CHAPTER 19 ELECTROCHEMISTRY

Transcription

1 CHAPTR 19 LCTROCHMISTRY 19.1 We fllw the steps are described i detail i Secti 19.1 f the text. (a) The prblem is give i iic frm, s cmbiig Steps 1 ad, the half-reactis are: xidati: Fe Fe 3+ reducti: H O H O Step 3: We balace each half-reacti fr umber ad type f atms ad charges. The xidati half-reacti is already balaced fr Fe atms. There are three et psitive charges the right ad tw et psitive charges the left, we add e electrs t the right side t balace the charge. Fe Fe 3+ + e Reducti half-reacti: we add e H O t the right-had side f the equati t balace the O atms. H O H O T balace the H atms, we add H + t the left-had side. H O + H + H O There are tw et psitive charges the left, s we add tw electrs t the same side t balace the charge. H O + H + + e H O Step 4: We w add the xidati ad reducti half-reactis t give the verall reacti. I rder t equalize the umber f electrs, we eed t multiply the xidati half-reacti by. (Fe Fe 3+ + e ) H O + H + + e H O Fe + H O + H + + e Fe 3+ + H O + e The electrs bth sides cacel, ad we are left with the balaced et iic equati i acidic medium. Fe + H O + H + Fe 3+ + H O The prblem is give i iic frm, s cmbiig Steps 1 ad, the half-reactis are: xidati: Cu Cu reducti: HNO 3 NO Step 3: We balace each half-reacti fr umber ad type f atms ad charges. The xidati half-reacti is already balaced fr Cu atms. There are tw et psitive charges the right, s we add tw electrs t the right side t balace the charge. Cu Cu + e

2 CHAPTR 19: LCTROCHMISTRY 549 Reducti half-reacti: we add tw H O t the right-had side f the equati t balace the O atms. HNO 3 NO + H O T balace the H atms, we add 3H + t the left-had side. 3H + + HNO 3 NO + H O There are three et psitive charges the left, s we add three electrs t the same side t balace the charge. 3H + + HNO 3 + 3e NO + H O Step 4: We w add the xidati ad reducti half-reactis t give the verall reacti. I rder t equalize the umber f electrs, we eed t multiply the xidati half-reacti by 3 ad the reducti half-reacti by. 3(Cu Cu + e ) (3H + + HNO 3 + 3e NO + H O) 3Cu + 6H + + HNO 3 + 6e 3Cu + NO + 4H O + 6e The electrs bth sides cacel, ad we are left with the balaced et iic equati i acidic medium. 3Cu + 6H + + HNO 3 3Cu + NO + 4H O (c) (d) 3CN + MO 4 + H O 3CNO + MO + OH 3Br + 6OH BrO 3 + 5Br + 3H O (e) Half-reactis balaced fr S ad I: xidati: reducti: S O 3 S4 O 6 I I Bth half-reactis are already balaced fr O, s we balace charge with electrs S O 3 S4 O 6 + e I + e I The electr cut is the same bth sides. We add the equatis, cacelig electrs, t btai the balaced equati. S O 3 + I S 4 O 6 + I 19. Strategy: We fllw the prcedure fr balacig redx reactis preseted i Secti 19.1 f the text. Sluti: (a) Step 1: The ubalaced equati is give i the prblem. M + H O MO + H O Step : The tw half-reactis are: M H O xidati MO reducti H O

3 550 CHAPTR 19: LCTROCHMISTRY Step 3: We balace each half-reacti fr umber ad type f atms ad charges. The xidati half-reacti is already balaced fr M atms. T balace the O atms, we add tw water mlecules the left side. M + H O MO T balace the H atms, we add 4 H + t the right-had side. M + H O MO + 4H + There are fur et psitive charges the right ad tw et psitive charge the left, we add tw electrs t the right side t balace the charge. M + H O MO + 4H + + e Reducti half-reacti: we add e H O t the right-had side f the equati t balace the O atms. H O H O T balace the H atms, we add H + t the left-had side. H O + H + H O There are tw et psitive charges the left, s we add tw electrs t the same side t balace the charge. H O + H + + e H O Step 4: We w add the xidati ad reducti half-reactis t give the verall reacti. Nte that the umber f electrs gaied ad lst is equal. M + H O MO + 4H + + e H O + H + + e H O M + H O + e MO + H + + e The electrs bth sides cacel, ad we are left with the balaced et iic equati i acidic medium. M + H O MO + H + Because the prblem asks t balace the equati i basic medium, we add e OH t bth sides fr each H + ad cmbie pairs f H + ad OH the same side f the arrw t frm H O. M + H O + OH MO + H + + OH Cmbiig the H + ad OH t frm water we btai: M + H O + OH MO + H O Step 5: Check t see that the equati is balaced by verifyig that the equati has the same types ad umbers f atms ad the same charges bth sides f the equati. This prblem ca be slved by the same methds used i part (a). Bi(OH) 3 + 3SO Bi + 3H O + 3SO 3

4 CHAPTR 19: LCTROCHMISTRY 551 (c) Step 1: The ubalaced equati is give i the prblem. Cr O 7 + C O 4 Cr 3+ + CO Step : The tw half-reactis are: C O 4 Cr O 7 xidati CO reducti Cr 3+ Step 3: We balace each half-reacti fr umber ad type f atms ad charges. I the xidati half-reacti, we first eed t balace the C atms. C O 4 CO The O atms are already balaced. There are tw et egative charges the left, s we add tw electrs t the right t balace the charge. C O 4 CO + e I the reducti half-reacti, we first eed t balace the Cr atms. Cr O 7 Cr 3+ We add seve H O mlecules the right t balace the O atms. Cr O 7 Cr H O T balace the H atms, we add 14H + t the left-had side. Cr O H + Cr H O There are twelve et psitive charges the left ad six et psitive charges the right. We add six electrs the left t balace the charge. Cr O H + + 6e Cr H O Step 4: We w add the xidati ad reducti half-reactis t give the verall reacti. I rder t equalize the umber f electrs, we eed t multiply the xidati half-reacti by 3. 3(C O 4 CO + e ) Cr O H + + 6e Cr H O 3C O 4 + Cr O H + + 6e 6CO + Cr H O + 6e The electrs bth sides cacel, ad we are left with the balaced et iic equati i acidic medium. 3C O 4 + Cr O H + 6CO + Cr H O Step 5: Check t see that the equati is balaced by verifyig that the equati has the same types ad umbers f atms ad the same charges bth sides f the equati. (d) This prblem ca be slved by the same methds used i part (c). Cl + ClO 3 + 4H + Cl + ClO + H O

5 55 CHAPTR 19: LCTROCHMISTRY Half-reacti (V) Mg (aq) + e Mg(s).37 Cu (aq) + e Cu(s) The verall equati is: Mg(s) + Cu (aq) Mg (aq) + Cu(s) 0.34 V (.37 V).71 V 19.1 Strategy: At first, it may t be clear hw t assig the electrdes i the galvaic cell. Frm Table 19.1 f the text, we write the stadard reducti ptetials f Al ad Ag ad apply the diagal rule t determie which is the ade ad which is the cathde. Sluti: The stadard reducti ptetials are: Ag + (1.0 M) + e Ag(s) Al 3+ (1.0 M) + 3e Al(s) 0.80 V 1.66 V Applyig the diagal rule, we see that Ag + will xidize Al. Ade (xidati): Cathde (reducti): Overall: Al(s) Al 3+ (1.0 M) + 3e 3Ag + (1.0 M) + 3e 3Ag(s) Al(s) + 3Ag + (1.0 M) Al 3+ (1.0 M) + 3Ag(s) Nte that i rder t balace the verall equati, we multiplied the reducti f Ag + by 3. We ca d s because, as a itesive prperty, is t affected by this prcedure. We fid the emf f the cell usig quati (19.1) ad Table 19.1 f the text. cell cathde ade + 3+ Ag /Ag Al /Al cell 0.80 V ( 1.66 V) +.46 V Check: The psitive value f shws that the frward reacti is favred The apprpriate half-reactis frm Table 19.1 are I (s) + e I (aq) ade 0.53 V Fe 3+ (aq) + e Fe (aq) cathde 0.77 V Thus ir(iii) shuld xidize idide i t idie. This makes the idide i/idie half-reacti the ade. The stadard emf ca be fud usig quati (19.1). cell cathde ade 0.77 V 0.53 V 0.4 V (The emf was t required i this prblem, but the fact that it is psitive cfirms that the reacti shuld favr prducts at equilibrium.) The halfreacti fr xidati is: H O(l) xidati (ade) O (g) + 4H + (aq) + 4e ade V

