GEN CHEM & QUAL ANALY
GEN CHEM & QUAL ANALY CHM 2046
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CHAPTER 20 ELECTRON TRANSFER AND ELECTROCHEMISTRY Section 201 Electrochemistry represents the interconversion of chemical energy and electrical energy Electrochemistry involves redox reactions because electrical energy flow of electrons results from loss of electrons oxidation at one place and gain of electrons reduction at another place Important questions 1 Can we use the energy released AG when a spontaneous reaction occurs to do work YES common eg car batteries barbecue etc 2 Can we cause a nonspontaneous reaction to occur by supplying energy YES We have already seen one way to accomplish 2 by supplying thermal energy raising the temperature if AS gt 0 Remember A6 AH TAS if AS gt 0 then TAS lt 0 and we can make a nonspontaneous reaction spontaneous by raising the temperature so that TAS term dominates But is there a convenient and general way capable of driving reactions in their nonspontaneous directions How about supplying electrical energy The answer is YES Electrochemistry brings together as a technique several topics we have already studied redox reactions equilibrium and thermodynamics Electrochemistry influences many areas of our lives from batteries car flashlights calculators watches and electric cars to electroplating 69 surgical instruments and many many others Definitions of Oxidation and Reduction Oxidation removal of one or more electrons Reduction addition of one or more electrons Oxidizing agent substance that is doing the oxidizing taking the electrons Reducing agent substance that is doing the reducing giving the electrons If something is oxidized its oxidation state ox state is increased If something is reduced its oxidation state decreases EXAMPLE Consider adding copper metal to a solution of silver nitrate AgNO3 the colorless solution turns blue and silver metal appears r Wilhh me l W l g ions The blue color ofthe solution is due to inasolulinn ufsiivernilme AgNDZ tosilv Wm r m r am er metal I5 oxidized lo upper ions Cuz Aftel several days Silver inns in solution Surface of topper wile FIGURE 201 The equation is Cus 2 Agaq gt Cu2aq 2 Ags For the reaction lefttoright Cus is the reducing agent giving electrons e39 to Agaq Agaq is the oxidizing agent taking e39 from Cus Cus is being oxidized iT is losing e39 ox sTaTe CuO increases To Cue Agaq is being reduced iT is gaining e39ox sTaTe Ag decreases To AgO For The reacTion righTTolefT Ags is The reducing agenT Cu2aq is The oxidizing agenT Ags is geTTing oxidized Cu2aq is geTTing reduced NoTe The oxidizing agenT is The subsTance being reduced The reducing agenT is The subsTance being oxidized Balancing Redox ReacTions We musT keep Track of The exchange of eecTrons among species in a redox reacTion The besT way is To consider The Two halfreacTionsquot occurring one describing The oxidaTion and The oTher The reducTion See Tebeook for some easier examples I will give a more difficulT one here EXAMPLE MnO439 aq 6204239 aq gt MnOZ s 603239 aq 1 WriTe The Two halfreacTions and decide ox sTaTe changes MnO439 aq gt MnOz 5 Mn 7 reduced To Mn 4 6204239 aq gt 603239 aq C 3 oxidized To C 4 2 Balance The aToms which change in oxidaTion sTaTe MnO439 aq gt MnOz s 6204239 aq gt 2 603239 aq 3 Add eecTrons To one side or oTher To balance ox sTaTe changes 3e39 MnO439 aq gt MnOZ s 6204239 aq gt 2603239 aq 2e39 A Balance the electrons gained and lost in the two halfreactions 6e39 2MnO439 aq gt 2Mn02 s 3 6204239 aq gt 6603239 aq 6 e39 01 Now add the half reactions electrons should not appear they MUST cancel 2 MnO439 aq 3 6204239 aq gt 2 MnOz s 6 603239 aq O Balance other atoms gexcept H and 0 if any that do not change in oxidation state None in this example l Now balance 0 by adding H20 as needed then balance H by adding H as needed ZHZO 2 MnO439aq 36204Z39aq gt 2 Mn02s 6 Cng39aq ZHZO 2MnO439aq 36204Z39aq gt 2 Mn02s 6 Cng39aq 4 H This is the FINAL BALANCED REDOX EQUATION 39f the reacton was done in acidic or netfra sou on 8 When reaction is in basic solution add as many OH39 to both sides as it takes to neutralize the acid H OHquot gt H20 Add 4 OH39 to each side giving 4OH39 ZHZO 2MnO439aq 36204Z39aq gt 2Mn02s 6Cng39aq 4 H20 and then cancel some waters OH39aq 2MnO439aq 36204Z39aq gt 2Mn02s 6Cng39aq 2H20l his is the FINAL BALANCED REDOX EQUATION in basic solution Example balance Cr207239 aq I39 aq gt Cr3 aq I2 5 in acidic soln SecTion 202 An Overview of Electrochemical Cells 1 A volTaic or galvanic cell uses a sponTaneous reacTion A6 lt 0 To generaTe elecTrical energy Thus reacTanTs have higher free energy Than producTs All baTTeries are volTaic cells SysTem does work on The surroundings 2 An elecTrolyTic cell uses elecTrical energy To drive a non sponTaneous reacTion A6 gt 0 Thus producTs have higher free energy Than reacTanTs eg meTal plaTing Surroundings do work on The sysTem VolTaic Cells Figure 201 see above Cu meTal in Ag soluTion This is a sponTaneous reacTion buT how can we use iT To do work Answer use This reacTion in a volTaic cell which separaTes The Two half reacTions and gives a flow of e39 beTween Them Figure 205 Voltmeter Cu anode Salt bridge contains NaNO3 Ag cathode Net reaction Cus 2 Agaq gt Cuzaq 2 Ags m Figure 205 The same redox reaction occurs but reagents are kept separate Cus 2Ag gt Cue 2Ag s I We find that the concentration of the Cue solution increases with time concentration of Ag solution decreases over time I We find mass of Cu rod decreases Silver metal is deposited on the silver rod Electrons flow through the external circuit and can do work eg can light a bulb run a heater etc Left compartment Anode Cus gt Cue 2 e39 Right compartment Cathode Ag equot gt Ag 5 Net reaction Cus 2Ag gt Cue 2Ag s Salt bridge serves to keep the solution neutral without the salt bridge the Cue solution would become more and more positively charged and the Ag solution would become more and more negatively charged and the reaction would stop uction occurs at the cathode red cat Notation for a Voltaic Cell shorthand way of describing cell Reaction Cu 5 2 Ag aq gt Cue aq 2 Ag 5 Cell notation Cu 5 I Cue aq II Ag aq I Ag 5 j k j Y Y anode cathode salt bridge 1 vertical bars I indicate a phase boundary 2 electrodes placed at far left and right If electrode not involved in halfreaction put it at end and show reagents in order they appear in halfreaction 69 2 I39 aq gt I2 s 2 e39 anode ie source of e39 ie e39 on right MnO439 aq 8 H aq 5 e39 gt an aq 4 H2O I cathode ie where e39 go and are used in a reduction ie e39 on left Electrodes are graphite platinum etc inert not involved in rxns graphite I I39 aq I I2 s II MnO439 aq H aq Mn2aq I graphite or Pt s I I39 aq I I2 s II MnO439 aq H aq Mn2aq I Pt s PROBLEM Cr bar dipping in CrN033 solution on one side of cell Ag bar in AgN03 solution on the other side and connected by salt bridge The Cr electrode is the negative one Draw diagram and give cell reaction and cell notation ANSWER Since Cr s I Cr3 aq is the negative electrode it is the anode Ag s I Ag aq is the positive electrode it is the M Cr s gt Cr3 aq 3 e39 anode source of e39 put on left of cell Ag aq e39 gt Ag s cathode where e39 are used put on right of cell Cr s 3 Ag aq gt Cr3 aq 3 Ag s and cell notation is Cr s I Cr3 aq II Ag aq I Ag s th and how long does a Voltaic Cell Work Why Because the rxn is spontaneous AG lt 0 How Long Until the reaction is no longer spontaneous ie until it reaches equilibrium A6 0 We say the cell is fully discharged flat and no more equots flow Section 203 Commercial Voltaic Cells read for yourSelves Section 204 Standard Electrochemical Potentials A reaction is spontaneous if AG lt 0 In a Voltaic cell this reaction drives the current equots from anode gt cathode In electrochemistry language we talk about the cell potential Ecen or electromotive force emf 396 the electrical potential difference or quotvoltagequot in everyday English between the anode and cathode units are volts can results from the spontaneous reaction related to A6 n No of electrons exchanged in balanced chemical reaction AG FEW F Faraday Constant 965 x104 JVmol e39 if AG lt 0 Ece gt 0 spontaneous Voltaic cell AG gt 0 Ece lt 0 nonspontaneous electrolytic cell A6 0 Ece 0 equilibrium Remember A6 wmaX AG and Ece tell us maximum work the cell can do Standard Cell Potentials Measured cell potential Eceu depends on concs At standard conditions 1 atm gas 1 M solquots Ecle Eocell E ceu standard cell potential at 25 C unless otherwise stated E ceu for ZnCu cell 110 V Zn5 Zn2aq1M Cu2aq1M Cu 5 How do we calculate E ceu We use thaIf cell for the two halfreactions HalfCell Electrode Potentials E halfce Each half of the cell has a halfcell potential that added together give the total Ece or E28 By convention halfcell potentials refer to halfreactions written as reductions Ile ox e39 gt red ox oxidized form red reduced form and this is the way they are listed in Tables such as Appendix M Reversing reaction direction changes the sign of the halfcell potential 50 for Zn 5 Cue aq gt an aq Cu s an aq 2 e39 gt Zn 5 E zn listed in App M like this Cue aq 2 e39 gt Cu 5 E cahode E CU listed like this In the cell the Zn halfreaction occurs in the opposite direction oxidation so we change the sign Zn 5 gt an aq 2 e39 E anode E 2n E ceu is the sum of the two halfcell rxns E0cell E0cathode annode EoCu EOZn Since the anode is always an oxid E anode E hahcrxn always Eocell Eocat annode NOTE the minus sign already takes into account the fact that the anode reaction occurs in the opposite direction from that listed in Appendix M so do NOT change the sign of numbers you take from Appendix M and plug into this equation Determining thafce with the Standard Hydrgen Electrode From above if we have tables of E hahcce values ie Appendix M we can calculate E ceu for any cell But where does Appendix M come from E ceu is related to A6 Remember 6 not known absolutely only differences AG Similarly E hahccen values cannot be known absolutely and are obtained by measurement of the difference with a reference electrode which we define as 000 V 39e ve FigHe is ve pov ev a difference between vav hafreac on and ve reference haf reac ovv Reference electrode standard hydrogen electrode SHE SHE Pt solid dipping in 1M aq strong acid through which 1atm H2 gas is bubbled This E hahcce is by convention given a value of 000 V 2 H aq 1M 2 e39 gt He g 1atm E mc 000 V So construct voltaic cell with SHE on one side and unknown on other gives unknown E hahcce since E ceu is known measured with a voltmeter E0cell Eocat Eano If SHE is the cathode E ceu 000 Eunk 39 E ceu Eunk If SHE is the anode E ceu E unk 000 39 E ceu E unk This way thalf cell for my halfcell can be obtained Calculations using thafcg Values Example 1 What is Eocell for Zn 5 I an aq II Cue aq I Cu 5 E ceu E caT E fmo 034 V 076 V 110 V NOTE use numbers straight from Appendix M Do not change sign E ceu 110 V Example 2 Consider The VolTaic cell driven by The overall reacTion Brg aq Zn 5 gt an aq 2 Br39 aq The E ce 183 V WhaT is E Br for The halfcell reacTion Brg aq 2 e39 gt 2 Br39 aq Look aT overall rxn Zn 5 gt an aq 2 e39 iT is The anode E0cell EocaT 39 Eoano Eoce Eogr Eozn V Eogr Eogr V E Br 183 V 076 V E Br 107 V Check E ce E ca E ano 107 V 076 V 183 V CorrecT Example 3 Consider The VolTaic cell driven by The overall reacTion Bra aq 2 V3 aq 2 H20 l gt 2 V02 aq 4 H aq 2 Br39 aq E ceu 139 V WhaT is E v for V02 gt V3 Look aT overall rxn Brg aq 2 equot gt 2 Br39 aq iT is The caThode E caT 107 V E ceu E caT E ano 139 V 107 V E anode 39 E anode 107 139 032 V E v 032 V RelaTive STrengIhs of Oxidizing and Reducing AgenTs E hahccen values represenT how easy or difficulT iT is To add elecTrons comparing E hahcceu gives us sTrengThs of oxidizingreducing agenTs Consider Cue aq 2 e39 gt Cu 5 E 034 V 2 H aq 2 equot gt H2 9 E 000 v an aq 2 e39 gt Zn 5 E 076 V The bigger more ve or less ve The E The easier is The reducTion The smaller more ve or less ve The E The harder is The reducTion Conversely The bigger more ve or less Ve The E The harder is The oxidaTion The smaller more ve or less ve The E The easier is The oxidaTion redLIcing sTrengTh Zn gt H2 gt CLI oxidizing sTrengTh Cu2 gt H39 gt Zn239 General Rule STrengTh of reducing agenT increases as E mimn becomes smaller down Table 201 or Appendix M STrengTh of oxidizing agenT increases as E hawxn becomes bigger LIp Table 201 and A p pend ix M lame 201 mmm magma v Mg 2 g 7 2F 1qu 7237 Hx lau2li laq2 2H0U 177 mm Sui am amp 5 H my 2 e 7 mm 9 2 man 165 Mum 39 5 Mllaq s a may a mom 131 Au WNi 3 e39 gt Aulsl 50 W 2 7gt m in 125 LnUy aqy14H aul5e ZCI aq7N0U 133 mm a H m 4 7 z mom H229 Elf29 2K an 108 N0112ql 4 Wm 3 g mom 4 2 my 095 an in mum 2e 7 u aq 2 DH in mm u my 7 2 e 39Hgll mm A i H e 7 m 15 mm monasng swim 2 e 392 Huh 3 D739 3 l e39 7 F anl g 077 Strength 0f 1 7 2 g 2 x W i 032 reducmg 2N0142 AON all 393 00 all 2 7 m H ems Svil laql 29 mm 393 0 my 5 H m 1 mi 2 7 ms 7 mm Pbtx 50 law 7 mm 39 F6a m a 7 2m pp 2 i mm 2 an q MW 3 Aim mam 2 a 7 mm ma I 30 K lzn e 77 m a mi 2 mm M Oxidizing agents Reducing agents Predicting Direction of Spontaneous Reactions E values are related to A6 we can use them to predict spontaneous direction in reactant or productfavored reactions ie small or large K 9 Cu 5 an aq Cu2aq Zn 5 j weaker weaker stronger stronger red ag ox ag ox ag red ag stronger redox agents win and push reaction towards the other side spontaneous direction is righttoleft ie K is small and the equilibrium lies far to the lefthandside ie reaction as written is reactant favored ie e39s grabbed by stronger oxid agent Cub converting it to Cu 5 ie electrons flow from the smaller E haham to the larger E hahcrxn M Negative numbers are smaller than positive numbers ie 8 is smaller than 2 This gives the NorthwestSouthwestquot rule described on page 968 Example 1 an aq 2 e39 gt Mn 5 E hahcm 118 V SnZ aq 2 e39 gt Sn 5 E hahcm 014 V Mn 5 SnZ aq LW5an aq Sn 5 K large Example 2 NOg39 aq 4 H aq 3 e39 gt NO g 2 H20 l E 096 V N2 g 5 H aq 4 e39 gt N2H5 aq E 023 V MnOZ s 4 H 2 e39 gt an aq 2 H20 l E 123 V Reducing strength N2H5 gt NO gt an Oxidizing strength MnOz gt NOg39 gt N2 Spontaneous Reactions ie productfavored 1 3 N2H5 aq 4 NOg39 aq H aq gt N2 g 4 NO 8 H20 I 2 2 NO g 3 MnOz s 4 H gt 2 NOg39 aq 3 an aq 2 H20 I 3 N2H5 aq 2 MnOz s 3 H gt NE g 2 an aq 4 H20 I E ceu 1 096 V 023 V 119 V E cen 2 123 V 096 V 027 V E cen 3 123 V 023 V 146 V An Application the Reaction of Metals with Acids Some metals eg Fe dissolve in acid eg HCI to give H2 g others eg Cu do not Why Consider M s 2 H aq gt M2 aq H2 g stronger stronger weaker weaker r ed agent ox agent ox agent r ed agent for39 the M s to react to give M2 aq and H2 g the r39xn has to be spontaneous Iefttor39ight Iar39ge K ie