Ch14-Note

Ch14-Note - Chapter 1-! Fundamentals of Eleelruchemistl'y...

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Unformatted text preview: Chapter 1-! Fundamentals of Eleelruchemistl'y page I Ch. 14 Fundamentals of Electrochemistry - Electrochemistry {oorrosion, galyanio and electrolytic cells} - Electrochemical deposition processes — electroplating electroless deposition [for adyanced integrated circuit interconnects and magnetic data storage thin-film heads] - Electrochemical energy systems (batteries, fuel cells, photovoltaios]. - Electrochemical analytical methods, e.g., potentiometryr and yoltammetry - Adyanced sensors, e.g., chem-FETs, chemfiETs - lylodern approaches, e.g., ultramicroelectrodes, modified electrodes, and scanning probe methods elec1rocr1errncal am" is Widely used lot of 1echniuues ane eleclrocnemucal Chapter l4 Fu ndamm tale of Eleclrnch :mlstry page 2 Electrochemistry Objectives *Ee familiar with the concept of oxidation state and the methods of assigning formal oxidation states to elements in a compound. *Understand the concept of oxidation and reduction, of half reactions and the use, derivation, manipulation and balancing of redox equations. *Be familiar with the operation of a galyanic {yoltaicjl cell and with the concepts of electrodes, salt bridges, half-cell reactions, net cell reaction and cell diagrams. *Eie familiar with the definition of the standard potential, E“, for a half reaction. -Describe the standard hydrogen electrode {SHE} and explain how its potential relates to other standard electrode potentials. Chapter l4 Fundamentals of E Il'fll‘lltl'l t-mistrj: page 3 Electrochemistry Objectives (cont) -Use tabulated E” values to predict the values of EN" for an oxidationl'reduction reaction and determine conditions under which certain species are liker to react. -Decide from E” and Egg” values whether or not a reaction will be spontaneous under a given set of conditions. 'Ouantitativelv and qualitativer predict the effect of varying conditions [concentrations and gas pressures] on values of Em”. *Know and be able to use equations that relate so, E”, Em”, and K. -Use Faradav’s laws to relate the quantity' of chemical change produced by a given amount of electric charge. -F’erform stoichiometric calculations based on oxidationireduction reactions. Flmptcr l4 Fundamentals ul‘ Eleclrnch :mlstry page 4 Basic Concepts I. Definitions Redox (reduction-oxidation} reactions involve the transfer of electrons from one species to another oxidation: removal of electrons reduction: addition of electrons oxidizing agent {oxidant} takes electrons {is is reduced} reducing agent treductant] gives electrons tie. is oxidized} Chapter 1-! Chaplcr 14 Fundamentals of Elt'clrucltumistl'y II. Balancing Redox Reactions Step 1. 1lllirite the unbalanced redox reaction. Step 2. Separate the reaction into its two half reactions (don't 1worry about electrons yet}. Step 3. Balance the atoms other than D and H in each half reaction Step 4. Add H20 to balance oxygens then add H' to balance hydrogens Step 5. Add electrons to one side of each half-reaction in order to make the net charge on both sides of the equation equal Step S. l‘itiultiolyr one or both half-reactions to equalize number of e' Step 'i’. Add the Mo half reactions together and cancel an other species Step S. Check. Verify that atoms and charge is the same on both sides Step El. If in basic solution, cancel out H+ using GH' See Appendix D F'u ndam an I als ol' Elecl ruch emisl r}- Redox reactions: Examples Example 1: HE‘S!2 + Fe?