6 CHAPTR 19: LCTROCHMISTRY 553 The species that ca xidize water t mlecular xyge must have a red mre psitive tha +1.3 V. Frm Table 19.1 f the text we see that ly Cl (g) ad MO 4 (aq) i acid sluti ca xidize water t xyge The verall reacti is: 5NO 3 (aq) + 3M (aq) + H O(l) 5NO(g) + 3MO 4 (aq) + 4H + (aq) cell cathde ade 0.96 V 1.51 V 0.55 V The egative emf idicates that reactats are favred at equilibrium. NO 3 will t xidize M t MO4 uder stadard-state cditis Strategy: cell is psitive fr a sptaeus reacti. I each case, we ca calculate the stadard cell emf frm the ptetials fr the tw half-reactis. Sluti: (a) cell cathde ade 0.40 V (.87 V).47 V. The reacti is sptaeus. (c) (d) 0.14 V 1.07 V 1.1 V. The reacti is t sptaeus. 0.5 V 0.80 V 1.05 V. The reacti is t sptaeus V 0.15 V 0.6 V. The reacti is sptaeus Frm Table 19.1 f the text, we cmpare the stadard reducti ptetials fr the half-reactis. The mre psitive the ptetial, the better the substace as a xidizig aget. (a) Au 3+ Ag + (c) Cd (d) O i acidic sluti Strategy: The greater the tedecy fr the substace t be xidized, the strger its tedecy t act as a reducig aget. The species that has a strger tedecy t be xidized will have a smaller reducti ptetial. Sluti: I each pair, lk fr the e with the smaller reducti ptetial. This idicates a greater tedecy fr the substace t be xidized. (a) Li H (c) Fe (d) Br 19.1 We fid the stadard reducti ptetials i Table 19.1 f the text. cell cathde ade 0.76 V (.37 V) 1.61 V cell l K V l K cell V

7 554 CHAPTR 19: LCTROCHMISTRY K e cell V ()(1.61 V) V K e K Strategy: The relatiship betwee the equilibrium cstat, K, ad the stadard emf is give by quati (19.5) f the text: cell (0.057 V / )l K. Thus, kwig (the mles f electrs trasferred) ad the equilibrium cstat, we ca determie cell. Sluti: The equati that relates K ad the stadard cell emf is: cell V l K We see i the reacti that Mg ges t Mg ad Z ges t Z. Therefre, tw mles f electrs are trasferred durig the redx reacti. Substitute the equilibrium cstat ad the mles f e trasferred ( ) it the abve equati t calculate. (0.057 V) l K (0.057 V) l( ) V I each case we use stadard reducti ptetials frm Table 19.1 tgether with quati (19.5) f the text. (a) cell cathde ade 1.07 V 0.53 V 0.54 V l K cell V K e cell V ()(0.54 V) K V e cell cathde ade 1.61 V 1.36 V 0.5 V ()(0.5 V) K V e (c) cell cathde ade 1.51 V 0.77 V 0.74 V (5)(0.74 V) K V e

8 CHAPTR 19: LCTROCHMISTRY (a) We break the equati it tw halfreactis: xidati (ade) Mg(s) Mg (aq) + e ade.37 V Pb (aq) + e reducti (cathde) Pb(s) cathde 0.13 V The stadard emf is give by cell cathde ade 0.13 V (.37 V).4 V We ca calculate ΔG frm the stadard emf. Δ G Fcell Δ G ()(96500 J/V ml)(.4 V) 43 kj/ml Next, we ca calculate K usig quati (19.5) f the text. r ad V cell l K l K cell V K e K ()(.4) e Tip: Yu culd als calculate K c frm the stadard free eergy chage, ΔG, usig the equati: ΔG RTl K c. We break the equati it tw halfreactis: Br (l) + e I (aq) reducti (cathde) Br (aq) cathde 1.07 V xidati (ade) I (s) + e ade 0.53 V The stadard emf is cell cathde ade 1.07 V 0.53 V 0.54 V We ca calculate ΔG frm the stadard emf. Δ G Fcell Δ G ()(96500 J/V ml)(0.54 V) 104 kj/ml Next, we ca calculate K usig quati (19.5) f the text. K e 0.057

9 556 CHAPTR 19: LCTROCHMISTRY K ()(0.54) e (c) This is wrked i a aalgus maer t parts (a) ad. cell cathde ade 1.3 V 0.77 V 0.46 V Δ G Fcell ΔG (4)(96500 J/V ml)(0.46 V) 178 kj/ml K e K (4)(0.46) e (d) This is wrked i a aalgus maer t parts (a),, ad (c). cell cathde ade 0.53 V ( 1.66 V).19 V Δ G Fcell ΔG (6)(96500 J/V ml)(.19 V) kj/ml K e K (6)(.19) e The half-reactis are: Fe 3+ (aq) + e Fe (aq) ade 0.77 V Ce 4+ (aq) + e Ce 3+ (aq) cathde 1.61 V Thus, Ce 4+ will xidize Fe t Fe 3+ ; this makes the Fe /Fe 3+ half-reacti the ade. The stadard cell emf is fud usig quati (19.1) f the text. cell cathde ade 1.61 V 0.77 V 0.84 V The values f ΔG ad K c are fud usig quatis (19.3) ad (19.5) f the text. Δ G Fcell (1)(96500 J/V ml)(0.84 V) 81 kj/ml l K cell V cell (1)(0.84 V) V V K c e e 10 14

10 CHAPTR 19: LCTROCHMISTRY Strategy: The relatiship betwee the stadard free eergy chage ad the stadard emf f the cell is give by quati (19.3) f the text: Δ G Fcell. The relatiship betwee the equilibrium cstat, K, ad the stadard emf is give by quati (19.5) f the text: cell (0.057 V / )l K. Thus, if we ca determie cell, we ca calculate ΔG ad K. We ca determie the cell f a hypthetical galvaic cell made up f tw cuples (Cu /Cu + ad Cu + /Cu) frm the stadard reducti ptetials i Table 19.1 f the text. Sluti: The half-cell reactis are: Ade (xidati): Cathde (reducti): Overall: Cu + (1.0 M) Cu (1.0 M) + e Cu + (1.0 M) + e Cu(s) Cu + (1.0 M) Cu (1.0 M) + Cu(s) cell cathde ade + + Cu /Cu Cu /Cu cell 0.5 V 0.15 V 0.37 V Nw, we use quati (19.3) f the text. The verall reacti shws that 1. Δ G Fcell ΔG (1)(96500 J/V ml)(0.37 V) 36 kj/ml Next, we ca calculate K usig quati (19.5) f the text. r ad cell l K V l K cell V K e K (1)(0.37) e e 6 10 Check: The egative value f ΔG ad the large psitive value f K, bth idicate that the reacti favrs prducts at equilibrium. The result is csistet with the fact that fr the galvaic cell is psitive If this were a stadard cell, the ccetratis wuld all be 1.00 M, ad the vltage wuld just be the stadard emf calculated frm Table 19.1 f the text. Sice cell emf's deped the ccetratis f the reactats ad prducts, we must use the Nerst equati [quati (19.8) f the text] t fid the emf f a stadard cell V l Q V [Z ] 1.10 V l [Cu ]