productfavored Halfreactions ar39e M2 aq 2 e39 gt M s E M 2 H aq 2 e39 gt He g E SHE For39 reaction to occur39 spontaneously the following must be true i M s must be a stronger reducing agent than H2 g ii H aq must be a stronger oxidizing agent than M2 aq the E haham for39 M E M must be smaller less positive or39 more negative than E haham for39 H E2 E M must be listed below H2 halfcell SHE in Table 201 and App M Co Mn Na Zn Mg AI etc dissolve in read with acid wher39eas Cu Hg Au Ag etc do not CHAPTER 19 ELECTRON TRANSFER AND ELECTROCHEMISTRY Electrochemistry represents the interconversion of chemical energy and electrical energy Electrochemistry involves redox reactions because electrical energy flow of electrons results from loss of electrons oxidation at one place and gain of electrons reduction at another place Important questions 1 Can we use the energy released AG when a spontaneous reaction occurs to do work YES common eg car batteries barbecue etc 2 Can we cause a nonspontaneous reaction to occur by supplying energy YES We have already seen one way to accomplish 2 by supplying thermal energy raising the temperature if AS gt 0 Remember A6 AH TAS 2 if AS gt 0 then TAS lt 0 and we can make a nonspontaneous reaction spontaneous by raising the temperature so that TAS term dominates But is there a convenient and general way capable of driving reactions in their nonspontaneous directions How about supplying electrical energy The answer is YES Electrochemistry brings together as a technique several topics we have already studied redox reactions equilibrium and thermodynamics Electrochemistry influences many areas of our lives from batteries car flashlights calculators watches and electric cars to electroplating 69 surgical instruments and many many others SECTION 191 REDOX REACTIONS REDUCTIONOXIDATION Oxidation removal of one or39 more electr39ons Reduction addition of one or39 more electrons Oxidizing agent substance that is doing the oxidizing taking the electrons Reducing agent substance that is doing the reducing giving the electrons If something is oxidized its oxidation state ox state is increased If something is reduced its oxidation state decr39eases Example Consider adding copper39 metal to a solution of silver39 nitrate AgN03 the color39less solution tur39ns blue and silver39 metal appear39s A clean piece of copper wire will be plaed Willi time the copper reduces Ag ions The blue color of the solution is due to in a solution of silver nitrate AgNOzt to silver metal crystals and the copper the presence of aqueous copperlll ions metal is oxidized to copper ions Cu 39 several days Silver ions in solution Surface of copper wire The equation is Cus 2 Agaq gt Cu2aq 2 Ags For39 the reaction lefttor39iqht Cus is the reducing agent giving electr39ons e39 to Agaq Agaq is the oxidizing agent taking e39 from Cus Cus is being oxidized it is losing e39 ox state CuO increases to Cue Agaq is being reduced it is gaining e39ox state Ag decreases to AgO For39 The r39eacTion r39ighTTolefT Ags is The r39educing agenT Cu2aq is The oxidizing agenT Ags is geTTing oxidized Cu2aq is geTTing r39educed NoTe The oxidizing agenT is The subsTance being reduced The r39educing agenT is The subsTance being oxidized Balancing Redox ReacTions We musT keep Track of The exchange of elecTr39ons among species in a r39edox r39eacTion The besT way is To consider39 The Two halfr39eacTionsquot occurring one descr39ibing The oxidaTion and The oTher39 The r39educTion See Tebeook for39 one example I will give anoTher39 difficulT one here Example MnO439 aq 6204239 aq gt MnOz s 603239 aq 1 Wr39iTe The Two halfr39eacTions and decide ox sTaTe changes MnO439 aq gt MnOZ s Mn7 r39educed To Mn 6204239 aq gt 603239 aq C3 oxidized To 64 2 Balance The aToms which change in oxidaTion sTaTe MnO439 aq gt MnOz s 6204239 aq gt 2 603239 aq 3 Add elecTr39ons To one side or39 oTher39 To balance ox sTaTe changes 3e39 MnO439 aq gt MnOz s 6204239 aq gt 2603239 aq 2e39 4 Balance The elecTr39ons gained and losT in The Two halfr39eacTions 6e39 2Mn0439 aq gt 2Mn02 s 3 6204239 aq gt 6603239 aq 6 e39 5 Now add the half reactions electrons should not appear they MUST cancel 2 MnO439 aq 3 6204239 aq gt 2 MnOz s 6 603239 aq 6 Balance other atoms except H and 0 if any that do not change in oxidation state None in this example 7 Now balance 0 by adding H20 as needed then balance H by adding H as needed ZHZO 2 MnO439aq 36204239aq gt 2 Mn02s 6 cog CK ZHZO 2MnO439aq 36204Z39aq gt 2 Mn02s 6 Cng39aq 4 H This is the FINAL BALANCED REDOX EQUATION 39f the reaction was done in acidic or netfra sou an 8 When reaction is in basic solution add as many OH to both sides as it takes to neutralize the acid H1 OH gt H20 Add 4 OH39 to each side giving 4OH39 ZHZO 2MnO439aq 36204239aq gt 2Mn02s 6Cng39aq 4 H20 and then cancel some waters 4OH39aq 2MnO439aq 36204Z39aq gt 2Mn02s 6Cng39aq 2H20l This is the FINAL BALANCED REDOX EQUATION in basic solution Example balance Cr207239 aq I39 aq gt Cr3 aq I2 5 in acidic soln Section 192 An Overview of Electrochemical Cells 1 A voltaic or galvanic cell uses a spontaneous reaction AG lt 0 to generate electrical energy Thus reactants have higher free energy than products All batteries are voltaic cells System does work on the surroundings 2 An electrolytic cell uses electrical energy to drive a non spontaneous reaction AG gt 0 Thus products have higher free energy than reactants eg metal plating Surroundings do work on the system GALVANIC OR VOLTAIC CELLS Question The previous figure see above showed Cu metal in a Ag soln this is a spontaneous rxn but how can we use it to do work Answer use This reacTion in a volTaic cell which separaTes The Two half reacTions and gives a flow of e39 beTween Them Figure 205 Figure 191 The same redox reacTion occurs buT reagenTs are kepT separaTe Cus 2Ag gt Cu2 2Ag s I We find ThaT The concenTraTion of The Cu2 soluTion increases wiTh Time concenTraTion of Ag soluTion decreases over Time I We find mass of Cu rod decreases Silver meTal is deposiTed on The silver rod ElecTrons flow Through The exTernal circuiT and can do work eg can lighT a bulb run a moTor or heaTer eTc LefT comparTmenT Anode Cus gt Cu2 2 e39 RighT comparTmenT CaThode Ag e39 gt Ag s NeT reacTion Cus 2Ag gt Cu2 2Ag s SalT bridge serves To keep The soluTion neuTral wiThouT The salT bridge The Cu2 soluTion would become more and more posiTively charged and The Ag soluTion would become more and more negaTively charged and The reacTion would very soon sTop uc on occurs aT The hode red caT Notation for a Voltaic Cell shorthand way of describing cell Reaction Cu 5 2 Ag aq gt Cue aq 2 Ag 5 Cell notation Cu 5 I Cue aq II Ag aq I Ag 5 J k j Y Y anode cathode salt bridge 1 vertical bars I indicate a phase boundary 2 electrodes placed at far left and right If electrode not involved in halfreaction put it at end and show reagents in order they appear in halfreaction 69 2 I39 aq gt I2 s 2 e39 anode source of e39 ie e39 on right of eq MnO439 aq 8 H aq 5 e39 gt an aq 4 H2O l cathode ie where e39 go and are used in a reduction ie e39 on left Electrodes are graphite platinum etc inert not involved in rxns graphite I I39 aq I I2 5 II MnO439 aq H aq Mn2aq I graphite or Pt 5 I I39 aq I I2 5 II MnO439 aq H aq Mn2aq I Pt 5 PROBLEM Cr bar dipping in CrN033 solution on one side of cell Ag bar in AgN03 solution on the other side and connected by salt bridge The Cr electrode is the negative one Draw diagram and give cell reaction and cell notation ANSWER Since Cr 5 I Cr3 aq is the negative electrode it is the anode Ag 5 Ag aq is the positive electrode it is the M Cr 5 gt Cr3 aq 3 e39 anode source of e39 put on left of cell Ag aq e39 gt Ag 5 cathode where e39 are used put on right of cell Cr 5 3 Ag aq gt Cr3 aq 3 Ag 5 and cell notation is Cr s I Cr3 aq II Ag aq I Ag 5 Why and how long does a Voltaic Cell Work Why Because the rxn is spontaneous AG lt 0 How Long Until the reaction is no longer spontaneous ie until it reaches equilibrium A6 0 We say the cell is fully discharged quotflatquot and no more equots flow Section 193 Standard Reduction Potentials A reaction is spontaneous if AG lt 0 In a Voltaic cell this reaction drives the current equots from anode gt cathode In electrochemistry language we talk about the cell potential Ecen or electromotive force emf ie the electrical potential difference or quotvoltagequot in everyday English between the anode and cathode units are volts EOE results from the spontaneous reaction related to A6 n No of electrons exchanged in balanced chemical reaction F Faraday Constant 965 x104 JVmol e39 A6 FEW if AG lt 0 Ecle gt 0 spontaneous Voltaic cell AG gt 0 Ecle lt 0 nonspontaneous electrolytic cell A6 0 Ecen 0 equilibrium Remember A6 wmaX AG and Ecle tell us maximum work the cell can do Standard Cell Potentials Measured cell potential Eceu depends on concs At standard conditions 1 atm gas 1 M concs E55 E ceu E ceu standard cell potential at 25 C unless otherwise stated Eocell for ZnCu cell 110 V M I Zn aq 1M II Cu aq 1M I Cu s How do we calculate E cen We use thalf cell for the two halfreactions HalfCell Electrode Potentials E halfce Each half of the cell has a halfcell potential that added together give the total Ecle or E28 By convention halfcell potentials refer to halfreactions written as reductions 39e ox e39 gt red ox oxidized form red reduced form and this is the way they are listed in tables such as Table 191 Reversing reaction direction changes the sign of the halfcell potential 50 for Zn 5 Cue aq gt an aq Cu s an aq 2 e39 gt Zn 5 E zn listed in Table 191 like this Cue aq 2 e39 gt Cu 5 E cahode E CU listed like this In The cell The Zn halfreacTion occurs in The opposiTe direcTion oxidaTion so we change The sign Zn 5 gt an aq 2 e39 E anode E zn E ceu is The sum of The Two halfcell rxns E0cell EocaThode annode EoCu EOZn Since The anode is always an oxid E anode E hahcrxn always o o o E cell E caT E anode IMPORTANT 1 The negaTive sign already Takes inTo accounT ThaT The anode rxn occurs in The opposiTe direcTion from ThaT in Table 191 so do M change The sign of numbers you Take from Table 191 To puT inTo The eqn 2 Do NOT mulTiply The E cammo by The coefficienTs in The balanced eqn JusT use Them as They are lisTed in Table 191 DeTermining thafcg wiTh The STandard Hydrgen ElecTrode From above if we have Tables of E hahcce values ie Table 191 we can calculaTe Eocell for any cell BuT where does Table 191 come from E ceu is relaTed To AG Remember 6 noT known absoluTely only differences AG Similarly E hahacen values cannoT be known absoluTely and are obTained by measuremenT of Their difference from a reference elecTrode which we define as 000 V 39e The Fig ee is The poTehTa difference beTweeh ThaT hafreacTon and The reference hafreacTon This is like quoTing a TemperaTure as The difference from The freezing poinT of waTer 0 C in The cenTigrade scale Reference elecTrode sTandard hydrogen elecTrode SHE SHE PT solid dipping in 1M aq sTrong acid Through which 1 aTm H2 gas is bubbled This E hahacen is by convenTion given a value of 000 V 2 2 Haq1M 2 e39 gt H2 g 1 aTm Eoref 000 V So construct voltaic cell with SHE on one side and unknown on other gives unknown E hahcce since E ceu is known measured with a voltmeter E0cell E0cat Eano If SHE is the cathode E ceu 000 Eunk E ceu Eunk If SHE is the anode E ceu E unk 000 E ceu E unk This way thalf cell for my halfcell can be obtained CALCULATIONS USING thafcg VALUES Example 1 What is Eocell for Zn 5 I an aq II Cue aq I Cu 5 E cen E caT E ano 034 V 076 V 110 V NOTE use numbers straight from Table 191 Do not change sign E ceu 110 V Example 2 Consider the Voltaic cell driven by the overall reaction Bra aq Zn 5 gt an aq 2 Br39 aq The E ceu 183 V What is E Br for the halfcell reaction Bra aq 2 e39 gt 2 Br39 aq Look at overall rxn Zn 5 gt an aq 2 e39 it is the anode E0cell E0cat 39 Eoano Eocen Eogr Eozn V Eogr Eogr V E Br 183 V 076 V E Br 107 V Check E cen E ca E ano 107 V 076 V 183 V Correct Example 3 Consider the Voltaic cell driven by the overall reaction Bra aq 2 V3 01 2 H20 0 2 VO2 aq 4 H aq 2 BF aq Eocell 139 V What is E V for V02 gt V3 Look at overall rxn Brg aq 2 e39 gt 2 Br39 aq it is the cathode E caT 107 V E ceu E ca E ano 139 V 107 V E anode 39 E anode 107 139 032 V E v 032 V Relative Strengths of Oxidizing and Reducing Agents E hahacen values represent how easy or difficult it is to add electrons comparing E hahaceu gives us strengths of oxidizingreducing agents Consider Cue aq 2 e39 gt Cu 5 E 034 V 2 H aq 2 e39 gt He g E 000 V an aq 2 e39 gt Zn s E 076 V The bigger more ve or less ve the E the easier is the reduction The smaller more ve or less ve the E the harder is the reduction Conversely The bigger more ve or less ve the E the harder is the oxidation The smaller more ve or less ve the E the easier is the oxidation reducing strength Zn gt H2 gt Cu oxidizing strength Cue gt H gt an General Rule Strength of reducing agent increases as E haham becomes smaller down Table 191 Strength of oxidizing agent increases as E haharxn becomes bigger up Table 191 increasing strength of oxidizing agents Increasing smngtli of oxidizing agaji Table 201 Standard Reduction Potentials in Aqueous Solution at 25 C Redudiun HalfReattion Equot V Mg 2 2 iv 2 F aq 287 Hoiaq2H aq2e 39ZHMi 177 Pb0s 4 so aq o 4 Mac 4 2e Plasma 2 H10 1685 Mun aq 8 H39aq 5 e v Mnquotaq 4 H700 151 AuWaq 3 e Ants 150 Iggy z Q 2 u aq 136 PO 0 14mm 69 2c aq7H0u 133 0g 4 H aq 4 e 7 2 man H229 3 2 e 2 3 sq 108 N0 1an 2 H qu 1 9 mom 4 z mow 096 on aq Hm 2 2 v i aq 2 DH an 089 Hgquotaqy Z Hum 0355 Aq 2q a i pig s 3 07 Hgmaq 2 e 2 H9 0739 Fequotaq 4 e 7 equotaq E 071 15 2 e 2 I iaq 0535 Mg 2 mom z e 4 0H 11 2 040 ul iaq 2 e Cum 2 033 Snquot aq 2 e Snquotaq g 015 H39mmu H1I 1 M in aq z e SnlsJ g 7014 lli aq 2 r Nils I 39 025 W39 an 9 v itquotaq 70255 PbSOJs 2 2 ms sn aq 0336 V dquotlaq 2 e cam o40 Fequotaq z e vFels 7044 Znquotaq 439 2239 7 Zns o753 2 H70 4 2 e 39 Hztg 2 DH aq 7032T7 M njaq 3 9 Als l56 Mcf iaq 2 e 7 Ngls 227 Na taq e v Nam 2714 K39mi a 7 Ks 2925 Li aq e 39 Lils 3045 L mum m ummm munlmi imimwulmuuh Oxidizing agents ed Reducing agents sanctions increasing strength of reducing agents E values are related to A6 we can use them to predict spontaneous direction in reactant or productfavored reactions ie small or large K Cu 5 weaker red ag weaker ox ag stronger ox ag 9 an aq 39gt Cu2aq Zn 5 9 stronger redag stronger redox agents win and push reaction towards the other side spontaneous direction is righttoleft ie K is very small and the equilibrium lies far to the lefthandside ie reaction as written is reactant favored ie e39s grabbed by stronger oxid agent Cub converting it to Cu s ie electrons flow from the smaller E haham to the larger E hahcrxn M Negative numbers are smaller than positive numbers