+ --‘:- Fe3+ + H20 {in acid} 2s + 2H' + HE'S“2 ———e SHED {reduction} 2(F32+ -__:r Fe3* + e") (oxidation) 2H+ + H202 + 2FeZ+ -—:- 2H2C} + 2Fe3* {total reaction} Example 2: Bi(DH)3 + Snflzz- --:=~ Snflf- + Si (in base] 2 {Se- + 3H‘ + BifDHh -> El + SHED } {reduction} SiHED + SnDEE' ---e San- + 2H‘ + 2e] {oxidation} sH+ + 2 etng + 3 H._,o + 3 SnCIf' m2» EBi + s Hgo + s sno32- + sH+ cancelling the H+ from both sides and 3 H20 yields: 2 EtiiOHh + 3 San‘ ---e 2Eii + 3 H20 + 3 San‘ (don't have to worry about the DH'} page 5 was it Chapter Id Fu ndamt-n tals cl‘ E lt-cl ruch urnistry Chemistry and Electricity Redox reactions work by the movement of electrons, which creates a current. We need to refresh our vocabulary and some basic concepts of electricity. Charge - electrical charge, q, is measured in coulombs {C}: charge of single electron is 1.602 x 10'” C one mole of electrons is 9.648 x 10“ C = Faradav's constant {F} q = n - F (ii = number of moles} Current - quantity of charge flowing each second through a circuit, measured in amperes {A}: 1 ampere = 1 coulombfsec page T Chapter I4 Fu ndarncnl als ul' Elm-J ruche mi slry Voltage, Work, and Free Energy Difference in electric potential between two points is measure of the work, W W: E as = r1 FE joules = volts ' coulombs The free energy change, ac. for a chemical reaction conducted reversiva at constant temperature and pressure equals the max. possible electrical work that can be done by the reaction on its surroundings: Hymn : The relation between the free energy difference and the electrical potential difference: .—. AG : -a FE Cine joule of energyr is gained or lost when one coulomb of charge is moved through a potential difference of one volt Past 3 Chapter I4 Fundamentals nl' Elutrnuhenrlstrgr page 9‘ Ohm’s Law Resistor Ohm's Law: E = ii? or i= E/R “39 fl The greater the 1lloltage, the more current will flow, the greater the resistance, the less current will flow. Power, P, is the work done per unit time: one Jls = one watt [W] P = El = iii? if the electrons involved in a radar reaction can be made to flow through an electric circuit, we can learn about the reaction by studyingI the circuit. Chapter II Fundamentals of Electron hemistrjr Electromotive Force (EMF) when} + Zn[s] —} Cu{s} + Zn2+{ao) Overall reaction is transfer of electrons from Zn to Guy. producing Cuts} and Zn“. Electron transfer corresponds to current flow; Requires a potential dlfi'erence Define E as the potential difference generated by a chemical reaction {electromotitie force, emf} Chap-tut 14 En ndammtnls ol‘Elcutruehcmlstry page 11 Potentiometer We can use electrochemical cells Anode (Oxidation) ' Oxidation: Collie} —} Cd Wan} + 26' ‘1 I Reduction: AQGHS) + e‘ —3» Ag[s) + «Cl—{ad} Duerall: 2 AgCI{S) + Cells} —} 2 Hats} + thtatfl + 2 CHEW} Chapter ]4 Fundamentals ofElcetroclIemistrJ-r Generally, we need a salt bridge Millivolts i3 " Potentiomemr Salt bridge KNUstfiq] Short Circuit! Cd{s) —> Cd2+lao) + 2e‘ 2Ag+{ao} + 2e‘—:— 2%{3} Overall. 2A9" + Cd(s} —> ZAgis} + Cid2+ Chapter l4 Fu ndamvntals ol' Elt-clrucht-mlstry Galvanic cells A galvanic cell consists of: 1. Two conductors {electrodes} immersed in electrolyte solutions - anode = electrode @ which oxidation occurs {Cd} 4 cathode = electrode @ which reduction occurs {Ag} 2. a salt bridge [to avoid mixing} - in this case KNOa 3. conduction of electricity from one electrolyte solution to another - occurs via migration of PC from the bridge in one direction and N0:- in the other 4. even though halt reactions are separated, a given up at Cd anode {to form Cd”) pass through the external circuit. reduce Aug" at Ag cathode 5. voltage measuring device [potential difference is developed by this cell, this potential difference is a measure of the driving force of the reaction) = potentiometer — two connectors are positive and negative {indicates direction of current} Ei. it the concentrations of reactants and products in the cell are far from equilibrium value, potential difference is very large i’. if concentration of reactants and products are at equilibrium, potential difference = D Chapter l-l Fu ndamvn tale of Elect rush vmistl'y Galvanic and electrolytic cells If concentrations are far from equilibrium, current flowing through circuit can be harvested - the principle of a battery Battery - electrochemical cell from which we receive energy galvanic or voltaic cell stores electricity — spontaneous chemical reaction to generate electricity We can also put a battery or power supply across cell and drive the electrochemical reaction — electrolytic cell — requires an external source of electrical energy — energy required because we force the reaction to proceed in the opposite {non— spontaneous] direction flmptcr 14 Fundsmmtsls ul‘ EIL-ctrnnhcmlstty Comparison . . . Voltaic Cell Electrolygjo Cell Potcntiom ctcr Salt bridge __ is It bridge l .. I “u ~—'}'Cflfi “4-Cli— 1.oo IvI zn{N03]2{3q] IDDI'vI Cutwflglgteg] HID rvi Zntucatztaqi 1-DUM CutNDaiatanl Anode = oxidation = negative Anode = oxidation = positive {negative charge supplied by reaction} {negative charge supplied externally} Cathode = reduction = positive Cathode = reduction = negative 'cell 3" 0 Egg" "5 U Chapler 14 F'u ndamcnlals of Elcclruch cmiailry page I15 ‘Ir'll'hytr does the sign change? Assigning the term “anode” and "cathode" is not based on the sign of the electrode, but on whether the chemical species is oxidized or reduced. Oxidation ALWAYS occurs at the anode {by definition]. 4 In a vottar'c cell. the anode liberates the electrons from solution and the electrode becomes negatively charged. : In an electrolytic cell. the electrons are withdrawn from the electrode, which therefore becomes positive. Reduction ALWAYS occurs at the cathode {by definition). * In a voltar'c cell. the cathode loses electrons to the solution and the electrode becomes positive. v In an etectrotyttc cell. the cathode has electrons pumped into it from the battery. and the electrode becomes negative. Anode - Cathode | atphabetr’cat | Oxidation ------ -- Reduction vowets consonants Fhsptcr l4 Fu misread tsls ol' Eleclrllch :mlstry Line Notation for electrochemical Cells Before we discuss half cells and the standard hydrogen electrode. we need to know this shorthand. Zn |ZnSO4{conc}| |CuSO4lconci | Cu 1. material anode is made of {anode is always on the left, by convention} . single line indicates phase boundary _ electrolyte contacting anode _ concentration of [3} . double line indicates salt bridge {separation between the 2 half cells] _ electrolyte contacting cathode _ concentration of {6} . single line indicates phase boundary _ material cathode is made of comhqmmhcom with anode on the left, cathode on the right] reduction is on the right Chaplcr 14 Fundamentals of Fleclruchemislry The standard hydrogen electrode — S.H.E. Since we can not measure a single "half—cell" potential, we must put two half cells together and measure the overall cell potential We must make sure that all half—cell potentials are measured relative to a common reference half cell — in general we use the standard hydrogen electrode - 5.H.E. [more below}. The 5.H.E. consists of 1. piece of Pt foil {coated with finely divided Pt} in contact with acidic solution 2. solution with [H"'] = 1, in which the Pt foil is immersed 3. stream of Hg with a partial pressure = 1 bar [= 1 atm) 4. half cell reaction is: Hgtg} ---:=- 2H+tao} + 2e Potential for this half cell reaction is defined E U at all temperatures “EL—g} (Jam—'1 J' '- [I bar =' ialIr-I '_- ' Pitsil Hztg.s=1iIH*taa.s=1i II Standard hydrogen electrode iS.H.E_, E”! G] page I T P3141: '3 Chapter 14 Fundammtnls ol' Elcclruth :mistry Given that SHE a D, we can now measure half cell potentials relative to SHE Pntcntiomctcr Suit bridge J ' F'ttsi | Hzifl'ufl =1} l Wiesel =11 N iii-913%?! =1} l fists} _Y—J Standard hydrogen electrode {S.H.E., a“: c} Chaplcr 14 F'u ndam en [sis of E'Icel ruch cmislry Conventions for the S.H.E. The S.H.E. is reversible and as such, can act as either an anode or a cathode. depending on the half cell with which it is coupled. So, when measuring the half-cell potential, you must specify.r whether the half cell is the cathode or the anode relative to S.H.E_ This is done by assigning a sign [+ or —. respectively) to the potential measured. This is arbitrarv, put the convention is: 1. Write all half-cell reactions as reductions ca2+ + an: —.~. Cd{s} 2n1+ + 2e' —> Znis] 2. When half cell is coupled to S.H.E. in an electrochemical cell, if the half—cell reaction proceeds as a reduction (i. e. as written}, half-cell is potential is positive. 3. When half cell is coupled to S.H.E., if the reaction proceeds as an oxidation, the potential is negative. Clmptcr 14 En ndammtals ul‘ ElL-ctrlrchcmlstty page 2| Half reactions, arranged by E” values {Table 14-1. tllrtlt-rt-tl t'L-rttn pult'ltli:ll~ fixidizing agent Reducing agent E9 [W [32ml + 2: :EF :3 2.8534] oytg} + 2H+ + 2.:- .=~ this) + HID E ems E :3 Mnoy + of + s:- .= Mt13+ + 4HIU E 1.5m . l1) = e sg+ + c‘ = Ago-l 5. was E {3112+ + s:- = Cum E {1.339 3 ~ '3 .-— ' fl a sac + s:- :Hztgl com a ; at ca2t + 2c :Cdm —o.4o2 IE 5 75 K' + e .——~ Hts: 4.9313 0 Li+ + e‘ : Lita} -3.U-=ltl For more, see Appendix H Chapter l4 Fundamentals of Elt-cl ruch t'mlfll'jf Standard Potentials What factors determine the magnitude of the cell potential? - The inherent driving force of reactions. All reactions move to minimize their free energy, so it a lot of energy is released when the reaction occurs. the potential will be large. This is expressed as the standard electrode potential. E“. These are conventionally listed as reduction reactions, so the potentials are generally called the standard reduction potential. [See Appendix H for a list} - Concentration. If concentration is high {SA a 1}, LeChatlier's principle would say the driving force is greater, therefore the potential is greater These concentration and E“ factors are taken into account using the Nernst Equation Chaplcr 14 Fundamflnlals ol' Eleclruchemislry page 23 The Nernst Equation Given a hypothetical half reaction: 0x + ne‘ —> Red Half-cell potential associated with this half cell is given by: s: so— E “1 [Rwl IKPIF/ lax where E” is the thermodynamic force for the reaction when the other half—cell reaction is SHE .9" NOTE: All of these E“ values (also known as the standard reduction potential} are based on writing the equation as a reduction Cl; + 2e" —} ECI’ E” = 1.35; veryr strong tendency to occur as a reduction (is. as written} Na+ + e‘ —> Hats] E“ = —2.?1; verv strong tendency to occurv as an oxidation (is. opposite direction as written) If we take the Nernst equation, convert it to base 1U. put in the actual values for R and F and assume we're at room temperature (293K). we get: rtossl‘h i" [Ralf .tt ; M [Ox] 2; F3313 E=E° —[ I: haprvr 1-! Fa ndam vn tale of E It'cl rush vmistry Calculation of Eng" To calculate Em" for a galvanic cell . . . - write both half reactions as reductions - calculate Efor the two half-cell reactions using Nernst equation - Set up the cell by arbitrarily assigning the left half cell as the anode, right half cell as the cathode {according to convention}. - Calculate Em”: tat— EEm - If Econ is positive, cell is set up correctly according to convention - If Ecell is negative. East and Ean assignments are backwards and must be reversed. Chaplcr 14 Chapter l-l Fu ndam an I als. ol' Elccl ruch cmisl rgi' Use of the Nernst Equation Example 1: ‘1. Calculate the half—cell potential for Zn electrode in 113'? M ZI'I{NU_.1}2 half reaction: 2n?