0 and Q = K Example: What is ΔG fr the aut-inizatin f water? Big Questin 2 9 350:53-002/80
Chapter 2 Electrchemistry Review: Reductin-Oxidatin Reactins (See Chapter 4) Driving frce: transfer/shift f electrns LEO the lin says GER Lss f Electrns is Oxidatin; Gain f Electrns is Reductin OIL RIG Oxidatin Is Lss; Reductin Is Gain ELMO Electrn Lss Means Oxidatin The substance that is xidized the reducing agent. The substance that is reduced the xidizing agent. A reactin is a redx reactin if the xidatin numbers f ne r mre atms changes. e.g., 2Zn(s) + O2(g) 2ZnO(s) ON f Zn changes frm 0 t +2 ON f O changes frm 0 t 2 This is a redx reactin, and Zn was xidized and was the reducing agent, and O2 was reduced and was the xidizing agent. Electrchemistry the study f interchange f chemical and electrical energy invlves redx reactins Gal: t be able t d useful things (wrk) by generating an electric current. Much f the experimental evidence fr thermdynamic cncepts arse frm electrchemical experiments! Experimentally, this reactin is spntaneus (i.e., ΔG is negative): Zn(s) + Cu 2+ (aq) Cu(s) + Zn 2+ (aq) The prblem is that the electrn transfer ccurs at the interface between the substances. T d useful electrical wrk, we need t frce the electrns t travel thrugh a wire; i.e., the reductin and xidatin prcesses must be separated! Big Questin 2 20 350:53-002/80
We need t add an apparatus that will allw electrical neutrality t be maintained in bth beakers withut allwing the cntents f the beakers t mix. Salt bridge allws ins t flw, maintaining electrical neutrality cntains a strng electrlyte in a gelatinus matrix r a prus/absrbent substance saturated with a strng electrlytes slutin, with the ends clsed with a glass frit. is an external surce f ins. Prus disk prevents slutins frm mixing significantly but allws ins t crss, maintaining charge balance. is an internal surce f ins. vltaic (galvanic) cell: device in which chemical energy is cnverted t electrical energy electrdes: metallic cnductrs that make electrical cntact with cntents f half-cells. ande: electrde at which xidatin ccurs cathde: electrde at which reductin ccurs In the salt bridge: anins flw tward/int ande cmpartment catins flw tward/int cathde cmpartment Big Questin 2 2 350:53-002/80
Cell Ptential, E (E ) measured in vlts vlt = jule energy per culmb charge transferred [ V = J C ] E CELL = E OVERALL = E OXIDATION + E REDUCTION = E CATHODE E ANODE E OXIDATION = E ANODE E REDUCTION = E CATHODE e.g., the tw tabulated reductins Zn 2+ + 2 e Zn E RED = 0.76 V and Cu 2+ + 2 e Cu E RED = +0.34 V Reverse Zn half-reactin: Zn Zn 2+ + 2 e Add t Cu half-reactin: Cu 2+ + 2 e Cu E OX = +0.76 V (= E ANODE = ( 0.76 V)) E RED = +0.34 V (= E CATHODE) Fr Zn + Cu 2+ Cu + Zn 2+ : E CELL = +.0 V We need a reference t measure half-cell E s Standard Hydrgen Electrde (SHE) (really a half-cell) Assigned exactly zer vlts H2(g) 2H + (aq) + 2e OR 2H + (aq) + 2e H2(g) Cell Shrthand Ntatin ande species xidized, slid if different xidatin prduct species reduced reductin prduct cathde slid, if different e.g., Fr Zn(s) + Cu 2+ (aq) Cu(s) + Zn 2+ (aq): Zn(s) Zn 2+ (aq) Cu 2+ (aq) Cu(s) e.g., Fr 2H + (aq) + Fe(s) H2(g) + Fe 2+ (aq): Fe(s) Fe 2+ (aq) H + (aq) H 2 (g) Pt(s) E and Reactin Spntaneity r Predicting Redx Reactins Redx reactins that have a psitive verall E are spntaneus as written. G = nfe G = nfe n = mles e transferred n, F always psitive F = Faraday cnstant G, G negative 96,485 cul/ml e if E, E psitive Big Questin 2 22 350:53-002/80
Example: Can acid xidize irn? Nernst Equatin Cell ptentials at ther-than-standard cnditins ECELL = E CELL RT nf ln Q R = 8.34 J ml K T in Kelvin n = mles e transferred F = 96,485 cul ml Q = reactin qutient Applicatin: Cncentratin Cells Same half-reactins, but different cncentratins. Cell will run s as t equalize cncentratins. Big Questin 2 23 350:53-002/80
Applicatin: Measurement f Equilibrium Cnstants ECELL = E CELL RT nf ln Q At equilibrium, E = 0 and Q = K, s ln K = nf E RT A Cuple Selected Redx/Electrchemistry Applicatins Crrsin (ingredients: ande + cathde + electrlyte + cnductr) Sme metals frm durable xide catings that prevent further xidatin. Many d nt, hwever, like irn.. Crrsin Preventin. Galvanizing cat steel with a layer f zinc Fe Fe 2+ + 2 e E OX = +0.44 V Zn Zn 2+ + 2 e E OX = +0.76 V ~Fig. 2.22B 2. Allying add metals that frm a durable xide cating, such as Cr r Ni 3. Cathdic Prtectin cnnect irn t a metal with a mre psitive E OX with a wire. used fr buried tanks/pipes, piers, lck gates, ship hulls, bridge decks sme cathdic prtectin uses external current t frce the prtected steel t be the cathde. Fe Fe 2+ + 2 e Mg Mg 2+ + 2 e E OX = +0.44 V E OX = +2.37 V Big Questin 2 24 350:53-002/80 I-70 Indianaplis
The advantage f cathdic prtectin is that it can halt the prgress f crrsin withut the remval f chlride-cntaminated cncrete. Crrsin requires an ande, a pint n the reinfrcing steel where ins are released. Cathdic prtectin is the applicatin f direct current such that the steel becmes cathdic t artificial andes lcated n the deck. These andes usually cnsist f sheets f thin wire mesh. A relatively small DC rectifier perating n AC line vltage and a cntrl panel are nrmally lcated beneath the bridge. Cathdic prtectin systems d nt need t perate 24 hurs per day t be beneficial. Therefre, they can be pwered by slar panels r in line with highway lighting systems. Cathdic prtectin systems shuld be cnsidered fr lcatins where traffic maintenance csts are very high and where a few years f additinal service between repairs wuld be advantageus. Surce: Indiana Department f Transprtatin Electrlytic Cells Using electric current t drive nnspntaneus reactin e.g., plating Cr(s) n irn Fe Fe 2+ + 2 e Cr 3+ + 3 e Cr E OX = +0.44 V E RED = 0.74 V e.g., Hw many grams f slid chrmium culd be depsited by running 0 amps f current thrugh a Cr 3+ slutin fr 2 hurs? 2 hr 60 min 60 s 0 cul ml e ml Cr 52.00 g Cr 3 g Cr hr min s 96485 cul 3 ml e ml Cr Big Questin 2 25 350:53-002/80