11 558 CHAPTR 19: LCTROCHMISTRY V V l V Hw did we fid the value f 1.10 V fr? Strategy: The stadard emf ( ) ca be calculated usig the stadard reducti ptetials i Table 19.1 f the text. Because the reactis are t ru uder stadard-state cditis (ccetratis are t 1 M), we eed Nerst's equati [quati (19.8) f the text] t calculate the emf () f a hypthetical galvaic cell. Remember that slids d t appear i the reacti qutiet (Q) term i the Nerst equati. We ca calculate ΔG frm usig quati (19.) f the text: ΔG F cell. Sluti: (a) The half-cell reactis are: Ade (xidati): Cathde (reducti): Overall: Mg(s) Mg (1.0 M) + e S (1.0 M) + e S(s) Mg(s) + S (1.0 M) Mg (1.0 M) + S(s) cell cathde ade S /S Mg /Mg cell 0.14 V (.37 V) Frm quati (19.8) f the text, we write: V l Q.3 V V l [Mg ] [S ] V V l.3 V We ca w fid the free eergy chage at the give ccetratis usig quati (19.) f the text. Nte that i this reacti,. ΔG F cell ΔG ()(96500 J/V ml)(.3 V) 430 kj/ml The half-cell reactis are: Ade (xidati): 3[Z(s) Z (1.0 M) + e ] Cathde (reducti): [Cr 3+ (1.0 M) + 3e Cr(s)] Overall: 3Z(s) + Cr 3+ (1.0 M) 3Z (1.0 M) + Cr(s) cell cathde ade 3+ Cr /Cr Z /Z cell 0.74 V ( 0.76 V) 0.0 V

12 CHAPTR 19: LCTROCHMISTRY 559 Frm quati (19.8) f the text, we write: V l Q V [Z ] l [Cr 3+ ] V (0.0085) 0.0 V l 0.04 V 6 (0.010) We ca w fid the free eergy chage at the give ccetratis usig quati (19.) f the text. Nte that i this reacti, 6. ΔG F cell ΔG (6)(96500 J/V ml)(0.04 V) 3 kj/ml The verall reacti is: Z(s) + H + (aq) Z (aq) + H (g) cell cathde ade 0.00 V ( 0.76 V) 0.76 V V [Z ] PH l + [H ] V (0.45)(.0) 0.76 V l 0.78 V (1.8) 19.3 Let s write the tw half-reactis t calculate the stadard cell emf. (Oxidati ccurs at the Pb electrde.) xidati (ade) Pb(s) Pb (aq) + e ade 0.13 V H + (aq) + e reducti (cathde) H (g) cathde 0.00 V H + (aq) + Pb(s) H (g) + Pb (aq) cell cathde ade 0.00 V ( 0.13 V) 0.13 V Usig the Nerst equati, we ca calculate the cell emf, V [Pb ] PH l + [H ] V (0.10)(1.0) 0.13 V l V (0.050) As writte, the reacti is t sptaeus uder stadard state cditis; the cell emf is egative. cell cathde ade 0.76 V 0.34 V 1.10 V

13 560 CHAPTR 19: LCTROCHMISTRY The reacti will becme sptaeus whe the ccetratis f zic(ii) ad cpper(ii) is are such as t make the emf psitive. The turig pit is whe the emf is zer. We slve the Nerst equati fr the [Cu ]/[Z ] rati at this pit V cell l Q V [Cu ] V l [Z ] [Cu ] l 85.6 [Z ] [Cu ] [Z ] 85.6 e I ther wrds fr the reacti t be sptaeus, the [Cu ]/[Z ] rati must be less tha Is the reducti f zic(ii) by cpper metal a practical use f cpper? All ccetrati cells have the same stadard emf: zer vlts. Mg (aq) + e reducti (cathde) Mg(s) cathde.37 V xidati (ade) Mg(s) Mg (aq) + e ade.37 V cell cathde ade.37 V (.37 V) 0.00 V We use the Nerst equati t cmpute the emf. There are tw mles f electrs trasferred frm the reducig aget t the xidizig aget i this reacti, s V l Q V [Mg ] l x [Mg ] red V 0.4 0V l V 0.53 What is the directi f sptaeus chage i all ccetrati cells? (a) The ttal charge passig thrugh the circuit is 8.5 C 3600 s h C 1s 1h Frm the ade half-reacti we ca fid the amut f hydrge. 4 mlh 1mle (9. 10 C) 0.48 ml H 4mle C

14 CHAPTR 19: LCTROCHMISTRY 561 The vlume ca be cmputed usig the ideal gas equati V RT (0.48 ml)(0.081 L atm/k ml)(98 K) L P 155 atm The charge passig thrugh the circuit i e miute is 8.5 C 60 s 510 C/mi 1s 1mi We ca fid the amut f xyge frm the cathde half-reacti ad the ideal gas equati. 510 C 1 ml e 1mlO ml O /mi 1 mi C 4mle V ml O (0.081 L atm/k ml)(98 K) RT P 1mi 1atm 0.03 L O /mi 0.03 L O 1.0 L air 0.16 L f air/mi 1 mi 0.0 L O We ca calculate the stadard free eergy chage, ΔG, frm the stadard free eergies f frmati, ΔGf usig quati (18.1) f the text. The, we ca calculate the stadard cell emf, The verall reacti is: C 3 H 8 (g) + 5O (g) 3CO (g) + 4H O(l) rx f f f 3 8 f cell Δ G 3 Δ G [CO ( g)] + 4 ΔG [H O( l)] { Δ G [C H ( g)] + 5 ΔG [O ( g)]}, frm ΔG. Δ Grx (3)( kj/ml) + (4)( 37. kj/ml) [(1)( 3.5 kj/ml) + (5)(0)] kj/ml We ca w calculate the stadard emf usig the fllwig equati: r Δ G cell F Δ G F cell Check the half-reactis p. 843 f the text t determie that 0 mles f electrs are trasferred durig this redx reacti. cell 3 ( J/ml) (0)(96500 J/V ml) 1.09 V Des this suggest that, i thery, it shuld be pssible t cstruct a galvaic cell (battery) based ay cceivable sptaeus reacti? Mass Mg 1 ml Mg 4.31 g Mg 1.00 F mle 1mlMg 1. g Mg

15 56 CHAPTR 19: LCTROCHMISTRY (a) The ly is preset i mlte BaCl are Ba ad Cl. The electrde reactis are: ade: Cl (aq) Cl (g) + e cathde: Ba (aq) + e Ba(s) This cathde half-reacti tells us that mles f e are required t prduce 1 mle f Ba(s). Strategy: Accrdig t Figure 19.0 f the text, we ca carry ut the fllwig cversi steps t calculate the quatity f Ba i grams. curret time culmbs ml e ml Ba g Ba This is a large umber f steps, s let s break it dw it tw parts. First, we calculate the culmbs f electricity that pass thrugh the cell. The, we will ctiue t calculate grams f Ba. Sluti: First, we calculate the culmbs f electricity that pass thrugh the cell. 1C 60s 0.50 A 30 mi C 1A s 1mi We see that fr every mle f Ba frmed at the cathde, mles f electrs are eeded. The grams f Ba prduced at the cathde are: 1ml 1mlBa 137.3gBa? g Ba ( C) e 0.64 g Ba 96,500 C mle 1 ml Ba The half-reactis are: Na + + e Na Al e Al Sice 1 g is the same idea as 1 t as lg as we are cmparig tw quatities, we ca write: 1ml 1g Na 1e 0.043mle.99 g Na 1ml 1g Al 3e 0.11mle 6.98 g Al It is cheaper t prepare 1 t f sdium by electrlysis The cst fr prducig varius metals is determied by the mles f electrs eeded t prduce a give amut f metal. Fr each reducti, let's first calculate the umber f ts f metal prduced per 1 mle f electrs (1 t g). The reductis are: Mg + e Al e Mg Al 1 ml Mg 4.31 g Mg 1 t t Mg/ml 1mlMg 5 ml g e 1 ml Al 6.98 g Al 1 t t Al/ml 1mlAl 5 3 ml g e e e Na + + e Na 1 ml Na.99 g Na 1 t t Na/ml 1mlNa 5 1 ml g e e