“ + 2e —-—> Zn 5 = E“ — [Uflfifilni log {mam} E = sass -{s.ossi2}ing{1lo.ci} E = 41.32 V What does this mean ? The reaction proceeds as an oxidation. therefore En is the anode This is the cell potential we would get lfwe assembled the cell: S.H.E.|IZn[NDG}2{1U'3 MHEn Any half-cell potential calculated using the Nernst Equation is an E for a hypothetical cell which has the half cell of interest as the cathode and the S.H_E. as [he anode. However, the negative sign on our example indicates that if assembled, the En electrode would be the anode. Fu ndarn en ml: of E Iecl rush emisrry Use of the Nernst Equation Example 2: Consider half cell that is the S.H.E.1 but PIIHE} and [W] are different! {F'ng = {15 atrn and [H‘] = ELD‘l M} So, reaction is still: 2H‘ + 2e ms H2 E = E“ - {c.cssei log {Pngli'IHT} = c - {UflfiSl‘Z} log {carom a} E:-lZl_1lIlL'5l3‘|uIIr This is how you calculate the half—cell potential, but we'd also like to do this for the whole cell (is. when you actuallyr have two half cells connected to each other}. Chaplcr 1-1 Fundamcnlals ul' Elmlruchemislq page 2? Examples Example 1: Calculate E m" for the following : 2n|znso4n .c MHICUSD4H .c Milcu anode on left: Zn2+ + 2e' ——=v Ents} E = 43.?53 V E:cln = -Ei.?E3 - {BESSIE} log {{1}![Zn2*]} Ean = $163 cathode on right Cu?“ + 2e ' -—-:r Cuts]: E = (133? U Emt = 0.33? - {D.fl53!2} log {{1ti{‘i}} EnEII = i133? "i! E coil = E En =c.ss?-to.?ss}=1.1cu 1531— a So1 cell reaction is spontaneous as written1 the anode reaction is in fact the oxidation and the cathode reaction is in fact a reduction reaction Chapter l4 Fundamentals of Elt-clruch t-mlstr} So. what if the cell were drawn in this way: Co | 0115044110 M} H anCMU .d M} | En Cu is now anode. Zn is cathode — if you were to follow the same steps. you find that the Efor the cell would be negative MUTE: cell potentials calculated in this manner are: -positive for galvanic cell — spontaneous, no need for excess power sources -negatiue for electrolytic cell - not spontaneous, needs added power source DO A BUNCH MORE EXAMPLES FDR PEACHCE!!! Chapter 1-1 Chapter 14 CH3C02H $ mace; + H+taqj This galvanic cell can be used to measure the pH of the left half cell, determine the Kg 01’ the acid. or the Km ef AQCI F'u ndam en tale of Flttl ruch cmislry Nernst Equation For a half reactien: M5; +Jtc—t—b .513 RT E 233;” HP Reaction quotient. Ct: Q: For a eemplete reaction: Witt: 19‘ 'Willt Lht: putt-Illqu Druid:le E — E. 1:: reducti m1 Relation of E“ and Hm .-"'tl 25‘”: Finding E“ from K: =[Ufl5915]10gK H Finding Kfrern ES: K- : IUHEWUIIEEHE Fu ndamm tale {if E Ititlrtlth :mistrjt' Using cells as chemical probes .t-I--—-— - {fit-=1)” _ II In: '-=lrr.m| CI-I3t302Htflflfifl M} Aha.“ A bufi‘er. CHacOQNa {flflflfifl M} int] F reatttjtm 1«ti-"rillten tut a page 30 AgCItsl é Ag+taqi + Cl'taq} His} | H2{1_flfl bar} | 0H3002H{U.t15[i M], eH3e02H3{n_nnen M} H CI"{G.1D M] | AgeltsH Agtej Chapler 14 F'u ndamenlals ol' Eleelruehcmislry page 3| In this osll. Em" = E — E— = 0.503 ‘0' The cathode {EX} process is 0.90115} —3 Figs} + Cl" {0.10 M} For which it}. = 0.222 '0' The anode {5;} process is EH‘iaoH? M} + 2e —> H: {9031.00 bar} For which ECl = 0 ‘0' So, - - 0.05010 [PHzl r. = .222— _ 1 -| I — _ | n.