ie 8 is smaller than 2 This gives the NWSE diagonal rulequot rule described on page 829 Example 1 S I39i 39aq2e gt Sn s E haham 014 V Mn2 aq 2 e svwresx Exam 118 v Mn 5 SnZ aq SP quot MnZ aq Sn 5 K large Example 2 NOg39 aq 4 H aq 3 e39 gt NO g 2 H20 l E 096 V N2 g 5 H aq 4 e39 gt N2H5 aq E 023 V MnOZ s 4 H 2 e39 gt an aq 2 H20 l E 123 V Reducing strength N2H5 gt NO gt an Oxidizing strength MnOz gt NOg39 gt N2 The Spontaneous Reactions ie productfavored are 1 3 N2H5 aq 4 N0339 aq H aq gt N2 g 4 NO 8 H20 l 2 2 NO g 3 MnOz s 4 H gt 2 NOg39 aq 3 an aq 2 H20 I 3 N2H5 aq 2 MnOz s 3 H gt N2 g 2 an aq 4 H20 l E ceu 1 096 V 023 V 119 V E cen 2 123 V 096 V 027 V E cen 3 123 V 023 V 146 V An Application The ReacTion of MeTals wiTh Acids Some meTals 69 Fe dissolve in acid 69 HCI To give H2 9 oThers 69 Cu do noT Why Consider M s 2 H aq gt M2 aq H2 9 sTronger sTronger weaker weaker red agenT ox agenT ox agenT red agenT for The M 5 To reacT To give M2 aq and H2 9 The rxn has To be sponTaneous lefT TorighT large K ie producT favored HalfreacTions are M2 aq 2 e39 gt M s E M 2 H aq 2 equot a H2 9 E SHE For reacTion To occur sponTaneously The following musT be True i M s musT be a sTronger reducing agenT Than H2 9 ii H aq musT be a sTronger oxidizing agenT Than M2 aq The thaIf rxn for M E M musT be smaller less posiTive or more negaTive Than E haham for H E2 E M musT be lisTed below H2 halfcell SHE in Table 191 Co Mn Na Zn Mg Al eTc dissolve in reacT wiTh acid whereas Cu Hg Au Ag eTc do noT GROUP 13 nsznpl Me rallic charac rer decreases moving righ r and we find rha r boron B is no r a me ral me ralloid 3203 is acidic cf LiZO is basic gives OHquot in wa rer Down group me rallic charac rer increases remainder are me rals bu r oxides of Al and Ga are ampho reric see below while Those of In and TI are basic Compare 2 3203 s 6 NaOH aq gt 2 NClgBOg aq 3 H20 l acid base sal r wa rer cf HCI g NaOH aq gt NaCl aq H20l in wa rer forms BOH3 or H3303 known as boric acid BOH3 2 H20 2 BOH439 H30 pKa 925 Al Ga oxides reac r wi rh bases as above bu r also wi rh acids They are ampho reric can behave as bo rh acids or bases Al203 S 6 gt 2 NClgAlOg 3 H2O acid base sal r wa rer Al203 S 3 H2504 gt Al25043 S 3 H2O base acid sal r wa rer InI Tl oxides are basic and reac r only wi rh acids II39I203 S 3 H2504 gt IH25043 S 3 H2O base acid sal r wa rer Bonding B compounds are covalen r Al are some rimes ionic and some rimes covalen r ionic charac rer increases down The group Oxidation States All form 3 but Tl also 1 In general when two oxidation states are possible lower one becomes more important down the group and its properties are more metallike Again B period 2 more different from the rest of group For example B forms many electrondeficient compounds stable but nevertheless reactive to Lewis bases gt attain an octet eg BFg F B has only 6 e39 in its outer valence FszF shell strong Lewis acid39 will accept electron pair from Lewis base eg BFg ZNHg gt F3B NH3 this is source of acidity of BOH3 H20 gt BOH439 H Uses Plentiful 3203 used in production of borosilicate glass BOH3 boric acid used as disinfectant eyewash insecticide Na2B4O5OH48HZO used in washing powders Boron hydrides BXHy very important class of compounds Aluminum Sulfate used in water purification dye industry antiperspirants etc Al203 used as a support for industrial catalysts chromatography supports etc BN compounds similar to analogous C I H C compounds eg borazene like benzene Hiy in C4 CH l l il B H H H V H Borazon BN has a Similar structure to Ll CH3 diamond and thus also very hard Diagonal Relationships Be with Al and B with Si l 39 Be B C Be and Al have similar properties Na Mg AI Si eg similar covalences in their compounds GROUP 14 nsznpz first group to show complete range of properties from nonmetal C through metalloids Si Ge to metals Sn and Pb shows up in properties such as melting points and AHfug nonmetals give strong covalent bonding higher melting points Table 142 Elements show important allotropes for the first time in C chemistry graphite 2D sheets diamond 3D network Cn molecular fullerenes eg C60 C70 C60 looks like a soccer ball eg diamond graphite very hard soft and greasy used as lubricant colorless black 3D 2D sheets insulator conductor Graphite is the standard state of C at 298 K and 1 atm Diamond formed at high T and P and interconverts to graphite at 298 K 1 atm but very slowly Oxidation states multiple ones now more common C nonmetal compounds all covalent except 6439 carbide ion eg Cagc Oxide C02 acidic almost all oxid states from 4 to 4 eg CH4 to C02 known 5i Ge metalloids essentially all compounds 4 ox state but some 2 5n Pb metals 4 covalent eg M02 2 ionic usually eg MO C usually fourcoordinate exception CEO Other elements show more exceptions eg SiFGZ39 GeClGZ39 SnOH6Z39 PbOH6239 but still usually fourcoordinate Why this difference between C and the rest C has ZsZZpZ outer configuration with no available d orbitals Remainder of elements have d orbitals they can use in bonding can form six bonds sp3d2 hybrids As we saw for the other groupsI there is a big difference between C and rest of the group Main differences i ability to form low down to 4 oxidation states ii forms multiple bonds to itself or other light elements N O eg C02 ie OCO acetone ie CH32CO iii forms stablecommon single bonds to itself catenationquot chain formation eg nbutane is HgCCHgCHgCHg containing a CCCC chain Include O N 5 etc and you have Organic Chemistry and Biochemistry and Life nbonding C is small forms strong enough 7 bonds to give stable compounds H H L R eg HC CH O C O C O RgtC2NH For the elements further down the group nbonding is much weaker and therefore usually prefers to form extra sigma 0 bonds 67 C02 vsSiOZ both are oxidation state 4 molecule Oquotquot 0 3D polymer oco vs 39 oSl0 strong C0 7 bond vs weak Si0 7 bond Therefore each Si C0 is 1 sigma 1 pi forms only four Si0 bonds each C has 2 sigma 2 pi Note C can even form further 7 bonding gt C20 1 sigma 2 pi bonds Let39s look further at CO carbon monoxide C02gCs2C0g AHgt0ASgt0 Nonspontaneous AG gt 0 at RT but spontaneous AG lt 0 at high T In fact favored at gt 700 C unless excess 02 present CO is a major pollutant from combustion processes at high T andor in 02 starved conditions when 2 C 02 gt 2 C0 rather than C02 Interesting difference C02 is acidic C0 is not Therefore C02 9 H20 l H2C03 aq carbonic acid but C0 9 H20 I no reaction C0 9 very toxic binds to hemoglobin irreversibly Diagonal Relationship B with Si GROUP 15 nsznp3 N P nonmetals As Sb metalloids Bi metal N again different from the rest All show 5 ox state 3 becomes more important down group NEN VS P4 P P P 1027 bonds P lgt or P P P P l P P P white P red P Oxid States N all from 5 to 3 common P As Sb 5 and 3 common Bi only 3 is common Lowest oxid state 3 reacts with H20 to give EH3 toxic except for39 NH3 CGgASZ s 6 H20 I gt 2 Ang g 3 CaOH2 aq N P oxides acidic As Sb amphoter39ic Bi basic Acidity decreases with lower39 oxid states Let39s look at the oxides of N many are known N20 NO N203 N02 N204 N205 1 2 3 4 4 5 nitr39ous nitric oxide oxide laughing gas In contrast only two Ncontaining acids are known learn the ous vs ic naming r39ulesll HNOZ is nitrous acid NOZ39 nitr39ite ion Flo N N is 3 oxid state 0 HN03 is nitric acid NOg39 nitr39ate ion HONO N is 5 oxid state 0 HN03 is a strong acid HNOZ weak K 71 x 10394 pKa 315 Similarly for P P4 s 3 02 g gt P406 5 oxid st 3 P4 s 5 02 g gt P4010 s oxid st 5 1 P406 H20 gt Hglzjog phosphorous acid dipr39otic acid Hill gt H2P0339 and HPng39 m POgg39 HO OH 2 P4010 H20 gt H3P04 phosphoric acid HO O tr39ipr39otic acid p gt HgPO439 gt HPO4Z39 gt PO4339 HO OH GROUP 16 nsznp4 O 5 Se nonmetals Te metalloid Po meta r39adioactive 0 too electr39onegative to show maximum oxidation state 6 S does then lower39 oxidation states again become more important Main oxidation state O2 S6 Se4 Te2 0 found everywhere as 02 and H20 5 also found as element others as miner39als 02 is 00 1 sigma 1 pi 5 occurs as 8 cyclic only sigma Oxides 502 and 503 gt H2503 and H2504 with water39 4 6 sulfur39ous acid sulfuric acid SeOZ and 5e03 gt H25e03 and H25e04 4 6 acid acid Hydrgen compounds H20 H25 H25e H2Te all 2 oxidation state Note usually bpts of compounds increase 100 H20 down a group M H2O is an exception due to bpt strong OHO hydrogenbonding in liquid each bond is 36 kJmol strong 5H5 etc much Awe weaker Similarly for Group 15 hydrogen compounds NH3 PH3 etc but there they are all gases at room temperature SE The high bpt of water 100 C at 1 atm is the reason that life on this planet is the way it is Liquid water allowed a place for life to develop Note H25 H25e H2Te very toxic GROUP 17 The Halgens nsznp5 F the most electronegative element only 1 oxidation state Others show range from maximum 7 to minimum 1 with 1 the most important A nonmetals acidic oxides Oxyacids eg Cl anion ox st HCIO4 perchloric acid perchlorate 7 HClOg chloric acid chlorate 5 HClO2 chlorous acid chlorite 3 HCIO hypochlorous acid hypochlorite 1 HCl hydrochloric acid chloride 1 Section 166 Slightly Soluble Ionic Salts Ionic salts are collections of cations M and anions X39 When an ionic salt dissolves in water it does so by the ions separating as they become surrounded by H20 molecules A very soluble ionic salt 69 NaCl dissolves in water completely giving Naaq and Cl39aq and there is no solid left However some salts are only slightly soluble and an equilibrium exists between dissolved and undissolved ie solid compound Consider the addition of PbSO4s lead sulfate to water PbSO4 s 2 sz aq 504239 aq At equilibrium the rate at which more PbSO4s dissolves ie the forward reaction is equal to the rate at which Pb2aq and SO4Z39aq ions come together to give PbSO4s the reverse reaction We say the solution is saturated ie the concs are as big as they can be If the reaction has not yet reached equilibrium Qc is Pb25042 QC PbSO4 The conc of a solid its density is a constant combine it with QC Qc Pb504 Qsp Pb2504239 QSP ionproduct expressionquot or solubility product expressionquot At equilibrium QP KSP K Pb2SO4Z39 quot Ksp solubility product constantquot or just quotsolubility product KSP like all other equilibrium constants only changes with temperature In general MPXq s 2 p Mn aq q XZ39 aq KSP MnPXz q We usually only consider sysTems aT equilibrium use KSP noT QSP Occasionally we will consider QSP see laTer Examples CuOH2 s 2 Cue aq 2 OH39 aq KSP Cu2OH39Z CaC03 s Ca2 aq 603239 aq KSP Ca2603239 Ca3PO42 s 2 3 Caz aq 2 PO4339 aq KSP Ca23PO43392 The greaTer is Ksp The more soluble is The subsTance eg PbSO4 KSP 16 x 10398 insoluble 60603 KSP 10 x 103910 more insoluble or less soluble FeOH2 KSP 41 x 103915 mosT insoluble or leasT soluble CalculaTions Involving SolubiliTy ProducTs Two Types Use KSP To find conc of dissolved ions Use concs To find KSP Make sure The equaTions are balanced Example 1 The solubiliTy of Aggcog is 0032 M aT 20 C WhaT is KSP 0f Answer This is a common Type of quesTion noTe ThaT we are Told The molar solubiliTy of Aggcog ie whaT concenTraTion will dissolve buT remember iT compleTely dissociaTe inTo ions Therefore Aggcog s 2 2Agaq Cng39aq iniT solid 0 0 change 0032 M 0064 M 0032 M equil solid 0064 M 0032 M KSP Ag2C03 006420032 K5 13 x 104 Example 2 The solubiliTy of Zn oxalaTe is 79 x 10393 M aT 18 C WhaT is iTs KSP Zn ox ZnZ aq oxz39 aq iniT solid 0 0 change 79 x 10393 M 79 x 10393 M 79 x 10393 M equil solid 79 x 10393 M 79 x 10393 M KSP Zn2ox 79 x103932 Ksp 62 x 105 Example 3 WhaT is The molar solubiliTy of smog KSP 54 x 1010 This Time we are given KSP and asked To find The solubiliTy 5rcogs 5r2aq coaz39mq iniT solid 0 0 change x X X equil solid x x KSP x2 x V54 gtlt103910 x 23 x 10395 M SolubiliTy of 5r603 is 23 x 10395 M Example 4 WhaT is The molar solubiliTy of CaOH2 in waTer39 KSP 65 x 10396 CaOH2 s Ca2aq 20H39aq iniT solid 0 0 change x x 2x equil solid x 2x KSP 65 x 10396 x2x2 4x3 careful I SolubiliTy of CaOH2 12 x 10392 M To obTain The f of a number learn To use The W buTTon on your calculaTor Q Take The log divide by x Then anTilog PracTice wiTh The number 8 iTs cuberooT is 2 SecTion 168 CommonIon EffecT on SolubiliTx The addiTion of a common ion ie one involved in a KSP reacTion decreases solubiliTy due To Le ChaTelier39s principle PbCrO4 s 2 sz aq CrO4Z39 aq KSP Pb2CrO4239 23 gtlt103913 Imagine This PbCrO4 sysTem aT equilibrium If we dissolve some Na26r04 s verx soluble in This souTion whaT will happen CrO4239 will increase because The Na26r04 will dissociaTe To give more Cr04239 Na26r04 s LO Z 2 Na aq CrO4Z39 aq The KSP for PbCrO4 Tells us ThaT Pb2CrO4239 is a consTanT if CrO4239 increases Pb2 musT decrease How can ThaT happen Some PbCrO4 s precipiTaTes from souTion 39e The equilibrium is shifTed To The lefT PbCrO4 s Pb2aq Cr04239aq 4 5hi add CrO4Z39 The equilibrium will shifT To The lefT ie more PbCrO4 s will form unTil Pb2newCrO4239new KSP 23 x 1013 The same thing will happen if we add a soluble source of sz eg PbN032 that will increase Pb2 ie addition of PbN032 to a solution of PbCrO4 will cause some more PbCrO4 s to precipitate from solution To understand this let39s make up an example that will use more convenient numbers Imagine a solid MX with KSP 100 MX s 2 NF aq Xquot aq at equilibrium M 10 M X39 10 M KSP 10x10 100 ie MXs will dissolve until MX39 10M because KSP MX39100 Now let39s add enough very soluble NaX to make new X39 20 M 39e we double X39 New MX39 1020 200 QSP 7t KSP Reaction no longer at equilibrium QSP gt KSP shifts to the left to return to equilibrium more MX 5 forms we say MX solid precipitatesquot decreasing M and X39 When M 10M X39 20M QSP 200 st KSP W1 9M X39 19M Q3P 171 2 KSP W W X39 18M QP 144 st KSP W1 7M X39 17M QP 119 7b KSP M 65M X39 165M QSP 10725 st KSP M 618M X39 1618M QSP 100 KSP reaction at equilibrium again New M 618 M new X39 1618 M QSP MX39 6181618 100 KSP Remember Le Chatelier39s principle tells us that if we do something to perturb a reaction at equilibrium the reaction will shift in the direction that allows it to reach equilibrium again Section 169 Effect of pH on Solubility Similarly to the previous section if a compound contains an anion that is the conjugate base of a weak acid addition of H30 from a strong acid increases its solubility Why Le Chatelier39s principle again Ca603 s 2 Caz aq 603239 aq Addition of H30 causes the 603239 to decrease 603239 aq H30 aq gt HCOg39 aq H20 Ca603 s 2 Caz aq 603239 aq Shift gt decrease 603239 When we decrease 603239 by adding H30 reaction will shift to the right to make more 603239 and restore equilibrium Note Some anions are such strong weak bases that even in pure water ie not acid they will give extensive reaction with water making the salt more soluble than you would expect from its KSP For example The 5239 aq anion