CHAPTER 21 ELECTROCHEMISTRY: CHEMICAL CHANGE AND ELECTRICAL WORK

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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 independently 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, spectatr ins are nt used t balance the half-reactins. Add spectatr ins t the balanced inic equatin t btain the balanced mlecular equatin. 1.8 Spntaneus reactins, ΔG sys < 0, take place in vltaic s, which are als called galvanic s. Nn-spntaneus reactins take place in electrlytic s and result in an increase in the free energy f the (ΔG 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 Assign xidatin numbers: H + (aq) + MnO 4 (aq) + 10 Cl (aq) Mn (aq) + 5 Cl (g) + 8 H O(l) a) T decide which reactant is xidized, lk at xidatin numbers. Cl is xidized because its xidatin number increases frm 1 in Cl t 0 in Cl. b) MnO 4 is reduced because the xidatin number f Mn decreases frm +7 in MnO 4 t + in Mn. 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 atms 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 fr a 1 charge n each side I (aq) I (s) + e add e t prducts fr a charge n each side 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 } r multiply by 3 t give 6 e 6 I (aq) 3 I (s) + 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 fr a 0 charge n each side SO 3 (aq) + H O(l) SO 4 (aq) + H + (aq) + e add e t prducts fr a charge n each side 1-

3 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)} r multiply by t give 6 e MnO 4 (aq) + 8 H + (aq) + 6 e MnO (s) + 4 H O(l) 3{SO 3 (aq) + H O(l) SO 4 (aq) + H + (aq) + e } r multiply by 3 t give 6 e 3 SO 3 (aq) + 3 H O(l) 3 SO 4 (aq) + 6 H + (aq) + 6 e 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) MnO 4 (aq) + H O( l) + 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: MnO 4 (aq) Mn (aq) Mn is balanced H O (aq) O (g) N ther elements t balance Balance O by adding H O: MnO 4 (aq) Mn (aq) + 4 H O(l) add 4 H O t prducts H O (aq) O (g) O is balanced 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 fr + n each side H O (aq) O (g) + H + (aq) + e add e t prducts fr 0 charge n each side 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)} r multiply by t give 10 e MnO 4 (aq) + 16 H + (aq) + 10 e Mn (aq) + 8 H O(l) 5{H O (aq) O (g) + H + (aq) + e } r multiply by 5 t give 10 e 5 H O (aq) 5 O (g) + 10 H + (aq) + 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. 1-3

4 1.13 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) balance Cr Cr O 7 (aq) Cr 3+ (aq) + 7 H O(l) balance O by adding H O Cr O 7 (aq) + 14 H + (aq) Cr 3+ (aq) + 7 H O(l) balance H by adding H + Cr O 7 (aq) + 14 H + (aq) + 6 e Cr 3+ (aq) + 7 H O(l) balance charge by adding 6 e Balance the xidatin half-reactin: Zn(s) Zn (aq) + e balance charge by adding e Add the tw half-reactins multiplying the xidatin half-reactin by 3 t equalize the electrns: Cr O 7 (aq) + 14 H + (aq) + 6 e Cr 3+ (aq) + 7 H O(l) 3 Zn(s) 3 Zn (aq) + 6 e 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 by adding H O MnO 4 (aq) + 4 H + (aq) MnO (s) + H O(l) balance H by adding H + MnO 4 (aq) + 4 H + (aq) + 3 e MnO (s) + H O(l) balance charge by adding 3e Balance the xidatin half-reactin: Fe(OH) (s) + H O(l) Fe(OH) 3 (s) balance O by adding H O Fe(OH) (s) + H O(l) Fe(OH) 3 (s) + H + (aq) balance H by adding H + Fe(OH) (s) + H O(l) Fe(OH) 3 (s) + H + (aq) + e balance charge by adding 1 e Add half-reactins after multiplying xidatin half-reactin by 3: MnO 4 (aq) + 4 H + (aq) + 3 e MnO (s) + H O(l) 3 Fe(OH) (s) + 3 H O(l) 3 Fe(OH) 3 (s) + 3 H + (aq) + 3 e 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) MnO 4 (aq) + H + (aq) + 3 Fe(OH) (s) + H O(l) MnO (s) + 3 Fe(OH) 3 (s) Add OH t bth sides t neutralize the H + and cnvert H + + OH H O: MnO 4 (aq) + H + (aq) + OH (aq) + 3 Fe(OH) (s) + H O(l) MnO (s) + 3 Fe(OH) 3 (s) + OH (aq) 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 by adding H O NO 3 (aq) + 1 H + (aq) N (g) + 6 H O(l) balance H by adding H + NO 3 (aq) + 1 H + (aq) + 10 e N (g) + 6 H O(l) balance charge by adding 10 e Balance the xidatin half-reactin: Zn(s) Zn (aq) + e balance charge by adding e 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) + 10 e N (g) + 6 H O(l) 5 Zn(s) 5 Zn (aq) + 10 e 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. 1-4

