2. Before we answer the question, here are four important terms relating to redox reactions and galvanic cells.

Transcription

1 CHAPTER SEVENTEEN ELECTROCHEMISTRY Fr Review 1. Electrchemistry is the study f the iterchage f chemical ad electrical eergy. A redx (xidati-reducti) reacti is a reacti i which e r mre electrs are trasferred. I a galvaic, a sptaeus redx reacti ccurs which prduces a electric curret. I a electrlytic, electricity is used t frce a sptaeus redx reacti t ccur.. Befre we aswer the questi, here are fur imprtat terms relatig t redx reactis ad galvaic s. a. Cathde: The electrde at which reducti ccurs. b. Ade: The electrde at which xidati ccurs. c. Oxidati half-reacti: The half-reacti i which electrs are prducts. I a galvaic, the xidati half-reacti always ccurs at the ade. d. Reducti half-reacti: The half-reacti i which electrs are reactats. I a galvaic, the reducti half-reacti always ccurs at the cathde. See Figures 17. ad 17. fr desigs f galvaic s. The electrde cmpartmet i which reducti ccurs is called the cathde ad the electrde cmpartmet i which xidati ccurs is called the ade. These cmpartmets have electrdes (a slid surface) immersed i a sluti. Fr a stadard, the sluti ctais the reactat ad prduct slutes ad gases that are i the balaced half-reactis. The slute ccetratis are all 1 M ad gas partial pressures are all 1 atm fr a stadard. The electrdes are cected via a wire ad a saltbridge cects the tw slutis. The purpse f the electrdes is t prvide a slid surface fr electr trasfer t ccur i the tw cmpartmets. Electrs always flw frm the ade (where they are prduced) t the cathde (where they are reactats). The salt bridge allws cuter is t flw it the tw cmpartmets t maitai electrical eutrality. Withut a salt bridge, sustaied electr flw ca ccur. I the salt bridge, ais flw it the ade t repleish the lss f egative charge as electrs are lst; catis flw it the cathde t balace the egative charge as electrs are trasferred it the cathde. The pull r drivig frce the electrs is called the ptetial (E ) r the electrmtive frce. The uit f electrical ptetial is the vlt (V) which is defied as 1 jule f wrk per culmb f charge trasferred. It is the ptetial that ca be used t d useful wrk. We haress the sptaeus redx reacti t prduce a ptetial which ca d useful wrk. 66

2 66 CHAPTER 17 ELECTROCHEMISTRY. The zer pit fr stadard reducti ptetials (E) is the stadard hydrge electrde. The half-reacti is: H + + e H. This half-reacti is assiged a stadard ptetial f zer, ad all ther reducti half-reactis are measured relative t this zer pit. Substaces less easily reduced tha H + have egative stadard reducti ptetials (E < 0), while substaces mre easily reduced tha H + have psitive stadard reducti ptetials (E > 0). The species mst easily reduced has the mst psitive E value; this is F. The least easily reduced species is Li + with the mst egative E value. Whe a reducti half-reacti is reversed t btai a xidati half-reacti, the sig f the reducti ptetial is reversed t give the ptetial fr the xidati half-reacti (E x = E red ). The species xidized are the prduct side f the reducti half-reactis listed i Table Li will have the mst psitive xidati ptetial [E x = E red = (.05 V) =.05 V, s Li is the mst easily xidized f the species. The species mst easily xidized is the best reducig aget. The wrst reducig aget is F because it has the mst egative xidati ptetial (E x =.87 V). Fr a sptaeus reacti at stadard cditis, E must be psitive (E = E red + E x > 0). Fr ay tw half-reactis, there is ly e way t maipulate them t cme up with a psitive E (a sptaeus reacti). Because the half-reactis d t deped hw may times the reacti ccurs, half-reactis are a itesive prperty. This meas that the value f E red r E x is t chaged whe the half-reactis are multiplied by itegers t get the electrs t crss ff. The lie tati f the stadard galvaic illustrated i Figure 17.5 wuld be: Z(s) Z + (aq) H (g) H + (aq) Pt r Z(s) Z + (1.0 M) H (1.0 atm) H + (1.0 M) Pt The duble lie represets the salt-bridge separatig the ade ad cathde cmpartmets. T the left f the duble lie are the pertiet ade cmpartmet ctets ad t the right are the pertiet cathde cmpartmet ctets. At each ed, the electrdes are listed; t the iside, the sluti ctets are listed. A sigle lie is used t separate the ctets f each cmpartmet wheever there is a phase chage. Here i the cathde cmpartmet, a sigle lie is used t separate H (g) frm H + (aq) ad t separate H + (aq) frm Pt (the electrde). Whe ccetratis ad partial pressures are t listed, they are assumed t be stadard (1.0 M fr slutes ad 1.0 atm fr gases). Fr s havig stadard ccetratis ad pressures, we always iclude the actual ccetratis ad pressures i the lie tati istead f the phases.. G = FE; G is the stadard free eergy chage fr the verall balaced reacti, is the umber f electrs trasferred i the verall balaced reacti, F is called the Faraday cstat (1 F = 96,85 culmbs f charge trasferred per mle f electrs), ad E is the stadard ptetial fr the reacti. Fr a sptaeus redx reacti, E is psitive while G rx is egative. The egative sig is ecessary t cvert the psitive E value fr a sptaeus reacti it a egative G rx. The superscript idicates stadard cditis. These are T = 5C, slute ccetratis f 1.0 M, ad gas partial pressures f

3 CHAPTER 17 ELECTROCHEMISTRY atm. Nte that is ecessary i rder t cvert the itesive prperty E it the extesive prperty, G. 5. E = E RT F l Q; At 5C, the Nerst equati is: E = E lg Q Nstadard cditis are whe slutes are t all 1.0 M ad/r partial pressures f gases are t all 1.0 atm. Nstadard cditis als ccur whe T 5C, Fr mst prblem slvig, T = 5C is usually assumed, hece the secd versi f the Nerst equati is mst fte used. E = ptetial at the cditis f the ; E = stadard ptetial; = umber f electrs trasferred i the verall reacti, ad Q is the reacti qutiet determied at the ccetratis ad partial pressures f the ctets. At equilibrium, E = 0 ad Q = K. At 5C, E = (/) lg K. The stadard ptetial allws calculati f the equilibrium cstat fr a reacti. Whe K < 1, the lg K term is egative, s E is egative ad G is psitive. Whe K > 1, the lg K term is psitive, s E is psitive ad G is egative. Frm the equati E = (/) lg K, the value f E allws calculati f the equilibrium cstat K. We say that E gives the equilibrium psiti fr a reacti. E is the actual ptetial at the cditis f the reacti. If E is psitive, the the reacti is sptaeus as writte (the frward reacti ca be used t make a galvaic t prduce a vltage). If E is egative, the frward reacti is t sptaeus at the cditis f, but the reverse reacti is sptaeus. The reverse reacti ca be used t frm a galvaic. E ca ly be used t determie sptaeity whe all reactats ad prducts are at stadard cditis (T = 5C, [ = 1.0 M, P = 1.0 atm). 6. Ccetrati : a galvaic i which bth cmpartmets ctai the same cmpets, but at differet ccetratis. All ccetrati s have E = 0 because bth cmpartmets ctai the same ctets. The drivig frce fr the is the differet i ccetratis at the ade ad cathde. The prduces a vltage as lg as the i ccetratis are differet. Equilibrium fr a ccetrati is reached (E = 0) whe the i ccetratis i the tw cmpartmets are equal. The et reacti i a ccetrati is: M a+ (cathde, x M) M a+ (ade, y M) E = 0 ad the Nerst equati is: E = E lg Q = a [M (ade) lg where a is the umber f a a [M (cathde) electrs trasferred. T register a ptetial (E > 0), the lg Q term must be a egative value. This ccurs whe M a+ (cathde) > M a+ (ade). The higher i ccetrati is always at the cathde ad the

