CHAPTER 21 ELECTROCHEMISTRY: CHEMICAL CHANGE AND ELECTRICAL WORK

Transcription

1 CHAPTR 1 LCTROCHMISTRY: CHMICAL CHANG AND LCTRICAL WORK 1.1 Oxidatin is the lss f electrns (resulting in a higher xidatin number), while reductin is the gain f electrns (resulting in a lwer xidatin number). In an xidatin-reductin reactin, electrns transfer frm the xidized substance t the reduced substance. The xidatin number f the reactant being xidized increases while the xidatin number f the reactant being reduced decreases. 1. An electrchemical prcess invlves electrn flw. At least ne substance must lse electrn(s) and ne substance must gain electrn(s) t prduce the flw. This electrn transfer is a redx prcess. 1.3 N, ne half-reactin cannt take place independent f the ther because there is always a transfer f electrns frm ne substance t anther. If ne substance lses electrns (xidatin half-reactin), anther substance must gain thse electrns (reductin half-reactin). 1.4 O - is t strng a base t exist in H O. The reactin O - + H O OH - ccurs. Only species actually existing in slutin can be present when balancing an equatin. 1.5 Multiply each half-reactin by the apprpriate integer t make e - lst equal t e - gained. 1.6 T remve prtns frm an equatin, add an equal number f hydrxide ins t bth sides t neutralize the H + and prduce water: H + (aq) + OH - (aq) H O(l). 1.7 N, add spectatr ins t the balanced inic equatin t btain the balanced mlecular equatin. 1.8 Spntaneus reactins, G sys < 0, take place in vltaic s, which are als called galvanic s. Nnspntaneus reactins take place in electrlytic s and result in an increase in the free energy f the ( G sys > 0). 1.9 a) True b) True c) True d) False, in a vltaic, the system des wrk n the surrundings. e) True f) False, the electrlyte in a prvides a slutin f mbile ins t maintain charge neutrality a) T decide which reactant is xidized, lk at xidatin numbers. Cl - is xidized because its xidatin number increases frm -1 t 0. b) MnO 4 - is reduced because the xidatin number f Mn decreases frm +7 t +. c) The xidizing agent is the substance that causes the xidatin by accepting electrns. The xidizing agent is the substance reduced in the reactin, s MnO 4 - is the xidizing agent. d) Cl - is the reducing agent because it lses the electrns that are gained in the reductin. e) Frm Cl -, which is lsing electrns, t MnO 4 -, which is gaining electrns. f) 8 H SO 4 (aq) + KMnO 4 (aq) + 10 KCl(aq) MnSO 4 (aq) + 5 Cl (g) + 8 H O(l) + 6 K SO 4 (aq) 1-1

2 1.11 CrO - (aq) + H O(l) + 6 ClO - (aq) CrO 4 - (aq) + 3 Cl (g) + 4 OH - (aq) a) The CrO - is the xidized species because Cr increases in xidatin state frm +3 t +6. b) The ClO - is the reduced species because Cl decreases in xidatin state frm +1 t 0. c) The xidizing agent is ClO - ; the xidizing agent is the substance reduced. d) The reducing agent is CrO - ; the reducing agent is the substance xidized. e) lectrns transfer frm CrO - t ClO -. f) NaCrO (aq) + 6 NaClO(aq) + H O(l) Na CrO 4 (aq) + 3 Cl (g) + 4 NaOH(aq) 1.1 a) Divide int half-reactins: ClO 3 (aq) Cl (aq) I (aq) I (s) Balance elements ther than O and H ClO 3 (aq) Cl (aq) chlrine is balanced I (aq) I (s) idine nw balanced Balance O by adding H O ClO 3 (aq) Cl (aq) + 3 H O(l) add 3 waters t add 3 O s t prduct I (aq) I (s) n change Balance H by adding H + ClO 3 (aq) + 6 H + (aq) Cl (aq) + 3 H O(l) add 6 H + t reactants I (aq) I (s) n change Balance charge by adding e - ClO 3 (aq) + 6 H + (aq) + 6 e - Cl (aq) + 3 H O(l) add 6 e - t reactants I (aq) I (s) + e - add e - t prducts Multiply each half-reactin by an integer t equalize the number f electrns ClO 3 (aq) + 6 H + (aq) + 6 e - Cl (aq) + 3 H O(l) multiply by 1 t give 6 e - 3{ I (aq) I (s) + e - } multiply by 3 t give 6 e - Add half-reactins t give balanced equatin in acidic slutin. ClO 3 (aq) + 6 H + (aq) + 6 I(aq) Cl (aq) + 3 H O(l) + 3 I (s) Check balancing: Reactants: 1 Cl Prducts: 1 Cl 3 O 3 O 6 H 6 H 6 I 6 I -1 charge -1 charge Oxidizing agent is ClO 3 and reducing agent is I. b) Divide int half-reactins: MnO 4 (aq) MnO (s) SO 3 (aq) SO 4 (aq) Balance elements ther than O and H MnO 4 (aq) MnO (s) Mn is balanced SO 3 (aq) SO 4 (aq) S is balanced Balance O by adding H O MnO 4 (aq) MnO (s) + H O(l) add H O t prducts SO 3 (aq) + H O(l) SO 4 (aq) add 1 H O t reactants Balance H by adding H + MnO 4 (aq) + 4 H + (aq) MnO (s) + H O(l) add 4 H + t reactants SO 3 (aq) + H O(l) SO 4 (aq) + H + (aq) add H + t prducts Balance charge by adding e - MnO 4 (aq) + 4 H + (aq) + 3 e - MnO (s) + H O(l) add 3 e - t reactants SO 3 (aq) + H O(l) SO 4 (aq) + H + (aq) + e - add e - t prducts Multiply each half-reactin by an integer t equalize the number f electrns {MnO 4 (aq) + 4 H + (aq) + 3 e - MnO (s) + H O(l)} multiply by t give 6 e - 3{SO 3 (aq) + H O(l) SO 4 (aq) + H + (aq) + e - } multiply by 3 t give 6 e - 1-

3 Add half-reactins and cancel substances that appear as bth reactants and prducts MnO 4 (aq) + 8 H + (aq) + 3 SO 3 (aq) + 3 H O(l) MnO (s) + 4 H O(l) + 3 SO 4 (aq) + 6 H + (aq) The balanced equatin in acidic slutin is: MnO 4 (aq) + H + (aq) + 3 SO 3 (aq) MnO (s) + H O(l) + 3 SO 4 (aq) T change t basic slutin, add OH t bth sides f equatin t neutralize H +. MnO 4 (aq) + H + (aq) + OH (aq) + 3 SO 3 (aq) MnO (s) + H O(l) + 3 SO 4 (aq) + OH (aq) Balanced equatin in basic slutin: MnO 4 (aq) + H O(l) + 3 SO 3 (aq) MnO (s) + 3 SO 4 (aq) + OH (aq) Check balancing: Reactants: Mn Prducts: Mn 18 O 18 O H H 3 S 3 S -8 charge -8 charge Oxidizing agent is MnO 4 and reducing agent is SO 3. c) Divide int half-reactins: MnO 4 (aq) Mn (aq) H O (aq) O (g) Balance elements ther than O and H - Mn is balanced Balance O by adding H O MnO 4 (aq) Mn (aq) + 4 H O(l) add 4 H O t prducts Balance H by adding H + MnO 4 (aq) + 8 H + (aq) Mn (aq) + 4 H O(l) add 8 H + t reactants H O (aq) O (g) + H + (aq) add H + t prducts Balance charge by adding e - MnO 4 (aq) + 8 H + (aq) + 5 e - Mn (aq) + 4 H O(l) add 5 e - t reactants H O (aq) O (g) + H + (aq) + e - add e - t prducts Multiply each half-reactin by an integer t equalize the number f electrns {MnO 4 (aq) + 8 H + (aq) + 5 e - Mn (aq) + 4 H O(l)} multiply by t give 10 e - 5{H O (aq) O (g) + H + (aq) + e - } multiply by 5 t give 10 e - Add half-reactins and cancel substances that appear as bth reactants and prducts MnO 4 (aq) + 16 H + (aq) + 5 H O (aq) Mn (aq) + 8 H O(l) + 5 O (g) + 10 H + (aq) The balanced equatin in acidic slutin MnO 4 (aq) + 6 H + (aq) + 5 H O (aq) Mn (aq) + 8 H O(l) + 5 O (g) Check balancing: Reactants: Mn Prducts: Mn 18 O 18 O 16 H 16 H +4 charge +4 charge Oxidizing agent is MnO - 4 and reducing agent is H O a) 3 O (g) + 4 NO(g) + H O(l) 4 NO 3 - (aq) + 4 H + (aq) Oxidizing agent is O and reducing agent: NO b) CrO 4 - (aq) + 8 H O(l) + 3 Cu(s) Cr(OH) 3 (s) + 3 Cu(OH) (s) + 4 OH - (aq) Oxidizing agent is CrO 4 - and reducing agent: Cu c) AsO 4 3- (aq) + NO - (aq) + H O(l) AsO - (aq) + NO 3 - (aq) + OH - (aq) Oxidizing agent is AsO 4 3- and reducing agent: NO a) Balance the reductin half-reactin: Cr O 7 (aq) Cr 3+ (aq) Cr O 7 (aq) Cr 3+ (aq) + 7 H O(l) Cr O 7 (aq) + 14 H + (aq) Cr 3+ (aq) + 7 H O(l) Cr O 7 (aq) + 14 H + (aq) + 6 e - Cr 3+ (aq) + 7 H O(l) balance Cr balance O balance H balance charge 1-3