5 1.15 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 by adding H O NO 3 (aq) + 4 H + (aq) NO(g) + H O(l) balance H by adding H + NO 3 (aq) + 4 H + (aq) + 3 e NO(g) + H O(l) balance charge by adding 3 e 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 by adding H O 4 Sb(s) + 6 H O(l) Sb 4 O 6 (s) + 1 H + (aq) balance H by adding H + 4 Sb(s) + 6 H O(l) Sb 4 O 6 (s) + 1 H + (aq) + 1 e balance charge by adding 1 e Multiply each half-reactin by an integer t equalize the number f electrns: 4{ NO 3 (aq) + 4 H + (aq) + 3 e NO(g) + H O(l)} multiply by 4 t give 1 e 1{4 Sb(s) + 6 H O(l) Sb 4 O 6 (s) + 1 H + (aq) + 1 e } multiply by 1 t give 1 e This gives: 4 NO 3 (aq) + 16 H + (aq) + 1 e 4 NO(g) + 8 H O(l) 4 Sb(s) + 6 H O(l) Sb 4 O 6 (s) + 1 H + (aq) + 1 e 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. b) Balance reductin half-reactin: BiO 3 (aq) Bi 3+ (aq) + 3 H O(l) balance O by adding H O BiO 3 (aq) + 6 H + (aq) Bi 3+ (aq) + 3 H O(l) balance H by adding H + BiO 3 (aq) + 6 H + (aq) + e Bi 3+ (aq) + 3 H O(l) balance charge t give +3 n each side Balance xidatin half-reactin: Mn (aq) + 4 H O(l) MnO 4 (aq) balance O by adding H O Mn (aq) + 4 H O(l) MnO 4 (aq) + 8 H + (aq) balance H by adding H + Mn (aq) + 4 H O(l) MnO 4 (aq) + 8 H + (aq) + 5 e balance charge t give + n each side Multiply each half-reactin by an integer t equalize the number f electrns: 5{BiO 3 (aq) + 6 H + (aq) + e Bi 3+ (aq) + 3 H O(l)} multiply by 5 t give 10 e {Mn (aq) + 4 H O(l) MnO 4 (aq) + 8 H + (aq) + 5 e } multiply by t give 10 e This gives: 5 BiO 3 (aq) + 30 H + (aq) + 10 e 5 Bi 3+ (aq) + 15 H O(l) Mn (aq) + 8 H O(l) MnO 4 (aq) + 16 H + (aq) + 10 e Add half-reactins. 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 by adding H O Pb(OH) 3 (aq) + 3 H + (aq) Pb(s) + 3 H O(l) balance H by adding H + Pb(OH) 3 (aq) + 3 H + (aq) + e Pb(s) + 3 H O(l) balance charge t give 0 n each side Balance the xidatin half-reactin: Fe(OH) (s) + H O(l) Fe(OH) 3 (s) balance O by adding H O Fe(OH) (s) + H O(l) Fe(OH) 3 (s) + H + (aq) balance H by adding H + Fe(OH) (s) + H O(l) Fe(OH) 3 (s) + H + (aq) + e balance charge t give 0 n each side 1-5