4 666 CHAPTER 17 ELECTROCHEMISTRY lwer i ccetrati is always at the ade. The magitude f the ptetial depeds the magitude f the differeces i i ccetratis betwee the ade ad cathde. The larger the differece i i ccetratis, the mre egative the lg Q term ad the mre psitive the ptetial. Thus, as the differece i i ccetratis betwee the ade ad cathde cmpartmets icrease, the ptetial icreases. This ca be accmplished by decreasig the i ccetrati at the ade ad/r by icreasig the i ccetrati at the cathde. Whe NaCl is added t the ade cmpartmet, Ag + reacts with Cl t frm AgCl(s). Addig Cl, lwers the Ag + ccetrati which causes a icrease i the ptetial. T determie K sp fr AgCl (K sp = [Ag + [Cl ), we must kw the equilibrium Ag + ad Cl ccetratis. Here, [Cl is give ad we use the Nerst equati t calculate the [Ag + at the ade. 7. As a battery discharges, E decreases, evetually reachig zer. A charged battery is t at equilibrium. At equilibrium, E = 0 ad ΔG = 0. We get wrk ut f a equilibrium system. A battery is useful t us because it ca d wrk as it appraches equilibrium. Bth fuel s ad batteries are galvaic s that prduce ptetials t d useful wrk. Hwever, fuel s, ulike batteries, have the reactats ctiuusly supplied ad ca prduce a curret idefiitely. The verall reacti i the hydrge-xyge fuel is H (g) + O (g) H O(l). The half-reactis are: e + O + H O OH cathde H + OH H O + e ade Utilizig the stadard ptetials i Table 17.1, E = 0.0 V V = 1. V fr the hydrge-xyge fuel. As with all fuel s, the H (g) ad O (g) reactats are ctiuusly supplied. See Figure fr a schematic f this fuel. 8. The crrsi f a metal ca be viewed as the prcess f returig metals t their atural state. The atural state f metals is t have psitive xidati umbers. This crrsi is the xidati f a pure metal (xidati umber = 0) it its is. Fr crrsi f ir t take place, yu must have: a. expsed ir surface a reactat b. O (g) a reactat c. H O(l) a reactat, but als prvides a medium fr i flw (it prvides the salt bridge) d. is t cmplete the salt bridge

5 CHAPTER 17 ELECTROCHEMISTRY 667 Because water is a reactat ad acts as a salt bridge fr crrsi, cars d t rust i dry air climates, while crrsi is a big prblem i humid climates. Saltig rads i the witer als icreases the severity f crrsi. The dissluti f the salt it is the surface f a metal icreases the cductivity f the aqueus sluti ad accelerates the crrsi prcess. Sme f the ways metals (ir) are prtected frm crrsi are listed belw. a. Pait: Cvers the metal surface s ctact ccurs betwee the metal ad air. This ly wrks as lg as the paited surface is t scratched. b. Durable xide catigs: Cvers the metal surface s ctact ccurs betwee the metal ad air. c. Galvaizig: Catig steel with zic; Z frms a effective xide catig ver steel; als, zic is mre easily xidized tha the ir i the steel. d. Sacrificial metal: Attachig a mre easily xidized metal t a ir surface; the mre active metal is preferetially xidized istead f ir. e. Allyig: Addig chrmium ad ickel t steel; the added Cr ad Ni frm xide catigs the steel surface. f. Cathdic prtecti: A mre easily xidized metal is placed i electrical ctact with the metal we are tryig t prtect. It is xidized i preferece t the prtected metal. The prtected metal becmes the cathde electrde, thus, cathdic prtecti. 9. A electrlytic uses electrical eergy t prduce a chemical chage. The prcess f electrlysis ivlves frcig a curret thrugh a t prduce a chemical chage fr which the ptetial is egative. Electrical wrk is used t frce a sptaeus reacti t ccur. The uits fr curret are amperes (A) which equal 1 culmb f charge per sec. curret (A) time (s) = culmbs f charge passed We use Faraday s cstat (F = 96,85 culmbs f charge per mle) t cvert culmbs f charge passed it mles f electrs passed. The half-reacti gives the mle rati betwee mles f electrs ad mles f metal prduced (r plated ut). Platig meas depsitig the eutral metal the electrde by reducig the metal is i sluti. I electrlysis, as with ay redx reacti, the reacti that ccurs first is the e mst favred thermdyamically. The reducti reacti mst favred thermdyamically has the largest, mst psitive E red value. The xidati reacti mst likely t ccur is the e with the largest, mst psitive E x value. Nte that fr electrlytic s that E < 0, s the E red ad E xvalues are cmmly egative. The half-reactis that ccur first as a curret is applied are the es with the least egative ptetials (which are the mst psitive ptetials). T predict the cathde half-reacti, write dw the half-reacti ad E red value fr all

6 668 CHAPTER 17 ELECTROCHEMISTRY species preset that ca be reduced. The cathde reacti that ccurs has the least egative (mst psitive) E red value. The same thig is de fr the ade; write dw everythig preset that ca be xidized; the species xidized has the least egative (mst psitive) E xvalue. Nte that we cmmly assume stadard cditis whe predictig which halfreactis ccur, s we ca use the stadard ptetials i Table Whe mlte salts are electrlyzed, there is ly e species preset that ca be xidized (the ai i simple salts) ad there is ly e species that ca be reduced (the cati i simple salts). Whe H O is preset as is the case whe aqueus slutis are electrlyzed, we must csider the xidati ad reducti f water as ptetial reactis that ca ccur. Whe water is preset, mre reactis ca take place, makig predictis mre difficult. Whe the vltage required t frce a chemical reacti t ccur is larger tha expected, this is called vervltage. The amut f vervltage ecessary t frce a reacti t ccur varies with the type f substace preset. Because f this, E values must be used cautiusly whe predictig the half-reactis that ccur. 10. Electrlysis is used t prduce may pure metals ad pure elemets fr cmmercial use. It als is used t purify metals as well as t plate ut thi catigs substaces t prvide prtecti frm crrsi ad t beautify bjects. Ather applicati f electrlysis is the chargig f batteries. Whe aqueus NaCl is electrlyzed, water, with its less egative reducti ptetial is preferetially reduced ver Na + is. Thus, the presece f water des t allw Na + is t be reduced t Na. I mlte NaCl, water is t preset, s Na + ca be reduced t Na. Purificati by electrlysis is called electrrefiig. See the text fr a discussi f the electrrefiig f cpper. Electrrefiig is pssible because f the selectivity f electrde reactis. The ade is made up f the impure metal. A ptetial is applied s just the metal f iterest ad all mre easily xidized metals are xidized at the ade. The metal f iterest is the ly metal plated at the cathde due t the careful ctrl f the ptetial applied. The metal is that culd plate ut at the cathde i preferece t the metal we are purifyig will t be i sluti, because these metals were t xidized at the ade. Review f Oxidati - Reducti Reactis 1. Oxidati: icrease i xidati umber; lss f electrs Reducti: decrease i xidati umber; gai f electrs 1. See Table. i Chapter f the text fr rules fr assigig xidati umbers. a. H (+1), O (), N (+5) b. Cl (1), Cu (+) c. O (0) d. H (+1), O (1) e. H(+1), O (), C (0) f. Ag (0)

7 CHAPTER 17 ELECTROCHEMISTRY 669 g. Pb (+), O (), S (+6) h. O (), Pb (+) i. Na (+1), O (), C (+) j. O (), C (+) k. (NH ) Ce(SO ) ctais NH + is ad SO is. Thus, cerium exists as the Ce + i. H (+1), N (), Ce (+), S (+6), O () l. O (), Cr (+) 15. The species xidized shws a icrease i xidati umbers ad is called the reducig aget. The species reduced shws a decrease i xidati umbers ad is called the xidizig aget. The pertiet xidati umbers are listed by the substace xidized ad the substace reduced. Substace Substace Redx? Ox. Aget Red. Aget Oxidized Reduced a. Yes H O CH CH (C, +) H O (H, +1 0) b. Yes AgNO Cu Cu (0 +) AgNO (Ag, +1 0) c. Yes HCl Z Z (0 +) HCl (H, +1 0) d. N; There is chage i ay f the xidati umbers. 16. See Chapter.10 f the text fr rules balacig xidati-reducti reactis. a. Cr Cr + + e NO NO H + + NO NO + H O e - + H + + NO - NO + H O Cr Cr + + e e - + H + + NO NO + H O H + (aq) + NO (aq) + Cr(s) Cr + (aq) + NO(g) + H O(l) b. (Al Al + + e ) 5 MO M + 8 H + + MO M + + H O (5 e H + + MO M + + H O) 5 Al 5 Al e 15 e - + H + + MO M H O H + (aq) + MO - (aq) + 5 Al(s) 5 Al + (aq) + M + (aq) + 1 H O(l)