“ {0 13059 5 D'EllC [U 2 09( [H+]2 Solving for [H*] = 1.0 N 10 a M and pH = 3.?4 And _ .. _ [CH3602'1[H*] _ [n.ppspms x10"] 3 - [CHSCOEH] _ [0-0500] = 1.0 x 10 5 Chapter III Fu ndamerl tale of EIeetruch umistn' page 31 Analyzing a eomplieated eell: measuring the formation constant (Kf) of Hg(EDTA)2‘ V + Potenlimneler ,-- f Hamil—E" Z.- {xflufl ) ||har‘:la1m|_ 5 all; bridge Solution prepared from 50.0 mL of 0.0100 M HgC|2 -" 40.0 mL of 00500 M EDTA 10.0 n1L of buffer, pH 0.00 .5____ _ . - Standard hydrogen electrode S.H.E. || ngeoTAf-(aq, pppspp M}, eomraqpmsu M) | ngr} 1il“l|'.‘l I'JIJI Mr mllon oomtand Mr mercury Edi-El eomplex ("hauler l4 Fundsmwtsh al‘ Elcclruch :mlstry page it First, sea mL u some M = sees mmole Hg“ 110.13 mL x 11050:] M = 2.0131 mmole EDTA Assume Kr is large. so HQIIEDTA}2_ forms e:.=.,=.~:eritia||3,.r quantitatively. Hg? + eons. —t Hgteoraf' aseo mmole Residual EDTA 1.50 mmole 'Iflt'lI mL of total solution. HgthDTAf' {1.0051313 M EDTA {1.0150 M What is H'th HQfEDTAJI' '? Egg" = E+ — 5—: E- = = D 50 Egg" = Eh _ Ft". E+.- HQ? "' 23- —3' HQH} E“ = [1.352 ‘e' Eiglélfcfirgiitnehhalnghthano cell standard stale fll'tl'll'y' e1 1 fl-“5915Iog( 12+ = cast it 2 [He 1 2+ _ 1%- [Hg 1 _ 2'4 x 10 M jualilicd to think EDT-fl con-Ioch forms quanlilal'rb'cltr And E. = {1.352 - New _ [HQCEDTME'] _ [HglEDTAF'] [0.00500] f- [Hsz'llEDTea'l- [He2+][EDTema][ev»]_ [2.4x1o"3][o.otso][2.ax1o'5] as = so.I x to“ 111(4— at pH 6.00 1.u'alue from Table 12-1 Assumption of Kr large quite good. 1|” Chapter l4 Fu ndalnun tale of E lt-slruch :mlstry Bioehemists use E‘" Whenever IrI+ appears in a redth reaction, E9 applies when AHEL its. pH=D. The formal potential for reduction at pI-I=TIr is called 5"“. E“ is found by writing the Nemst equation for the desired half reaction and grouping together all terms except the logarithm conleining the formal concentrations of the reactant and product. EC' is this combination of terms, evaluated at [it-{=1 ChBPh-T 1" Fu nflamcnlals ul‘ Elculrucl'l :mislry Pal-W 35 5" far variuus biulugical reflux cnuples 'T-HIIIE 14-2 Hh'lllln'liflll |lii[l'I1Ill-l1"- III hirlhmiml inn-nu“ Hem-mm h“ Wt If" 1h CI: + d“ ' + 4e EHE-fl H.314 HUME- Ftl'_ + r' Fr:' --'[I.'-"-"l +fl.'-'T| II + 21: r-ll +£|LSJ£ H1535 Cylmhmnm n :Fc'”! 1- c ' = ty'nxh mm: r: (Fr:l “.I - {III-“Ell +fl1‘lfl filly] + Ill' + 2c' = llllllilJ *{IJfiEIfi +1131 {TymhnmI-r: -r' HT" ] +- r: = L'_'|-'I|.1'|'ll'l1-II11'.' r [FL-2' I — +£1.34 lfi-flichlumplrnnlindnplrnnl + 1H' + 24: :Itduccd Ifi-dichlmphemimlnphem —r +{I.12 IHIj'dI'E‘I-I'IWWMIE * le' + 1.3' :mcnrt‘me + [1:11 *HJ'JH +H.H.'EH Fumunlr + 1H' + 1:" T‘surrinalr --*L|I.43-3 +{IflJ-I h‘lflfl'lflfll'll'.‘ blue + 2H' + It :r'rcduE-L‘d PlThdllEi +{|I.'_"'rJ-1"_ +flfll] filymylal: + ZII' + Eu' =glymlnt¢ — flflfilfl UHuJLm'EIthL' + 1H" + 2:" = maluln- -r-IIII.]3-EII --III.1I|}1 win-+311: + EII' 1'- 1:: =|.ifl.'l.':1ll.‘. +1124 —{I.1‘:1'[I Rih-nl'lauin + 2H' + 21: : rchH:1:-I.l ril'nf'lavin — —{I.1flfl Fin + 2H' + 2:: = FAD-H: -r --[l.1|"=|' {filllTflTl'l'lli'lll-E-Sja + Ill' + 2:: #1 gluwhinne-El! — 4L1": Sm'runlmT + 1c" .— Icumul'ranin: T' —'[|.13l-5 —{IJE‘J {quash + _H‘ + 1: .—- achuisn — -n.Ju NAD' 1- ll' 'r 1¢"=‘.'~i.-‘I.DEI {LIES -IIJ.J|’."EI HADP' + II' ' Ec'.-—-T-'.-'I.DFI! -- "11.314- Cystin: + lli' + 2-: =2cysltlm — -fl.J-’lfl Mam-mm: + 2H' + 2: =I-B-h3dnn1rl'ruu'nu: — -fl.J+l~fi Kat-thine * EJ-l' + 2:: =' hymxanthine + Hlfl — --fl..1"'| 21-I' + 2e- ;- IL ELIIH'I —-fl.4l4 Glumnutt + H" + 2.5 :“gluruss: + H13 — 4-4.1 sui- + I: + 2H' .—- sfli- + Hru ' — 4:454 230-“; - 2r + arr =53D§ + EEIJD — "+3.52"! ...
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Ch14-Note - Chapter 1-! Fundamentals of Eleelruchemistl'y...

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