16 CHAPTR 19: LCTROCHMISTRY 563 Ca + e Ca 1 ml Ca g Ca 1 t t Ca/ml 1mlCa 5 ml g e e Nw that we kw the ts f each metal prduced per mle f electrs, we ca cvert frm $155/t Mg t the cst t prduce the give amut f each metal. (a) Fr alumium : 5 $ t Mg 1 ml e 10.0 ts Al 1tMg 6 1 ml t Al e 3 $ (c) Fr sdium: 5 $ t Mg 1 ml e 30.0 ts Na 1tMg 5 1 ml t Na Fr calcium: e 5 $ t Mg 1 ml e 50.0 ts Ca 1tMg 5 1 ml t Ca e 3 $ $ Fid the amut f xyge usig the ideal gas equati 1atm 755 mmhg (0.076 L) PV 760 mmhg ml O RT (0.081 L atm/k ml)(98 K) Sice the half-reacti shws that e mle f xyge requires fur faradays f electric charge, we write 3 4 F ( ml O ) 0.01 F 1mlO (a) The halfreacti is: H O(l) O (g) + 4H + (aq) + 4e First, we ca calculate the umber f mles f xyge prduced usig the ideal gas equati. O PV RT (1.0 atm)(0.84 L) ml O (0.081 L atm/ml K)(98 K) O Sice 4 mles f electrs are eeded fr every 1 mle f xyge, we will eed 4 F f electrical charge t prduce 1 mle f xyge. 4 F? F ml O 0.14 F 1mlO

17 564 CHAPTR 19: LCTROCHMISTRY The halfreacti is: Cl (aq) Cl (g) + e The umber f mles f chlrie prduced is: Cl PV RT 1atm 750 mmhg (1.50 L) 760 mmhg ml Cl (0.081 L atm/ml K)(98 K) Cl Sice mles f electrs are eeded fr every 1 mle f chlrie gas, we will eed F f electrical charge t prduce 1 mle f chlrie gas. F? F ml Cl 0.11 F 1mlCl (c) The halfreacti is: S (aq) + e S(s) The umber f mles f S(s) prduced is 1mlS? ml S 6.0 g S ml S g S Sice mles f electrs are eeded fr every 1 mle f S, we will eed F f electrical charge t reduce 1 mle f S is t S metal. F? F ml S 0.10 F 1mlS The half-reactis are: Cu (aq) + e Cu(s) The mass f cpper prduced is: Br (aq) Br (l) + e 3600 s 1 C 1 ml e 1 ml Cu g Cu 4.50 A 1 h 1h 1A s 96500C mle 1mlCu 5.33 g Cu The mass f brmie prduced is: 3600 s 1 C 1 ml e 1mlBr 159.8gBr 4.50 A 1 h 1 h 1 A s C mle 1 ml Br 13.4 g Br 19.5 (a) The halfreacti is: Ag + (aq) + e Ag(s)

18 CHAPTR 19: LCTROCHMISTRY 565 Sice this reacti is takig place i a aqueus sluti, the prbable xidati is the xidati f water. (Neither Ag + r NO 3 ca be further xidized.) H O(l) O (g) + 4H + (aq) + 4e (c) The half-reacti tells us that 1 mle f electrs is eeded t reduce 1 ml f Ag + t Ag metal. We ca set up the fllwig strategy t calculate the quatity f electricity (i C) eeded t depsit 0.67 g f Ag. grams Ag ml Ag ml e culmbs 1 ml Ag 1 ml e C 0.67 g Ag g Ag 1 ml Ag 1mle C The half-reacti is: C + e C 1 ml C ml e C.35 g C g C 1 ml C 1mle C (a) First fid the amut f charge eeded t prduce.00 g f silver accrdig t the halfreacti: Ag + (aq) + e Ag(s) 1 ml Ag 1 ml e C 3.00 g Ag C g Ag 1 ml Ag 1mle The halfreacti fr the reducti f cpper(ii) is: Cu (aq) + e Cu(s) Frm the amut f charge calculated abve, we ca calculate the mass f cpper depsited i the secd cell. 3 1mle 1mlCu 63.55gCu ( C) g Cu C mle 1 ml Cu We ca calculate the curret flwig thrugh the cells usig the fllwig strategy. Culmbs Culmbs/hur Culmbs/secd Recall that 1 C 1 A s The curret flwig thrugh the cells is: 3 1h 1 ( A s) A 3600 s 3.75 h

19 566 CHAPTR 19: LCTROCHMISTRY The half-reacti fr the xidati f chlride i is: Cl (aq) Cl (g) + e First, let's calculate the mles f e flwig thrugh the cell i e hur. 1 C 3600 s 1 ml e 1500 A ml e 1 A s 1 h C Next, let's calculate the hurly prducti rate f chlrie gas (i kg). Nte that the ade efficiecy is 93.0%. 1 ml Cl kg Cl 93.0% ml e 1.84 kg Cl /h mle 1 ml Cl 100% Step 1: Balace the halfreacti. Cr O 7 (aq) + 14H + (aq) + 1e Cr(s) + 7H O(l) Step : Calculate the quatity f chrmium metal by calculatig the vlume ad cvertig this t mass usig the give desity. Vlume Cr thickess surface area Vlume Cr ( m mm) 0.5 m 1000 mm.5 10 m Cvertig t cm 3, cm 3 (.5 10 m ).5 cm 0.01 m Next, calculate the mass f Cr. Mass desity vlume g Mass Cr.5 cm 18 g Cr 3 1cm Step 3: Fid the umber f mles f electrs required t electrdepsit 18 g f Cr frm sluti. The halfreacti is: Cr O 7 (aq) + 14H + (aq) + 1e Cr(s) + 7H O(l) Six mles f electrs are required t reduce 1 ml f Cr metal. But, we are electrdepsitig less tha 1 mle f Cr(s). We eed t cmplete the fllwig cversis: g Cr ml Cr ml e 1mlCr 6mle? faradays 18 g Cr.1 ml e 5.00 g Cr 1 ml Cr Step 4: Determie hw lg it will take fr.1 mles f electrs t flw thrugh the cell whe the curret is 5.0 C/s. We eed t cmplete the fllwig cversis: ml e culmbs secds hurs

20 CHAPTR 19: LCTROCHMISTRY ,500 C 1 s 1 h?h.1 ml e.3h 1mle 5.0 C 3600 s Wuld ay time be saved by cectig several bumpers tgether i a series? The quatity f charge passig thrugh the sluti is: 1C 60s 1mle A 5.0 mi ml e 1 A s 1 mi C Sice the charge f the cpper i is +, the umber f mles f cpper frmed must be: 1mlCu 3 ( ml e ) ml Cu mle The uits f mlar mass are grams per mle. The mlar mass f cpper is: g ml 63.1g/ml Based the half-reacti, we kw that e faraday will prduce half a mle f cpper. Cu (aq) + e Cu(s) First, let s calculate the charge (i C) eeded t depsit g f Cu. 1C (3.00 A)(304 s) 91 C 1A s We kw that e faraday will prduce half a mle f cpper, but we d t have a half a mle f cpper. We have: 1mlCu g Cu ml g Cu We calculated the umber f culmbs (91 C) eeded t prduce ml f Cu. Hw may culmbs will it take t prduce mles f Cu? This will be Faraday s cstat. 91 C ml Cu C 1 F ml Cu The umber f faradays supplied is: 1mlAg 1mle 1.44 g Ag ml e g Ag 1 ml Ag Sice we eed three faradays t reduce e mle f X 3+, the mlar mass f X must be: 0.10 g X 3 ml ml e e 1mlX 7.1 g/ml