8 670 CHAPTER 17 ELECTROCHEMISTRY c. (Ce + + e Ce + ) 6 CH OH CO H O + CH OH CO + 6 H + H O + CH OH CO + 6 H e 6 Ce e 6 Ce + H O + CH OH CO + 6 H e H O(l) + CH OH(aq) + 6 Ce + (aq) 6 Ce + (aq) + CO (g) + 6 H + (aq) d. PO PO (H O + PO PO + H + + e ) MO MO ( e - + H + + MO MO + H O) H O + PO PO + 6 H e 6 e H + + MO - MO + H O H + + MO + PO PO + MO + H O Nw cvert t a basic sluti by addig OH - t bth sides. H + + OH H O the reactat side. After cvertig H + t OH, simplify the verall equati by crssig ff e H O each side f the reacti. The verall balaced equati is: H O(l) + MO (aq) + PO (aq) PO (aq) + MO (s) + OH (aq) e. Mg Mg(OH) OCl Cl H O + Mg Mg(OH) + H + + e e - + H + + OCl Cl + H O H O + Mg Mg(OH) + H + + e e + H + + OCl Cl - + H O OCl (aq) + H O(l) + Mg(s) Mg(OH) (s) + Cl (aq) The fial verall reacti des t ctai H +, s we are de. f. H CO HCO Ag(NH ) + Ag + NH H O + H CO HCO + 5 H + + e (e - + Ag(NH ) + Ag + NH ) H O + H CO HCO + 5 H + + e e + Ag(NH ) + Ag + 8 NH Ag(NH ) + + H O + H CO HCO + 5 H + + Ag + 8 NH Cvert t a basic sluti by addig 5 OH t bth sides (5 H OH 5 H O). The, crss ff H O bth sides, which gives the verall balaced equati: 5 OH (aq) + Ag(NH ) + (aq) + H CO(aq) HCO (aq) + H O(l) + Ag(s) +

9 CHAPTER 17 ELECTROCHEMISTRY 671 Questis 8 NH (aq) 17. Magesium is a alkalie earth metal; Mg will xidize t Mg +. The xidati state f hydrge i HCl is +1. T be reduced, the xidati state f H must decrease. The bvius chice fr the hydrge prduct is H (g) where hydrge has a zer xidati state. The balaced reacti is: Mg(s) + HCl(aq) MgCl (aq) + H (g). Mg ges frm the 0 t the + xidati state by lsig tw electrs. Each H atm ges frm the +1 t the 0 xidati state by gaiig e electr. Sice there are tw H atms i the balaced equati, the a ttal f tw electrs are gaied by the H atms. Hece, tw electrs are trasferred i the balaced reacti. Whe the electrs are trasferred directly frm Mg t H +, wrk is btaied. I rder t haress this reacti t d useful wrk, we must ctrl the flw f electrs thrugh a wire. This is accmplished by makig a galvaic which separates the reducti reacti frm the xidati reacti i rder t ctrl the flw f electrs thrugh a wire t prduce a vltage. 18. Galvaic s use sptaeus redx reactis t prduce a vltage. The key is t have a verall psitive E value whe maipulatig the half-reactis. Fr ay tw half-reactis, the half-reacti with the mst psitive reducti ptetial will always be the cathde reacti. Fr egative ptetials, this will be the half-reacti with the stadard reducti ptetial clsest t zer. The remaiig half-reacti (the e with the mst egative E red ) will be reversed ad becme the ade half-reacti (E x = E red ). This cmbiati will always yield a psitive verall stadard ptetial which ca be used t ru a galvaic. 19. A extesive prperty is e that depeds directly the amut f substace. The free eergy chage fr a reacti depeds whether 1 ml f prduct is prduced r ml f prduct is prduced r 1 milli ml f prduct is prduced. This is t the case fr ptetials which d t deped the amut f substace. The equati that relates G t E is G = FE. It is the term that cverts the itesive prperty E it the extesive prperty G. is the umber f mles f electrs trasferred i the balaced reacti that G is assciated with. 0. E = E lg Q A ccetrati has the same ade ad cathde ctets; thus, E = 0 fr a ccetrati. N matter which half-reacti yu chse, the ppsite half-reacti is ccurrig i the ther. The drivig frce t prduce a vltage is the lg Q term i the Nerst equati. Q is determied by the ccetrati f is i the ade ad cathde cmpartmets. The larger the differece i ccetratis, the larger the lg Q term ad the larger the vltage prduced. Therefre, the drivig frce fr ccetrati s is the differece i i ccetratis betwee the cathde ad ade cmpartmets. Whe the i ccetratis are equal, Q = 1 ad lg Q = 0, ad vltage is prduced.

10 67 CHAPTER 17 ELECTROCHEMISTRY 1. A ptetial hazard whe jump startig a car is the pssibility fr the electrlysis f H O(l) t ccur. Whe H O(l) is electrlyzed, the prducts are the explsive gas mixture f H (g) ad O (g). A spark prduced durig jump startig a car culd igite ay H (g) ad O (g) prduced. Grudig the jumper cable far frm the battery miimizes the risk f a spark earby the battery where H (g) ad O (g) culd be cllectig.. Metals crrde because they xidize easily. Referecig Table 17.1, mst metals are assciated with egative stadard reducti ptetials. This meas the reverse reactis, the xidati half-reactis, have psitive xidati ptetials idicatig they xidize fairly easily. Ather key pit is that the reducti f O (which is a reactat i crrsi red prcesses) has a mre psitive E tha mst f the metals (fr O, E = 0.0 V). This meas that whe O is cupled with mst metals, the reacti will be sptaeus sice E > 0, s crrsi ccurs. The ble metals (Ag, Au, ad Pt) all have stadard reducti ptetials greater tha that f O. Therefre, O is t capable f xidizig these metals at stadard cditis. Nte: the stadard reducti ptetial fr Pt Pt + + e is t i Table As expected, its reducti ptetial is greater tha that f O ( E = 1.19 V).. Yu eed t kw the idetity f the metal s yu kw which mlar mass t use. Yu eed t kw the xidati state f the metal i i the salt s the ml f electrs trasferred ca be determied. Ad fially, yu eed t kw the amut f curret ad the time the curret was passed thrugh the electrlytic. If yu kw these fur quatities, the the mass f metal plated ut ca be calculated.. Alumium is fud i ature as a xide. Alumium has a great affiity fr xyge s it is extremely difficult t reduce the Al + is i the xide t pure metal. Oe ptetial way is t try t disslve the alumium xide i water i rder t free up the is. Eve if alumium is wuld g it sluti, water wuld be preferetially reduced i a electrlytic. Ather way t mbilize the is is t melt the alumium xide. This is t practical because f the very high meltig pit f alumium xide. The key discvery was fidig a slvet that wuld t be mre easily reduced tha Al + is (as water is). The slvet discvered by Hall ad Herult (separately) was Na AlF 6. A mixture f Al O ad Na AlF 6 has a meltig pit much lwer tha that f pure Al O. Therefre, Al + i mbility is easier t achieve, makig it pssible t reduce Al + t Al. Pt red

11 CHAPTER 17 ELECTROCHEMISTRY 67 Exercises Galvaic Cells, Cell Ptetials, Stadard Reducti Ptetials, ad Free Eergy 5. A typical galvaic diagram is: e - e - Ade (xidati) Salt Bridge catis ais ~~~~~~~ ~~~~~~ ~~~~~~~ ~~~~~~ Cathde (reducti) The diagram fr all s will lk like this. The ctets f each half- cmpartmet will be idetified fr each reacti, with all slute ccetratis at 1.0 M ad all gases at 1.0 atm. Fr Exercises 17.5 ad 17.6, the flw f is thrugh the salt bridge was t asked fr i the questis. If asked, hwever, catis always flw it the cathde cmpartmet, ad ais always flw it the ade cmpartmet. This is required t keep each cmpartmet electrically eutral. a. Table 17.1 f the text lists balaced reducti half-reactis fr may substaces. Fr this verall reacti, we eed the Cl t Cl - reducti half-reacti ad the Cr + t Cr O 7 xidati half-reacti. Maipulatig these tw half-reactis gives the verall balaced equati. (Cl + e Cl ) 7 H O + Cr + Cr O H e 7 H O(l) + Cr + (aq) + Cl (g) Cr O 7 (aq) + 6 Cl (aq) + 1 H + (aq) The ctets f each cmpartmet are: Cathde: Pt electrde; Cl bubbled it sluti, Cl i sluti Ade: Pt electrde; Cr +, H +, ad Cr O 7 i sluti We eed a reactive metal t use as the electrde i each case, sice all the reactats ad prducts are i sluti. Pt is a cmm chice. Ather pssibility is graphite.