4 Balance the xidatin half-reactin: Zn(s) Zn (aq) + e - balance charge Add the tw half-reactins multiplying the xidatin half-reactin by 3 t equalize the electrns. Cr O 7 (aq) + 14 H + (aq) + 3 Zn(s) Cr 3+ (aq) + 7 H O(l) + 3 Zn (aq) Oxidizing agent is Cr O 7 and reducing agent is Zn. b) Balance the reductin half-reactin: MnO 4 (aq) MnO (s) + H O(l) balance O MnO 4 (aq) + 4 H + (aq) MnO (s) + H O(l) balance H MnO 4 (aq) + 4 H + (aq) + 3 e - MnO (s) + H O(l) balance charge Balance the xidatin half-reactin Fe(OH) (s) + H O(l) Fe(OH) 3 (s) balance O Fe(OH) (s) + H O(l) Fe(OH) 3 (s) + H + (aq) balance H Fe(OH) (s) + H O(l) Fe(OH) 3 (s) + H + (aq) + e balance e - Add half-reactins after multiplying xidatin half-reactin by 3. MnO 4 (aq) + 4 H + (aq) + 3 Fe(OH) (s) + 3 H O(l) MnO (s) + H O(l) + 3 Fe(OH) 3 (s) + 3 H + (aq) Add OH - t bth sides t neutralize the H + and cnvert H + + OH - H O MnO 4 (aq) + 3 Fe(OH) (s) + H O(l) MnO (s) + 3 Fe(OH) 3 (s) + OH (aq) Oxidizing agent is MnO - 4 and reducing agent is Fe(OH). c) Balance the reductin half-reactin: NO 3 (aq) N (g) balance N NO 3 (aq) N (g) + 6 H O(l) balance O NO 3 (aq) + 1 H + (aq) N (g) + 6 H O(l) balance H NO 3 (aq) + 1 H + (aq) + 10 e - N (g) + 6 H O(l) balance charge Balance the xidatin half-reactin: Zn(s) Zn (aq) + e - balance charge Add the half-reactins after multiplying the reductin half-reactin by 1 and the xidatin half-reactin by 5. NO 3 (aq) + 1 H + (aq) + 5 Zn(s) N (g) + 6 H O(l) + 5 Zn (aq) Oxidizing agent is NO 3 and reducing agent is Zn a) 3 BH 4 - (aq) + 4 ClO 3 - (aq) 3 H BO 3 - (aq) + 4 Cl - (aq) + 3 H O(l) Oxidizing agent is ClO 3 - and reducing agent: BH 4 - b) CrO 4 - (aq) + 3 N O(g) + 10 H + (aq) Cr 3+ (aq) + 6 NO(g) + 5 H O(l) Oxidizing agent is CrO 4 - and reducing agent: N O c) 3 Br (l) + 6 OH - (aq) BrO 3 - (aq) + 5 Br - (aq) + 3 H O(l) Oxidizing agent is Br and reducing agent is Br a) Balance the reductin half-reactin: NO 3 (aq) NO(g) + H O(l) balance O NO 3 (aq) + 4 H + (aq) NO(g) + H O(l) balance H NO 3 (aq) + 4 H + (aq) + 3 e - NO(g) + H O(l) balance charge Balance xidatin half-reactin: 4 Sb(s) Sb 4 O 6 (s) balance Sb 4 Sb(s) + 6 H O(l) Sb 4 O 6 (s) balance O 4 Sb(s) + 6 H O(l) Sb 4 O 6 (s) + 1 H + (aq) balance H 4 Sb(s) + 6 H O(l) Sb 4 O 6 (s) + 1 H + (aq) + 1 e - balance charge Multiply reductin half-reactin by 4 and add half-reactins. Cancel cmmn reactants and prducts. 4 NO 3 (aq) + 16 H + (aq) + 4 Sb(s) + 6 H O(l) 4 NO(g) + 8 H O(l) + Sb 4 O 6 (s) + 1 H + (aq) Balanced equatin in acidic slutin: 4 NO 3 (aq) + 4 H + (aq) + 4 Sb(s) 4 NO(g) + H O(l) + Sb 4 O 6 (s) Oxidizing agent is NO - 3 and reducing agent is Sb. 1-4

5 b) Balance reductin half-reactin: BiO 3 (aq) Bi 3+ (aq) + 3 H O(l) balance O BiO 3 (aq) + 6 H + (aq) Bi 3+ (aq) + 3 H O(l) balance H BiO 3 (aq) + 6 H + (aq) + e - Bi 3+ (aq) + 3 H O(l) balance charge Balance xidatin half-reactin: Mn (aq) + 4 H O(l) MnO 4 (aq) balance O Mn (aq) + 4 H O(l) MnO 4 (aq) + 8 H + (aq) balance H Mn (aq) + 4 H O(l) MnO 4 (aq) + 8 H + (aq) + 5 e - balance H Multiply reductin half-reactin by and xidatin half-reactin by t transfer 10 e - in the verall reactin. Cancel H O and H + in reactants and prducts. 5 BiO 3 (aq) + 30 H + (aq) + Mn (aq) + 8 H O(l) 5 Bi 3+ (aq) + 15 H O(l) + MnO 4 (aq) + 16 H + (aq) Balanced reactin in acidic slutin: 5 BiO 3 (aq) + 14 H + (aq) + Mn (aq) 5 Bi 3+ (aq) + 7 H O(l) + MnO 4 (aq) BiO - 3 is the xidizing agent and Mn is the reducing agent. c) Balance the reductin half-reactin: Pb(OH) 3 (aq) Pb(s) + 3 H O(l) balance O Pb(OH) 3 (aq) + 3 H + (aq) Pb(s) + 3 H O(l) balance H Pb(OH) 3 (aq) + 3 H + (aq) + e - Pb(s) + 3 H O(l) balance charge Balance the xidatin half-reactin Fe(OH) (s) + H O(l) Fe(OH) 3 (s) balance O Fe(OH) (s) + H O(l) Fe(OH) 3 (s) + H + (aq) balance H Fe(OH) (s) + H O(l) Fe(OH) 3 (s) + H + (aq) + e - balance charge Multiply xidatin half-reactin by and add tw half-reactins. Cancel H O and H +. Pb(OH) 3 (aq) + 3 H + (aq) + Fe(OH) (s) + H O(l) Pb(s) + 3 H O(l) + Fe(OH) 3 (s) + H + (aq) Add OH - t bth sides t neutralize H +. Pb(OH) 3 (aq) + H + (aq) + OH (aq) + Fe(OH) (s) Pb(s) + H O(l) + Fe(OH) 3 (s) + OH (aq) Balanced reactin in basic slutin: Pb(OH) 3 (aq) + Fe(OH) (s) Pb(s) + Fe(OH) 3 (s) + OH (aq) Pb(OH) 3 is the xidizing agent and Fe(OH) is the reducing agent a) 1[ OH - (aq) + NO (g) NO - 3 (aq) + H O(l) + 1 e - ] 1[NO (g) + 1 e - NO - (aq)] Overall: NO (g) + OH - (aq) NO - 3 (aq) + NO - (aq) + H O(l) Oxidizing agent is NO and reducing agent is NO. b) 4[Zn(s) + 4 OH - (aq) Zn(OH) - 4 (aq) + e - ] 1[NO - 3 (aq) + 6 H O(l) + 8 e - NH 3 (aq) + 9 OH - (aq) Overall: 4 Zn(s) + 16 OH - (aq) + NO - 3 (aq) + 6 H O(l) + 8 e - 4 Zn(OH) - 4 aq) + NH 3 (aq) + 9 OH - (aq) + 8 e - 4 Zn(s) + 7 OH - (aq) + NO - 3 (aq) + 6 H O(l) 4 Zn(OH) - 4 (aq) + NH 3 (aq) Oxidizing agent is NO - 3 and reducing agent is Zn. c) 3[8 H S(g) S 8 (s) + 16 H + (aq) + 16 e - ] 16[3 e - + NO - 3 (aq) + 4 H + (aq) NO(g) + H O(l)] Overall: 4 H S(g) + 16 NO - 3 (aq) + 64 H + (aq) + 48 e - 3 S 8 (s) + 48 H + (aq) + 48 e NO(g) + 3 H O(l) 4 H S(g) + 16 NO - 3 (aq) + 16 H + (aq) 3 S 8 (s) + 16 NO(g) + 3 H O(l) Oxidizing agent is NO - 3 and reducing agent is H S. 1-5