21 568 CHAPTR 19: LCTROCHMISTRY First we ca calculate the umber f mles f hydrge prduced usig the ideal gas equati. H PV RT H 1atm 78 mmhg (0.845 L) 760 mmhg (0.081 L atm/k ml)(98 K) ml The umber f faradays passed thrugh the sluti is: F ml H F 1mlH (a) The half-reactis are: H (g) H + (aq) + e The cmplete balaced equati is: Ni (aq) + e Ni(s) Ni (aq) + H (g) Ni(s) + H + (aq) Ni(s) is belw ad t the right f H + (aq) i Table 19.1 f the text (see the half-reactis at 0.5 ad 0.00 V). Therefre, the sptaeus reacti is the reverse f the abve reacti, that is: Ni(s) + H + (aq) Ni (aq) + H (g) The half-reactis are: + MO 4 (aq) + 8H (aq) + 5e M (aq) + 4H O Cl (aq) Cl (g) + e The cmplete balaced equati is: MO 4 (aq) + 16H + (aq) + 10Cl (aq) M (aq) + 8H O + 5Cl (g) I Table 19.1 f the text, Cl (aq) is belw ad t the right f MO 4 (aq); therefre the sptaeus reacti is as writte. (c) The half-reactis are: Cr(s) Cr 3+ (aq) + 3e The cmplete balaced equati is: Z (aq) + e Z(s) Cr(s) + 3Z (aq) Cr 3+ (aq) + 3Z(s) I Table 19.1 f the text, Z(s) is belw ad t the right f Cr 3+ (aq); therefre the sptaeus reacti is the reverse f the reacti as writte The balaced equati is: Cr O Fe + 14H + Cr Fe H O The remaider f this prblem is a sluti stichimetry prblem. The umber f mles f ptassium dichrmate i 6.0 ml f the sluti is: ml ml ml KCrO ml sl

22 CHAPTR 19: LCTROCHMISTRY 569 Frm the balaced equati it ca be see that 1 mle f dichrmate is stichimetrically equivalet t 6 mles f ir(ii). The umber f mles f ir(ii) xidized is therefre 4 6mlFe 3 7 1mlCrO7 ( ml Cr O ) ml Fe Fially, the mlar ccetrati f Fe is: ml L ml/l M Fe The balaced equati is: 5SO (g) + MO 4 (aq) + H O(l) 5SO 4 (aq) + M (aq) + 4H + (aq) The mass f SO i the water sample is give by ml KMO4 5 ml SO g SO 7.37 ml 1000 ml sl ml KMO 1 ml SO g SO The balaced equati is: MO 4 + 5Fe + 8H + M + 5Fe H O First, let s calculate the umber f mles f ptassium permagaate i 3.30 ml f sluti ml ml ml KMO 1000 ml sl Frm the balaced equati it ca be see that 1 mle f permagaate is stichimetrically equivalet t 5 mles f ir(ii). The umber f mles f ir(ii) xidized is therefre 4 5mlFe 3 4 1mlMO4 ( ml MO ).6 10 ml Fe 4 The mass f Fe xidized is: g Fe mass Fe (.6 10 ml Fe ) 0.16 g Fe 1mlFe Fially, the mass percet f ir i the re ca be calculated. mass f ir mass % Fe 100% ttal mass f sample 0.16 g %Fe 100% 45.1% 0.79 g (a) The balaced equati is: MO 4 + 5H O + 6H + 5O + M + 8H O

23 570 CHAPTR 19: LCTROCHMISTRY The umber f mles f ptassium permagaate i ml f the sluti is ml ml ml f KMO ml sl Frm the balaced equati it ca be see that i this particular reacti mles f permagaate is stichimetrically equivalet t 5 mles f hydrge perxide. The umber f mles f H O xidized is therefre 4 5mlHO 3 ( ml MO 4) ml HO mlmo4 The mlar ccetrati f H O is: ml [HO ] ml/l M L (a) The halfreactis are: + (i) MO 4 (aq) + 8H (aq) + 5e M (aq) + 4H O(l) (ii) C O 4 (aq) CO (g) + e We cmbie the half-reactis t cacel electrs, that is, [ equati (i)] + [5 equati (ii)] + MO 4 (aq) + 16H (aq) + 5C O 4 (aq) M (aq) + 10CO (g) + 8H O(l) We ca calculate the mles f KMO 4 frm the mlarity ad vlume f sluti ml KMO ml KMO ml KMO ml sl We ca calculate the mass f xalic acid frm the stichimetry f the balaced equati. The mle rati betwee xalate i ad permagaate i is 5:. 4 5 ml HCO g HCO ( ml KMO 4 4) g HCO4 ml KMO4 1 ml HCO4 Fially, the percet by mass f xalic acid i the sample is: g % xalic acid 100% 5.40% 1.00 g ΔG Cell Reacti > 0 < 0 sptaeus < 0 > 0 sptaeus 0 0 at equilibrium The balaced equati is: MO 4 + 5C O H + M + 10CO + 8H O Therefre, ml MO 4 reacts with 5 ml C O 4

24 CHAPTR 19: LCTROCHMISTRY ml MO4 5 Mles f MO 4 reacted 4. ml ml MO ml sl Recgize that the mle rati f Ca t C O 4 is 1:1 i CaC O 4. The mass f Ca i 10.0 ml is: 5 5 ml Ca g Ca 3 ( ml MO 4 ) g Ca mlmo4 1mlCa Fially, cvertig t mg/ml, we have: g Ca 1000 mg 10.0 ml 1 g 0.31 mg Ca /ml bld The slubility equilibrium f AgBr is: AgBr(s) Ag + (aq) + Br (aq) By reversig the secd give half-reacti ad addig it t the first, we btai: Ag(s) Ag + (aq) + e ade 0.80 V AgBr(s) + e Ag(s) + Br (aq) cathde 0.07 V AgBr(s) Ag + (aq) + Br (aq) cell cathde ade 0.07 V 0.80 V 0.73 V At equilibrium, we have: V l[ag + ][Br ] V V lk sp 1 l K sp 8.4 K sp (Nte that this value differs frm that give i Table 16. f the text, sice the data quted here were btaied frm a studet's lab reprt.) (a) The halfreactis are: H + (aq) + e H (g) ade 0.00 V Ag + (aq) + e Ag(s) cathde 0.80 V cell cathde ade 0.80 V 0.00 V 0.80 V The sptaeus cell reacti uder stadard-state cditis is: Ag + (aq) + H (g) Ag(s) + H + (aq)

25 57 CHAPTR 19: LCTROCHMISTRY (c) Usig the Nerst equati we ca calculate the cell ptetial uder stadard-state cditis V [H ] l [Ag + ] P H (i) The ptetial is: V ( ) 0.80 V l 0.9 V (1.0) (1.0) (ii) The ptetial is: V ( ) 0.80 V l 1.10 V (1.0) (1.0) (d) Frm the results i part (c), we deduce that this cell is a ph meter; its ptetial is a sesitive fucti f the hydrge i ccetrati. ach 1 uit icrease i ph causes a vltage icrease f V (a) If this were a stadard cell, the ccetratis wuld all be 1.00 M, ad the vltage wuld just be the stadard emf calculated frm Table 19.1 f the text. Sice cell emf's deped the ccetratis f the reactats ad prducts, we must use the Nerst equati [quati (19.8) f the text] t fid the emf f a stadard cell V l Q V [Mg ] 3.17 V l + [Ag ] V V l [0.10] 3.14 V First we calculate the ccetrati f silver i remaiig i sluti after the depsiti f 1.0 g f silver metal Ag rigially i sluti: ml Ag 1L L ml Ag Ag depsited: Ag remaiig i sluti: 1ml 1.0 g Ag ml Ag g ( ml Ag) ( ml Ag) ml Ag ml [Ag ] M L The verall reacti is: Mg(s) + Ag + (aq) Mg (aq) + Ag(s) We use the balaced equati t fid the amut f magesium metal sufferig xidati ad disslvig. 1mlMg 3 ( ml Ag) ml Mg mlag