12 67 CHAPTER 17 ELECTROCHEMISTRY b. Cu + + e Cu Mg Mg + + e Cu + (aq) + Mg(s) Cu(s) + Mg + (aq) Cathde: Cu electrde; Cu + i sluti; Ade: Mg electrde; Mg + i sluti 6. Referece Exercise 17.5 fr a typical galvaic diagram. The ctets f each half- cmpartmet are idetified belw with all slute ccetratis at 1.0 M ad all gases at 1.0 atm. a. Referece Table 17.1 fr the balaced half-reactis. 5 e + 6 H + + IO 1/ I + H O (Fe + Fe + + e ) 5 6 H + + IO + 5 Fe + 5 Fe + + 1/ I + H O r 1 H + (aq) + IO - (aq) + 10 Fe + (aq) 10 Fe + (aq) + I (aq) + 6 H O(l) Cathde: Pt electrde; IO, I ad H SO (H + surce) i sluti. Nte: I (s) wuld make a pr electrde sice it sublimes. Ade: Pt electrde; Fe + ad Fe + i sluti b. (Ag + + e Ag) Z Z + + e - Z(s) + Ag + (aq) Ag(s) + Z + (aq) Cathde: Ag electrde; Ag + i sluti; Ade: Z electrde; Z + i sluti 7. T determie E fr the verall reacti, we must add the stadard reducti ptetial t the stadard xidati ptetial (E Ered Ex). Referece Table 17.1 fr values f stadard reducti ptetials. Remember that Ex Ered ad that stadard ptetials are t multiplied by the iteger used t btai the verall balaced equati. 5a. 5b. 8. 6a. 6b. E E E E E E Cl Cl Cr Cr O 7 E E Cu Cu Mg Mg E E IO I Fe Fe E E Ag Ag Z Z = 1.6 V + (1. V) = 0.0 V = 0. V +.7 V =.71 V = 1.0 V + (0.77 V) = 0. V = 0.80 V V = 1.56 V

13 CHAPTER 17 ELECTROCHEMISTRY Referece Exercise 17.5 fr a typical galvaic desig. The ctets f each half- cmpartmet are idetified belw with all slute ccetratis at 1.0 M ad all gases at 1.0 atm. Fr each pair f half-reactis, the half-reacti with the largest (mst psitive) stadard reducti ptetial will be the cathde reacti, ad the half-reacti with the smallest (mst egative) reducti ptetial will be reversed t becme the ade reacti. Oly this cmbiati gives a sptaeus verall reacti, i.e., a reacti with a psitive verall stadard ptetial. Nte that i a galvaic as illustrated i Exercise 17.5, the catis i the salt bridge migrate t the cathde, ad the ais migrate t the ade. a. Cl + e Cl E = 1.6 V Br - Br + e E = 1.09 V Cl (g) + Br - (aq) Br (aq) + Cl (aq) E = 0.7 V The ctets f each cmpartmet are: Cathde: Pt electrde; Cl (g) bubbled i, Cl i sluti Ade: Pt electrde; Br ad Br - i sluti b. ( e + H + + IO IO + H O) 5 E = 1.60 V ( H O + M + MO + 8 H e ) E = 1.51 V 10 H IO + 8 H O + M + 5 IO + 5 H O + MO + 16 H + E = 0.09 V This simplifies t: H O(l) + 5 IO (aq) + M + (aq) 5 IO (aq) + MO (aq) + 6 H + (aq) E = 0.09 V Cathde: Pt electrde; IO, IO, ad H SO (as a surce f H + ) i sluti Ade: Pt electrde; M +, MO ad H SO i sluti 0. Referece Exercise 17.5 fr a typical galvaic desig. The ctets f each half- cmpartmet are idetified belw, with all slute ccetratis at 1.0 M ad all gases at 1.0 atm. a. H O + H + + e H O E = 1.78 V H O O + H + + e E = 0.68 V H O (aq) H O(l) + O (g) E = 1.10 V Cathde: Pt electrde; H O ad H + i sluti Ade: Pt electrde; O (g) bubbled i, H O ad H + i sluti b. (Fe + + e Fe) E = 0.06 V

14 676 CHAPTER 17 ELECTROCHEMISTRY (M M + + e ) E = 1.18 V Fe + (aq) + M(s) Fe(s) + M + (aq) E = 1.1 V Cathde: Fe electrde; Fe + i sluti; Ade: M electrde; M + i sluti 1. I stadard lie tati, the ade is listed first ad the cathde is listed last. A duble lie separates the tw cmpartmets. By cveti, the electrdes are the eds with all slutes ad gases twards the middle. A sigle lie is used t idicate a phase chage. We als icluded all ccetratis. 5a. Pt Cr + (1.0 M), Cr O 7 (1.0 M), H + (1.0 M) Cl (1.0 atm) Cl (1.0 M) Pt 5b. Mg Mg + (1.0 M) Cu + (1.0 M) Cu 9a. Pt Br (1.0 M), Br (1.0 M) Cl (1.0 atm) Cl (1.0 M) Pt 9b. Pt M + (1.0 M), MO (1.0 M), H + (1.0 M) IO (1.0 M), H + (1.0 M),. 6a. Pt Fe + (1.0 M), Fe + (1.0 M) IO (1.0 M), H + (1.0 M), I (1.0 M) Pt 6b. Z Z + (1.0 M) Ag + (1.0 M) Ag IO (1.0 M) Pt 0a. Pt H O (1.0 M), H + (1.0 M) O (1.0 atm) H O (1.0 M), H + (1.0 M) Pt 0b. M M + (1.0 M) Fe + (1.0 M) Fe. Lcate the pertiet half-reactis i Table 17.1, ad the figure which cmbiati will give a psitive stadard ptetial. I all cases, the ade cmpartmet ctais the species with the smallest stadard reducti ptetial. Fr part a, the cpper cmpartmet is the ade, ad i part b, the cadmium cmpartmet is the ade. a. Au + + e Au E = 1.50 V (Cu + Cu + + e ) E = V Au + (aq) + Cu + (aq) Au(s) + Cu + (aq) E = 1. V b. (VO + + H + + e VO + + H O) E = 1.00 V Cd Cd + + e - E = 0.0 V VO + (aq) + H + (aq) + Cd(s) VO + (aq) + H O(l) + Cd + (aq) E = 1.0 V. a. (H O + H + + e H O) E = 1.78 V Cr H O Cr O H e E = 1. V H O (aq) + Cr + (aq) + H O(l) Cr O 7 (aq) + 8 H + (aq) E = 0.5 V b. ( H + + e H ) E = 0.00 V