6 1.18 a) Balance reductin half-reactin: MnO 4 (aq) Mn (aq) + 4 H O(l) balance O MnO 4 (aq) + 8 H + (aq) Mn (aq) + 4 H O(l) balance H MnO 4 (aq) + 8 H + (aq) + 5 e - Mn (aq) + 4 H O(l) balance charge Balance xidatin half-reactin: As 4 O 6 (s) 4 AsO 3 4 (aq) balance As As 4 O 6 (s) + 10 H O(l) 4 AsO 3 4 (aq) balance O As 4 O 6 (s) + 10 H O(l) 4 AsO 3 4 (aq) + 0 H + (aq) balance H As 4 O 6 (s) + 10 H O(l) 4 AsO 3 4 (aq) + 0 H + (aq) + 8 e - balance charge Multiply reductin half-reactin by 8 and xidatin half-reactin by 5 t transfer 40 e - in verall reactin. Add the half-reactins and cancel H O and H +. 5 As 4 O 6 (s) + 8 MnO 4 (aq) + 64 H + (aq) + 50 H O(l) 0 AsO 3 4 (aq) + 8 Mn (aq) + 3 H O(l) H + (aq) Balanced reactin in acidic slutin: 5 As 4 O 6 (s) + 8 MnO 4 (aq) + 18 H O(l) 0 AsO 3 4 (aq) + 8 Mn (aq) + 36 H + (aq) - Oxidizing agent is MnO 4 and reducing agent is As 4 O 6. b) The reactin gives nly ne reactant, P 4. Since bth prducts cntain phsphrus, divide the half-reactins s each include P 4 as the reactant. Balance reductin half-reactin: P 4 (s) 4 PH 3 (g) balance P P 4 (s) + 1 H + (aq) 4 PH 3 (g) balance H P 4 (s) + 1 H + (aq) + 1 e - 4 PH 3 (g) balance charge Balance xidatin half-reactin: P 4 (s) 4 HPO 3 (aq) balance P P 4 (s) + 1 H O(l) 4 HPO 3 (aq) balance O P 4 (s) + 1 H O(l) 4 HPO 3 (aq) + 0 H + (aq) balance H P 4 (s) + 1 H O(l) 4 HPO 3 (aq) + 0 H + (aq) + 1 e - balance charge Add tw half-reactins and cancel H +. P 4 (s) + 1 H + (aq) + 1 H O(l) 4 HPO 3 (aq) + 4 PH 3 (g) + 0 H + (aq) Balanced reactin in acidic slutin: P 4 (s) + 1 H O(l) 4 HPO 3 (aq) + 4 PH 3 (g) + 8 H + (aq) r P 4 (s) + 6 H O(l) HPO 3 (aq) + PH 3 (g) + 4 H + (aq) P 4 is bth the xidizing agent and reducing agent. c) Balance the reductin half-reactin: MnO 4 (aq) MnO (s) + H O(l) balance O MnO 4 (aq) + 4 H + (aq) MnO (s) + H O(l) balance H MnO 4 (aq) + 4 H + (aq) + 3 e - MnO (s) + H O(l) balance charge Balance xidatin half-reactin: CN (aq) + H O(l) CNO (aq) balance O CN (aq) + H O(l) CNO (aq)+ H + (aq) balance H CN (aq) + H O(l) CNO (aq)+ H + (aq) + e - balance charge Multiply the xidatin half-reactin by 3 and reductin half-reactin by t transfer 6 e - in verall reactin. Add tw half-reactins. Cancel the H O and H +. MnO - 4 (aq) + 3 CN - (aq) + 8 H + (aq) + 3 H O(l) MnO (s) + 3 CNO - (aq) + 6 H + (aq) + 4 H O(l) Add OH - t bth sides t neutralize H + and frm H O. MnO - 4 (aq) + 3 CN - (aq) + H O(l) MnO (s) + 3 CNO - (aq) + H O(l) + OH - (aq) Balanced reactin in basic slutin: MnO - 4 (aq) + 3 CN - (aq) + H O(l) MnO (s) + 3 CNO - (aq) + OH - (aq) Oxidizing agent is MnO - 4 and reducing agent is CN

7 1.19 a) SO 3 - (aq) + OH - (aq) SO 4 - (aq) + H O(l) + e - Cl (g) + e - Cl - (aq) Overall: SO 3 - (aq) + OH - (aq) + Cl (g) SO 4 - (aq) + Cl - (aq) + H O(l) Oxidizing agent is Cl and reducing agent is SO 3 -. b) 7[Fe(CN) 6 3- (aq) + 1 e - Fe(CN) 6 4- (aq)] 1[Re(s) + 8 OH - (aq) ReO 4 - (aq) + 4 H O(l) + 7 e - ] Overall: 7 Fe(CN) 6 3- (aq) + Re(s) + 8 OH - (aq) + 7 e - 7 Fe(CN) 6 4- (aq) + ReO 4 - (aq) + 4 H O(l) + 7 e - Oxidizing agent is Fe(CN) 6 3- and reducing agent is Re. c) [MnO 4 - (aq) + 8 H + (aq) + 5 e - Mn (aq) + 4 H O(l)] 5[HCOOH(aq) CO (g) + H + (aq) + e - ] Overall: MnO 4 - (aq) + 5 HCOOH(aq) + 16 H + (aq) + 10 e - Mn (aq) + 5 CO (g) + 8 H O(l) + 10 H + (aq) + 10 e - MnO 4 - (aq) + 5 HCOOH(aq) + 6 H + (aq) Mn (aq) + 5 CO (g) + 8 H O(l) Oxidizing agent is MnO 4 - and reducing agent is HCOOH Fe (aq) + MnO 4 - (aq) + 8 H + (aq) Mn (aq) + 5 Fe 3+ (aq) + 4 H O(l) 1.1 a) Balance reductin half-reactin: NO - 3 (aq) NO (g) + H O(l) balance O NO - 3 (aq) + H + (aq) NO (g) + H O(l) balance H NO - 3 (aq) + H + (aq) + e - NO (g) + H O(l) balance charge Balance xidatin half-reactin: Au(s) + 4 Cl - (aq) AuCl - 4 (aq) balance Cl Au(s) + 4 Cl - (aq) AuCl - 4 (aq) + 3 e - balance charge Multiply reductin half-reactin by 3 and add half-reactins. Au(s) + 3 NO - 3 (aq) + 4 Cl - (aq) + 6 H + (aq) AuCl - 4 (aq) + 3 NO (g) + 3 H O(l) b) Oxidizing agent is NO - 3 and reducing agent is Au. c) The HCl prvides chlride ins that cmbine with the unstable gld in t frm the stable in, AuCl a) A is the ande because by cnventin the ande is shwn n the left. b) is the cathde because by cnventin the cathde is shwn n the right. c) C is the salt bridge prviding electrical cnnectin between the tw slutins. d) A is the ande, s xidatin takes place there. Oxidatin is the lss f electrns, meaning that electrns are leaving the ande. e) is assigned a psitive charge because it is the cathde. f) gains mass because the reductin f the metal in prduced the metal. 1.3 Unless the xidizing and reducing agents are physically separated, the redx reactin will nt generate electrical energy. This electrical energy is prduced by frcing the electrns t travel thrugh an external circuit. 1.4 The purpse f the salt bridge is t maintain charge neutrality by allwing anins t flw int the ande cmpartment and catins t flw int the cathde cmpartment. 1.5 An active electrde is a reactant r prduct in the reactin, whereas an inactive electrde is neither a reactant nr a prduct. An inactive electrde is present nly t cnduct electricity when the half- reactin des nt include a metal. Platinum and graphite are cmmnly used as inactive electrdes. 1-7