26 CHAPTR 19: LCTROCHMISTRY 573 The amut f magesium rigially i sluti was ml 0.88 L ml 1L The ew magesium i ccetrati is: 3 [( ) + ( )] ml 0.88 L M The ew cell emf is: V l Q V V l ( ) 3.13 V 19.7 The vervltage f xyge is t large eugh t prevet its frmati at the ade. Applyig the diagal rule, we see that water is xidized befre fluride i. F (g) + e F (aq).87 V O (g) + 4H + (aq) + 4e H O(l) 1.3 V The very psitive stadard reducti ptetial idicates that F has essetially tedecy t uderg xidati. The xidati ptetial f chlride i is much smaller (1.36 V), ad hece Cl (g) ca be prepared by electrlyzig a sluti f NaCl. This fact was e f the majr bstacles prevetig the discvery f flurie fr may years. HF was usually chse as the substace fr electrlysis, but tw prblems iterfered with the experimet. First, ay water i the HF was xidized befre the fluride i. Secd, pure HF withut ay water i it is a cductr f electricity (HF is a weak acid!). The prblem was fially slved by disslvig KF i liquid HF t give a cductig sluti The cell vltage is give by: V [Cu ] l dilute [Cu ] ccetrated V l V We ca calculate the amut f charge that 4.0 g f MO ca prduce. 1 ml ml e C g MO C g ml MO 1mle Sice a curret f e ampere represets a flw f e culmb per secd, we ca fid the time it takes fr this amut f charge t pass.

27 574 CHAPTR 19: LCTROCHMISTRY A C/s 3 1s 1h ( C) C 3600 s.5 10 h The tw electrde prcesses are: ade: H O(l) O (g) + 4H + (aq) + 4e cathde: 4H O(l) + 4e H (g) + 4OH (aq) The amut f hydrge frmed is twice the amut f xyge. Ntice that the sluti at the ade will becme acidic ad that the sluti at the cathde will becme basic (test with litmus paper). What are the relative amuts f H + ad OH frmed i this prcess? Wuld the slutis surrudig the tw electrdes eutralize each ther exactly? If t, wuld the resultig sluti be acidic r basic? Sice this is a ccetrati cell, the stadard emf is zer. (Why?) Usig the Nerst equati, we ca write equatis t calculate the cell vltage fr the tw cells. (1) RT RT [Hg ]sl A cell l Q l F F [Hg ]sl B () cell + RT RT [Hg ]sl A l Q l F 1 F + [Hg ]sl B I the first case, tw electrs are trasferred per mercury i ( ), while i the secd ly e is trasferred ( 1). Nte that the ccetrati rati will be 1:10 i bth cases. The vltages calculated at 18 C are: (1) () cell (8.314 J/K ml)(91 K) l (96500 J V ml ) V cell (8.314 J/K ml)(91 K) l (96500 J V ml ) V Sice the calculated cell ptetial fr cell (1) agrees with the measured cell emf, we cclude that the mercury(i) i exists as Hg i sluti Accrdig t the fllwig stadard reducti ptetials: O (g) + 4H + (aq) + 4e H O I (s) + e I (aq) 1.3 V 0.53 V we see that it is easier t xidize the idide i tha water (because O is a strger xidizig aget tha I ). Therefre, the ade reacti is: I (aq) I (s) + e The sluti surrudig the ade will becme brw because f the frmati f the triidide i: I + I (s) I 3 (aq) The cathde reacti will be the same as i the NaCl electrlysis. (Why?) Sice OH is a prduct, the sluti arud the cathde will becme basic which will cause the phelphthalei idicatr t tur red.

28 CHAPTR 19: LCTROCHMISTRY We begi by treatig this like a rdiary stichimetry prblem (see Chapter 3). Step 1: Calculate the umber f mles f Mg ad Ag +. The umber f mles f magesium is: 1mlMg 1.56 g Mg ml Mg 4.31 g Mg The umber f mles f silver i i the sluti is: ml Ag 1L L ml Ag Step : Calculate the mass f Mg remaiig by determiig hw much Mg reacts with Ag +. The balaced equati fr the reacti is: Ag + (aq) + Mg(s) Ag(s) + Mg (aq) Sice yu eed twice as much Ag + cmpared t Mg fr cmplete reacti, Ag + is the limitig reaget. The amut f Mg csumed is: + 1mlMg ml Ag ml Mg + mlag The amut f magesium remaiig is: 4.31 g Mg ( ) ml Mg 1.44 g Mg 1mlMg Step 3: Assumig cmplete reacti, calculate the ccetrati f Mg is prduced. Sice the mle rati betwee Mg ad Mg is 1:1, the ml f Mg frmed will equal the ml f Mg reacted. The ccetrati f Mg is: ml [Mg ] M L Step 4: We ca calculate the equilibrium cstat fr the reacti frm the stadard cell emf. cell cathde ade 0.80 V (.37 V) 3.17 V We ca the cmpute the equilibrium cstat. K e cell ()(3.17) K e

29 576 CHAPTR 19: LCTROCHMISTRY Step 5: T fid equilibrium ccetratis f Mg ad Ag +, we have t slve a equilibrium prblem. Let x be the small amut f Mg that reacts t achieve equilibrium. The ccetrati f Ag + will be x at equilibrium. Assume that essetially all Ag + has bee reduced s that the iitial ccetrati f Ag + is zer. Ag + (aq) + Mg(s) Ag(s) + Mg (aq) Iitial (M): Chage (M): +x x quilibrium (M): x ( x) K [Mg ] + [Ag ] ( x) ( x) We ca assume x ( x) ( x ) (x) x M [Ag + ] x M [Mg ] x M Weigh the zic ad cpper electrdes befre peratig the cell ad re-weigh afterwards. The ade (Z) shuld lse mass ad the cathde (Cu) shuld gai mass (a) Sice this is a acidic sluti, the gas must be hydrge gas frm the reducti f hydrge i. The tw electrde reactis ad the verall cell reacti are: ade: Cu(s) Cu (aq) + e cathde: H + (aq) + e H (g) Cu(s) + H + (aq) Cu (aq) + H (g) Sice g f cpper was csumed, the amut f hydrge gas prduced is: 1mlCu 1mlH g Cu ml H g Cu 1 ml Cu At STP, 1 mle f a ideal gas ccupies a vlume f.41 L. Thus, the vlume f H at STP is: V H 3.41 L ( ml H ) 1ml 0.06 L

30 CHAPTR 19: LCTROCHMISTRY 577 Frm the curret ad the time, we ca calculate the amut f charge: 1C A ( s) C 1A s Sice we kw the charge f a electr, we ca cmpute the umber f electrs. 3 1 e ( C) e C Usig the amut f cpper csumed i the reacti ad the fact that ml f e are prduced fr every 1 mle f cpper csumed, we ca calculate Avgadr's umber e 1 ml Cu ml Cu ml e /ml e I practice, Avgadr's umber ca be determied by electrchemical experimets like this. The charge f the electr ca be fud idepedetly by Millika's experimet The reacti is: Al e Al First, let's calculate the umber f culmbs f electricity that must pass thrugh the cell t depsit 60. g f Al. 1 ml Al 3 ml e C 60. g Al 6.98 g Al 1 ml Al 1mle C The time (i mi) eeded t pass this much charge is: 5 1A s 1 1mi 4 t mi ( C) mi 1 C 0.35 A 60 s 19.8 (a) We ca calculate ΔG frm stadard free eergies f frmati. Δ G Δ Gf (N ) + 6 ΔGf (HO) [4 Δ Gf (NH 3) + 3 ΔGf (O )] ΔG 0 + (6)(37. kj/ml) [(4)(16.6 kj/ml) + 0] ΔG kj/ml The half-reactis are: 4NH 3 (g) N (g) + 1H + (aq) + 1e 3O (g) + 1H + (aq) + 1e 6H O(l) The verall reacti is a 1-electr prcess. We ca calculate the stadard cell emf frm the stadard free eergy chage, ΔG. Δ G Fcell cell kj 1000 J Δ G 1ml 1kJ F (1)(96500 J/V ml) 1.17 V