a weak base but one of the strongest weak bases known is y unstable in water and undergoes 100 o l a basedissociation reaction with water to give HS39 aq and OH39 aq 5239 aq H20 l gt HS39 aq OH39 aq Kb 10 x 105 v large I the solubility equilibrium when PbS s is dissolved in water is PbS s H20 l 2 sz aq H539 aq OH39 aq Ksp Pb2H5 OH39 Section 167 Precipitation Reactions If we mix two solutions each containing one ion of a sparingly soluble ionic salt then we might see some solid of that salt form Such a formation of a solid from a clear solution is known as precipitation When will we see a solid ie Can we predict what will happen when we mix two solutions YES Calculate QSP and compare with KSP a If QSP KSP reaction at equilibrium solution is saturated no solid precipitate forms b If Qsp gt KSP reaction not at equilibrium of ions in solution too high solid precipitate forms until QSP KSP c If QSP lt KSP reaction not at equilibrium of ions in solution too low no solid precipitate forms Example Will a precipitate form if 0100 L of 030 M CaN032 is mixed with 0200 L of 0060 M NaF Method Calculate QSP and compare with KSP First What is the sparingly soluble salt NaN03 like all Na and N0339 salts is very soluble in water it must be CaFg CaFg s Caz aq 2 F39 aq KSP Ca2F392 need Ca2 and F39 a We start with 030 M CaN032 030 M Ca2 100 dissociation b We start with 0060 M NaF 0060 M F39 100 dissociation g we are mixing two solutions Therefore both will be diluted Remember MiVi Mfo i initial 1 final Cabf C021 i 030 M0100 L V 0200 L 0100 L Ca2f 010 M F39f W vf 0300 L ij 0040 M QsP Ca2F392 01000402 16 gtlt10394 KSP 32 x10quot11 Q3P gt KSP Can s precipitates until QSP KSP Section 1610 Complex Ions and Solubility AgCI s is only slightly soluble and it does NOT become more soluble if we lower the pH because CI39 is the conjugate base of a strong acid and therefore does not bind the added H BUT AgCI 5 becomes much more soluble if we add NH3 to the solution Why because the NH3 reacts with the Ag aq Lewis acidbase reaction to give a complex ion AgNH32 Consider what happens as two steps AgCI s 2 Ag aq Cl39 aq KSP18 x 103910 v small Ag aq 2 NH3 aq 2 AgNH32 Kc Kform 16 x 107 The equilibrium constant of a reaction which is the formation of a complex ion is called the formation constantquot Kfopm So the overall reaction that occurs when AgCI is dissolved in water with NH3 also present is AgCI s 2 NH3 aq 2 AgNH32 Cl39 aq Km The net Kc for this Km is the product of the KSP and Kform ie the K39s of the individual two steps Km KSP x Kform 18 x 10391O16 x 107 29 x 10393 therefore the solubility of the AgCI s is indeed much greater when NH3 is also present since Km gtgt KSP by 10 million times I CHAPTER 19 ELECTROCHEMISTRY We follow the steps are described in detail in Section 191 of the text a The problem is given in ionic form so combining Steps 1 and 2 the halfreactions are 2 Fe3 H202 gt H20 oxidation Fe reduction Step 3 We balance each halfreaction for number and type of atoms and charges The oxidation halfreaction is already balanced for Fe atoms There are three net positive charges on the right and two net positive charges on the left we add one electrons to the right side to balance the charge 2 3 7 Fe gtFee Reduction halfreaction we add one H20 to the righthand side of the equation to balance the O atoms H202 gt 2H2O To balance the H atoms we add 2H4r to the lefthand side H202 2H 2Hzo There are two net positive charges on the left so we add two electrons to the same side to balance the charge H202 2H 2e ZHZO Step 4 We now add the oxidation and reduction halfreactions to give the overall reaction In order to equalize the number of electrons we need to multiply the oxidation halfreaction by 2 2Fe2 gt Fe3 e7 H202 2H 2e ZHZO 2Fe2 H202 2H 2e a 2Fe3 2Hzo 2e The electrons on both sides cancel and we are left with the balanced net ionic equation in acidic medium 2Fe2 H202 2H gt 2Fe3 2H20 b The problem is given in ionic form so combining Steps 1 and 2 the halfreactions are Cu Cuz HN03 NO oxidation reduction Step 3 We balance each halfreaction for number and type of atoms and charges The oxidation halfreaction is already balanced for Cu atoms There are two net positive charges on the right so we add two electrons to the right side to balance the charge Cu gt Cu2 267 CHAPTER 19 ELECTROCHEMISTRY 549 Reduction halfreaction we add two H20 to the righthand side of the equation to balance the O atoms HNO3 gt NO ZHZO To balance the H atoms we add 3HJr to the lefthand side 3H HN03 NO 2Hzo There are three net positive charges on the left so we add three electrons to the same side to balance the charge 3H HN03 36 a NO 2Hzo Step 4 We now add the oxidation and reduction halfreactions to give the overall reaction In order to equalize the number of electrons we need to multiply the oxidation halfreaction by 3 and the reduction halfreaction by 2 3Cu Cu 26 23H HNO3 3e NO 2H20 3Cu 6H 2HNO3 6e 3Cu 2N0 4HZO 6e The electrons on both sides cancel and we are left with the balanced net ionic equation in acidic medium 3Cu 6H 2HNO3 3Cu2 2N0 4H20 c 3CN 2Mn04 H20 3CNO 2Mn02 20H d 3Br2 6OH Bro 5Br 3H20 e Halfreactions balanced for S andl oxidation 2520327 gt 40527 reduction 12 gt 217 Both halfreactions are already balanced for 0 so we balance charge with electrons 252032 a 40527 2e 12 267 gt 217 The electron count is the same on both sides We add the equations canceling electrons to obtain the balanced equation 2s2032 12 S4052 21 Strategy We follow the procedure for balancing redox reactions presented in Section 191 of the text Solution 3 Step 1 The unbalanced equation is given in the problem Mn H202 Mnoz H20 Step 2 The two halfreactions are WE oxidation WOZ H202 reduction H20 550 CHAPTER 19 ELECTROCHEMISTRY Step 3 We balance each halfreaction for number and type of atoms and charges The oxidation halfreaction is already balanced for M1 atoms To balance the O atoms we add two water molecules on the left side Mn 2Hzo M102 To balance the H atoms we add 4 H4r to the righthand side 2 Mi 2HzO M102 4H There are four net positive charges on the right and two net positive charge on the left we add two electrons to the right side to balance the charge Mn 2Hzo Mnoz 4H 2e Reduction halfreaction we add one HzO to the righthand side of the equation to balance the O atoms H202 2HzO To balance the H atoms we add 2H4r to the lefthand side H202 2H 2Hzo There are two net positive charges on the left so we add two electrons to the same side to balance the charge H202 2H 2e 2Hzo Step 4 We now add the oxidation and reduction halfreactions to give the overall reaction Note that the number of electrons gained and lost is equal Mn 2Hzo Mnoz 4H 2e H202 2H 2e 2Hzo Mn H202 2e gt MO 2H 2e The electrons on both sides cancel and we are left with the balanced net ionic equation in acidic medium Mn H202 Mnoz 2H Because the problem asks to balance the equation in basic medium we add one OH7 to both sides for each Hit and combine pairs of Hir and OH7 on the same side of the arrow to form HzO Mn H202 20H M102 2H 20H Combining the H4r and OH7 to form water we obtain Mn H202 20H Mn02 2H20 Step 5 Check to see that the equation is balanced by verifying that the equation has the same types and numbers of atoms and the same charges on both sides of the equation b This problem can be solved by the same methods used in part a 2BiOH3 3Sn02 2Bi 3H20 3Sn03 CHAPTER 19 ELECTROCHEMISTRY 551 c Step 1 The unbalanced equation is given in the problem Crzo czo Cr3 co Step 2 The two halfreactions are c204 oxidation COZ 27 39 3 cr207 reduction Cr Step 3 We balance each halfreaction for number and type of atoms and charges In the oxidation halfreaction we first need to balance the C atoms c2042 2coz The O atoms are already balanced There are two net negative charges on the left so we add two electrons to the right to balance the charge c2042 2coz 2e In the reduction halfreaction we first need to balance the Cr atoms Crzo72 2Cr3 We add seven HzO molecules on the right to balance the O atoms Crzo72 2Cr3 7HzO To balance the H atoms we add l4HJr to the lefthand side Crzo7z 14H 2Cr3 7HzO There are twelve net positive charges on the left and six net positive charges on the right We add six electrons on the left to balance the charge Crzo72 14H 6e 2Cr3 7HzO Step 4 We now add the oxidation and reduction halfreactions to give the overall reaction In order to equalize the number of electrons we need to multiply the oxidation halfreaction by 3 3c2042 2coz 2e Crzo72 14H 6e 2Cr3 7HzO 3c2042 Crzo72 14H 6e 6C0 2Cr3 7HzO 6e The electrons on both sides cancel and we are left with the balanced net ionic equation in acidic medium 3C2042 Cr2072 14H 6C0 2Cr3 7H20 Step 5 Check to see that the equation is balanced by verifying that the equation has the same types and numbers of atoms and the same charges on both sides of the equation d This problem can be solved by the same methods used in part c 2Cl 2ClO3 4H gt C12 2C102 2H20 552 1911 1912 1913 1914 CHAPTER 19 ELECTROCHEMISTRY Halfreaction E V Mg2aq 2e7 gt Mgs 7237 Cu2aq 267 gt Cus 034 The overall equation is Mgs Cu2aq gt Mg2aq Cus E 034V77237V 271V Strategy At first it may not be clear how to assign the electrodes in the galvanic cell From Table 191 of the text we write the standard reduction potentials of Al and Ag and apply the diagonal rule to determine which is the anode and which is the cathode Solution The standard reduction potentials are Ag10Me7 a Ags E 080V A1310M3e Als E 7166V Applying the diagonal rule we see that AgJr will oxidize Al Als A1310M 36 3Ag10 M 36 a 3Ags Anode oxidation Cathode reduction Overall Als 3Ag10 M A1310 M 3Ags Note that in order to balance the overall equation we multiplied the reduction of AgJr by 3 We can do so because as an intensive property E is not affected by this procedure We find the emf of the cell using Equation 191 and Table 191 of the text Egell Egathode Egnode EggAg T BillyA1 E3811 080 V 7 7166 V 246 V Check The positive value of E shows that the forward reaction is favored The appropriate halfreactions from Table 191 are 12s 267 gt 217aq Fe3aq ei gt Fe2aq Ede 053 v Egamode 077 v Thus iron111 should oxidize iodide ion to iodine This makes the iodide ioniodine halfreaction the anode The standard emf can be found using Equation 191 EEC Egamode 7 Ede 077 v 7 053 v 024 v The emf was not required in this problem but the fact that it is positive confirms that the reaction should favor products at equilibrium The halfireaction for oxidation is 2HzOl M 02g 4Haq 46 Egmde 123 v 1915 1916 1917 1918 1921 CHAPTER 19 ELECTROCHEMISTRY 553 The species that can oxidize water to molecular oxygen must have an Efed more positive than 123 V From Table 191 of the text we see that only C12g and Mn04aq in acid solution can oxidize water to oxygen The overall reaction is 5No3 aq 3anaq 2HzOl a 5NOg 3Mno4 aq 4Haq E3 Egamode7 Ede 096 v 7151v 7055 v The negative emf indicates that reactants are favored at equilibrium N037 will not oxidize 1W12 to M1047 under standardstate conditions Strategy E8611 is positive for a spontaneous reaction In each case we can calculate the standard cell emf from the potentials for the two halfreactions Egell Egathode 7 Egnode Solution 21 Equot 7040 V7 7287 V 247 V The reaction is spontaneous b E 7014 V 7 107 V 121 V The reaction is not spontaneous c E 7025 V 7 080 V 7105 V The reaction is not spontaneous d Equot 077 V 7 015 V 062 V The reaction is spontaneous From Table 191 of the text we compare the standard reduction potentials for the halfreactions The more positive the potential the better the substance as an oxidizing agent a Au3 b Ag c Cd d 02 in acidic solution Strategy The greater the tendency for the substance to be oxidized the stronger its tendency to act as a reducing agent The species that has a stronger tendency to be oxidized will have a smaller reduction potential Solution In each pair look for the one with the smaller reduction potential This indicates a greater tendency for the substance to be oxidized a Li c Fe b H2 d Br We find the standard reduction potentials in Table 191 of the text Egen Egamode7 Ede 7076 v 7 7237 v 161 v 0 00257 V Ecell nK n In K quotEgell 00257 v 554 1922 1923 CHAPTER 19 ELECTROCHEMISTRY quotEceu K e00257v 2161 V K e 00257v K3x1054 Strategy The relationship between the equilibrium constant K and the standard emf is given by Equation 195 of the text E2611 00257 Vn an Thus knowing n the moles of electrons transferred and the equilibrium constant we can determine E2611 Solution The equation that relates K and the standard cell emf is E2611 an 7 00257 V n We see in the reaction that Mg goes to Mg2 and Zn2 goes to Zn Therefore two moles of electrons are transferred during the redox reaction Substitute the equilibrium constant and the moles of ei transferred n 2 into the above equation to calculate E0 7 00257 V1nK n E 0368 V 7 00257 V1n269 x10 2 In each case we use standard reduction potentials from Table 191 together with Equation 195 of the text a Egen E amodei Ede 107 V 7 053 V 054 V 1n K quotEgell 00257 V quotE3611 K 600257V 2054 V K e 00257V 2x 1018 b Egen Egamodei Ede 161 V 7136 V 025 V 2025 V K e 00257V 3x108 c Egen E amodei Egmde 151 V7 077 V 074 V 5074 V K e 00257V 3x1062 CHAPTER 19 ELECTROCHEMISTRY 555 a We break the equation into two half7reactions Mgs oxidation anode Mg2 MD 267 Pb2aq 267 reduction cathode Pbs 1924 Ede 7237 v Egamode 7013 V The standard emf is given by E3611 Egamode7 Ede 7013 V 7 7237 V 224 V We can calculate AGO from the standard emf AGquot 7 nFEgell AG 7296500 JVmol224 V 7432 kJmol Next we can calculate K using Equation 195 of the text E2611 0027 V an 1n K quotEgell 00257 V L K 600257 2224 K e 00257 5x 1075 Tip You could also calculate K from the standard free energy change AGO using the equation AGO 7RTanc b We break the equation into two half7reactions Egamode 107 V BUG 267 reduction cathode 2Br7aq 2raq M 12s2e Egmde 053 V The standard emf is EEC Egamode7 Egrme 107 V 7 053 V 054 V We can calculate AGO from the standard emf AGquot 7nFEgell AG 7296500 JVmol054 V 7104kJmol Next we can calculate K using Equation 195 of the text L K 600257 556 CHAPTER 19 ELECTROCHEMISTRY 2054 K 6 00257 2x 1018 c This is worked in an analogous manner to parts a and b E3611 Egamode 7 Ede 123 V 7 077 V 046 V AGquot 7nFEgen AGquot 74965001Vmo1o46V 717s kJmol L K 600257 4046 K 6 00257 1gtlt 1031 d This is worked in an analogous manner to parts a b and c EEC Egamode 7 Ede 053 V 7 7166 V 219 V AGquot 7nFEgen AG 76965001Vmo1219V 127gtlt 103kJmol L K 600257 6219 K e 00257 8x 10211 1925 The halfreactions are Fe3aq ei 7 Fe2aq Ede 077 V Ce4aqei 7 Ce3aq Egamode 161 V Thus Ce4 Will oxidize Fe2 to F e3 this makes the F e2F e3 halfreaction the anode The standard cell emf is found using Equation 191 of the text Egg Egamodf Egmde 161 V 7 077 V 084 V The