15 CHAPTER 17 ELECTROCHEMISTRY 677 (Al Al + + e ) E = 1.66 V 6 H + (aq) + Al(s) H (g) + Al + (aq) E = 1.66 V 5. a. (5 e + 8 H + + MO M + + H O) E = 1.51 V ( I I + e ) 5 E = -0.5 V 16 H + (aq)+ MO (aq) + 10 I (aq) 5 I (aq)+ M + (aq) + 8 H O(l) E = 0.97 V This reacti is sptaeus at stadard cditis because E > 0. b. (5 e + 8 H + + MO M + + H O) E = 1.51 V ( F F + e ) 5 E =.87 V 16 H + (aq) + MO (aq) + 10 F (aq) 5 F (aq) + M + (aq) + 8 H O(l) E = -1.6 V This reacti is t sptaeus at stadard cditis because E < a. H H + + e E = 0.00 V H + e H E =. V H (g) H + (aq) + H (aq) E =. V Nt sptaeus b. Au + + e Au E = 1.50 V (Ag Ag + + e ) E = 0.80 V Au + (aq) + Ag(s) Au(s) + Ag + (aq) E = 0.70 V Sptaeus 7. Cl + e Cl E = 1.6 V (ClO ClO + e ) E = 0.95 V ClO (aq) + Cl (g) ClO (aq) + Cl (aq) E = 0.1 V = 0.1 J/C ΔG = FE = ( ml e )(96,85 C/ml e )(0.1 J/C) = J = 79 kj 8. a. ( H + + NO + e NO + H O) E = 0.96 V (M M + + e ) E = 1.18 V M(s) + 8 H + (aq) + NO (aq) NO(g) + H O(l) + M + (aq) E =.1 V ( e - + H + + IO - IO - + H O) 5 E = 1.60 V (M + + H O MO + 8 H e ) E = 1.51 V 5 IO (aq) + M + (aq) + H O(l) 5 IO (aq) + MO (aq) + 6 H + (aq) E = 0.09 V b. Nitric acid xidati (see abve fr E ):

16 678 CHAPTER 17 ELECTROCHEMISTRY ΔG = FE (6 ml e )(96,85 C/ml e )(.1 J/C) = J = 10 kj Peridate xidati (see abve fr E ): ΔG = (10 ml e - )(96,85 C/ml e )(0.09 J/C)(1 kj/1000 J) = 90 kj 9. Because the s are at stadard cditis, w max = ΔG = ΔG = 17. fr the balaced verall equatis ad fr E. FE. See Exercise a. w max = ( ml e )(96,85 C/ml e )(1. J/C) = J = 88 kj b. w max = ( ml e )(96,85 C/ml e )(1.0 J/C) = J = 70. kj 0. Because the s are at stadard cditis, w max = ΔG = ΔG = 17. fr the balaced verall equatis ad fr E. FE. See Exercise a. w max = (6 ml e )(96,85 C/ml e )(0.5 J/C) = J = 60 kj b. w max = (6 ml e )(96,85 C/ml e )(1.66 J/C) = J = 961 kj 1. CH OH(l) + / O (g) CO (g) + H O(l) ΔG = (-7) + (-9) [-166 = 70 kj The balaced half-reactis are: H O + CH OH CO + 6 H e ad O + H + + e H O Fr / ml O, 6 mles f electrs will be trasferred ( = 6). ΔG = FE, E = G F (6 mlee ( 70,000J) Δ )(96,85C / mle = 1.1 J/C = 1.1 V ). Fe + + e Fe E = 0. V = 0. J/C ΔG = FE = ( ml e )(96,85 C/ml e )(0. J/C)(1 kj/1000 J) = 85 kj f, Fe 85 kj = 0 ΔG 0 G f, Fe [, Δ = 85 kj We ca get G f, Fe Δ tw ways. Csider: Fe + + e Fe + E = 0.77 V ΔG = (1 ml e)(96,85 C/ml e - )(0.77 J/C) = 7,00 J = 7 kj Fe + Fe + + e ΔG = 7 kj Fe Fe + + e ΔG = 85 kj Fe Fe + + e ΔG = 11 kj, Δ = 11 kj/ml G f, Fe r csider: Fe + + e Fe E = 0.06 V

17 CHAPTER 17 ELECTROCHEMISTRY 679 ΔG = ( ml e )(96,85 C/ml e )(0.06 J/C) = 10,00 J 10. kj f, Fe 10. kj = 0 ΔG 0 G f, Fe [, Δ = 10. kj/ml; Rud-ff errr explais the 1 kj discrepacy.. Gd xidizig agets are easily reduced. Oxidizig agets are the left side f the reducti half-reactis listed i Table We lk fr the largest, mst psitive stadard reducti ptetials t crrespd t the best xidizig agets. The rderig frm wrst t best xidizig agets is: K + < H O < Cd + < I < AuCl < IO E (V) Gd reducig agets are easily xidized. The reducig agets are the right side f the reducti half-reactis listed i Table The best reducig agets have the mst egative stadard reducti ptetials (E ) r the mst psitive stadard xidati ptetials, E x (= E ). The rderig frm wrst t best reducig agets is: F < H O < I < Cu + < H < K E (V) a. H + + e H E = 0.00 V; Cu Cu + + e E = -0. V E = 0. V; N, H + cat xidize Cu t Cu + at stadard cditis ( E < 0). b. Fe + + e Fe + E = 0.77 V; I I + e E = -0.5 V E = = 0. V; Yes, Fe + ca xidize I t I. c. H H + + e E = 0.00 V; Ag + + e Ag E = 0.80 V E = 0.80 V; Yes, H ca reduce Ag + t Ag at stadard cditis ( E > 0). d. Fe + Fe + + e E = V; Cr + + e Cr + E = V E = = 1.7 V; N, Fe + cat reduce Cr + t Cr + at stadard cditis. 6. Cl + e Cl E = 1.6 V Ag + + e Ag E = 0.80 V Pb + + e Pb E = 0.1 V Z + + e Z E = V Na + + e Na E =.71 V a. Oxidizig agets (species reduced) are the left side f the abve reducti halfreactis. Of the species available, Ag + wuld be the best xidizig aget sice it has the largest E value. Nte that Cl is a better xidizig aget tha Ag +, but it is t e f the chices listed. b. Reducig agets (species xidized) are the right side f the reducti half-reactis. Of the species available, Z wuld be the best reducig aget sice it has the largest E value.

18 680 CHAPTER 17 ELECTROCHEMISTRY c. SO + H + + e H SO + H O E = 0.0 V; SO ca xidize Pb ad Z at stadard cditis. Whe SO is cupled with these reagets, E is psitive. d. Al Al + + e E = 1.66 V; Al ca reduce Ag + ad Z + at stadard cditis sice E > a. Br - Br + e E = 1.09 V; Cl - Cl + e E = 1.6 V; E > 1.09 V t xidize Br - ; E < 1.6 V t t xidize Cl ; Cr O 7, O, MO, ad IO are all pssible sice whe all f these xidizig agets are cupled with Br -, E > 0, ad whe cupled with Cl, E < 0 (assumig stadard cditis). b. M M + + e E = 1.18; Ni Ni + + e E = 0. V; Ay xidizig aget with 0. V > E > 1.18 V will wrk. PbSO, Cd +, Fe +, Cr +, Z + ad H O will be able t xidize M but t Ni (assumig stadard cditis). 8. a. Cu + + e Cu E = 0. V; Cu + + e Cu + E = 0.16 V; T reduce Cu + t Cu but t reduce Cu + t Cu +, the reducig aget must have a stadard xidati ptetial E x = E ) betwee 0. V ad 0.16 V (s E is psitive ly fr the Cu + t Cu reducti). The reducig agets (species xidized) are the right side f the halfreactis i Table The reagets at stadard cditis which have E x (=E ) betwee 0. V ad 0.16 V are Ag (i 1.0 M Cl ) ad H SO. b. Br + e Br - E = 1.09 V; I + e - I E = 0.5 V; Frm Table 17.1, VO +, Au (i 1.0 M Cl - ), NO, ClO, Hg +, Ag, Hg, Fe +, H O ad MO are all capable at stadard cditis f reducig Br t Br - but t reducig I t I. Whe these reagets are cupled with Br, E > 0, ad whe cupled with I, E < ClO + H O + e OH + Cl E = 0.90 V NH + OH N H + H O + e E = 0.10 V ClO (aq) + NH (aq) Cl (aq) + N H (aq) + H O(l) E = 1.00 V Because E is psitive fr this reacti, at stadard cditis ClO ca sptaeusly xidize NH t the smewhat txic N H. 50. Tl + + e Tl + E = 1.5 V I I + e E = V Tl + + I Tl + + I E = 0.70 V I sluti, Tl + ca xidize I t I. Thus, we expect TlI t be thallium(i) triidide. The Nerst Equati 51. H O + H + + e H O E = 1.78 V