8 1.6 a) The metal A is being xidized t frm the metal catin. T frm psitive ins, an atm must always lse electrns, s this half-reactin is always an xidatin. b) The metal in B is gaining electrns t frm the metal B, s it is displaced. c) The ande is the electrde at which xidatin takes place, s metal A is used as the ande. d) Acid xidizes metal B and metal B xidizes metal A, s acid will xidize metal A and bubbles will frm when metal A is placed in acid. The same answer results if strength f reducing agents is cnsidered. The fact that metal A is a better reducing agent than metal B indicates that if metal B reduces acid, then metal A will als reduce acid. 1.7 a) If the zinc electrde is negative, xidatin takes place at the zinc electrde: Zn(s) Zn (aq) + e - Reductin half-reactin: Sn (aq) + e - Sn(s) Overall reactin: Zn(s) + Sn (aq) Zn (aq) + Sn(s) b) Vltmeter e e Zn Salt bridge Sn () (+) 1 M Zn Anin flw Catin flw 1 M Sn 1.8 a) (red half-rxn) Ag + (aq) + 1 e - Ag(s) (x half-rxn) Pb(s) Pb (aq) + e - (verall rxn) Ag + (aq) + Pb(s) Ag(s) + Pb (aq) b) Vltmeter e e Pb Salt bridge Ag () (+) 1 M Pb Anin flw Catin flw 1 M Ag + 1-8

9 1.9 a) lectrns flw frm the ande t the cathde, s frm the irn half- t the nickel half-, left t right in the figure. By cnventin, the ande appears n the left and the cathde n the right. b) Oxidatin ccurs at the ande, which is the electrde in the irn half-. c) lectrns enter the reductin half-, the nickel half- in this example. d) lectrns are cnsumed in the reductin half-reactin. Reductin takes place at the cathde, nickel electrde. e) The ande is assigned a negative charge, s the irn electrde is negatively charged. f) Metal is xidized in the xidatin half-, s the irn electrde will decrease in mass. g) The slutin must cntain nickel ins, s any nickel salt can be added. 1 M NiSO 4 is ne chice. h) KNO 3 is cmmnly used in salt bridges, the ins being K + and NO - 3. Other salts are als acceptable answers. i) Neither, because an inactive electrde culd nt replace either electrde since bth the xidatin and the reductin half-reactins include the metal as either a reactant r a prduct. j) Anins will mve tward the half- in which psitive ins are being prduced. The xidatin half- prduces Fe, s salt bridge anins mve frm right (nickel half-) t left (irn half-). k) Oxidatin half-reactin: Fe(s) Fe (aq) + e - Reductin half-reactin: Ni (aq) + e - Ni(s) Overall reactin: Fe(s) + Ni (aq) Fe (aq) + Ni(s) 1.30 a) The electrns flw left t right. b) Reductin ccurs at the electrde n the right. c) lectrns leave the frm the left side. d) The zinc electrde generates the electrns. e) The cbalt electrn has the psitive charge. f) The cbalt electrde increases in mass. g) The ande electrlyte culd be 1 M Zn(NO 3 ). h) One pssible pair wuld be K + and NO 3 -. i) Neither electrde culd be replaced because bth electrdes are part f the reactin. j) The catins mve frm left t right t maintain charge neutrality. k) Reductin: C (aq) + e - C(s) Oxidatin: Zn(s) Zn (aq) + e - Overall: Zn(s) + C (aq) C(s) + Zn (aq) 1.31 a) The cathde is assigned a psitive charge, s the irn electrde is the cathde. Reductin half-reactin: Fe (aq) + e - Fe(s) Oxidatin half-reactin: Mn(s) Mn (aq) + e - Overall reactin: Fe (aq) + Mn(s) Fe(s) + Mn (aq) b) Vltmeter e e Mn Salt bridge Fe () (+) 1 M Mn Anin flw Catin flw 1 M Fe 1-9

10 1.3 a) (red half-rxn) Cu (aq) + e - Cu(s) (x half-rxn) Ni(s) Ni (aq) + e - (verall rxn) Cu (aq) + Ni(s) Cu(s) + Ni (aq) b) e Vltmeter e Ni Salt bridge Cu () (+) 1 M Ni Anin flw Catin flw 1 M Cu 1.33 a) Al is xidized, s it is the ande and appears first in the ntatin: Al(s) Al 3+ (aq) Cr 3+ (aq) Cr(s) b) Cu is reduced, s Cu is the cathde and appears last in the ntatin. The xidatin f SO des nt include a metal, s an inactive electrde must be present. Hydrgen in must be included in the xidatin half. Pt SO (g) SO 4 - (aq), H + (aq) Cu (aq) Cu(s) 1.34 a) Mn(s) + Cd (aq) Mn (aq) + Cd(s) b) 3 Fe(s) + NO 3 - (aq) + 8 H + (aq) 3 Fe (aq) + NO(g) + 4 H O(l) 1.35 An islated reductin r xidatin ptential cannt be directly measured. Hwever, by assigning a standard half ptential t a particular half-reactin, the standard ptentials f ther half-reactins can be determined relative t this reference value. The standard reference half- is a standard hydrgen electrde defined t have an value f V A negative with > 0. indicates that the reactin is nt spntaneus, G > 0. The reverse reactin is spntaneus 1.37 Similar t ther state functins, the sign f changes when a reactin is reversed. Unlike G, H and S, is an intensive prperty, the rati f energy t charge. When the cefficients in a reactin are multiplied by a factr, the values f G, H and S are multiplied by the same factr. Hwever, des nt change because bth the energy and charge are multiplied by the factr and their rati remains unchanged a) Divide the balanced equatin int reductin and xidatin half-reactins and add electrns. Add water and hydrxide in t the half-reactin that includes xygen. Oxidatin: Se - (aq) Se(s) + e - Reductin: SO - 3 (aq) + 3 H O(l) + 4 e - S O - 3 (aq) + 6 OH - (aq) b) = ande - cathde = V V = -0.9 V 1-10

11 1.39 a) (red half-rxn) O 3 (g) + H + (aq) + e - O (g) + H O(l) (x half-rxn) Mn (aq) + H O(l) MnO (s) + 4 H + (aq) + e - b) = zne - manganese manganese = zne - =.07 V V = 1.3 V 1.40 a) The greater the reductin ptential, the greater the strength as an xidizing agent. When placed in rder f decreasing strength as xidizing agents: Br > Fe 3+ > Cu. b) When placed in rder f increasing strength as xidizing agents: Ca < Ag + < Cr O a) When placed in rder f decreasing strength as reducing agents: SO > MnO > PbSO 4 b) When placed in rder f increasing strength as reducing agents: Hg < Sn < Fe 1.4 a) Oxidatin: C(s) C (aq) + e - - = 0.8 V Reductin: H + (aq) + e - H (g) = 0.00 V Overall reactin: C(s) + H + (aq) C (aq) + H (g) = 0.8 V Reactin is spntaneus under standard state cnditins because is psitive. b) Oxidatin: {Mn (aq) + 4 H O(l) MnO - 4 (aq) + 8 H + (aq) + 5 e - } - = V Reductin: 5{Br (l) + e - Br - (aq)} = V Overall: Mn (aq) + 5 Br (l) + 8 H O(l) MnO - 4 (aq) + 10 Br - (aq) + 16 H + (aq) = V Reactin is nt spntaneus under standard state cnditins with < 0. c) Oxidatin: Hg (aq) Hg (aq) + e - - = -0.9 V Reductin: Hg (aq) + e - Hg(l) = V Overall: Hg (aq) Hg (aq) + Hg(l) = V r Hg (aq) Hg (aq) + Hg(l) Negative indicates reactin is nt spntaneus under standard state cnditins a) Cl (g) + Fe (aq) Cl - (aq) + Fe 3+ (aq) = Cl - 3+ Fe = 1.36 V - (0.77 V) = 0.59 V The reactin is spntaneus. b) Mn (aq) + H O(l) + C 3+ (aq) MnO (s) + 4 H + (aq) + C (aq) = 3+ C - MnO = 1.8 V - (1.3 V) = 0.59 V The reactin is spntaneus. c) 3 AgCl(s) + NO(g) + H O(l) 3 Ag(s) + 3 Cl - (aq) + NO - 3 (aq) + 4 H + (aq) = AgCl - NO3 - = 0. V - (0.96 V) = V The reactin is nnspntaneus. 1-11