values of AG0 and K are found using Equations 193 and 195 of the text AG 7nFEgen 7196500 JVmol084 V 81 kJmol K quotEgell 00257 V quotEden 1084 V Kc 600257V e 00257V 2X 1014 1926 1929 CHAPTER 19 ELECTROCHEMISTRY 557 Strategy The relationship between the standard free energy change and the standard emf of the cell is given by Equation 193 of the text AGO K and the standard emf is given by Equation 195 of the text E2611 00257 Vnan Thus if we can 7 nFEgell The relationship between the equilibrium constant determine E2611 we can calculate AG0 and K We can determine the E2611 of a hypothetical galvanic cell made up of two couples CuerCuJr and CuCu from the standard reduction potentials in Table 191 of the text Solution The halfcell reactions are Cu10M Cu210Me7 Cu101ei gt Cus Anode oxidation Cathode reduction Overall 2Cu10 111 gt Cu210 111 Cus o ESell Egathode T Egnode Egutcu 7 EcuZ rju r E38quot 052 V 7 015 V 037 V Now we use Equation 193 of the text The overall reaction shows that n 1 AGquot inFEgell AGquot 7196500 JVmol037 V 36 kJmol Next we can calculate K using Equation 195 of the text E2611 00257 V an n or 1n K quotEgell 00257 V and L K 600257 1037 K e 00257 e144 2x 106 Check The negative value of AG0 and the large positive value of K both indicate that the reaction favors products at equilibrium The result is consistent with the fact that E for the galvanic cell is positive If this were a standard cell the concentrations would all be 100 M and the voltage would just be the standard emf calculated from Table 191 of the text Since cell emfs depend on the concentrations of the reactants and products we must use the Nernst equation Equation 198 of the text to find the emf of a nonstandard cell 00257 V n EE an anti ln Cult E 110V7 00257 V 2 558 1930 CHAPTER 19 ELECTROCHEMISTRY 025 015 E 110V7 002257 V In E 109V How did we find the value of 110 V for Equot Strategy The standard emf E can be calculated using the standard reduction potentials in Table 191 of the text Because the reactions are not run under standardstate conditions concentrations are not 1 111 we need Nernst39s equation Equation 198 of the text to calculate the emf E of a hypothetical galvanic cell Remember that solids do not appear in the reaction quotient Q term in the Nernst equation We can calculate AG from E using Equation 192 of the text AG inFEcen Solution 21 The halfcell reactions are Anode oxidation Mgs gt Mg210 111 2e7 Cathode reduction Sn210 111 267 gt Sns Overall Mgs Snz10I gt Mg2101 Sns Ecell Ecathode Eanode Esn2Sn T EMgZMg Egg 7014 V 7 7237 V 223 V From Equation 198 of the text we write 7 00257 V E E an n 2 E E0 7 00257 V llg2 7 Sn 1 E 223 vimmw 223 v 2 0035 We can now find the free energy change at the given concentrations using Equation 192 of the text Note that in this reaction n 2 AG inFEcen AG 7296500 JVmol223 V 430 kJmol b The halfcell reactions are Anode oxidation Cathode reduction Ov erall 3Zns an10M 26 2Cr310M 36 Crs 3Zns 2Cr310 M a 3an10 M 2Crs Eell Egathode Egnode BlahCr T EZHZ rZn Eggquot 7074 v 7 7076 V 002 V CHAPTER 19 ELECTROCHEMISTRY 559 From Equation 198 of the text we write 7 00257 V n EE an 7 00257 V 1nant3 n cr3 2 EE 3 E 7 002 V 7 00257 V1n 00085 2 004V 6 0010 We can now find the free energy change at the given concentrations using Equation 192 of the text Note that in this reaction n 6 AG inFEcen AG 7696500 JVmol004 V 723 kJmol 1931 The overall reaction is Zns 2Haq 7 Zn2aq Hzg Eggquot Egamodf Ede 000 V 7 7076 V 076 V 7 00257 V1nZnZPHZ n H 2 EE E076V7 OJSV 00257 V 1n045 2 18 1932 Let s write the two halfreactions to calculate the standard cell emf Oxidation occurs at the Pb electrode Pbs M Pb2aq 7 2e Egmde 7013 V 2Haq 77 267 reduction cathode H2g EgamOde 000 V 2HaqPbs 7 Hzg Pb2aq Egg Egamode 7 Egmde 000 V 7 7013 V 013 V Using the Nernst equation we can calculate the cell emf E 2 7 00257 V 1nPb lPHZ n H2 EE 7 013 V7 00257 Vln0101 7 0083V 2 0050 1933 As written the reaction is not spontaneous under standard state conditions the cell emf is negative EEC Egamodf Ede 7076 V 7 034 V 7110 V 560 CHAPTER 19 ELECTROCHEMISTRY The reaction will become spontaneous when the concentrations of Zincll and copperH ions are such as to make the emf positive The turning point is when the emf is zero We solve the Nernst equation for the Cu2 Zn2 ratio at this oint P 00257 V Ecell O T an n 2 0 ililovi 00257 V1nCu 2 ant Cult ln Z 2 7 7856 n l 2 Cu 1 856 67X10 38 ZnZW In other words for the reaction to be spontaneous the Cu2an ratio must be less than 67 X 10738 Is the reduction of zincll by copper metal a practical use of copper 1934 All concentration cells have the same standard emf zero volts Mg2aq 267 reduction cathode Mgs EgamOde 7237 V Mgs oxidation anode Mg2aq 267 Egrmde 7237 V EEC Egamode 7 Ede 237 v 7 237 V 000 v We use the Nernst equation to compute the emf There are two moles of electrons transferred from the reducing agent to the oxidizing agent in this reaction so n 2 7 00257 v E EO an 71 2 E E0 7 00257 Vln Mg2 0x n Mg lred E owmm 0010V 2 053 What is the direction of spontaneous change in all concentration cells 1937 a The total charge passing through the circuit is 85Cgtlt 3600s ls l 30hgtlt 92gtlt104C From the anode halfreaction we can find the amount of hydrogen 2 mol H2 X lmol e7 92 X 104 C X 4 mol 5 96500 c 048 mol H2 1938 1945 CHAPTER 19 ELECTROCHEMISTRY 561 The volume can be computed using the ideal gas equation V 7 nRT 7 048 mol00821LatmKmol298 K P 0076 L 155 atm b The charge passing through the circuit in one minute is 8395 c X 510 Cmin l s 1 mm We can find the amount of oxygen from the cathode halfreaction and the ideal gas equation 510 C 1min lmol e X 1 mol 02 96500 C gtlt 13 gtlt1073 mol Ozmin 4 mol 6 73 V nRT 13x10 mol Oz00821LatmKmol298 10 003 Ozmm P l min 1 atm wx I39OL alr 016 L of airmin 1 mm 020 L 02 We can calculate the standard free energy change AGO from the standard free energies of formation AGE using Equation 1812 of the text Then we can calculate the standard cell emf E2611 from AGquot The overall reaction is C3H8g 5028 gt 3C02g 4HZOl AGroxn 3AGFC028 4AG leOll AGFC3H88 5AG 028 AG 373944 kJmol 472372 kJmol 7 17235 kJmol 50 721085 kJmol We can now calculate the standard emf using the following equation AGquot inFEgell 0139 o iAGO E cell nF Check the halfreactions onp 843 of the text to determine that 20 moles of electrons are transferred during this redox reaction 7721085 gtlt103Jmol Ecell 109 V 2096500 JVmol Does this suggest that in theory it should be possible to construct a galvanic cell battery based on any conceivable spontaneous reaction 1 mol Mg X 2431 g Mg lmol Mg MassMg 100 F X 122 g Mg 2 mol 6 562 1946 1947 1948 CHAPTER 19 ELECTROCHEMISTRY a The only ions present in molten BaClz are Ba2 and C17 The electrode reactions are anode 2Cliaq Clzg 267 cathode Ba2aq 267 Bas This cathode halfreaction tells us that 2 moles of ei are required to produce 1 mole of Bas 1 Strategy According to Figure 1920 of the text we can carry out the following conversion steps to calculate the quantity of Ba in grams currentgtlttime gt coulombs gt molei gt mol Ba gt gBa This is a large number of steps so let s break it down into two parts First we calculate the coulombs of electricity that pass through the cell Then we will continue on to calculate grams of Ba Solution First we calculate the coulombs of electricity that pass through the cell 050Axix lAs x 30 min 90gtlt102C l min We see that for every mole ofBa formed at the cathode 2 moles of electrons are needed The grams ofBa produced at the cathode are lmole lmol Ba 1373gBa gtltgtlt 064 g Ba 96500 C 2 mol 6 lmol Ba gBa 90gtlt102Cgtlt NaJr ei gt Na A13 3e A1 The half re actions are Since 1 g is the same idea as 1 ton as long as we are comparing two quantities we can write lgNaxlixle 0043mole 2299gNa lgAlx x3ei 0llmole 2698gAl It is cheaper to prepare 1 ton of sodium by electrolysis The cost for producing various metals is determined by the moles of electrons needed to produce a given amount of metal For each reduction let39s first calculate the number of tons of metal produced per 1 mole of electrons 1 ton 9072 X 105 g The reductions are Mg22e7 Mg MXMXLHS l340gtlt 10 5 ton Mgmol e 2mole lmolMg 9072gtlt10 g A133e Al M let Ons 9913x10 6 ton Almol e 3mole lmolAl 9072gtlt10 g NaJr ei Na M gtlt M xlt Ons 2534 X 10 5 ton Namol e lmol e lmolNa 9072gtlt 10 g CHAPTER 19 ELECTROCHEMISTRY 563 lmol Ca 4008 g Ca 1 ton gtlt gtlt 2209 X 10 5 ton Camol e 2 mol e 1molCa 9072gtlt105 g Ca2 2e7 Ca Now that we know the tons of each metal produced per mole of electrons we can convert from 155ton Mg to the cost to produce the given amount of each metal a For aluminum 155 X 1340x10 5 ton Mg X lmol e X 100 tons A1 210 lt 103 lton Mg 1 mol e 9913 gtlt10 6 ton Al b For sodium 155 X l340gtlt 10 5 ton Mg X lmol e X 300 tons Na 246 x 103 lton Mg 1 mol e 2534 gtlt1075 ton Na c For calcium 155 X l340gtlt10 5tonMg X lmole X 500 tons Ca 470 x 103 1 ton Mg 1 mol e 2209 gtlt10 5 ton Ca 1949 Find the amount of oxygen using the ideal gas equation 755 mmHg X i 0076 L PV 760 mmHg 3 n 3lgtlt10 molOz RT 00821 L atmKmol298 K Since the halfreaction shows that one mole of oxygen requires four faradays of electric charge we write F 3lgtlt10 3molO gtlt 4 0012F 2 lmol 02 1950 a The halfireaction is 2H20l 02g4Haq4e First we can calculate the number of moles of oxygen produced using the ideal gas equation 7 PV quot02 7 E no M 0034 mo 02 2 00821 LatmmolK298 K Since 4 moles of electrons are needed for every 1 mole of oxygen we will need 4 F of electrical charge to produce 1 mole of oxygen F 0034 mol Ozgtlt L 014 F lmol Oz 564 CHAPTER 19 ELECTROCHEMISTRY b The halfireaction is 2Cliaq Clzg 2e7 The number of moles of chlorine produced is PV quotc1Z E 750 mmHg X i 150 L 760 mmHg 00605 mol c12 00821 LatmmolK298 K quotC12 Since 2 moles of electrons are needed for every 1 mole of chlorine gas we Will need 2 F of electrical charge to produce 1 mole of chlorine gas 7 F 00605 mol Clzgtlt L 0121 F 1 mol C12 c The halfireaction is Sn2aq 2e Sns The number of moles of Sns produced is mol Sn 60 g Sngtlt m 0051mol Sn 1187 g Sn Since 2 moles of electrons are needed for every 1 mole of Sn we will need 2 F of electrical charge to reduce 1 mole of Sn2 ions to Sn metal F 0051 mol Sn gtlt L 010 F 1 mol Sn 1951 The halfreactions are Cu2aq 267 gt Cus 2Br7aq gt Brzl 267 The mass of copper produced is 4ISOAgtlt1hgtlt3600sgtlt 1C leole gtltlmolCugtlt6355gCu533gcu 1h 1As 96500C Zmole lmolCu The mass of bromine produced is 7 1 113 1598 B 450Axlhx xix x x 134gBr2 lh lAs 96500C 2m01 e lmol Brz 1952 a The halfireactionis Agaq e Ago CHAPTER 19 ELECTROCHEMISTRY 565 b Since this reaction is taking place in an aqueous solution the probable oxidation is the oxidation of water Neither AgJr nor N037 can be further oxidized 2H20l 02g4Haq4e c The halfreaction tells us that 1 mole of electrons is needed to reduce 1 mol of AgJr to Ag metal We can set up the following strategy to calculate the quantity of electricity in C needed to deposit 067 g of Ag grams Ag gt mol Ag gt mol ei gt coulombs 067gAgx x xm 60x102C 1079 g Ag lmol Ag 1 mol 6 1953 The halfreaction is Coz 267 gt Co lmol Co X 2 mole X 96500 C 770x 103C 235 g Co gtlt 5893 g Co lmol Co 1 mol e 1954 a First find the amount of charge needed to produce 200 g of silver according to the halfireaction AgYaq e Ago lmol Ag X lmol e X 96500 C l79gtlt103C 1079gAg lmolAg lmolei 200g Ag X The halfireaction for the reduction of copperH is Cu2aq 267 Cus From the amount of charge calculated above we can calculate the mass of copper deposited in the second cell lmole lmol Cu 6355g Cu X X 0589 g Cu 96500 C 2 mol 6 1 mol Cu 179 gtlt103 C X b We can calculate the current owing through the cells using the following strategy Coulombs gt Coulombshour gt Coulombssecond Recall thatl C 1 As The current owing through the cells is lh gtlt 0133 A 3600 s 375 h l79gtlt103Asgtlt 566 1955 1956 CHAPTER 19 ELECTROCHEMISTRY The halfreaction for the oxidation of chloride ion is 2Cliaq gt Clzg 267 First let39s calculate the moles of ei owing through the cell in one hour 1500Agtlt gtlt lAs lh 3600 s X 1 mol 6 96500 C 5596 mole Next let39s calculate the hourly production rate of chlorine gas in kg Note that the anode efficiency is 930 1 mol C12 X 007090 kg C12 X 930 1 mol C12 100 5596 mol e7gtlt 184 kg C12h 2 mol 6 Step 1 Balance the halfireaction Crzo72 aq 14Haq 12e 2Crs 7HZOl Step 2 Calculate the quantity of chromium metal by calculating the volume and converting this to mass using the given density Volume Cr thickness gtlt surface area Volume Cr 10 gtlt10 megtlt l mx 025 m2 25 gtlt1076m3 1000 mm Converting to cm3 3 25 x10 6m3gtlt 25 cm3 001 m Next calculate the mass of Cr Mass density gtlt volume Mass Cr 25 cm3 X 73919g 18g Cr 1 cm3 Step 3 Find the number of moles of electrons required to electrodeposit 18 g of Cr from solution The half reaction is Crzo72 aq 14Haq 12e 2Crs 7H20l Six moles of electrons are required to reduce 1 mol of Cr metal But we are electrodepositing less than 1 mole of Crs We need to complete the following conversions gCr gt molCr gt molei lmol Cr 6 mol 6 5200 g Cr lmol Cr faradays 18 g Cr gtlt 21 mol 6 Step 4 Determine how long it will take for 21 moles of electrons to flow through the cell when the current is 250 Cs We need to complete the following conversions mol ei gt coulombs gt seconds gt hours CHAPTER 19 ELECTROCHEMISTRY 567 96500C ls lh gtlt gtlt h2lmole gtlt lmole 250C 3600s Would any time be saved by connecting several bumpers together in a series 1957 The quantity of charge passing through the solution is 0750AX1C X 60 s gtltlmole lAs lmin 96500C gtlt 250 min 117 X lO zmol e Since the charge of the copper ion is 2 the number of moles of copper formed must be lmol Cu 117x10 2 mol e x 585x10 3 mol Cu The units of molar mass are grams per mole The molar mass of copper is 0369 g 73 631gmol 585 X 10 mol 1958 Based on the halfreaction we know that one faraday will produce half a mole of copper Cu2aq 2e7 gt Cus First let s calculate the charge in C needed to deposit 0300 g of Cu 1c 9l2C 300A 304sgtlt 7 lt x 1M We know that one faraday will produce half a mole of copper but we don t have a half a mole of copper We have 0300 g Cu gtlt 472 gtlt10 3 mol 6355 g Cu We calculated the number of coulombs 912 C needed to produce 472 X 1073 mol of Cu How many coulombs will it take to produce 0500 moles of Cu This will be Faraday s constant x 0500 mol Cu 966 x 104C 1 F 472 gtlt107 mol Cu 1959 The number of faradays supplied is lmol Ag X lmol e 00133 mol 6 1079 g Ag lmol Ag l44g Ag gtlt Since we need three faradays to reduce one mole ofX3 the molar mass ofX must be 0l20g X X 3mol e 271 gmol 00133 mol e lmolX 568 1960 1961 1962 CHAPTER 19 ELECTROCHEMISTRY First we can calculate the number of moles of hydrogen produced using the ideal gas equation 7 PV E 1 atm 782 