19 CHAPTER 17 ELECTROCHEMISTRY 681 (Ag Ag + + e ) E = V H O (aq) + H + (aq) + Ag(s) H O(l) + Ag + (aq) E = 0.98 V a. A galvaic is based sptaeus redx reactis. At stadard cditis, this reacti prduces a vltage f 0.98 V. Ay chage i ccetrati that icreases the tedecy f the frward reacti t ccur will icrease the ptetial. Cversely, ay chage i ccetrati that decreases the tedecy f the frward reacti t ccur (icreases the tedecy f the reverse reacti t ccur) will decrease the ptetial. Usig Le Chatelier s priciple, icreasig the reactat ccetratis f H O ad H + frm 1.0 M t.0 M will drive the frward reacti further t right (will further icrease the tedecy f the frward reacti t ccur). Therefre, E will be greater tha E. b. Here, we decreased the reactat ccetrati f H + ad icreased the prduct ccetrati f Ag + frm the stadard cditis. This decreases the tedecy f the frward reacti t ccur which will decrease E as cmpared t E (E < E ). 5. The ccetratis f Fe + i the tw cmpartmets are w 0.01 M ad M. The drivig frce fr this reacti is t equalize the Fe + ccetratis i the tw cmpartmets. This ccurs if the cmpartmet with M Fe + becmes the ade (Fe will be xidized t Fe + ) ad the cmpartmet with the 0.01 M Fe + becmes the cathde (Fe + will be reduced t Fe). Electr flw, as always fr galvaic s, ges frm the ade t the cathde, s electr flw will g frm the right cmpartmet ([Fe + 7 = 1 10 M) t the left cmpartmet ([Fe + = 0.01 M). 5. Fr ccetrati s, the drivig frce fr the reacti is the differece i i ccetratis betwee the ade ad cathde. I rder t equalize the i ccetratis, the ade always has the smaller i ccetrati. The geeral setup fr this ccetrati is: Cathde: Ag + (x M) + e Ag E = 0.80 V Ade: Ag Ag + (y M) + e E = 0.80 V Ag + (cathde, x M) Ag + (ade, y M) E = 0.00 V E = E [Ag lg Q lg 1 [Ag ade cathde Fr each ccetrati, we will calculate the ptetial usig the abve equati. Remember that the ade always has the smaller i ccetrati. a. Bth cmpartmets are at stadard cditis ([Ag + = 1.0 M), s E = E = 0 V. N vltage is prduced sice reacti ccurs. Ccetrati s ly prduce a vltage whe the i ccetratis are t equal. b. Cathde =.0 M Ag + ; Ade = 1.0 M Ag + ; Electr flw is always frm the ade t the cathde, s electrs flw t the right i the diagram.

20 68 CHAPTER 17 ELECTROCHEMISTRY E = lg [Ag [Ag ade cathde 1.0 lg = V 1.0 c. Cathde = 1.0 M Ag + ; Ade = 0.10 M Ag + ; Electrs flw t the left i the diagram. E = lg [Ag [Ag ade cathde 0.10 lg = V d. Cathde = 1.0 M Ag + ; Ade = M Ag + ; Electrs flw t the left i the diagram E = lg = 0.6 V 1.0 e. The i ccetratis are the same, thus lg ([Ag + ade /[Ag + cathde ) = lg (1.0) = 0 ad E = 0. N electr flw ccurs. 5. As is the case fr all ccetrati s, E = 0, ad the smaller i ccetrati is always i the ade cmpartmet. The geeral Nerst equati fr the Ni Ni + (x M) Ni + (y M) Ni ccetrati is: E = E [Ni lg Q lg [Ni ade cathde a. Bth cmpartmets are at stadard cditis ([Ni + = 1.0 M), ad E = N electr flw ccurs. E = 0 V. b. Cathde =.0 M Ni + ; Ade = 1.0 M Ni + ; Electr flw is always frm the ade t the cathde, s electrs flw t the right i the diagram. E = lg [Ni [Ni ade cathde 1.0 lg = V.0 c. Cathde = 1.0 M Ni + ; Ade = 0.10 M Ni + ; Electrs flw t the left i the diagram. E = lg = 0.00 V d. Cathde = 1.0 M Ni + ; Ade = M Ni + ; Electrs flw t the left i the diagram. E = lg = 0.1 V e. Because bth ccetratis are equal, lg (.5/.5) = lg 1.0 = 0 ad E = 0. N electr flw ccurs. 55. = fr this reacti (lead ges frm Pb Pb + i PbSO ).

21 CHAPTER 17 ELECTROCHEMISTRY 68 E = E 1 lg [H [HSO 1 =.0 V lg (.5) (.5) E =.0 V V =.1 V 56. Cr O H e Cr H O E = 1. V (Al Al + + e ) E = 1.66 Cr O H + + Al Al + + Cr H O E =.99 V E = E [Al [Cr lg Q, E =.99 V lg 6 [Cr O [H 7 1, (0.0) (0.15).01 =.99 lg Q 1 (0.55)[H.7 10 [H 1 = 6(0.0).7 10 lg 1 [H.0 10 = 0.01, [H + 1 = 0.7, [H + = 0.9 = 0.9 M, ph = -lg(0.9) = Cu + + e Cu E = 0. V Z Z + + e E = 0.76 V Cu + (aq) + Z(s) Z + (aq) + Cu(s) E = 1.10 V Because Z + is a prduct i the reacti, the Z + ccetrati icreases frm 1.00 M t 1.0 M. This meas that the reactat ccetrati f Cu + must decrease frm 1.00 M t 0.80 M (frm the 1:1 ml rati i the balaced reacti). E = E lg Q 1.10V [Z lg [Cu E = 1.10 V lg = 1.10 V V = 1.09 V 58. (Pb + + e Pb) E = -0.1 V (Al Al + + e ) E = 1.66 V Pb + (aq) + Al(s) Pb(s) + Al + (aq) E = 1.5 V Frm the balaced reacti, whe the Al + has icreased by 0.60 ml/l (Al + is a prduct i the sptaeus reacti), the the Pb + ccetrati has decreased by / (0.60 ml/l) = 0.90 M. [Al (1.60) E = 1.5 V lg 1.5 lg 6 [Pb 6 (0.10)

22 68 CHAPTER 17 ELECTROCHEMISTRY E = 1.5 V 0.0 V = 1.50 V 59. Cu + (aq) + H (g) H + (aq) + Cu(s) P H = 1.0 atm ad [H + = 1.0 M: E = E = 0. V 0.00 V = 0. V; = ml electrs E 1 lg [Cu 1 a. E = 0. V lg = 0. V 0.11V = 0. V.5 10 b V = 0. V 1 1 lg, lg =.91, [Cu + = [Cu [Cu Nte: Whe determiig expets, we will carry extra sigificat figures = M 60. Ni + (aq) + Al(s) Al + (aq) + Ni(s) = 6 ml electrs fr this reacti. E = = 1. V; [Al a. E = 1. V lg 6 [Ni (7. 10 = 1. lg 6 (1.0) ) E = 1. V (-0.0 V) = 1.7 V [Al b. 1.6 V = 1. V lg 6 (1.0) [Al = 10, [Al + 10 =. 10 M 61. Cu + (aq) + H (g) H + (aq) + Cu(s) P H = 1.0 atm ad [H + = 1.0 M: E =, lg [Al + = 19.9 E = 0. V 0.00 V = 0. V; = E 1 lg [Cu Use the K sp expressi t calculate the Cu + ccetrati i the. Cu(OH) (s) Cu + (aq) + OH (aq) K sp = = [Cu + [OH Frm prblem, [OH = 0.10 M, s: [Cu = (0.10) 19 = M E = E 1 lg [Cu = 0. V 1 lg = = 0.16 V Because E < 0, the frward reacti is t sptaeus, but the reverse reacti is sptaeus. The Cu electrde becmes the ade ad E = 0.16 V fr the reverse reacti. The reacti is: H + (aq) + Cu(s) Cu + (aq) + H (g). 6. Ni + (aq) + Al(s) Al + (aq) + Ni(s) E = -0. V V = 1. V; = 6