6 Multiply each half-reactin by an integer t equalize the number f electrns: 1{Pb(OH) 3 (aq) + 3 H + (aq) + e Pb(s) + 3 H O(l)} multiply by 1 t give e {Fe(OH) (s) + H O(l) Fe(OH) 3 (s) + H + (aq) + e } multiply be t give e This gives: Pb(OH) 3 (aq) + 3 H + (aq) + e Pb(s) + 3 H O(l) Fe(OH) (s) + H O(l) Fe(OH) 3 (s) + H + (aq) + e Add the 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) Pb(OH) 3 (aq) + H + (aq) + Fe(OH) (s) Pb(s) + H O(l) + Fe(OH) 3 (s) Add ne 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) Pb(OH) 3 (aq) + H O(l) + 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 + 16 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 a) Balance reductin half-reactin: MnO 4 (aq) Mn (aq) + 4 H O(l) balance O by adding H O MnO 4 (aq) + 8 H + (aq) Mn (aq) + 4 H O(l) balance H by adding H + MnO 4 (aq) + 8 H + (aq) + 5 e Mn (aq) + 4 H O(l) balance charge by adding 5 e 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 by adding H O As 4 O 6 (s) + 10 H O(l) 4 AsO 3 4 (aq) + 0 H + (aq) balance H by adding H + As 4 O 6 (s) + 10 H O(l) 4 AsO 3 4 (aq) + 0 H + (aq) + 8 e balance charge by adding 8 e 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 + : 8 MnO 4 (aq) + 64 H + (aq) + 40 e 8 Mn (aq) + 3 H O(l) 5 As 4 O 6 (s) + 50 H O(l) 0 AsO 3 4 (aq) H + (aq) + 40 e 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

7 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 by adding H + P 4 (s) + 1 H + (aq) + 1 e 4 PH 3 (g) balance charge by adding 1 e 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 by adding H O P 4 (s) + 1 H O(l) 4 HPO 3 (aq) + 0 H + (aq) balance H by adding H + P 4 (s) + 1 H O(l) 4 HPO 3 (aq) + 0 H + (aq) + 1 e balance charge by adding 1 e Add tw half-reactins and cancel H + : P 4 (s) + 1 H + (aq) + 1 e 4 PH 3 (g) P 4 (s) + 1 H O(l) 4 HPO 3 (aq) + 0 H + (aq) + 1 e 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 by adding H O MnO 4 (aq) + 4 H + (aq) MnO (s) + H O(l) balance H by adding H + MnO 4 (aq) + 4 H + (aq) + 3 e MnO (s) + H O(l) balance charge by adding 3 e Balance xidatin half-reactin: CN (aq) + H O(l) CNO (aq) balance O by adding H O CN (aq) + H O(l) CNO (aq)+ H + (aq) balance H by adding H + CN (aq) + H O(l) CNO (aq)+ H + (aq) + e balance charge by adding e 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) + 8 H + (aq) + 6 e MnO (s) + 4 H O(l) 3 CN (aq) + 3 H O(l) 3 CNO (aq)+ 6 H + (aq) + 6 e 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) MnO 4 (aq) + 3 CN (aq) + H + (aq) MnO (s) + 3 CNO - (aq) + H O(l) Add OH t bth sides t neutralize H + and frm H O: MnO - 4 (aq) + 3 CN - (aq) + H + (aq) + OH (aq) MnO (s) + 3 CNO (aq) + H O(l) + OH - (aq) 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 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) 3 6 (aq) + 1 e Fe(CN) 4 6 (aq)] 1[Re(s) + 8 OH (aq) ReO 4 (aq) + 4 H O(l) + 7 e ] Overall: 7 Fe(CN) 3 6 (aq) + Re(s) + 8 OH (aq) + 7 e 7 Fe(CN) 4 6 (aq) + ReO 4 (aq) + 4 H O(l) + 7 e Oxidizing agent is Fe(CN) 3 6 and reducing agent is Re. 1-7

8 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 by adding H O NO 3 (aq) + H + (aq) NO (g) + H O(l) balance H by adding H + NO 3 (aq) + H + (aq) + e NO (g) + H O(l) balance charge t give 0 n each side 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 t 4 n each side Multiply each half-reactin by an integer t equalize the number f electrns: 3{NO 3 (aq) + H + (aq) + e NO (g) + H O(l)} Multiply by 3 t give 3 e 1{ Au(s) + 4 Cl (aq) AuCl 4 (aq) + 3 e } Multiply be 1 t give 3 e This gives: 3 NO 3 (aq) + 6 H + (aq) + 3 e 3 NO (g) + 3 H O(l) Au(s) + 4 Cl (aq) AuCl 4 (aq) + 3 e 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 prduces the slid metal which plates ut n. 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. 1.6 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. 1-8

9 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 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-). 1-9

10 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 1.3 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) Vltmeter e e Ni Salt bridge Cu () (+) 1 M Ni Anin flw Catin flw 1 M Cu 1-10