12 1.44 a) Oxidatin: Ag(s) Ag + (aq) + e - - = V Reductin: Cu (aq) + e - Cu(s) = V Overall: Ag(s) + Cu (aq) Ag + (aq) + Cu(s) = V The reactin is nt spntaneus. b) Oxidatin: Cd(s) Cd (aq) + e - - = V Reductin: Cr O - 7 (aq) + 14 H + (aq) + 6 e - Cr 3+ (aq) + 7 H O(l) = V Overall: Cr O - 7 (aq) + 3 Cd(s) + 14 H + (aq) Cr 3+ (aq) + 3 Cd (aq) + 7 H O(l) The reactin is spntaneus. = V c) Oxidatin: Pb(s) Pb (aq) + e - - = V Reductin: Ni (aq) + e - Ni(s) = -0.5 V Overall: Pb(s) + Ni (aq) Pb (aq) + Ni(s) The reactin is nt spntaneus a) Cu + (aq) + PbO (s) + 4 H + (aq) + SO - 4 (aq) Cu (aq) + PbSO 4 (s) + H O(l) = PbO - Cu = 1.70 V - (0.15 V) = 1.55 V The reactin is spntaneus. b) H O (aq) + Ni (aq) H + (aq) + O (g) + Ni(s) = Ni - O = -0.5 V - (0.68 V) = V The reactin is nnspntaneus. c) 3 Ag + (aq) + MnO (s) + 4 OH - (aq) 3Ag(s) + MnO - 4 (aq) + H O(l) = Ag + - MnO4 - = 0.80 V - (0.59 V) = 0.1 V The reactin is spntaneus Adding (1) and () t give a spntaneus reactin invlves cnverting (1) t xidatin: 3 N O 4 (g) + Al(s) 6 NO - (aq) + Al 3+ (aq) = V - (-1.66 V) =.53 V Reverse (1), and then add (1) and (3) t give a spntaneus reactin. Al(s) + 3 SO - 4 (aq) + 3 H O(l) Al 3+ (aq) + 3 SO - 3 (aq) + 6 OH - (aq) = 0.93 V - (-1.66 V) =.59 V Reverse (), then add () and (3) fr the spntaneus reactin: SO - 4 (aq) + NO - (aq) + H O(l) SO - 3 (aq) + N O 4 (g) + OH - (aq) = 0.93 V V = 0.06 V Rank xidizing agents (substance being reduced) in rder f increasing strength: Al 3+ < N O 4 < SO 4 - Rank reducing agents (substance being xidized) in rder f increasing strength: SO 3 - < NO - < Al = -0.1 V 1-1

13 N O(g) + 6 H + (aq) + Cr(s) 3 N (g) + 3 H O(l) + Cr 3+ (aq) = 1.77 V - (-0.74 V) =.51 V Oxidizing agents: N O > Cr 3+ ; reducing agents: Cr > N 3 Au + (aq) + Cr(s) 3 Au(s) + Cr 3+ (aq) = 1.69 V - (-0.74 V) =.43 V Oxidizing agents: Au + > Cr 3+ ; reducing agents: Cr > Au N O(g) + H + (aq) + Au(s) N (g) + H O(l) + Au + (aq) = 1.77 V - (1.69 V) = 0.08 V Oxidizing agents: N O > Au + ; reducing agents: Au > N Oxidizing agents: N O > Au + > Cr 3+ ; reducing agents: Cr > Au > N 1.48 Spntaneus reactin results when () is reversed and added t (1): HClO(aq) + Pt(s) + H + (aq) Cl (g) + Pt (aq) + H O(l) = 1.63 V V = 0.43 V Spntaneus reactin results when (3) is reversed and added t (1): HClO(aq) + Pb(s) + SO - 4 (aq) + H + (aq) Cl (g) + PbSO 4 (s) + H O(l) = 1.63 V - (-0.31 V) = 1.94 V Spntaneus reactin results when (3) is reversed and added t (): Pt (aq) + Pb(s) + SO - 4 (aq) Pt(s) + PbSO 4 (s) = 1.0 V - (-0.31 V) = 1.51 V Order f increasing strength as xidizing agent: PbSO 4 < Pt < HClO Order f increasing strength as reducing agent: Cl < Pt < (Pb + SO 4 - ) 1.49 S O - 8 (aq) + I - (aq) SO - 4 (aq) + I (s) =.01 V - (0.53 V) = 1.48 V Oxidizing agents: S O - 8 > I ; reducing agents: I - - > SO 4 Cr O - 7 (aq) + 14 H + (aq) + 6 I - (aq) Cr 3+ (aq) + 7 H O(l) + 3 I (s) = 1.33 V - (0.53 V) = 0.80 V Oxidizing agents: Cr O - 7 > I ; reducing agents: I - > Cr 3+ 3 S O - 8 (aq) + Cr 3+ (aq) + 7 H O(l) 6 SO - 4 (aq) + Cr O - 7 (aq) + 14 H + (aq) =.01 V - (1.33 V) = 0.68 V Oxidizing agents: S O 8 - > Cr O 7 - ; reducing agents: Cr 3+ > SO 4 - Oxidizing agents: S O 8 - > Cr O 7 - > I ; reducing agents: I - > Cr 3+ > SO Metal A + Metal B salt slid clred prduct n metal A Cnclusin: Prduct is slid metal B. A is better reducing agent than B. Metal B + acid gas bubbles Cnclusin: Prduct is H gas prduced as result f reductin f H +. B is better reducing agent than acid. Metal A + Metal C salt n reactin Cnclusin: C must be a better reducing agent than A. Since C is a better reducing agent than A, which is a better reducing agent than B and B reduces acid, then C wuld als reduce acid t frm H bubbles. The rder f strength f reducing agents is: C > A > B. 1-13

14 1.51 a) Cpper metal is cating the irn. b) The xidizing agent is Cu and the reducing agent is Fe. c) Yes, this reactin, being spntaneus, may be made int a vltaic. d) Cu (aq) + Fe(s) Cu(s) + Fe (aq) e) = Cu - Fe = 0.34 V - (-0.44 V) = 0.78 V 1.5 = V lg Q n K and G = -nf a) When Q / K < 1, > 0 and G < 0. When Q / K = 1, = 0 and G = 0. When Q / K > 1, < 0 and G > 0. b) Only when Q / K < 1 will the reactin prceed spntaneusly and be able t d wrk At the negative (ande) electrde, xidatin ccurs s the verall reactin is A(s) + B + (aq) A + (aq) + B(s) with Q = [A + ] / [B + ]. a) The reactin prceeds t the right because with > 0 (vltaic ), the spntaneus reactin ccurs. As the perates, [A + ] increases and [B + ] decreases. b) decreases because the reactin takes place t apprach equilibrium, = 0. c) and are related by the Nernst equatin: = - (RT / nf)ln([a + ] / [B + ]). = when (RT / nf)ln([a + ] / [B + ]) = 0. This ccurs when ln([a + ] / [B + ]) = 0. Recall that e 0 = 1, s [A + ] must equal [B + ] fr t equal. d) Yes, it is pssible fr t be less than when [A + ] > [B + ] a) xamine the Nernst equatin: = RT nf lg Q.303 RT.303 RT = lg K - lg Q nf nf.303 RT K.303 RT K = lg = - lg nf Q nf Q If Q / K < 1, will decrease with a decrease in temperature. If Q / K > 1, will increase (becme less negative) with a decrease in temperature..303 RT [ Active in at ande] b) = - lg nf [ Active in at cathde] will decrease as the cncentratin f an active in at the ande increases. c) will increase as the cncentratin f an active in at the cathde increases. d) will increase as the pressure f a gaseus reactant in the cathde cmpartment increases In a cncentratin, the verall reactin takes place t decrease the cncentratin f the mre cncentrated electrlyte. The mre cncentrated electrlyte is reduced, s it is in the cathde cmpartment. 1-14

15 1.56 The equilibrium cnstant can be fund by cmbining tw equatins: G = -nf and G = -RT ln K, We get: ln K = nf n r lg K = RT a) = cathde - ande = 0.80 V - (-0.5 V) = 1.05 V; electrns are transferred. n 1.05V ( ) lg K = = = (unrunded) V K = x = 3 x b) = V - (-0.44 V) = V; 6 electrns are transferred. n 6( 0.30 V) lg K = = = (unrunded) V K = x = 4 x a) Al(s) + 3 Cd (aq) Al 3+ (aq) + 3 Cd(s) n = 6 = 3+ Cd - Al 1.58 a) lg K = = V - (-1.66 V) = 1.6 V n = 61.6V ( ) V = (unrunded) K = x = 6 x b) I (s) + Br - (aq) Br (l) + I - (aq) n = = - I Br = 0.53 V - (1.07 V) = V n ( 0.54 V) lg K = = = (unrunded) V K = x = 6 x b) = V V = V; electrns transferred. n ( 1.98 V) lg K = = = (unrunded) V K = x = 1 x = 1.36 V V = 0.9 V; electrns transferred. lg K = n = 0.9V ( ) V K = x 10 9 = 6 x 10 9 = (unrunded) 1.59 a) Cr(s) + 3 Cu (aq) Cr 3+ (aq) + 3 Cu(s) n = 6 = 3+ Cu - Cr lg K = = 0.34 V - (-0.74 V) = 1.08 V n = 61.08V ( ) V K = x = 3 x = (unrunded) 1-15