23 CHAPTER 17 ELECTROCHEMISTRY 685 [Al [Al E E lg, 1.8 V = 1. V lg [Ni 6 (1.0) lg [Al + = 9.59, [Al + = , [Al + 0 = M Al(OH) (s) Al + (aq) + OH (aq) K sp = [Al + [OH ; Frm the prblem, [OH = M. 0 K sp = ( ) ( ) = Cathde: M + + e M(s) E= 0.1 V Ade: M(s) M + + e E = 0.1 V M + (cathde) M + (ade) E = 0.00 V E = 0. V = 0.00 V lg [M + ade = [M lg [M ade cathde, 0. = (0.) = 1.89, [M + 15 ade = M 0 ade.0591 [M lg 1.0 Because we started with equal umbers f mles f SO ad M +, [M + = [SO at equilibrium. K sp = [M + [SO - = ( ) = a. Ag + (x M, ade) Ag + (0.10 M, cathde); Fr the silver ccetrati, E = 0.00 (as is always the case fr ccetrati s) ad = 1. E = 0.76 V = 0.00 [Ag lg 1 [Ag ade cathde 0.76 = - lg [ ade ade Ag 0.10 [Ag, 0.10 b. Ag + (aq) + S O (aq) Ag(S O ) (aq) = , [Ag + 1 ade = M K = [Ag(S [Ag O) [SO = (0.050) = See Exercises 17.5, 17.7, ad 17.9 fr balaced reactis ad stadard ptetials. Balaced reactis are ecessary t determie, the mles f electrs trasferred. 5a. 7 H O + Cr + + Cl Cr O Cl + 1 H + E = 0.0 V = 0.0 J/C ΔG = FE = (6 ml e )(96,85 C/ml e )(0.0 J/C) = J = 0 kj

24 686 CHAPTER 17 ELECTROCHEMISTRY E E lg Q: At equilibrium, E = 0 ad Q = K, s: E lg K, lg K = E 6(0.0) =.05, K = = 1 10 Nte: Whe determiig expets, we will rud ff t the crrect umber f sigificat figures after the calculati is cmplete i rder t help elimiate excessive rud-ff errr. 5b. ΔG = ( ml e )(96,85 C/ml e )(.71 J/C) = J = 5 kj lg K = (.71) = , K = a. ΔG = ( ml e )(96,85 C/ml )(0.7 J/C) = J = 5 kj lg K = (0.7) = 9.1, K = b. ΔG = (10 ml e )(96,85 C/ml e )(0.09 J/C) = J = -90 kj lg K = 10(0.09) = 15., K = ΔG = FE; E lg K, lg K = E 6a. ΔG = (10 ml e )(96,85 C/ml e )(0. J/C) = J = -10 kj lg K = 10(0.) = 7.76, K = = b. ΔG = ( ml e )(96,85 C/ml e )(1.56 J/C) = J = 01 kj lg K = (1.56) = 5.79, K = a. ΔG = ( ml )(96,85 C/ml e )(1.10 J/C) = J = 1 kj lg K = (1.10) = 7.5, K = b. ΔG = (6 ml e )(96,85 C/ml e )(1.1 J/C) = J = 660. kj

25 CHAPTER 17 ELECTROCHEMISTRY 687 lg K = 6(1.1) =115.76, K = Cu + + e Cu E = 0. V Fe Fe + + e E = 0. V Fe + Cu + Cu + Fe + E = 0.78 V Fr this reacti, K = E = [Fe [Cu lg K, lg K =, s let s slve fr K t determie the equilibrium i rati. (0.78) = 6.0, K = 68. Ni + + e Ni E = 0. V S S + + e E = 0.1 V Ni ++ + S S + + Ni E = 0.09 V [Fe [Cu = = Let s calculate the i rati whe this reacti is at equilibrium (E = 0). E = [S [Ni lg K, 0.09 = ( 0.09)/ = 10 = 9 10 [S lg [Ni The reacti is at equilibrium whe the [S + /[Ni + rati is equal t A reacti shifts t the right (which is what we wat i rder t prduce a vltage) whe Q < K sp. The miimum rati ecessary t make this reacti sptaeus is a [S + /[Ni + rati just less tha Oly whe the rati is belw 9 10 will E be psitive. 69. a. Pssible reacti: I (s) + Cl (aq) I (aq) + Cl (g) E = 0.5 V 1.6 V = 0.8 V This reacti is t sptaeus at stadard cditis because E < 0. N reacti ccurs. b. Pssible reacti: Cl (g) + I (aq) I (s) + Cl (aq) E = 0.8 V; This reacti is sptaeus at stadard cditis because E > 0. The reacti will ccur. Cl (g) + I (aq) I (s) + Cl (aq) E = 0.8 V = 0.8 J/C ΔG = FE = ( ml e - )(96,85 C/ml e - )(0.8 J/C) = J = -160 kj E = lg K, lg K = E (0.8) = 7.75, K = =

26 688 CHAPTER 17 ELECTROCHEMISTRY c. Pssible reacti: Ag(s) + Cu + (aq) Cu(s) + Ag + (aq) ccurs. E = 0.6 V; N reacti d. Fe + ca be xidized r reduced. The ther species preset are H +, SO, H O, ad O frm air. Oly O i the presece f H + has a large eugh stadard reducti ptetial t xidize Fe + t Fe + (resultig i E > 0). All ther cmbiatis, icludig the pssible reducti f Fe +, give egative ptetials. The sptaeus reacti is: Fe + (aq) + H + (aq) + O (g) Fe + (aq) + H O(l) E = = 0.6 V ΔG = lg K = FE = ( ml e )(96,85 C/ml e )(0.6 J/C)(1 kj/1000 J) = 180 kj (0.6) = 1.1, K = a. Cu + + e Cu E = 0.5 V Cu + Cu + + e E = 0.16 V Cu + (aq) Cu + (aq) + Cu(s) E = 0.6 V; Sptaeus ΔG = FE = (1 ml e )(96,85 C/ml e )(0.6 J/C) =,700 J = -5 kj E = lg K, lg K = E 1(0.6) = 6.09, K = = b. Fe + + e Fe E = 0. V (Fe + Fe + + e ) E = 0.77 V Fe + (aq) Fe + (aq) + Fe(s) E = 1.1 V; Nt sptaeus c. HClO + H + + e HClO + H O E = 1.65 V HClO + H O ClO - + H + + e E = 1.1 V HClO (aq) ClO (aq) + H + (aq) + HClO(aq) E = 0. V; Sptaeus ΔG = FE = ( ml e )(96,85 C/ml e )(0. J/C) = 8,900 J = 85 kj E (0.) lg K = = 1.89, K = a. Au + + e Au E = 1.50 V (Tl Tl + + e ) E = 0. V Au + (aq) + Tl(s) Au(s) + Tl + (aq) E = 1.8 V b. ΔG = FE = ( ml e )(96,85 C/ml e )(1.8 J/C) = J = 5 kj

27 CHAPTER 17 ELECTROCHEMISTRY 689 lg K = E (1.8) = 9.01, K = = c. E = 1.8 V [Tl ( ) lg = 1.8 lg [Au E = (-0.0) =.0 V 7. (Cr + Cr + + e ) C + + e C Cr + + C + Cr + + C E = lg K = lg ( ) = 0.0 V E = E [Cr lg [Cr [C = 0.0 V (.0) lg (0.0) (0.0) = V ΔG = FE = ( ml e )(96,85 C/ml e )(0.151 J/C) = J = 9.1 kj 7. The K sp reacti is: FeS(s) Fe + (aq) + S (aq) K = K sp. Maipulate the give equatis s whe added tgether we get the K sp reacti. The we ca use the value f E fr the reacti t determie K sp. FeS + e Fe + S E = 1.01 V Fe Fe + + e E = 0. V Fe(s) Fe + (aq) + S (aq) E = 0.57 V lg K sp = E ( 0.57) = 19.9, K sp = = Al + + e Al E = 1.66 V Al + 6 F AlF e E =.07 V Al + (aq) + 6 F (aq) AlF 6 (aq) E = 0.1 V K =? lg K = E (0.1) = 0.81, K = = NO is a spectatr i. The reacti that ccurs is Ag + reactig with Z. (Ag + + e Ag) E = 0.80 V Z Z + + e E = 0.76 V