mmHg gtlt 0845 L 760 mmHg nHZ 00355 mol 00821 L atmK mol298 K The number of faradays passed through the solution is 00355 mol H2gtlt 00710 F 2 1 mol H2 Hzg a 2Haq2ei Nizaq 26 Nis NizaqHzg Nis 2Haq a The halfreactions are The complete balanced equation is Nis is below and to the right of Haq in Table 191 of the text see the halfreactions at 7025 and 000 V Therefore the spontaneous reaction is the reverse of the above reaction that is Nis 2Haq Nizaq Hzg b The halfreactions are M1047aq 8Haq 567 gt M12aq 4HzO 2Cliaq gt Clzg 267 The complete balanced equation is 2Mno4 aq16Haq 10Cliaq 2Mn2aq 8HZO 5Clzg In Table 191 of the text Cliaq is below and to the right of M1047aq39 therefore the spontaneous reaction is as written c The halfreactions are Crs gt Cr3aq 367 Zn2aq 267 gt Zns 2Crs 3Zn2aq gt 2Cr3aq 3Zns The complete balanced equation is In Table 191 of the text Zns is below and to the right of Cr3aq39 therefore the spontaneous reaction is the reverse of the reaction as written The balanced equation is Cr2072 6 Fe2 14H 2Cr3 6Fe3 7H20 The remainder of this problem is a solution stoichiometry problem The number of moles of potassium dichromate in 260 mL of the solution is 00250 mol 260 mL gtlt 1000 mL soln 650 gtlt10 4 mol KZCrZO7 CHAPTER 19 ELECTROCHEMISTRY 569 From the balanced equation it can be seen that 1 mole of dichromate is stoichiom etrically equivalent to 6 moles of ironH The number of moles of ironH oxidized is therefore 6 mol Fe2 2 390gtlt 10 3m61Fezt lmol CrzOf 650 X 104 mol Crzo r gtlt Finally the molar concentration of Fe2 is 390 gtlt1073 mol 3 0156 molL 0156 M Fe2 250gtlt10 L 1963 The balanced equation is 5502g 2Mno4 aq 2H20l 5so4z aq 2anaq 4Haq The mass of 02 in the water sample is given by 000800 mol KlVan4 X 5 mol 502 X 6407 g 802 1000 mL soln 2 mol KlVan4 1 mol 502 737 mL gtlt 944 x 10393 g s02 1964 The balanced equation is Mno4 5Fe2 8H Mn 5Fe3 4HzO First let s calculate the number of moles of potassium permanganate in 2330 mL of solution 00194 mol 2330 mL gtlt 452 gtlt1074 mol KMnO4 1000 mL soln From the balanced equation it can be seen that 1 mole of permanganate is stoichiometrically equivalent to 5 moles of ironH The number of moles of ironH oxidized is therefore 5 mol Fe2 1 mol M10 452 gtlt10 4 mol Mnog gtlt 226 gtlt10 3 mol Fe The mass of Fe2 oxidized is 5585 g Fe mass Fe2 226 X 10 3 mol Fe2 X 1 mol Fe2 0126 g Felt Finally the mass percent of iron in the ore can be calculated f mass Fe m X 100 total mass of sample 0l26g Fe gtlt100 451 g 1965 a The balanced equation is 2M1047 5HzOz 6HJr 502 2MB SHZO 570 CHAPTER 19 ELECTROCHEMISTRY b The number ofmoles of potassium permanganate in 3644 mL of the solution is 001652 mol 3644 mL gtlt 6020 gtlt1074 mol of KMnO4 1000 mL soln From the balanced equation it can be seen that in this particular reaction 2 moles of permanganate is stoichiometrically equivalent to 5 moles of hydrogen peroxide The number of moles of H202 oxidized is therefore 5 mol H202 6020 gtlt10 4 mol Mnog gtlt 1505 gtlt10 3 mol H202 2 mol M10 The molar concentration of H202 is 73 11202 W 00602 molL 00602 M 250 X 10 3 L 1966 a The halfireactions are i Mno4 aq 8Haq 5e M127 4 4H2Ol ii czo mq 2CO2g 4 2e We combine the halfreactions to cancel electrons that is 2 X equation i 5 X equation ii 2Mno4 aq 16Haq 5C2O427aqgt 2anaq l0CO2g 8H2Ol b We can calculate the moles of KlVan4 from the molarity and volume of solution 00100 1 KWO 240 mL KMnO4x L 240 X 10 4molKM104 1000 mL soln We can calculate the mass of oxalic acid from the stoichiometry of the balanced equation The mole ratio between oxalate ion and permanganate ion is 52 240 gtlt1074 mol KMnO4 X M X W 00540 g HZCZO4 2 mol KWO4 lmol HZCZO4 Finally the percentby mass of oxalic acid in the sample is oxalic acid X 100 540 1967 E AG 3997 Rear nn gt 0 lt 0 spontaneous lt 0 gt 0 nonspontaneous 0 0 at equilibrium 1968 The balanced equation is 2Mno4 5c2042 16H 2Mn 10CO2 8H2O Therefore 2 mol M1047 reacts with 5 mol C204 CHAPTER 19 ELECTROCHEMISTRY 571 956 gtlt10 4 mol M101 1000 mL soln Moles ofMiOZ reacted 242 mL gtlt 231 X 10 5 mol M10 Recognize that the mole ratio of Ca2 to C204 is 11 in CaCzO4 The mass of Ca2 in 100 mL is 5 mol Ca2 X 4008 g Ca2 231 X 10 5 mol MnO X 4 7 2 2 mol MnO4 lmol Ca 231gtlt10 3g Ca2 Finally converting to mgmL we have 0231 mg Ca2mL blood 231gtlt10 3g Ca2 X1000 mg 100 mL g 1969 The solubility equilibrium of AgBr is AgBrs Agaq Br7aq By reversing the second given halfreaction and adding it to the first we obtain Ags Agaqe Egmde 080 v AgBrs ei gt Ags Br7aq Egamode 007 V AgBrs Agaq Br7aq E2611 Egamodei Egnmde 007 V 7 080 V 7073 V At equilibrium we have E E0 7 00257 VlnAgBr n 0 7073 Viman 1 5P ansp 7284 KSp 5 gtlt 10 13 Note that this value differs from that given in Table 162 of the text since the data quoted here were obtained from a student39s lab report 1970 a The halfireactions are 2Haq2e Hzg Egmde 000 v AgYaq e a Ags Esamode 080 v E38quot Egamodei Egrme 080 V 7 000 V 080 V b The spontaneous cell reaction under standardstate conditions is 2AgaqH2g 2Ags 2Haq 572 CHAPTER 19 ELECTROCHEMISTRY c Using the Nernst equation we can calculate the cell potential under nonstandardstate conditions 2 7 00257 V1n H E E T 7 Ag 1 PHZ i The potential is 72 2 E 080V7 0390257 Vln I39OXZIO 092 V 2 10 10 ii The potential is 75 2 E 080V7 00257Vn10gtlt10 LIOV 2 10210 d From the results in part c we deduce that this cell is a pH meter its potential is a sensitive function of the hydrogen ion concentration Each 1 unit increase in pH causes a voltage increase of 0060 V 1971 a If this were a standard cell the concentrations would all be 100 M and the voltage would just be the standard emf calculated from Table 191 of the text Since cell emf s depend on the concentrations of the reactants and products we must use the Nernst equation Equation 198 of the text to find the emf of a nonstandard cell E E07 00257 van n 2 E 317 V7 00257 VlnMg 2 2 Ag E 317 V7 00257 v1n 0102 2 010 E 314V b First we calculate the concentration of silver ion remaining in solution after the deposition of 120 g of silver meta 0100 mol Ag Ag originally in solution 1 L X 0346 L 346 X 107211161 Ag 1mol A de os1ted 120 A X g P g g 107 9 g 111gtlt10 2molAg Ag remaining in solution 346 X 1072 mol Ag 7 111 X 1072 mol Ag 235 X 1072 mol Ag 7 235gtlt10 2mol 679 gtlt10 ZM 0346 L A5 The overall reaction is Mgs 2Agaq gt Mg2aq 2Ags We use the balanced equation to find the amount of magnesium metal suffering oxidation and dissolving M 555 gtlt 107311161 Mg g 111 107211161 X Aggtlt 2molA CHAPTER 19 ELECTROCHEMISTRY 573 The amount of magnesium originally in solution was 0288 L X m 288 gtlt10 2 mol l L The new magnesium ion concentration is 555 x1073 288 x1072mol 0119 M 0288 L The new cell emfis E E07 00257 Van n E 317V7Mln 01 313V 679 gtlt10 22 7 1972 The overvoltage of oxygen is not large enough to prevent its formation at the anode Applying the diagonal rule we see that water is oxidized before uoride ion Fzg2e 2F aq E 287v 02g4Haq4e 2HzOl E 123 v The very positive standard reduction potential indicates that F7 has essentially no tendency to undergo oxidation The oxidation potential of chloride ion is much smaller 7136 V and hence Clzg can be prepared by electrolyzing a solution of NaCl This fact was one of the major obstacles preventing the discovery of uorine for many years HF was usually chosen as the substance for electrolysis but two problems interfered with the experiment First any water in the HF was oxidized before the uoride ion Second pure HF without any water in it is a nonconductor of electricity HF is a weak acid The problem was finally solved by dissolving KF in liquid HF to give a conducting solution 1973 The cell voltage is given by 7 00257 v In 31121 ld ute 2 E E0 2 cu lconcentrated 0 7 00257 v 2 l 0035 V 0080 n 12 1974 We can calculate the amount of charge that 40 g of M102 can produce 1mol X m 8694 g 2 mol M102 1mol e 40gMiOzx 444gtlt103C Since a current of one ampere represents a ow of one coulomb per second we can find the time it takes for this amount of charge to pass 574 1975 1976 1977 CHAPTER 19 ELECTROCHEMISTRY 00050A 00050 Cs 1s 1h gtlt 25gtlt102h 00050 c 3600 s 444 gtlt103 C X 2H20l 029g 4Haq 4 4e 4HzOl 467 gt 2Hzg 4OH7aq anode cathode The two electrode processes are The amount of hydrogen formed is twice the amount of oxygen Notice that the solution at the anode will become acidic and that the solution at the cathode will become basic test with litmus paper What are the relative amounts of H4r and 0H7 formed in this process Would the solutions surrounding the two electrodes neutralize each other exactly If not would the resulting solution be acidic or basic Since this is a concentration cell the standard emf is zero Why Using the Nernst equation we can write equations to calculate the cell voltage for the two cells RT RT 1 E i ln i ln cell nF Q 2 F Hggir soln A Hgsoln B RT In Hgir soln A RT 2 Emu i an 7 nF 1 F Hg soln B In the first case two electrons are transferred per mercury ion n 2 while in the second only one is transferred n 1 Note that the concentration ratio will be 110 in both cases The voltages calculated at 18 C are 78314 IKmol291K1n101 1 1 00289 v 296500 I v mol 1 Ecell 48314 Kmol291K1n101 00577 v 196500 I V lmol l 2 Ecell Since the calculated cell potential for cell 1 agrees with the measured cell emf we conclude that the mercury1 ion exists as Hg2 in solution According to the following standard reduction potentials Ozg 4Haq 4e a ZHZO E 123 V 12s 2e 217aq E 053 V we see that it is easier to oxidize the iodide ion than water because 02 is a stronger oxidizing agent than 12 Therefore the anode reaction is 217aq gt 12s 267 The solution surrounding the anode will become brown because of the formation of the triiodide ion 1 120 9 1370111 The cathode reaction will be the same as in the NaCl electrolysis Why Since OH7 is a product the solution around the cathode will become basic which will cause the phenolphthalein indicator to turn red 1978 CHAPTER 19 ELECTROCHEMISTRY 575 We begin by treating this like an ordinary stoichiometry problem see Chapter 3 Step 1 Calculate the number of moles of Mg and Ag The number of moles ofmagnesium is 156 g Mg X M 00642 mol Mg 2431 g Mg The number of moles of silver ion in the solution is i W X 01000 L 00100 mol Ag Step 2 Calculate the mass of Mg remaining by determining how much Mg reacts with Ag The balanced equation for the reaction is 2Agaq Mgs me Mngaq Since you need twice as much Agir compared to Mg for complete reaction AgJr is the limiting reagent The amount ofMg consumed is 00100 mol Ag X M 000500 mol Mg Ag The amount of magnesium remaining is 006427 000500 molM gtlt 144 M g 1 1 g g mo Step 3 Assuming complete reaction calculate the concentration of Mg2 ions produced Since the mole ratio between Mg and Mg2 is 11 the mol of Mg2 formed Will equal the mol of Mg reacted The concentration of Mg is 7 000500 mol Mg20 0100 L 00500 M Step 4 We can calculate the equilibrium constant for the reaction from the standard cell emf E3611 Egamode 7 Ede 080 V 7 7237 V 317 V We can then compute the equilibrium constant quotE2611 K 600257 2317 K 6 00257 lgtlt 10107 576 CHAPTER 19 ELECTROCHEMISTRY Step 5 To find equilibrium concentrations of Mg2 and Ag we have to solve an equilibrium problem Let x be the small amount of Mg2 that reacts to achieve equilibrium The concentration of Aglr will be 2x at equilibrium Assume that essentially all AgJr has been reduced so that the initial concentration of Aglr is zero 2Agaq Mgs 2Ags Mngaq Initial M 00000 00500 Change 1V1 2x 7x Equilibrium 1V1 2x 00500 7 x K Mgzu Ag 12 1X10107 00500 x 202 We can assume 00500 7 x m 00500 lgtlt10107 m m 202 2x2 amp37 00500 gtlt10 107 lgtlt 10 7109 7110 202 500gtlt 10 500gtlt 10 2x 7gtlt 10 55M Ag 2x 7 x 10 55M Mg2 00500 7 x 00500M 1979 Weigh the zinc and copper electrodes before operating the cell and reweigh afterwards The anode Zn should lose mass and the cathode Cu should gain mass 1980 21 Since this is an acidic solution the gas must be hydrogen gas from the reduction of hydrogen ion The two electrode reactions and the overall cell reaction are anode Cus Cu2aq 267 cathode 2Haq 267 Hzg Cus 2Haq Cu2aq Hzg Since 0584 g of copper was consumed the amount of hydrogen gas produced is lmol Cu X lmole 0584 g Cu gtlt 6355 g Cu lmol Cu 920 x10 3molH2 At STP 1 mole of an ideal gas occupies a volume of 2241 L Thus the volume of Hz at STP is 2241 L lmol 0206 L VH2 920x10 3molH2gtlt 1981 1982 CHAPTER 19 ELECTROCHEMISTRY 577 b From the current and the time we can calculate the amount of charge 1c gtltl52gtlt103s l79gtlt103C lAs 118 Agtlt Since we know the charge of an electron we can compute the number of electrons l e 112x1022e 16022x10 19c 179 gtlt103 C X Using the amount of copper consumed in the reaction and the fact that 2 mol of ei are produced for every 1 mole of copper consumed we can calculate Avogadro39s number 112x1022e lmol Cu 920 gtlt10 3 mol Cu 609 x 1023 lmol e 2 mol 6 In practice Avogadro39s number can be determined by electrochemical experiments like this The charge of the electron can be found independently by lLillikan s experiment The reaction is Al3 367 gt Al First let39s calculate the number of coulombs of electricity that must pass through the cell to deposit 602 g of A1 602 g Al X M X 2698 g Al 3 mol 6 X 96500 C 646x105 c lmol Al 1 mol 6 The time inmin needed to pass this much charge is l l gtlt gtlt 0352 A 306 x 104 min 5 tmin 646 X 10 C X 60 S A l C a We can calculate AGO from standard free energies of formation AGquot ZAG N2 6AG H20 7 MAG N39H3 SAG 02 AG 0 672372 kJmol 7 47166 kJmol 0 AG 13568 kJmol b The halfreactions are 4N39H3g 2Nzg12Haq 12e 30zg12Haq 12e 6HzOl The overall reaction is a lZelectron process We can calculate the standard cell emf from the standard free energy change AGO AGquot inFEgell 7 713568 kJ X 1000 J o 7 G0 1 mol 1 k1 Ecen 117 V A nF 12965001Vm61 578 1983 1984 CHAPTER 19 ELECTROCHEMISTRY Cathode Anode Au3aq 367 gt Aus 2HzOl a 02g 4Haq 46 21 First from the amount of gold deposited we can calculate the moles of 02 produced Then using the ideal gas equation we can calculate the volume of Oz 926 g Au XM w w 003525mO102 1970 g Au lmol Au 4 mol 6 003525 1 00821L39altm 296K quot0 RT 39 me I molK V02 ZT 0872 L 747 mmHg X i 760 mmHg 1 Current 1 M 15 r 200hx60 x S 720gtlt103s l h lm1n lmol Au 3mol e 965OOC lA X X X Q 926gAux S l36gtlt104As 1970gAu lmol Au C 1 mol e7 1 Current I 4 wwisszamp s 720 X 103 s The reduction of Agir to Ag metal is Agaq ei Ag We can calculate both the moles of Ag deposited and the moles of Au deposited mol Ag 264g Agx 245gtlt lO zmol Ag 1079g Ag mol Au l6lgAugtltM 8l7gtlt10 3mol Au 1970gAu We do not know the oxidation state of Au ions so we will represent the ions as Au If we divide the mol of Ag by the mol of Au we can determine the ratio of Agir reduced