16 b) Sn(s) + Pb (aq) Sn (aq) + Pb(s) n = = Pb - Sn = V - (-0.14 V) = 0.01V n lg K = K =.1767 = = 0.01V ( ) V = (unrunded) 1.60 Substitute J / C fr V. a) G = -nf = -( ml e - ) (96485 C / ml e - ) (1.05 J / C) = x 10 5 = -.03 x 10 5 J b) G = -nf = -(6 ml e - ) (96485 C / ml e - ) (-0.30 J / C) = x 10 5 = 1.73 x 10 5 J 1.61 Substitute J / C fr V. a) G = -nf = -(6 ml e - ) (96485 C / ml e - ) (1.6 J / C) = x 10 5 = -7.9 x 10 5 J b) G = -nf = -( ml e - ) (96485 C / ml e - ) (-0.54 J / C) = x 10 5 = 1.0 x 10 5 J 1.6 Substitute J / C fr V. a) G = -nf = -( ml e - ) (96485 C / ml e - ) (-1.98 J / C) = x 10 5 = -3.8 x 10 5 J b) G = -nf = -( ml e - ) (96485 C / ml e - ) (0.9 J / C) = x 10 4 = -5.6 x 10 4 J 1.63 Substitute J / C fr V. a) G = -nf = -(6 ml e - ) (96485 C / ml e - ) (1.08 J / C) = x 10 5 = -6.5 x 10 5 J b) G = -nf = -( ml e - ) (96485 C / ml e - ) (0.01 J / C) = x 10 3 = - x 10 3 J 1.64 Find G frm the fact that G = -RT ln K. Then use G value t find, and frm G = -nf. T = (73 + 5)K = 98 K G = -RT ln K = -(8.314 J/ml K) (98 K) ln (5.0 x 10 3 ) = x 10 4 = -.1 x 10 4 J = - G / nf = -( x 10 4 J) / (1 ml e - ) (96485 C / ml e - )[V / (J / C)] = = 0. V 1.65 Find G frm the fact that G = -RT ln K. Then use G value t find, and frm G = -nf. T = (73 + 5)K = 98 K G = -RT ln K = -(8.314 J/ml K) (98 K) ln (5.0 x 10-6 ) = x 10 4 = 3.0 x 10 4 J = - G / nf = -( x 10 4 J) / (1 ml e - ) (96485 C / ml e - )[V / (J / C)] = = V n 1.66 Use lg K = and G = -nf. T = (73 + 5)K = 98 K = (0.059 / n) lg K = (0.059 / ) lg 75 = = V G = -RT ln K = -(8.314 J/ml K) (98 K) ln (75) = x 10 4 = -1.1 x 10 4 J n 1.67 Use lg K = and G = -nf. T = (73 + 5)K = 98 K = (0.059 / n) lg K = (0.059 / ) lg = = V G = -RT ln K = -(8.314 J/ml K) (98 K) ln (0.075) = x 10 3 = 6.4 x 10 3 J 1-16

17 1.68 The reactin is: Cu (aq) + H (g) Cu(s) + H + (aq) = + - Cu H + = 0.34 V V = 0.34 V = H lg n Cu [ H ] Fr a standard hydrgen electrde [H + ] = 1.0 M and [H ] = 1.0 atm 0.5 V = 0.34 V lg Cu (0.5 V V) (- / 0.059) = lg = (unrunded) Cu 1.0 [Cu ] = x 10-4 = 9 x 10-4 M 1.69 The reactin is: Pb (aq) + Mn(s) Pb(s) + Mn (aq) = Pb - Mn = V - (-1.18 V) = 1.05 V = Mn lg n Pb 0.4 V = 1.05 V [ 1.3] lg Pb + [ 1.3] (0.4 V V) (- / 0.059) = lg = (unrunded) Pb + [Pb ] = x 10 - = 7 x 10 - M 1.70 The spntaneus reactin (vltaic ) is Ni (aq) + C(s) Ni(s) + C (aq) with = -0.5 V - (-0.8 V) = 0.03 V a) Use the Nernst equatin: = - (0.059 V / n) lg Q = 0.03 V - (0.059 V / ) lg ([C ] / [Ni ]) = 0.03 V - (0.059 V / ) lg (0.0 M / 0.80 M) = V = 0.05 V b) Frm part (a), ntice that an increase in [C ] leads t a decrease in ptential. Therefre, the cncentratin f cbalt in must increase further t bring the ptential dwn t 0.03 V. Thus, the new cncentratins will be [C ] = 0.0 M + x and [Ni ] = 0.80 M - x 0.03 V = 0.03 V - (0.059 V / ) lg [(0.0 + x) / ( x)] x = 0.0 [Ni ] = = 0.60 M c) At equilibrium = 0.00, t decrease the ptential t 0.00, [C ] increases and [Ni ] decreases V = 0.03 V - (0.059 V / ) lg [(0.0 + x) / ( x)] x = (unrunded) [C ] = = = 0.91 M [Ni ] = = = 0.09 M 1-17

18 1.71 The spntaneus reactin (vltaic ) is Cd (aq) + Mn(s) Cd(s) + Mn (aq) with = V - (-1.18 V) = 0.78 V. a) Use the Nernst equatin: = - (0.059 V / n) lg Q. = 0.78 V - (0.059 V / ) lg ([Mn ] / [Cd ]) = 0.78 V - (0.059 V / ) lg (0.090 M / M) = V = 0.77 V b) Fr the [Cd ] t decrease frm M t M, a change f M, the [Ni ] must increase by the same amunt, frm M t M. = 0.78 V - (0.059 V / ) lg (0.100 M / M) = V = 0.77 V c) Increase the manganese and decrease the cadmium by equal amunts V = 0.78 V - (0.059 V / ) lg ([Mn ] / [Cd ]) ([Mn ] / [Cd ]) = x 10 4 (unrunded) [Mn ] + [Cd ] = M [Cd ] = x 10-6 M (unrunded) [Mn ] = M - [Cd ] = M d) At equilibrium = V = 0.78 V - (0.059 V / ) lg ([Mn ] / [Cd ]) ([Mn ] / [Cd ]) = x 10 6 (unrunded) [Mn ] + [Cd ] = M [Cd ] = x 10-8 = 7 x 10-8 M [Mn ] = M - [Cd ] = M 1.7 The verall reactin prceeds t increase the 0.10 M H + cncentratin and decrease the.0 M H + cncentratin. Therefre, half- A is the ande because it has the lwer cncentratin. + H P H(cathde) ( ) 0.10 ( 0.50 ) ande Q fr the equals = = (unrunded) + H P (.0) ( 0.90) H(ande) cathde = 0.00 V - (0.059 V / ) lg ( ) = = V 1.73 Sn (0.87 M) Sn (0.13 M) Half- B is the cathde. = 0.00V - (0.059 V / ) lg (0.13 / 0.87) = = 0.04 V 1.74 lectrns flw frm the ande, where xidatin ccurs, t the cathde, where reductin ccurs. The electrns always flw frm the ande t the cathde, n matter what type f The electrdes are separated by an electrlyte paste, which fr the rdinary dry battery cntains ZnCl, NH 4 Cl, MnO, starch, graphite, and water A D-sized battery is much larger than an AAA-sized battery, s the D-sized battery cntains a greater amunt f the cmpnents. The ptential, hwever, is an intensive prperty and des nt depend n the amunt f the cmpnents. (Nte that amunt is different frm cncentratin.) The ttal amunt f charge a battery can prduce des depend n the amunt f cmpnents, s the D-sized battery prduces mre charge than the AAA-sized battery a) Alkaline batteries = (6.0 V) (1 alkaline battery / 1.5 V) = 4 alkaline batteries. b) Vltage = (6 Ag batteries) (1.6 V / Ag battery) = 9.6 V c) The usual 1 vlt car battery cnsists f six vlt s. If tw s are shrted nly fur s remain. Vltage = (4 s) ( V / ) = 8 V 1.78 The Tefln spacers keep the tw metals separated s the cpper cannt cnduct electrns that wuld prmte the crrsin f the irn skeletn. Oxidatin f the irn by xygen causes rust t frm and the metal t crrde. 1-18