11 1.33 In ntatin, the xidatin cmpnents f the ande cmpartment are written n the left f the salt bridge and the reductin cmpnents f the cathde cmpartment are written t the right f the salt bridge. A duble vertical line separates the ande frm the cathde and represents the salt bridge. A single vertical line separates species f different phases. Ande Cathde a) Al is xidized, s it is the ande and appears first in the ntatin. There is a single vertical line separating the slid metals frm their slutins. 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 indicates that the reactin is nt spntaneus, ΔG > 0. The reverse reactin is spntaneus with > Similar t ther state functins, the sign f changes when a reactin is reversed. Unlike ΔG, ΔH 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 ΔG, ΔH 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 cathde ande = = 0.57 V 0.35 V = 0.9 V 1.39 a) (red halfrxn) O 3 (g) + H + (aq) + e O (g) + H O(l) (x halfrxn) Mn (aq) + H O(l) MnO (s) + 4 H + (aq) + e b) = zne manganese manganese = zne =.07 V 0.84 V = 1.3 V 1.40 The greater (mre psitive) the reductin ptential, the greater the strength as an xidizing agent. a)frm Appendix D: Fe 3+ (aq) + e Fe (aq) = 0.77 V Br (l) + e Br (aq) = 1.07 V Cu (aq) + e Cu(s) = 0.34 V When placed in rder f decreasing strength as xidizing agents: Br > Fe 3+ > Cu. b) Frm Appendix D: Ca (aq) + e Ca(s) =.87 V Cr O 7 (aq) + 14H + (aq) 6e Cr 3+ (aq) + 7H O(l) = 1.33 V Ag + (aq) + e Ag(s) = 0.80 V + When placed in rder f increasing strength as xidizing agents: Ca < Ag < Cr O

12 1.41 a) When placed in rder f decreasing strength as reducing agents: SO > MnO > PbSO4 b) When placed in rder f increasing strength as reducing agents: Hg < Sn < Fe cathde ande 1.4 = values are fund in Appendix D. Spntaneus reactins have > 0. 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) = 0.00 V (0.8 V) = 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) = 1.07 V 1.51 V = 0.44 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) r Hg (aq) Hg (aq) + Hg(l) = 0.85 V 0.9 V = 0.07 V Negativ e 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) = 0.74 V The reactin is nnspntaneus. cathde ande 1.44 = values are fund in Appendix D. Spntaneus reactins have > 0. a) Oxidatin: {Ag(s) Ag + (aq) + e } = 0.80 V Reductin: Cu (aq) + e Cu(s) = V Overall: Ag(s) + Cu (aq) Ag + (aq) + Cu(s) = V 0.80 V = 0.46 V The reactin is nt spntaneus. b) Oxidatin: 3{Cd(s) Cd (aq) + e } = 0.40 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) = V (0.40 V) = The reactin is spntaneus. 1-1

13 c) Oxidatin: Reductin: Pb(s) Pb (aq) + e Ni (aq) + e Ni(s) = 0.13 V = 0.5 V Overall: Pb(s) + Ni (aq) Pb (aq) + Ni(s) = 0.5 V (0. 13 V) = 0.1 V 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) = 0.93 V The reactin is nnspntaneus. c) 3 Ag + (aq) + MnO (s) + 4 OH (aq) 3Ag(s) + MnO 4 (aq) + H O(l) = + Ag MnO 4 = 0.80 V (0.59 V) = 0.1 V The reactin is spntaneus. Al(s) + 3 SO + 3 H 1.46 Spntaneus reactins have > 0. All three reactins are written as reductins. When tw half-reactins are paired, ne half-reactin must be reversed and written as an xidatin. Reverse the half-reactin that will result in a psitive value f. Adding (1) and () t give a spntaneus reactin invlves cnverting (1) t xidatin: Oxidatin: {Al(s) Al 3+ (aq) + 3 e } = 1.66 V Reductin: 3{ N O4(g) + e NO (aq)} = V 3 N O 4 (g) + Al( s) 6 NO l 3+ (aq) + A (aq) = V (1.66 V) =.53 V Oxidizing agents: N O 4 > Al 3+ ; reducing agents: Al > NO Adding (1) and (3) t give a spntaneus reactin invlves cnverting (1) t xidatin: Oxidati n: {Al(s) Al 3+ (aq) + 3 e } =1.66 V Reducti n: 3{SO 4 (aq) + H e SO 3 (aq) + OH O(l) + (aq)} = 0.93 V 4 (aq) l) Al 3+ (aq) + 3 SO 3 (aq) + 6 OH O( (aq) = 0.93 V (1.66 V) =.59 V 3+ Oxidizing agents: SO 4 > Al ; reducing agents: Al > SO 3 Add ing () and (3) t give a spntaneus reactin invlves cnverting t xidatin: Oxidatin: NO (aq) N O 4 (g) + e = V Reductin: SO 4 (aq) + H O(l) + e SO 3 (aq) + OH (aq) = 0.93 V SO 4 (aq) + NO (aq) + H O(l) SO 3 (aq) + N O 4 (g) + OH (aq) = 0.93 V V = V educing agents: NO Oxidizing agents: SO 4 > N O 4 ; r > SO 3 Rank xidizing agents (substance being reduced) in rder f increasing strength: Al 3+ < NO 4 < SO4 Rank reducing agents (su bstance being xidized) in rder f increasing strength: SO 3 < NO < Al 1-13