compared to Au reduced 245 X 10 2 mol Ag 7 3 817x10 3molAu That is the same number of electrons that reduced the Ag ions to Ag reduced only onethird the number of moles of the Aunir ions to Au Thus each Aunir required three electrons per ion for every one electron for Ag The oxidation state for the gold ion is 3 the ion is Au Au3aq 3e7 Au CHAPTER 19 ELECTROCHEMISTRY 579 1985 Heating the garage Will melt the snow on the car which is contaminated with salt The aqueous salt will hasten corrosion 1986 We reverse the first halfireaction and add it to the second to come up with the overall balanced equation ngz 2Hg2 2e Ede 092 V ng 26 2Hg Egamode 085 V 2Hg22 2Hg2 2Hg EEC 085 V 7 092 V 7007 V Since the standard cell potential is an intensive property ng aq ngYaq ng Es 7 007 v We calculate AGO from E0 AG inFEO 7l96500 JVmol7007 V 68 kJmol The corresponding equilibrium constant is ngt Hg ti We calculate K from AGO AGO iRTan K 7 768x103Jmol 8314 JK mol298 K K 0064 1987 a Anode ZFT gt F2g267 Cathode 2H2e H2g Overall 2H2F a H2gFzg b KF increases the electrical conductivity What type of electrolyte is HF D The Kit is not reduced c V Calculating the moles of F2 3600sgtlt lC gtltlmole xlmole lh lAs 96500C Zmole 502Agtlt15hgtlt l40molF2 Using the ideal gas law 7 nRT 7 140 mol00821LatmKmol297 K P 12 atm V 28 x 103 L 1988 The reactions for the electrolysis of NaClaq are Anode 2Cliaq Clzg 267 Cathode 2HzOl 267 gt Hzg 20Hiaq Overall 2HzOl 2Cliaq Hzg Clzg 20Hiaq 580 1989 1990 CHAPTER 19 ELECTROCHEMISTRY From the pH of the solution we can calculate the OH7 concentration From the 0H7 we can calculate the moles of OH7 produced Then from the moles of 0H7 we can calculate the average current used pH 1224 pOH 1400 71224 176 0H 174 gtlt10TZM The moles of OH7 produced are 72 Wx 0300 L 522 x10 3mol OH From the balanced equation it takes 1 mole of ei to produce 1 mole of OH7 ions 1 mol 6 96500 C 522X10 3molOH gtlt 504 C 1mol OH lmol 6 Recall that 1 C 1 As 504Cgtlt1AISgtlt1m mgtlt 14A 1 C 60 s 600 min a Anode Cus gt Cu2aq 267 Cathode Cu2aq 267 gt Cus The overall reaction is Cus gt Cus Cu is transferred from the anode to cathode b Consulting Table 191 of the text the Zn will be oxidized but Zn2 will not be reduced at the cathode Ag will not be oxidized at the anode c The moles of Cu 1000 g Cu X M 157 mol Cu 6355 g Cu The coulombs required 157 mol Cu gtlt X W 303 gtlt106 C 1mol Cu 1 mol e 6 The time required 7s 160 gtlt105s 189 A 160gtlt105sgtlt 1hr 444 hr 3600 s The reaction is Ptnne7 Pt Thus we can calculate the charge of the platinum ions by realizing that n mol of ei are required per mol of Pt formed CHAPTER 19 ELECTROCHEMISTRY 581 The moles ofPt formed are 909 g Pt gtlt 1L1 00466 mol Pt 1951 g Pt Next calculate the charge passed in C c 200hx x 180gtlt104C 1h 1s Convert to moles of electrons 4 lmol e mol e 180gtlt 10 Cgtlt 0187 mole 96500 C We now know the number ofmoles of electrons 0187 mol e7 needed to produce 00466 mol of Pt metal We can calculate the number of moles of electrons needed to produce 1 mole of Pt metal m 401 mol eTmol Pt 00466 mol Pt Since we need 4 moles of electrons to reduce 1 mole of Pt ions the charge on the Pt ions must be 4 1991 Using the standard reduction potentials found in Table 191 Cd2aq2e Cds E 7040 V Mg2aq2e a Mgs E 7237 V Thus Cd2 will oxidize Mg so that the magnesium halfreaction occurs at the anode 2 2 Mgs Cd aq gt Mg aq Cds EEC Egamode 7 Ede 7040 V 7 7237 V 197 V e gt 67 Cd salt cathode anOde I I l bridge I l I 1992 The halfireaction for the oxidation of water to oxygen is ZHZOU oxidation anode 02g4Haq 467 Knowing that one mole of any gas at STP occupies a volume of 2241 L we find the number ofmoles of oxygen mol 426 L 02 X 2241 L 0190 mol Oz 582 1993 1994 1995 CHAPTER 19 ELECTROCHEMISTRY Since four electrons are required to form one oxygen molecule the number of electrons must be 4 mol e 23 7 0190 mol 02x w lmol Oz 1 mol 458 x1023 The amount of charge passing through the solution is 1C X 3600s lAs 600Agtlt gtlt340h 734gtlt104C We find the electron charge by dividing the amount of charge by the number of electrons 734gtlt104C 23 160 x 10 19Ce 458 X 10 e In actual fact this sort of calculation can be used to find Avogadro39s number not the electron charge The latter can be measured independently and one can use this charge together with electrolytic data like the above to calculate the number of objects in one mole See also Problem 1980 a Aus 3HNO3aq 4HClaq gt HAuCl4aq 3HzOl 3NOzg b To increase the acidity and to form the stable complex ion AuCl47 Cells of higher voltage require very reactive oxidizing and reducing agents which are difficult to handle From Table 191 of the text we see that 592 V is the theoretical limit of a cell made up of LiLi and FzF electrodes under standardstate conditions Batteries made up of several cells in series are easier to use The overall cell reaction is 2Agaq Hzg gt 2Ags 2Haq We write the Nernst equation for this system E E 7 00257 van n 00257 v Ht2 2 E0080V 2 Ag lPHZ The measured voltage is 0589 V and we can find the silver ion concentration as follows 0589 v 080 v 7 Mm 2 Agz 1 ln 2 1642 Ag 2 l4gtlt107 Ag Ag 27gtlt1074M CHAPTER 19 ELECTROCHEMISTRY 583 Knowing the silver ion concentration we can calculate the oxalate ion concentration and the solubility product constant czoi Ag 135x10 4M Ksp Agt2czo42 27x 10 42135 x1074 98x 10 12 1996 The halfreactions are Zns 4OH7aq 7 ZnOH427aq 267 Ede 7136 V anaq72e 7 Zns Egamode 7076 V Zn2aq 7 4OH aq 7 ZnOH427aq E3611 7076 V 7 7136 V 060 V E2611 00257 V1rle n It 2060 Kf 600257 6 00257 2 x 1020 1997 The half reactions are HzOzaq 7 02g2Haq 2e Egmde 068 V HzOzaq 7 2Haq 7 2e 7 2HzOl Egamode 177 V 2HzOzaq 7 ZHZOU 7 02g EEC Egamode 7 Ede 177 V 7 068 V 109 V Thus products are favored at equilibrium H202 is not stable it disproportionates 1998 21 Since electrons ow from X to SHE E for X must be negative Thus E for Y must be positive b Y2t72e 7 Y Egamode 034 V X 7 X 72e Egmde 7025 V X 737 7 XZ7 Y Eggquot 034 V 7 7025 V 059 V 1999 a The half reactions are Sn4aq72e 7 Snzaq Egamode 013 V 2Tls 7 Tlaq 7 e Egmde 7034 V Sn4aq 7 2Tls 7 Sn2aq 7 2T1aq E3611 Egamode 7 Egmde 013 V 7 7034 V 047 V b V 0 RT Ecell Ean o47 v WM 296500 K s x 1015 584 CHAPTER 19 ELECTROCHEMISTRY 2 c E E 7 mw 047 7 00592v 041 V 2 10 19100 21 Gold does not tarnish in air because the reduction potential for oxygen is insufficient to result in the oxidation of gold 02 4H 4e 2Hzo Egamode 123 V That is E2611 Egamode 7 Ede lt 0 for either oxidation by 02 to AuJr or Au3 Egg 123 v7150 V lt 0 or E3611 123 v7169 V lt 0 b 3Au e a Au Egamode 169 V Au a Au33e Egmde 150 V 3Au a 2AuAu3 Egg 169 V 7150 V 019 V Calculating AG AGO 7nFE 7396500 JVmol0l9 V 7550 kJmol For spontaneous electrochemical equations AGO must be negative Thus the disproportionation occurs spontaneously c Since the most stable oxidation state for gold is Au3 the predicted reaction is 2Au 3F2 2AuF3 19101 It is mercury ion in solution that is extremely hazardous Since mercury metal does not react with hydrochloric acid the acid in gastric juice it does not dissolve and passes through the human body unchanged Nitric acid not part of human gastric juices dissolves mercury metal see Problem 19113 if nitric acid were secreted by the stomach ingestion of mercury metal would be fatal 2 7 2 3 19102 The balanced equation is 5Fe M104 8H M1 5Fe 4 H20 Calculate the amount of ironH in the original solution using the mole ratio from the balanced equation 00200 mol KMno4 X 5 mol Fe 230 mL gtlt 000230 mol Fe2 1000 mL soln 1 mol KlVan4 The concentration of ironll must be Fey 000230 mol 00920 M 00250 L The total iron concentration can be found by simple proportion because the same sample volume 250 mL and the same KlVan4 solution were used 400 mL KlVan4 gtlt 00920 M 0160 M 230 mL KlVan4 Feltotal CHAPTER 19 ELECTROCHEMISTRY 585 Fey Fe1ma17Fe2 00680M Why are the two titrations with permanganate necessary in this problem 19103 Viewed externally the anode looks negative because of the ow of the electrons from Zn gt Zn2 267 toward the cathode 1n solution anions move toward the anode because they are attracted by the Zn2 ions surrounding the anode 19104 From Table 191 of the text Hzozaq2Haq 26 2HzOl Egamode 177 v HzOzaq 02g 2Haq 26 Egmde 068 V 2HzOzaq a 2HzOl Ozg EEC Egamodei Egmde 177 v 7 068 V 109 v Because E is positive the decomposition is spontaneous 19105 21 The overall reaction is Pb PbOz H2504 gt 2PbSO4 2HZO 129 g 1n1t1al mass of H2504 724 mL X X 0380 355 g 1 mL 119 g F1nal mass oszSO4 724 mL X 1 L X 0260 224 g m Mass of H2504 reacted 355 g 7 224 g 131 g 1mol 9809 g 134 mol Moles ofHZSO4 reacted 131g gtlt Q 134 molHZSO4gtlt x m 129 x 105C 2 mol H2504 1 mol 6 5 b t 9 w 576gtlt103s 160hr I 224 A 19106 21 unchanged b unchanged c squared d doubled e doubled 19107 21 Spoon cathode Pt anode 4 AgN03 solution 586 CHAPTER 19 ELECTROCHEMISTRY 1079 g lmol 185 X 10 3A 088 X lmol 196500 c b t 7 427gtlt104s 119 hr 19108 F g 2H aq 2e ZHF g 2 E E0 ElnPH Fz 2F PF2H With increasing H E Will be larger F2 Will become a stronger oxidizing agent 19109 Advantages 19110 Pb a Pb22e 19111 a At the anode Mg a No startup problems b much quieter c no pollution smog d more energy efficient in the sense that when the car is not moving for example at a traffic light no electricity is consumed Disadvantages a Driving range is more limited than automobiles b total mass of batteries is appreciable c production of electricity needed to charge the batteries leads to pollution Ede 7013 v 2H2e a H Egamode 000 v Pb2H a Pb2Hz E3611 000 v7 7013 V 013 v pH 160 H 10 13960 00st RT 1sz JPH E E 7 1n 2 nF Ht2 00257 v 0035PH 0 013 7 1n 2 2 00252 026 7 0035PHZ 00257 7 00252 PH2 44 X 102 atm Mg a Mg2 2e Also Mg ZHCl gt MgClz H2 At the cathode Cu 2H4r 267 gt Hz A 5quot V c V 19112 21 CHAPTER 19 ELECTROCHEMISTRY 587 The solution does not turn blue After all the HCl has been neutralized the white precipitate is Mg2 on MgOHzs The halfreactions are Anode Zn gt Zn2 267 Cathode 02 267 gt 027 Overall Zn Oz gt ZnO To calculate the standard em f we first need to calculate AGO for the reaction From Appendix 3 of the text we write AGquot AG ZnO 7AG Zn AG OZ AGquot 73182 kJmol 7 00 AGquot 73182 kJmol AGquot inFEquot 73182 X 103 Jmol 7296500 JVmolE E 165V b We use the following equation c V V RT E E 7 ln nF Q E 165V700257v1n 1 POZ E 165V700257v1n 1 021 E 165 V7 0020 V E 163 V Since the free energy change represents the maximum work that can be extracted from the overall reaction the maximum amount of energy that can be obtained from this reaction is the free energy change To calculate the energy density we multiply the free energy change by the number of moles of Zn present in 1 kg of Zn 1 mol Zn 6539 g Zn 3182 kJ 1000 g Zn 1 mol Zn 487 x 103 kJ Zn 1 kg Zn kg energy density One ampere is l Cs The charge drawn every second is given by nF charge nF 21 X 105 c n96500 Cmol e n 22 mol ei 588 19113 21 CHAPTER 19 ELECTROCHEMISTRY From the overall balanced reaction we see that 4 moles of electrons will reduce 1 mole of Oz therefore the number ofmoles of Oz reduced by 22 moles of electrons is m 055 mol 02 m mol Oz 22 mol e X 4 The volume of oxygen at 10 atm partial pressure can be obtained by using the ideal gas equation 7 nRT 7 055 mol0082lLatmmolK298 K 2 P 10 atm V0 13L Since air is 21 percent oxygen by volume the volume of air required every second is Vairl3LOzgtlt 62L0fair 100 air 21 Oz HCl First we write the halfreactions Oxidation 2Hgl Hg22lA4 2e7 Reduction 2Hl A42ei Hzl atm Overall 2Hgl 2H1M ngzm M H20 atm The standard emf E0 is given by EC Egathode T Egnode E 0 7 085 V E 085V We omit the subscript cell because this reaction is not carried out in an electrochemical cell Since E is negative we conclude that mercury is not oxidized by hydrochloric acid under standardstate conditions HNO3 The reactions are Oxidation 32Hgl ngzl A4 2e7 Reduction 2NO37l A4 4Hl A4 367 gt NOl atm 2HzOl Overall 6Hgl 2NO37l A4 8Hl A4 3ngzl A4 2NOl atm 4HzOl Thus E0 Egathode T Egnode EO 096V7 085V E 011V Since E is positive products are favored at equilibrium under standardstate conditions The test tube on the left contains HNO3 and Hg HNO3 can oxidize Hg and the product NO reacts with oxygen to form the brown gas NOz CHAPTER 19 ELECTROCHEMISTRY 589 19114 We can calculate AGEXH using the following equation AGroxn EnAGE products 7 ZmAGf reactants AG 0 0 7 l72938 kJmol 0 2938 kJmol Next we can calculate E using the equation AGquot 7nFE We use a more accurate value for Faraday39s constant 2938 gtlt 103 Jmol 7l964853 JVmolE E 7305V 19115 21 First we calculate the coulombs of electricity that pass through the cell 1 C X 3600 s l As h 022Agtlt x316h 25gtlt104C We see that for every mole of Cu formed at the cathode 2 moles of electrons are needed The grams of Cu produced at the cathode are lmol e lmol Cu 6355 g Cu gtlt gtlt 82 g Cu 96500 C 2 mol e lmol Cu gCu 25gtlt104Cgtlt b 82 g of Cu is 013 mole of Cu The moles of Cu2 in the original solution are 2 0218 L X 0218 mol Cult 1 L soln The mole ratio between Cu2 and Cu is 11 so the moles of Cu2 remaining in solution are moles Cu2 remaining 0218 mol 7 013 mol 0088 mol Cuer The concentration of Cu2 remaining is 0088 mol Cult Cu2 040 M 0218 L soln 19116 First we need to calculate E2611 then we can calculate K from the cell potential H2g 2Haq2e Egmde 000 V 2HzOl 2e Hzg 2OH Egamode 7083 v 2HzOl 7 2Haq 20H aq EEC 7083 v 7 000 v 7083 v We want to calculate K for the reaction HzOl 7 Haq OH7aq The cell potential for this reaction will be the same as the above reaction but the moles of electrons transferred n will equal one 590 19117 19118 CHAPTER 19 ELECTROCHEMISTRY 7 00257 v Egell lan n In K quotEgell W 00257 V L KW 600257 10083 KW e 00257 e42 1x10 14 H202 Fuel Cell Pros 1 Almost inexhaustible supplies of Hz from H20 and 02 from air 2 There is no pollution because the product generated is H20 3 The generation of electricity is not restricted by second law efficiency as in combustion reactions 4 There is no thermal pollution Cons I A large input of energy is required to generate Hz 2 Storage of hydrogen gas is problematic 3 Electrocatalysts are expensive 4 The impact of appreciable leakage of Hz into the atmosphere is uncertain How will Hz affect climate and O3 in the stratosphere Coal red Power Station Pros 1 2 Large deposits of coal Technology available at acceptable cost Use of coal and oil are interchangeable Cons I The emission pollution to the environment C02 NOX SOz Hg etc leads to global warming acid rain and smog Combustion reactions cause thermal pollution The generation of electricity by combustion reactions is restricted by second law ef ciency Mning coal is unsafe and damaging to the environment a V8 4 a lAh 1Agtlt 3600s 3600 C b Anode Pb 5042 a PbSO4 2e Two moles of electrons are produced by 1 mole of Pb Recall that the charge