19 1.79 Bridge supprts rust mre rapidly at the water line due t the presence f large cncentratins f bth O and H O Chrmium, like irn, will crrde thrugh the frmatin f a metal xide. Unlike the irn xide, which is relatively prus and easily cracks, the chrmium xide frms a prtective cating, preventing further crrsin Sacrificial andes are metals with less than that fr irn, V, s they are mre easily xidized than irn. a) (aluminum) = Yes, except aluminum resists crrsin because nce a cating f its xide cvers it, n mre aluminum crrdes. Therefre, it wuld nt be a gd chice. b) (magnesium) = -.37 V. Yes, magnesium is apprpriate t act as a sacrificial ande. c) (sdium) = -.71 V. Yes, except sdium reacts with water, s it wuld nt be a gd chice. d) (lead) = V. N, lead is nt apprpriate t act as a sacrificial ande because its value is t high. e) (nickel) = -0.5 V. N, nickel is inapprpriate as a sacrificial ande because its value is t high. f) (zinc) = V. Yes, zinc is apprpriate t act as a sacrificial ande. g) (chrmium) = V. Yes, chrmium is apprpriate t act as a sacrificial ande. 1.8 a) Oxidatin ccurs at the left electrde (ande). b) lemental M frms at the right electrde (cathde). c) lectrns are being released by ins at the left electrde. d) lectrns are entering the at the right electrde Cd (aq) + Cr(s) 3 Cd(s) + Cr 3+ (aq) = V - (-0.74 V) = 0.34 V T reverse the reactin requires 0.34 V with the in its standard state. A 1.5 V supplies mre than enugh ptential, s the cadmium metal xidizes t Cd and chrmium plates ut The half- values are different than the half- values because in pure water, the [H + ] and [OH - ] are 1.0 x 10-7 M rather than the standard-state value f 1 M The xidatin number f nitrgen in the nitrate in, NO 3 -, is +5 and cannt be xidized further since nitrgen has nly five electrns in its uter level. In the nitrite in, NO -, n the ther hand, the xidatin number f nitrgen is +3, s it can be xidized t the +5 state Due t the phenmenn f vervltage, the prducts predicted frm a cmparisn f electrde ptentials are nt always the actual prducts. When gases (such as H (g) and O (g)) are prduced at metal electrdes, there is an vervltage f abut 0.4 t 0.6V mre than the electrde ptential indicates. Due t this, if H r O is the expected prduct, anther species may be the true prduct a) At the ande, brmide ins are xidized t frm brmine (Br ). b) At the cathde, sdium ins are reduced t frm sdium metal (Na) a) At the negative electrde (cathde) barium ins are reduced t frm barium metal (Ba). b) At the psitive electrde (ande), idide ins are xidized t frm idine (I ) ither idide ins r fluride ins can be xidized at the ande. The in that mre easily lses an electrn will frm. Since I is less electrnegative than F, I - will mre easily lse its electrn and be xidized at the ande. The prduct at the ande is I gas. The idine is a gas because the temperature is high t melt the salts. ither ptassium r magnesium ins can be reduced at the cathde. Magnesium has greater inizatin energy than ptassium because magnesium is lcated up and t the right f ptassium n the peridic table. The greater inizatin energy means that magnesium ins will mre readily add an electrn (be reduced) than ptassium ins. The prduct at the cathde is magnesium (liquid) Liquid strntium (Sr) frms at the negative electrde, and gaseus brmine (Br ) frms at the psitive electrde. 1-19

20 1.91 Brmine gas frms at the ande because the electrnegativity f brmine is less than that f chlrine. Calcium metal frms at the cathde because its inizatin energy is greater than that f sdium. 1.9 Liquid calcium frms at the negative electrde and chlrine gas frms at the psitive electrde Cpper and brmine can be prepared by electrlysis f their aqueus salts because their half- ptentials are mre psitive than the ptential fr the electrlysis f water with vervltage, abut 1 V fr reductin f water and abut 1.4 V fr the xidatin f water Strntium is t electrpsitive t frm frm the electrlysis f an aqueus slutin. The elements that electrlysis will separate frm an aqueus slutin are gld, tin, and chlrine Idine, zinc, and silver can be prepared by electrlysis f their salt slutins because water is less readily xidized than idide ins and less readily reduced than zinc ins and silver ins lectrlysis will separate bth irn and cadmium frm a aqueus slutin a) Pssible xidatins: H O(l) O (g) + 4 H + (aq) + 4 e - - = -1.4 V with vervltage F - F (g) + e - - = -.87 V The xidatin f water prduces xygen gas (O ), and hydrnium ins (H 3 O + ) at the ande. Pssible reductins: H O(l) + e - H (g) + OH - (aq) = -1 V with vervltage Li + (aq) + e - Li(s) = V The reductin f water prduces H gas and OH - at the cathde. b) Pssible xidatins: H O(l) O (g) + 4 H + (aq) + 4 e - - = -1.4 V with vervltage The xidatin f water prduces xygen gas (O ), and hydrnium ins (H 3 O + ) at the ande. Pssible reductins: H O(l) + e - H (g) + OH - (aq) = -1 V with vervltage Sn (aq) + e - Sn(s) = V SO - 4 (aq) + 4 H + (aq) + e - SO (g) + H O(l) = V (apprximate) The ptential fr sulfate reductin is estimated frm the Nernst equatin using standard state cncentratins and pressures fr all reactants and prducts except H +, which in pure water is 1 x 10-7 M. = 0.0 V - (0.059 / ) lg [1 / (1 x 10-7 ) 4 ] = = V The mst easily reduced in is Sn, s tin metal frms at the cathde a) Slid zinc (Zn) frms at the cathde and liquid brmine (Br ) frms at the ande. b) Slid cpper (Cu) frms at the cathde and bth xygen gas (O ) and aqueus hydrgen ins (H + ) frm at the ande a) Pssible xidatins: H O(l) O (g) + 4 H + (aq) + 4 e - - = -1.4 V with vervltage The xidatin f water prduces xygen gas (O ), and hydrnium ins (H 3 O + ) at the ande. Pssible reductins: H O(l) + e - H (g) + OH - (aq) = -1 V with vervltage Cr 3+ (aq) + 3 e - Cr(s) = V NO - 3 (aq) + 4 H + (aq) + 3 e - NO(g) + H O(l) = V (apprximate) The ptential fr nitrate reductin is estimated frm the Nernst equatin using standard state cncentratins and pressures fr all reactants and prducts except H +, which in pure water is 1 x 10-7 M. = 0.96 V - (0.059 / ) lg [1 / (1 x 10-7 ) 4 ] = = 0.13 V The mst easily reduced in is NO - 3, s NO gas is frmed at the cathde. 1-0

21 b) Pssible xidatins: H O(l) O (g) + 4 H + (aq) + 4 e - - = -1.4 V with vervltage Cl - (aq) Cl (g) + e - - = V The xidatin f chlride ins t prduce chlrine gas ccurs at the ande. Pssible reductins: H O(l) + e - H (g) + OH - (aq) = -1 V with vervltage Mn (aq) + e - Mn(s) = V It is easier t reduce water than t reduce manganese ins, s hydrgen gas and hydrxide ins frm at the cathde a) Slid irn (Fe) frms at the cathde, and slid idine (I ) frms at the ande. b) Gaseus hydrgen (H ) and aqueus hydrxide in (OH - ) frm at the cathde, while gaseus xygen (O ) and aqueus hydrgen ins (H + ) frm at the ande Mg + e - Mg a) (35.6 g Mg) (1 ml Mg / 4.31 g Mg) ( ml e - / 1 ml Mg) = =.93 ml e - b) ( ml e - ) (96485 C / ml e - ) = x 10 5 =.83 x 10 5 culmbs x 10 C 1 h A c) = = 31.4 A.50 h 3600 s C s 1.10 Na e - Na a) (15 g Na) (1 ml Na /.99 g Na) (1 ml e - / 1 ml Na) = = 9.35 ml e - b) ( ml e - ) (96485 C / ml e - ) = x 10 5 = 9.0 x 10 5 culmbs x 10 C 1 h A c) = = 6.4 A 9.50 h 3600 s C s Ra + e - Ra In the reductin f radium ins, Ra, t radium metal, the transfer f tw electrns ccurs. 1 ml e 1 ml Ra 6 g Ra ( 15 C) C mle = = 0.8 g Ra 1 ml Ra Al e - Al In the reductin f aluminum ins, Al 3+, t aluminum metal the transfer f three electrns ccurs. 1 ml e 1 ml Al 6.98 g Al ( 305 C) C 3mle = = g Al 1 ml Al Zn + e - Zn 1 ml Zn ml e C 1 1 A Time = ( 85.5 g Zn) g Zn 1 ml Zn 1mle 3.0 A C s = x 10 4 = 1.10 x 10 4 secnds Ni + e - Ni Time = ( 1.63 g Ni) 1 ml Ni ml e C 1 1 A = = 391 secnds g Ni 1 ml Ni 1mle 13.7 A C s 1-1