14 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 + 3+ Oxidizing agents: Au > Cr ; 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: NO > Au + ; reducing agents: Au > N 3+ Oxidizing agents: N O > Au + > Cr ; reducing agents: Cr > Au > N 1.48 Spntaneus reactins have > 0. All three reactins are written as reductins. When tw half-reactins are paired, ne half-reactin must be reversed and written as an xidatin. Reverse the half-reactin that will result in a psitive value f. Adding (1) and () t give a spntaneus reactin invlves cnverting () t xidatin: Oxidatin: Pt(s) Pt (aq) + e = 1.0 V Reducti n: HCl O(aq) + H + (aq) + e Cl (g) + H O(l) = 1.63 V HC lo (aq) + Pt(s) + H + (aq) Cl (g) + Pt (aq) + H O(l) = 1.63 V 1.0 V = 0.43 V Oxidizing agents: HClO > Pt ; reducing agents: Pt > Cl Adding (1) and (3) t give a spntaneus reactin invlves cnverting (3) t xidatin: Oxidatin: Pb(s) + SO 4 (aq) PbSO 4 (s) + e + Reductin: HClO(aq) + H (aq) + e Cl (g) + H O(l) = 0.31 V = 1.63 V 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 Oxidizing agents: HClO > PbSO 4 ; reducing agents: Pb > Cl Adding () and (3) t give a spntaneus reactin invlves cnverting (3) t xidatin: Oxidatin: Pb(s) + SO4 (aq) PbSO 4 (s) + e Reductin: Pt (aq) + e Pt(s) = 0.31 V = 1.0 V Pt (aq) + Pb(s) + SO4 (aq) Pt( s) + PbSO4(s) = 1.0 V (0.31 V) = 1.51 V Oxidizing agents: Pt > PbSO 4 ; reducing agents: Pb > Pt 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: SO8 > I ; reducing agents: I > SO4 CrO7 (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 xidizin agents: O 3+ O g Cr 7 > I ; reducing ag ents: I > Cr 3+ 3 S (aq l) O (aq Cr O (aq) + 14 H + O 8 ) + Cr (aq) + 7 H O( 6 S 4 ) + 7 (aq) =.01 V (1.33 V) = 0.68 V Oxidizing agents: SO > CrO7 ; reducing agents: Cr 3+ > SO4 8 Oxidizing agents: S O 8 > Cr O 7 > I ; reducing agents: I > Cr 3+ > SO

15 1.50 Metal A + Metal B salt slid clred prduct n metal A Cn clusin: Prduct is slid metal B. B is underging reductin and plating ut n A. A is better reducing agent than B. Metal B + acid gas bubbles Cnclusin: Prduct is H as result f reductin f H + gas prduced. B is better reducing agent than acid. Metal A + Metal C salt n reactin Cnclusin: C is nt underging reductin. 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 r educe acid t frm H bubbles. The rder f strength f reducing agents is: C > A > B 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 ( V) = 0.78 V 1.5 = V lg Q n K and ΔG = nf a)when Q / K < 1, the reactin is preceding t the right; > 0 and ΔG < 0 and the reactin is spntaneus. When Q / K = 1, the reactin is at equilibrium; = 0 and ΔG = 0. When Q / K > 1, the reactin is preceding t the left; < 0 and ΔG > 0 and the reactin is nt spntaneus. 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 ce ll), 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 RT ln K nf ln Q K RT ln = Q nf RT nf ln Q = RT Q ln nf K 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. RT [ Active in at ande] b) = Active in at cathde nf ln [ ] 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. 1-15