of 1 mol ei is 96500 C We can set up the following conversions to calculate the capacity of the battery mol Pb gt molei gt coulombs gt ampere hour 1 h 3600 s 1 mol Pb X 2 mol 6 X 96500 C 2072 g Pb lmol Pb 104 Ah 406nggtlt 374gtlt105Cgtlt 1 mol 6 This amperehour cannot be fully realized because the concentration of H2504 keeps decreasing c E2811 170 V 7 7031 V 201 V From Table 191 ofthe text CHAPTER 19 ELECTROCHEMISTRY 591 AGquot 7nFE AGquot 7296500 JVmol201 V 388x 105Jmol Spontaneous as expected 19119 First we start with Equation 193 of the text AGO 7nFE Then we substitute into Equation 1810 of the text AGO AHO 7 TASO 7nFE AHO 7 TASO E0 7AH TAS nF The reaction is Zns Cu2aq 7 Zn2aq Cus We use data in Appendix 3 of the text to calculate AHO and ASquot AH AH Cus AH Zn aq 7 AHFZns AH Cu aq AHO 0 71524 kJmol 7 0 6439 kJmol 72168 kJmol AS S Cus S anaq 7 S Zns S Cu2aq AS 333 JKmol 710648 JKmol 7 416 JKmol7 996 JKmol 7152 JKmol At 298 K 25 C 7 72168 kJ X10001 298 K 7152 J 1 mol l kJ K mol E 1100 V camed to 3 dec1mal places 296500 JV mol At 353 K 80 C 211681k gtlt 10T 353 K7152 K J I E m 39m0 1095 V 296500 JVmol This calculation shows that E is practically independent of temperature In practice E does decrease more noticeably with temperature The reason is that we have assumed AH and ASquot to be independent of temperature which is technically not correct 19120 The surface area of an open cylinder is 21trh The surface area of the culvert is 21t0900 m400 m X 2 for both sides of the iron sheet 452 m2 Converting to units of cmz 2 452 m2gtlt1010 452gtlt106cm2 m 592 CHAPTER 19 ELECTROCHEMISTRY The volume of the Zn layer is C 0200 mm x1 m X 452x106cm2 904gtlt104cm3 m m The mass of Zn needed is 645 gtlt105 g Zn 904 gtlt104 cm3 gtlt 73914 g 1 C1113 anJr 267 gt Zn Q 645gtlt105g Znx xmxm l90gtlt109C 6539 g Zn lmol Zn 1m01 e l Il C X l V 9 Total energy 652 x109 I 095 lt eff1c1ency Cost 652 x109 J gtlt 1ka gtlt 1h gtlt 012 217 1000 3600 s lkwh s 19121 21 The halfcell reactions are anode 217 gt 12 267 cathode 2HzO 2e7 gt Hz 2OH7 779 mmHg X i 122 L PV 760 mmHg nHZ E Liam 005094 mol 00821 j299 K molK Q 005094 mol gtlt x w 983 x 103C 1 101 H2 lmol e b QIr 3 I 2 WXIAS 1302s 217min 1 755A 1c c The White precipitate is MgOH2 Mg aq2OH aqgt MgOH2S From Table 162 of the text we see that MgOH2 is insoluble Ksp 12 X 10711 From the cathode half cell reaction 71 2nHZ OH CHAPTER 19 ELECTROCHEMISTRY 593 2005094 mol OH 0900 L 0ll3M llgz 0200M We assume that all of the 0H7 ions are converted to MgOH2 hydroxide ion is the limiting reagent Mg2 20H MgOHz Because the mole ratio between OH7 and MgOH2 is 21 the moles of MgOH2 produced is 005094 mole The mass of MgOH2 produced is 5833 OH 005094 mol MgOH2 gtlt 297 g Mg0H2 1 mol MgOH2 19122 It might appear that because the sum of the first two halfreactions gives Equation 3 E is given by E1 E 033 V This is not the case however because emf is not an extensive property We cannot set E E1 On the other hand the Gibbs energy is an extensive property so we can add the separate Gibbs energy changes to obtain the overall Gibbs energy change AG AG1 AG Substituting the relationship AG0 7 inFEO we obtain n3FE anE10 nzFEE n1E1 anE quot3 E n1 2 n2 1 andn33 E 27044 V l077 V 0037 V 19123 21 The reaction is Zn Cu2 gt Zn2 Cu Using the Nernst equation E0 7 00257 020 E ln 110 V 2 020 i If NH3 is added to the CuSO4 solution Cu 4NH3 CuNH342 The concentration of copper ions Cu2 decreases so the ln term becomes greater than 1 and E decreases 594 CHAPTER 19 ELECTROCHEMISTRY ii If NH3 is added to the ZnSO4 solution an 4NH3 ZnNH34Z The concentration of zinc ions Zn2 decreases so the ln term becomes less than 1 andE increases b After addition of 250 mL of 30 MN39H3 Cu 4NH3 a CuNH342 Assume that all Cu2 becomes CuN39H342 CuNH342 010M N113 M44014 110M E E0 7 00257lnZn2 2 cm 068 v 110 viwm 03920 2 Cu211 Cuz 13x10 15M c NH 2 Kr u 34 l 7 010 53x1013 7 Cu2NH34 7 13 X 10 15114 Note this value differs somewhat from that listed in Table 164 of the text 19124 First calculate the standard emf of the cell from the standard reduction potentials in Table 191 of the text Then calculate the equilibrium constant from the standard emf using Equation 195 of the text E3611 Egamodei Ede 034 V 7 7076 V 110 V quotE12211 an 00257 V quotE3611 2010 V K 600257v e 00257v K 2 x 1037 The very large equilibrium constant means that the oxidation of Zn by Cu2 is virtually complete 19125 21 From Table 191 of the text we see that Al can reduce Agir applying the diagonal rule Al gt Al3 3e Ag e7 Ag A1 3Ag A13 3Ag CHAPTER 19 ELECTROCHEMISTRY 595 b A NaHC03 solution is basic The surplus 0H7 ions will convert Al3 ions to AlOH3 which precipitates out of solution Otherwise the Al ions formed can from A1203 which increases the tenacious oxide layer over the aluminum foil thus preventing further electrochemical reaction The A1203 layer is responsible for preventing aluminum beverage cans from corroding c Heating the solution speeds up the process and drives air oxygen out of solution which minimizes A1203 formation d HCl reacts with Ang 2HCl Ang a HzS 2AgCl The HZS gas has an unpleasant odor and in fact is poisonous Also this method removes Ag from the spoon as AgCl The method described in part a does not Another example of calculating whether a reaction is spontaneous Consider 609 2 H2g gt CH30H Question is the reaction spontaneous lefttoright at 298 K Answer calculate AS SYS A5 rxn AS surr and thus A5 univ i Calculate AS Sys from the Table of 5 values Appendix 3 AS SYS1mol1268JKmol1mol1979JKmol2mo1310JKmo AS SYS 3311 JK Note sign agrees with prediction ii Calculate AH sys from Table of AH f values Appendix 3 AH SyS 1mol2387kJmol 1mo1105 kJmol 2mo0 kJmo AH SYS 4282 kJ Now AS SW AH SyST As m 1282 kJ298K1ooo JkJ 4302 JK iii Calculate A5 umv AS SYS AS SW 3311JK 4302 JK As um 991 JK Sectio 18 5 Gibbs Free e 6 To avoid having to consider AS SYS and AS SW all the time a new thermodynamic quantity was introduced in the 19H1 century called the Gibbs Free Energy 6 A Since ASH v ASSYS ASSN and ASSN A Hg Asuniv A5s s ys Y T TASuniv TASSYS AHSYS and TASW AHSYS TASSYS AGSYS AGSYS AHSYS TASSS or usually just as A6 AH TAS Since AGSYS TASUiv and ASWgt 0 for a spontaneous reaction AG lt 0 ie negative for a spontaneous rxn AG gt 0 ie positive for a nonspontaneous rxn A6 0 for a process at equilibrium This is the most convenient form of the 2quot 39 Law of Thermodynamics ie saying AGSYS lt 0 for a spontaneous reaction is the same as saying Asuniv gt O Free Energy is a state function and AG standard free energy change kJmol Note since H values not available 6 values also are not and we can only calculate AG values Appendix 3 lists AGOform values for this Calculating A6 rxn AG SYS for reactions Example Calculate AG M for 4 KClOa s gt 3 K00 s KCI 5 Method 1 Calculate AH rxn and A5 m use AG AH TAS AH W 3mol4335 kJmol1mo4359 kJmo4mol3912kJmo AH rxn 1716 kJ exothermic A5 m 3mo1510 JmolK1mol827 JmolK 4mo1430 JmolK AS rxn 363 JK 363 x10393 kJK A6 m AH rxn TAS rxn 1716 kJ 298K363 x 10393 kJK AG rxn 161 kJ spontaneous Method 2 Use AG f values of products and reactants AG f standard free energy of formation free energy change to make substance from its component elements Like for AH f the AG f 0 for an element in its standard state AG rxn EmAG f products 2nAG f reactants AG rxn 1 mol 4083 kJmol 3 mol 3042 kJmol 4 mol 2899 kJmol AG rxn 1613 kJ same as Method 1 Slight discrepancies sometimes due to experimental sig figserrors in AH f 5 and AG f values in Appendix 3 The Meaning of AG For a spontaneous process A6 maximum work w the system can do For a nonspontaneous process A6 the minimum work that must be done to the system to make the reaction happen A6 wmax Since A6 AH TAS 39AH A6 TAS AH wmax TAS AH heat transferred A6 wmax portion usable to do work TAS portion lost as entropy change i e part of the energy released AH goes into an entropy change and is not available to do work Consider burning of octane in a car engine C8H18l12 02 9 gt 8 C02 9 9 H20 9 Very exothermic AH lt 0 and AS gt 0 12 moles gas gt 17 moles gas AH A6 TAS wmuX TAS AG is portion of AH usable to do work drive car wheels TAS portion lost as increased entropy of car engine temperature and surrounding air not usable for work m A6 wmax in practice we cannot get maximum amount because of further losses to eg friction Fraction of AG used determines the efficiency of the process no process is 100 efficient Summam If AG lt 0 process is spontaneous and can do work on the surroundings If AG gt 0 process is nonspontaneous and will not occur unless surroundings do work on it If A6 0 reaction is at equilibrium and can no longer do work E ect 0 Te erature on Reaction 5 ontaneit M exothermic reactions are also spontaneous because the AH contribution to A6 is much greater than the entropy contribution TAS ie the neqative AH helps make AG neqative However in many cases the T can change the importance of the TAS term Let39s look at the possibilities and assume AH and A5 are approximately independent of T which they are if no phase change occurs Remember AG AH TA AH A5 TA5 AG Conclusion AG lt 0 at all T r39xn spontaneous at all T AG gt 0 at all T r39xn nonspontaneous at all T or AGlt0atowT spontaneous at low T A6 gt 0 at high T nonspontaneous at high T or AGgt0atowT nonspontaneous at low T A6 lt 0 at high T spontaneous at high T Examples 2 H202 gt 2 H20 02 AH lt 0 AS gt 0 Spontaneous at all T 302g gt203g AHgt0ASlt0 nonspontaneous at all T AHgt0ASgt0 nonspontaneous at lower T spontaneous at higher T Na 5 Clz g gt 2 NaCl 5 AH lt 0 AS lt 0 spontaneous at lower T nonspontaneous at higher T Consider 2 502 g 02 g 2 2 503 g at 298 K AG 1416 kJ AH 1984 kJ AS 1879 JK AG lt 0 at 298 K rxn is spontaneous at 298 C But AH lt 0 and AS lt 0 expect it to be nonspont at high enough T Question Is it still spontaneous at 900 C ie T 1173 K Answer AG AH TAS 1984 kJ 1173 K1879 J39K10393 kJJ39 AG 220 kJ nonspontaneous at 200 C Te erature at which Reaction beco es 5 ontaneous When the signs of AH and A5 are the same the sign of AG changes with T At which T will this happen Crossover from A6 lt 0 to A6 gt 0 or viceversa occurs when AG 0 AH TASquot ie when AH TA5 Example CuZO s C s gt 2 Cu 5 CO g AH 581 kJ and AS 165 JK At 298 K A6 581 kJ298 K165 JK10393 kJJ A6 89 kJ nonspontaneous Crossover to spontaneous when A6 0 ie AH TAS 39AHO 39As k E 165JK10 3kJJ 79 C spontaneous above 352 K Coupling Reactions to Drive a NonSpontaneous Reaction The negative A6 of a spontaneous reaction can be used to drive another reaction in its nonspontaneous direction Example Making copper metal from CuZO 1 CuaO s gt 2 Cu 5 02 9 At 375 K A6375 140 kJ nonspontaneous However at 375 K 2 C s 02 9 a CO 9 A6375 1438 kJ spontaneous Couple combine reactions 1 and 2 ie add together CuZO s C s gt 2 Cu 5 CO g A6375 38 kJ spontaneous mixing Cs with Cu20s allows rxn 2 to drive rxn 1 We can drive a nonspontaneous reaction by combining it with a sufficiently spontaneous one Similarly reversing the rusting of Fe to get Fe s 1 2 Fezog s gt 4 Fe 5 3 02 g AG 1487 kJ 2 2 co g 02 g a 2 cog g AG 5144 kJ 239 6 co g 3 02 g gt 6 cog g AG 1543 kJ Use 2 to drive 1 2 Fe203 s 6 CO g gt 4 Fe 5 6 602 g AG 56 kJ39 spontaneous Sect 186 Free Energy and Equilibrium Let39s now be more general for all ran not just 100 7o ones The magnitude of A6 of a reaction controls the magnitude of K quot Ii A In Chapter 18 we learn A a B spontaneous direction if AG lt 0 B gt A is the nonspontaneous direction Consider reaction A gt B Q In Chapter 14 we learned Q lt K is the spontaneous direction for lefttoright where Q massaction expression How are the two rules related lt 139 sponlaneous39 AG lt 0 for rxn left gt right ie A gt B gt 1 nonspontaneous AG gt 0 1 equilibrium A6 0 NDNDKD How are QK and AG mathematically related A6 RT ln QK Note consistency with above ln1 0 39 AGRTnQRTln K This relationship tells us that the AG is the difference in free energy between the initial state and final equilibrium state At equilibrium Q K A6 0 Thus AG depends on the difference between Q and K When all initial concs are 1 M or all gas pressures 1 atm we have standard conditions Q 1 an 0andAGAG AG RT In K Extremely important relationship between AG and K Note Small change in A6 can give mg change in K Since most reactions do not begin with reactants in standard conditions we can calculate A6 for any conditions using AGAG RTan This allows us to calculate A6 for any initial conc39s if we know AG which we can calculate from Appendix 3 for example Let39s Practice using these equations Example 1 2 502 g 02 2 2 503 g Question Is it thermodynamically good to run this reaction at a higher T ie does equilibrium lie more to right so we get more product Answer We need to compare the K39s at 298 K and 973 K AG values at the two T39s are for balanced reaction as written AGozgs AG 973 1212 kJ Use these to calculate K at each T At 298 K AG RT In K In K AG RT o 1416kJmox1000JKJ quotAG MT l 831JmoK298K l 572 K e57392 7 x 1024 very big o 1212kJmox1000JkJ ML AG quot2T 831J moK973K K el3950 45 m big 150 Reaction much more to the right at lower temperature ie more products at lower T Exarmoe2 Question If we have two vessels each containing 050 atm 02 0010 atm Oz 010 atm 503 but one vessel is at 298 K and the other at 973 K in which direction will the reaction proceed to reach equilibrium at each T Answer Calculate Q and compare to K at each T P5203 0102 Q P5202332 0502oo10 40 At 298 K K7gtlt1024 Qltlt K 39 reaction spontaneously shifts very much to the right At 973 K K45 zQzK reaction moves only very slightly to the right Finally let39s calculate A6 at each T AGAG RTan A39l39 K39 A6298 AGozga RT ll39l Q 1416 x 103J8314 JK298Kln40 4382 x103 J 4382 kJ AT 973 K39 A6973 A60973 RT ln Q 1212 x 103J8314JK973Kln40 091 x 103339 091 kJ Since AG RT In K and A6 AG RT ln Q we can summarize on a figure how the equilibrium position is determined by the AG and how the reaction has to proceed to get to equilibrium Difference in free energy of pure reactants and pure products AG ltO 0 lt K Spontaneous reaction Increasing free energy 6 gt when 0 9E K 0 K Reactants Equilibrium Products only mixture only Course of reaction gt Remember AG is the change in G from reactants to products IfAG is negative ie AG lt 0 that means the free energy is decreasing on going from reactants to products This is thus a spontaneous process think of this as the reaction will spontaneously go 39downhill39 with respect to 6 and thus 39uphill39 is the non spontaneous direction Note that going from 100 products to equilibrium also occurs spontaneously ie is 39downhill39 in G 1 The spontaneous direction is towards equilibrium i e downhill in 6 Le A6 negative 2 The reaction never lies 100 to one side or the other
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