31 578 CHAPTR 19: LCTROCHMISTRY Cathde: Au 3+ (aq) + 3e Au(s) Ade: H O(l) O (g) + 4H + (aq) + 4e (a) First, frm the amut f gld depsited, we ca calculate the mles f O prduced. The, usig the ideal gas equati, we ca calculate the vlume f O. 1mlAu 3mle 1mlO 9.6 g Au ml O g Au 1 ml Au 4mle V O Latm ( ml) (96 K) O RT ml K P 1atm 747 mmhg 760 mmhg 0.87 L Curret ( I ) charge ( Q) t (s) t Q 60 mi 60 s 3.00 h s 1h 1mi 1 ml Au 3 ml e C 1 A s g Au A s g Au 1 ml Au 1mle 1 C Curret ( I ) 4 charge ( Q) A s t (s) s 1.89 A The reducti f Ag + t Ag metal is: Ag + (aq) + e Ag We ca calculate bth the mles f Ag depsited ad the mles f Au depsited. 1mlAg? ml Ag.64 g Ag ml Ag g Ag 1mlAu 3? ml Au 1.61 g Au ml Au g Au We d t kw the xidati state f Au is, s we will represet the is as Au +. If we divide the ml f Ag by the ml f Au, we ca determie the rati f Ag + reduced cmpared t Au + reduced ml Ag ml Au 3 That is, the same umber f electrs that reduced the Ag + is t Ag reduced ly e-third the umber f mles f the Au + is t Au. Thus, each Au + required three electrs per i fr every e electr fr Ag +. The xidati state fr the gld i is +3; the i is Au 3+. Au 3+ (aq) + 3e Au

32 CHAPTR 19: LCTROCHMISTRY Heatig the garage will melt the sw the car which is ctamiated with salt. The aqueus salt will haste crrsi We reverse the first halfreacti ad add it t the secd t cme up with the verall balaced equati Hg Hg + e ade V Hg + e Hg cathde V Hg Hg + Hg cell 0.85 V 0.9 V 0.07 V Sice the stadard cell ptetial is a itesive prperty, We calculate ΔG frm. Hg (aq) Hg (aq) + Hg(l) cell 0.07 V ΔG F (1)(96500 J/V ml)(0.07 V) 6.8 kj/ml The crrespdig equilibrium cstat is: [Hg ] K [Hg ] We calculate K frm ΔG. ΔG RTl K l K J/ml (8.314 J/K ml)(98 K) K (a) Ade F F (g) + e Cathde H + + e H (g) Overall: H + + F H (g) + F (g) KF icreases the electrical cductivity (what type f electrlyte is HF(l))? The K + is t reduced. (c) Calculatig the mles f F 3600 s 1 C 1 ml e 1mlF 50 A 15 h 140 ml F 1 h 1 A s C ml Usig the ideal gas law: RT (140 ml)(0.081 L atm/k ml)(97 K) 3 V.8 10 L P 1. atm e The reactis fr the electrlysis f NaCl(aq) are: Ade: Cl (aq) Cl (g) + e Cathde: H O(l) + e H (g) + OH (aq) Overall: H O(l) + Cl (aq) H (g) + Cl (g) + OH (aq)

33 580 CHAPTR 19: LCTROCHMISTRY Frm the ph f the sluti, we ca calculate the OH ccetrati. Frm the [OH ], we ca calculate the mles f OH prduced. The, frm the mles f OH we ca calculate the average curret used. ph 1.4 poh [OH ] M The mles f OH prduced are: ml L ml OH 1L Frm the balaced equati, it takes 1 mle f e t prduce 1 mle f OH is. 3 1 ml C (5. 10 ml OH ) e 504 C 1mlOH 1mle Recall that 1 C 1 A s 1A s 1mi C 1.4 A 1C 60s 6.00mi (a) Ade: Cu(s) Cu (aq) + e Cathde: Cu (aq) + e Cu(s) The verall reacti is: Cu(s) Cu(s) Cu is trasferred frm the ade t cathde. Csultig Table 19.1 f the text, the Z will be xidized, but Z will t be reduced at the cathde. Ag will t be xidized at the ade. (c) The mles f Cu: The culmbs required: 1mlCu 1000 g Cu 15.7 ml Cu g Cu ml e C ml Cu C 1mlCu 1ml e The time required: C 5? s s 18.9 A 5 1hr ( s) 44.4 hr 3600 s The reacti is: Pt + + e Pt Thus, we ca calculate the charge f the platium is by realizig that ml f e are required per ml f Pt frmed.

34 CHAPTR 19: LCTROCHMISTRY 581 The mles f Pt frmed are: Next, calculate the charge passed i C. Cvert t mles f electrs. 1mlPt 9.09 g Pt ml Pt g Pt 3600 s.50 C 4 C.00 h C 1h 1s 4 1mle? ml e ( C) ml e C We w kw the umber f mles f electrs (0.187 ml e ) eeded t prduce ml f Pt metal. We ca calculate the umber f mles f electrs eeded t prduce 1 mle f Pt metal ml e ml Pt 4.01 ml e /ml Pt Sice we eed 4 mles f electrs t reduce 1 mle f Pt is, the charge the Pt is must be Usig the stadard reducti ptetials fud i Table 19.1 Cd (aq + e Cd(s) Mg (aq) + e Mg(s) 0.40 V.37 V Thus Cd will xidize Mg s that the magesium half-reacti ccurs at the ade. Mg(s) + Cd (aq) Mg (aq) + Cd(s) cell cathde ade 0.40 V (.37 V) 1.97 V e e Mg ade salt bridge Cd cathde 19.9 The halfreacti fr the xidati f water t xyge is: H O(l) xidati (ade) O (g) + 4H + (aq) + 4e Kwig that e mle f ay gas at STP ccupies a vlume f.41 L, we fid the umber f mles f xyge. 4.6 L O 1ml ml O.41 L

35 58 CHAPTR 19: LCTROCHMISTRY Sice fur electrs are required t frm e xyge mlecule, the umber f electrs must be: 3 4 ml e e ml O mlO 1ml The amut f charge passig thrugh the sluti is: 1 C 3600 s A 3.40 h C 1A s 1h We fid the electr charge by dividig the amut f charge by the umber f electrs. e C e C/ e I actual fact, this srt f calculati ca be used t fid Avgadr's umber, t the electr charge. The latter ca be measured idepedetly, ad e ca use this charge tgether with electrlytic data like the abve t calculate the umber f bjects i e mle. See als Prblem (a) Au(s) + 3HNO 3 (aq) + 4HCl(aq) HAuCl 4 (aq) + 3H O(l) + 3NO (g) T icrease the acidity ad t frm the stable cmplex i, AuCl Cells f higher vltage require very reactive xidizig ad reducig agets, which are difficult t hadle. (Frm Table 19.1 f the text, we see that 5.9 V is the theretical limit f a cell made up f Li + /Li ad F /F electrdes uder stadard-state cditis.) Batteries made up f several cells i series are easier t use The verall cell reacti is: Ag + (aq) + H (g) Ag(s) + H + (aq) We write the Nerst equati fr this system V l Q V [H ] V l + [Ag ] P H The measured vltage is V, ad we ca fid the silver i ccetrati as fllws: V V 0.80 V l + [Ag ] 1 l [Ag ] 1 + [Ag ] [Ag + ] M

36 CHAPTR 19: LCTROCHMISTRY 583 Kwig the silver i ccetrati, we ca calculate the xalate i ccetrati ad the slubility prduct cstat [CO 4 ] [Ag ] M K sp [Ag + ] [C O 4 ] ( ) ( ) The half-reactis are: Z(s) + 4OH (aq) Z(OH) 4 (aq) + e ade 1.36 V Z (aq) +e Z(s) cathde 0.76 V Z (aq) + 4OH (aq) Z(OH) 4 (aq) cell 0.76 V ( 1.36 V) 0.60 V cell V l K f ()(0.60) K f e0.057 e The half reactis are: H O (aq) O (g) + H + (aq) + e ade 0.68 V H O (aq) + H + (aq) + e H O(l) cathde 1.77 V H O (aq) H O(l) + O (g) cell cathde ade 1.77 V 0.68 V 1.09 V Thus, prducts are favred at equilibrium. H O is t stable (it disprprtiates) (a) Sice electrs flw frm X t SH, fr X must be egative. Thus fr Y must be psitive. Y + e Y cathde 0.34 V X X + e ade 0.5 V X + Y X + Y cell 0.34 V ( 0.5 V) 0.59 V (a) The half reactis are: S 4+ (aq) + e S (aq) cathde 0.13 V Tl(s) Tl + (aq) + e ade 0.34 V S 4+ (aq) + Tl(s) S (aq) + Tl + (aq) cell cathde ade 0.13 V ( 0.34 V) 0.47 V cell RT l K F (8.314)(98) 0.47 V l K ()(96500) K