22 1.107 a) The sdium sulfate makes the water cnductive, s the current will flw thrugh the water t cmplete the circuit, increasing the rate f electrlysis. Pure water, which cntains very lw (10-7 M) cncentratins f H + and OH -, cnducts electricity very prly. b) The reductin f H O has a mre psitive half-ptential than the reductin f Na + ; the xidatin f H O is the nly reductin pssible because SO 4 - cannt be xidized under these cnditins. In ther wrds, it is easier t reduce H O than Na + and easier t xidize H O than SO Idine is frmed first since its xidatin ptential ( x (I - )) is mre psitive than the xidatin ptential ( x (Br - )) f brmide Zn + e - Zn C 3600 s 4 h 1 ml e 1 ml Zn g Zn Mass = ( A) s (.00 day) A 1 h 1 day C mle 1 ml Zn = = 44. g Zn Mn (aq) + H O(l) MnO (s) + 4 H + (aq) + e - Time = ( 1.00 kg MnO ) 10 g ml e C 1 A 1 h 1 kg g MnO 1 ml MnO 5.0 A C 3600 s s 3 1mlMnO 1mle = = 4.7 hurs The MnO prduct frms at the ande, since the half-reactin is an xidatin The half-reactin is: H + (aq) + e - H (g) a) First, find the mles f hydrgen gas. 6 ( 10.0 atm)(.5 x 10 L) n = PV / RT = = x 10 6 ml H (unrunded) L atm (( ) K) ml K Then, find the culmbs knwing that there are tw electrns transferred per ml f H. ( 6 ml e C x 10 mle H ) 1mlH = x 10 1mle 11 =.0 x culmbs b) Remember that 1 V equals 1 J / C. (1.4 J / C) ( x C) = x =.4 x J c) ( x J) (1 kj / 10 3 J) (1 kg / 4.0 x 10 4 kj) = x 10 3 = 6.1 x 10 3 kg 1.11 a) The half reactins are: H O(l) + Zn(s) ZnO(s) + H + (aq) + e - and e - + H + (aq) + MnO (s) Mn(OH) (s). mles f electrns flw per mle f reactin. 1mlZn 1 ml MnO g MnO b) Mass f MnO = (.50 g Zn) = = 3.3 g MnO g Zn 1 ml Zn 1 ml MnO 1mlZn 1 ml HO 18.0 g HO Mass f H O = (.50 g Zn) = = g H O g Zn 1 ml Zn 1 ml HO c) Ttal mass f reactants =.50 g Zn g MnO g H O = = 6.51 g 1 ml Zn ml e culmbs d) Charge = (.50 g Zn) g Zn 1 ml Zn 1mle = x 10 3 = 7.38 x 10 3 culmbs e) An alkaline battery cnsists f mre than just reactants. The case, electrlyte paste, cathde, absrbent, and unreacted reactants (less than 100% efficient) als cntribute t the mass f an alkaline battery. 1-

23 1.113 The reactin is: Ag + (aq) + e - Ag(s) Mass Ag = g g = g Ag prduced 1 ml Ag 1 ml e culmbs Culmbs = ( g Ag) g Ag 1 ml Ag 1mle = = 14.5 culmbs Frm the current 70.0% f the mles f prduct will be cpper and 30.0% zinc. Assume a current f exactly 100 culmbs. The amunt f current used t generate cpper wuld be (70.0% / 100%) (100 C) = 70.0 C, and the amunt f current used t generate zinc wuld be (30.0% / 100%) (100 C) = 30.0 C. The half-reactins are: Cu (aq) + e - Cu(s) and Zn (aq) + e - Zn(s). 1 ml e 1 ml Cu g Cu Mass cpper = ( 70.0 C) C mle = g Cu (unrunded) 1 ml Cu 1 ml e 1 ml Zn g Zn Mass zinc = ( 30.0 C) C mle = g Zn (unrunded) 1 ml Zn Mass % cpper = g Cu x 100% = = 69.4% Cu g Sample ( + ) Vltaic Cell lectrlytic Cell a) G is negative G is psitive b) Oxidatin Oxidatin c) Reductin Reductin d) (-) (+) e) Ande Ande The reactin is: Au 3+ (aq) + 3 e - Au(s) a) Mles f gld plated = (vlume f gld) (density f gld) (1 / mlar mass f gld) The vlume f the gld is the vlume f a cylinder (see the inside back cver f the text) V = πr h cm 10 m 1 cm 19.3 g Au 1 ml Au ( π)( ) ( 0.0 mm ) 3 1 mm 10 m = ml Au (unrunded) cm g Au 3 ml e C A 1 1 h 1 day Time = ( ml Au) 1 ml Au 1mle C A 3600 s 4 h s = = 13 days b) The time required dubles nce fr the secnd earring f the pair and dubles again fr the secnd side, thus it will take fur times as lng as ne side f ne earring. Time = (4) ( days) = = 5. x 10 1 days c) Start by multiplying the mles f gld frm part (a) by fur t get the mles fr the earrings. Cnvert this mles t grams, then t try unces, and finally t dllars g Au 1 Try Ounce $30 Cst = ( 4)( ml Au) = = $310 1 ml Au g Try Ounce 1-3

24 1.117 a) The half-reactin is: H O(l) O (g) + 4 H + (aq) + 4 e - Determine the mles f xygen frm the ideal gas equatin. Use the half-reactin and the current t cnvert the mles f xygen t time. ( )( ) n = PV / RT = 99.8 kpa 10.0 L 3 1atm 10 Pa L atm x 10 Pa 1kPa (( 73 8) K) + ml K = ml O (unrunded) 4 ml e C A 1 1 min Time = ( ml O ) 1mlO C 1mle 1.3A 60s s = x 10 3 =.0 x 10 3 min b) The balanced chemical equatin is: H O(l) H (g) + O (g) The mles f xygen determined previusly and this chemical equatin leads t the mass f hydrgen. Mass H = ( ml O ) ( ml H / 1 ml O ) (.016 g H / 1 ml H ) = = 1.61 g H Crrsin is an xidatin prcess. This wuld be favred by the metal being in cntact with the mist grund. T cunteract this, electrns shuld flw int the rails and away frm the verhead wire, s the verhead wire shuld be cnnected t the psitive terminal The half-reactins and the reactin are: Zn(s) + OH - (aq) ZnO(s) + H O(l) + e - Ag O(s) + H O(l) + e - Ag(s) + OH - (aq) Zn(s) + Ag O(s) ZnO(s) + Ag(s) The key is the mles f zinc. Frm the mles f zinc, the mles f electrns and the mles f Ag O may be fund. Mles Zn = (16.0 g Zn) (80% / 100%) (1 ml Zn / g Zn) = ml Zn (unrunded) The 80% is assumed t have tw significant figures. ml e 96485C A 1 ma 1 1 h 1 day a) Time = ( ml Zn) C 3 1 ml Zn 1mle 10 A 4.8 ma 3600 s 4 h s = = 91 days 100% ml Ag 95% g Ag b) Mass Ag = ( ml Zn) 80% 1 ml Zn 100% = = 50. g Ag 1 ml Ag c) Cst = ( g Ag) (1 try z / g) ($5.50 g / try z) = = $ Cu (aq) + e - Cu(s) C 3600 s 1 ml e 1 ml Cu g Cu Theretical amunt f cpper = ( 5.8 A) s ( 10 h) A 1 h C mle 1 ml Cu = g Cu (unrunded) Actual Yield 53.4 g Cu fficiency = x 100% = x100% = = 78% Theretical Yield g Cu The final assumes that 10 hurs has tw significant figures a) Mlten lectrlysis b) Aqueus lectrlysis Ande prduct Cl (g) Cl (g) Cathde prduct Na(l) H (g) and OH - (aq) Species reduced Na + (l) H O(l) Species xidized Cl - (l) Cl - (aq) 1-4