28 690 CHAPTER 17 ELECTROCHEMISTRY Ag + + Z Ag + Z + E = 1.56 V lg K = E (1.56), K = = CuI + e Cu + I E CuI =? Cu Cu + + e E = 0.5 V CuI(s) Cu + (aq) + I (aq) E = E 0.5 V Fr this verall reacti, K = K sp = : CuI E = 1 lg K sp = lg ( ) = V 1 Electrlysis E = 0.71 V = E CuI 0.5, E CuI = 0.19 V 77. a. Al + + e Al; ml e are eeded t prduce 1 ml Al frm Al mlAl mle 96,85C g Al 6.98g Al mlal mle b. 1.0 g Ni 1ml Ni 58.69g Ni mle ml Ni 96,85C mle 1 s C 1 s C = s = s = 0. hurs c. 5.0 ml Ag 1mle mlag 96,85C mle 1 s C =.8 10 s = 1. hurs 78. The xidati state f bismuth i BiO + is + because xyge has a xidati state i this i. Therefre, mles f electrs are required t reduce the bismuth i BiO + t Bi(s) g Bi A = 15C s 1mlBi 09.0 g Bi a C 60s mi mle mlbi 60mi h 1mlC mle 1mle 96,85C 96,85C mle 1 s 5.0 C = 55 s = 9. mi = C f charge passed i 1 hur 58.9 g C mlc = 16 g C b C 1mle 96,85C 1mlHf mle g Hf mlhf = 5 g Hf

29 CHAPTER 17 ELECTROCHEMISTRY 691 c. I - I + e - ; C 1mle 96,85C 1mlI mle 5.8 g I mli = 71 g I d. CrO (l) Cr 6+ + O ; 6 ml e are eeded t prduce 1 ml Cr frm mlte CrO C 1mle 96,85C 1mlCr 6 mle 5.00 g Cr mlcr =.9 g Cr 80. Al is i the + xidati i Al O, s ml e are eeded t cvert Al + it Al(s)..00 h 60mi h 60s mi s 6 C 1mle 96,85C 1mlAl mle 6.98 g Al mlal = g.00c 1mle 1mlM s = ml M where M = ukw metal s 96,85C mle Mlar mass = 0.107g M mlm 09g ml ; The elemet is bismuth. 8. Alkalie earth metals frm + is, s ml f e are trasferred t frm the metal, M. ml M = 78 s 5.00C s 1mle 96,85C 1mlM mle 1 mle 96,85C = ml M mlar mass f M = 0.71g M =. g/ml; MgCl was electrlyzed mlm 8. F is prduced at the ade: F - F + e -.00 h 60mi h 60s mi 10.0 C s 1mle = 0.76 ml e 96,85C 0.76 ml e 1 mlf RT = 0.7 ml F ; PV = RT, V = mle P ( 0.7ml)(0.0806Latm/ K ml)(98k) = 9.1 L F 1.00atm K is prduced at the cathde: K + + e K 0.76 ml e 1mlK mle 9.10g K mlk = 9. g K 8. The half-reactis fr the electrlysis f water are:

30 69 CHAPTER 17 ELECTROCHEMISTRY ( e + H O H + OH ) H O H + + O + e H O(l) H (g) + O (g) Nte: H + + OH H O ad = fr this reacti as it is writte. 60s.50C 1mle mlh 15.0 mi = ml H mi s 96,85C mle At STP, 1 mle f a ideal gas ccupies a vlume f. L (see Chapter 5 f the text) ml H.L = 0.6 L = 6 ml H mlh ml H 1mlO mlh.l = 0.11 L = 11 ml O mlo g C6H8N 1h 1mi 1mlC6H8N mle 96,85C 85. h 60mi 60s 108.1g C H N mlc H N mle = C/s r a curret f A 86. Al + + e Al; ml e are eeded t prduce Al frm Al g 1mlAl mle 96,85C 000 lb Al = C f electricity eeded lb 6.98g mlal mle C h 1h 60mi 1mi = C/s = A 60s mi 60 s.00c 1mle 1mlAg = 18 s; 18 s = ml Ag mi s 96,85C mle [Ag + = ml Ag + /0.50 L = M L ml Pt + /L = ml Pt + T plate ut 99% f the Pt +, we will prduce ml Pt mle 96,85C 1s 1mlAg ml Pt mlpt mle.00c mle = 80 s 89. Au + + e Au E = 1.50 V Ni + + e Ni E = 0. V

31 CHAPTER 17 ELECTROCHEMISTRY 69 Ag + + e Ag E = 0.80 V Cd + + e Cd E = 0.0 V H O + e H + OH E = -0.8 V Au(s) will plate ut first sice it has the mst psitive reducti ptetial, fllwed by Ag(s), which is fllwed by Ni(s), ad fially Cd(s) will plate ut last sice it has the mst egative reducti ptetial f the metals listed. Water will t iterfere with the platig prcess. 90. T begi platig ut Pd: E = 0.6 [Cl lg [PdCl = 0.6 (1.0) lg 0.00 E = 0.6 V V = 0.57 V Whe 99% f Pd has plated ut, [PdCl = 0.00 = M. 100 (1.0) E = 0.6 lg = 0.6 V 0.11V = 0.51 V.0 10 (1.0) T begi Pt platig: E = 0.7 V lg 0.00 = = 0.68 V (1.0) Whe 99% f Pt plated: E = 0.7 lg.0 10 = = 0.6 V (1.0) T begi Ir platig: E = 0.77 V lg = = 0.7 V 0.00 (1.0) Whe 99% f Ir plated: E = 0.77 lg.0 10 = = 0.70 V Yes, because the rage f ptetials fr platig ut each metal d t verlap, we shuld be able t separate the three metals. The exact ptetial t apply depeds the xidati reacti. The rder f platig will be Ir(s) first, fllwed by Pt(s) ad fially Pd(s) as the ptetial is gradually icreased. 91. Reducti ccurs at the cathde, ad xidati ccurs at the ade. First, determie all the species preset, the lk up pertiet reducti ad/r xidati ptetials i Table 17.1 fr all these species. The cathde reacti will be the reacti with the mst psitive reducti ptetial, ad the ade reacti will be the reacti with the mst psitive xidati ptetial. a. Species preset: Ni + ad Br ; Ni + ca be reduced t Ni, ad Br ca be xidized t Br (frm Table 17.1). The reactis are:

32 69 CHAPTER 17 ELECTROCHEMISTRY Cathde: Ni + + e Ni E = 0. V Ade: Br - Br + e E = 1.09 V b. Species preset: Al + ad F ; Al + ca be reduced, ad F ca be xidized. The reactis are: Cathde: Al + + e Al E = 1.66 V Ade: F F + e E =.87 V c. Species preset: M + ad I ; M + ca be reduced, ad I ca be xidized. The reactis are: Cathde: M + + e M E = 1.18 V Ade: I I + e E = 0.5 V 9. These are all i aqueus slutis, s we must als csider the reducti ad xidati f H O i additi t the ptetial redx reactis f the is preset. Fr the cathde reacti, the species with the mst psitive reducti ptetial will be reduced, ad fr the ade reacti, the species with the mst psitive xidati ptetial will be xidized. a. Species preset: Ni +, Br ad H O. Pssible cathde reactis are: Ni + + e Ni E = 0. V H O + e H + OH E = 0.8 V Because it is easier t reduce Ni + tha H O (assumig stadard cditis), Ni + will be reduced by the abve cathde reacti. Pssible ade reactis are: Br - Br + e E = 1.09 V H O O + H + + e E = 1. V Because Br - is easier t xidize tha H O (assumig stadard cditis), Br will be xidized by the abve ade reacti. b. Species preset: Al +, F ad H O; Al + ad H O ca be reduced. The reducti ptetials are E = 1.66 V fr Al + ad E = -0.8 V fr H O (assumig stadard cditis). H O will be reduced at the cathde ( H O + e H + OH ). F ad H O ca be xidized. The xidati ptetials are E =.87 V fr F ad E = 1. V fr H O (assumig stadard cditis). Frm the ptetials, we wuld predict H O t be xidized at the ade ( H O O + H + + e ). c. Species preset: M +, I ad H O; M + ad H O ca be reduced. The pssible cathde reactis are: M + + e M E = 1.18 V H O + e H + OH E = 0.8 V