37 584 CHAPTR 19: LCTROCHMISTRY (c) (1.0)(10.0) l ( ) V 0.41 V (1.0) (a) Gld des t tarish i air because the reducti ptetial fr xyge is isufficiet t result i the xidati f gld. O + 4H + + 4e H O cathde 1.3 V That is, cell cathde ade < 0, fr either xidati by O t Au + r Au 3+. r cell 1.3 V 1.50 V < 0 cell 1.3 V 1.69 V < 0 3(Au + + e Au) cathde 1.69 V Au Au e ade 1.50 V 3Au + Au + Au 3+ cell 1.69 V 1.50 V 0.19 V Calculatig ΔG, ΔG F (3)(96,500 J/V ml)(0.19 V) 55.0 kj/ml Fr sptaeus electrchemical equatis, ΔG must be egative. Thus, the disprprtiati ccurs sptaeusly. (c) Sice the mst stable xidati state fr gld is Au 3+, the predicted reacti is: Au + 3F AuF It is mercury i i sluti that is extremely hazardus. Sice mercury metal des t react with hydrchlric acid (the acid i gastric juice), it des t disslve ad passes thrugh the huma bdy uchaged. Nitric acid (t part f huma gastric juices) disslves mercury metal (see Prblem ); if itric acid were secreted by the stmach, igesti f mercury metal wuld be fatal The balaced equati is: 5Fe + MO 4 + 8H + M + 5Fe H O Calculate the amut f ir(ii) i the rigial sluti usig the mle rati frm the balaced equati ml KMO4 5mlFe 3.0 ml ml Fe 1000 ml sl 1 ml KMO4 The ccetrati f ir(ii) must be: ml [Fe ] M L The ttal ir ccetrati ca be fud by simple prprti because the same sample vlume (5.0 ml) ad the same KMO 4 sluti were used ml KMO [Fe] 4 ttal M ml KMO4 M

38 CHAPTR 19: LCTROCHMISTRY 585 [Fe 3+ ] [Fe] ttal [Fe ] M Why are the tw titratis with permagaate ecessary i this prblem? Viewed exterally, the ade lks egative because f the flw f the electrs (frm Z Z + e ) tward the cathde. I sluti, ais mve tward the ade because they are attracted by the Z is surrudig the ade Frm Table 19.1 f the text. H O (aq) + H + (aq) + e H O(l) cathde 1.77 V H O (aq) O (g) + H + (aq) + e ade 0.68 V H O (aq) H O(l) + O (g) cell cathde ade 1.77 V (0.68 V) 1.09 V Because is psitive, the decmpsiti is sptaeus (a) The verall reacti is: Pb + PbO + H SO 4 PbSO 4 + H O Iitial mass f H SO 4 : Fial mass f H SO 4 : 1.9 g 74 ml g 1mL 1.19 g 74 ml g 1mL Mass f H SO 4 reacted 355 g 4 g 131 g 1ml Mles f HSO 4 reacted 131 g 1.34 ml g Q ml e C 1.34 ml HSO4 mlhso 4 1mle C t 5 Q C s 1.60 hr.4 A I (a) uchaged uchaged (c) squared (d) dubled (e) dubled (a) Pt ade Sp (cathde) AgNO 3 sluti

39 586 CHAPTR 19: LCTROCHMISTRY t 1 ml C g Q g 1 ml s I A 11.9 hr F (g) + H + (aq) + e HF (g) RT PHF l F + PF [H ] With icreasig [H + ], will be larger. F will becme a strger xidizig aget Advatages: (a) N start-up prblems, much quieter, (c) plluti (smg), (d) mre eergy efficiet i the sese that whe the car is t mvig (fr example at a traffic light), electricity is csumed. Disadvatages: (a) (c) Drivig rage is mre limited tha autmbiles, ttal mass f batteries is appreciable, prducti f electricity eeded t charge the batteries leads t plluti Pb Pb + e ade 0.13 V H + + e H cathde 0.00 V Pb + H + Pb + H cell 0.00 V ( 0.13 V) 0.13 V ph 1.60 [H + ] M RT F [Pb ] PH l + [H ] V (0.035) P l (0.035) P l P H atm H H (a) At the ade (Mg): Mg Mg + e Als: Mg + HCl MgCl + H At the cathde (Cu): H + + e H

40 CHAPTR 19: LCTROCHMISTRY 587 (c) The sluti des t tur blue. After all the HCl has bee eutralized, the white precipitate is: Mg + OH Mg(OH) (s) (a) The half-reactis are: Ade: Z Z + e Cathde: 1 O + e O Overall: Z + 1 O ZO T calculate the stadard emf, we first eed t calculate ΔG fr the reacti. Frm Appedix 3 f the text we write: 1 Δ G ΔGf(ZO) [ Δ Gf(Z) + ΔG f(o )] ΔG 318. kj/ml [0 + 0] ΔG 318. kj/ml ΔG F J/ml ()(96,500 J/V ml) 1.65 V We use the fllwig equati: RT l Q F V V l P O V V l V 0.00 V 1.63 V (c) (d) Sice the free eergy chage represets the maximum wrk that ca be extracted frm the verall reacti, the maximum amut f eergy that ca be btaied frm this reacti is the free eergy chage. T calculate the eergy desity, we multiply the free eergy chage by the umber f mles f Z preset i 1 kg f Z kj 1 ml Z 1000 g Z 3 eergy desity kj/kg Z 1 ml Z g Z 1 kg Z Oe ampere is 1 C/s. The charge draw every secd is give by F. charge F C (96,500 C/ml e ). ml e

41 588 CHAPTR 19: LCTROCHMISTRY Frm the verall balaced reacti, we see that 4 mles f electrs will reduce 1 mle f O ; therefre, the umber f mles f O reduced by. mles f electrs is: 1mlO ml O. ml e 0.55 ml O 4mle The vlume f xyge at 1.0 atm partial pressure ca be btaied by usig the ideal gas equati. V O RT (0.55 ml)(0.081 L atm/ml K)(98 K) P (1.0 atm) 13 L Sice air is 1 percet xyge by vlume, the vlume f air required every secd is: V air 100% air 13 L O 1% O 6 L f air (a) HCl: First, we write the half-reactis. Oxidati: Hg(l) Hg (1 M) + e Reducti: H + (1 M) + e H (1 atm) Overall: Hg(l) + H + (1 M) Hg (1 M) + H (1 atm) The stadard emf,, is give by cathde ade V 0.85V (We mit the subscript cell because this reacti is t carried ut i a electrchemical cell.) Sice is egative, we cclude that mercury is t xidized by hydrchlric acid uder stadard-state cditis. HNO 3 : The reactis are: Oxidati: 3[Hg(l) Hg (1 M) + e ] Reducti: [NO 3 (1 M) + 4H + (1 M) + 3e NO(1 atm) + H O(l) Overall: Thus, 6Hg(l) + NO 3 (1 M) + 8H + (1 M) 3Hg (1 M) + NO(1 atm) + 4H O(l) cathde ade 0.96V 0.85V 0.11V Sice is psitive, prducts are favred at equilibrium uder stadard-state cditis. The test tube the left ctais HNO 3 ad Hg. HNO 3 ca xidize Hg ad the prduct NO reacts with xyge t frm the brw gas NO.