16 1.55 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. nf n 1.56 The equilibrium cnstant can be fund by using ln K = r lg K =. Use values frm t RT calculate and then calculate K. a) Oxidatin: Ni(s) Ni (aq) + e =0.5 V Reductin: {Ag + (aq) + 1 e Ag(s)} = 0.80 V Ni(s) + Ag + (aq) Ni (aq) + Ag(s) = = 0.80 V (0.5 V) = 1.05 V; electrns are transferred. cathde ande lg K = ( ) n 1.05V = = (unrund ed) V K = x = 3 x 10 K = b) Oxidatin: 3{Fe(s) Fe (aq) + e } = 0.44 V Reductin: {Cr 3+ (aq) + 3 e Cr(s)} = 0.74 V = = 0.74 V (0.44 V) = 0.30 V; 6 electrns are transferred. cathde lg K = ande n ( ) K = K = x = 4 x = V = (unrunded) V 1.57 a) Al(s) + 3 Cd (aq) Al 3+ (aq) + 3 Cd(s) n = 6 = 3+ Cd Al = 0.40 V (1.66 V) = 1.6 V n 61.6V ( ) lg K = = = (unrunded) V K = x = 5 x b) I (s) + Br ( aq) Br (l) + I (aq) n = = I Br = 0.53 V (1.07 V) = 0.54 V n ( 0.54 V) lg K = = = (unrunded) V K = x = 6 x nf n 1.58 The equilibrium cn stant can be fund by using ln K = r lg K =. Use values t calculate RT and then calculate K. a) Oxidatin: {Ag(s) Ag + (aq) + 1 e } = 0.80 V Reductin: Mn (aq) + e Mn(s) = 1.18 V Ni(s) + Ag + (aq) Ni (aq) + Ag(s) = cathde and e = 1.18 V (0.80 V) = 1. 98V; electrns are transferred. n ( 1.98 V) lg K = = = (unrunded) V K = 10 K = x = 1 x

17 b) Oxidatin: Br (aq) Br (l) + e = 1.07 V Reductin: Cl (g) + e Cl (aq) = 1.36 V Br (aq) + Cl (g) Br (l) + Cl (aq) = 1.36 V 1.07 V = 0.9 V; electrns transferred. l g K = n ( ) K = 10 K = x 10 9 = 6 x V = = (unrunded) V 1.59 a) Cr( s) + 3 Cu (aq) Cr 3+ (aq) + 3 Cu(s) 3+ = Cu Cr = 0.34 V (0.74 V) = 1.08 V n = 6 n 61.08V ( ) lg K = = = (unrunded) V K = x = 3 x b) S n(s) + Pb ( aq) Sn (aq) + Pb(s) n = = Pb Sn = 0.13 V (0.14 V) = 0.01V n lg K = K =.1767 = = 0.01V ( ) V = (unrunded) 1.60 Substitute J / C fr V. a) ΔG = nf = ( ml e ) (96485 C / ml e ) (1.05 J / C) = x 10 5 =.03 x 10 5 J b) ΔG = 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) ΔG = nf = (6 ml e ) (96485 C / ml e ) (1.6 J / C) = x 10 5 = 7.9 x 10 5 J b) ΔG = nf = ( ml e ) (96485 C / ml e ) (0. 54 J / C) = x 10 5 = 1.0 x 10 5 J 1.6 Substitu te J / C fr V. a) ΔG = nf = ( ml e ) (96485 C / ml e ) (1.98 J / C) = x 10 5 = 3.8 x 10 5 J b) ΔG = 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) ΔG = nf = (6 ml e ) (96485 C / ml e ) (1.08 J / C) = 6.58 x 10 5 = 6.5 x 10 5 J b) ΔG = nf = ( ml e ) (96485 C / ml e ) (0.01 J / C) = x 10 3 = x 10 3 J 1.64 Find ΔG frm the fact that ΔG = RT ln K. Then use ΔG value t find frm ΔG = nf. T = (73 + 5)K = 98 K ΔG = RT ln K = (8.314 J/ml K) (98 K) ln (5.0 x 10 4 ) = x 10 4 =.7 x 10 4 J 4 ΔG x 10 J 1 V = - = - - (1 ml e )(96485 C/ml e ) 1 J/C = = 0.8 V nf 1.65 Find ΔG frm the fact that ΔG = RT ln K. Then use ΔG value t find, and frm ΔG = nf. T = (73 + 5)K = 98 K ΔG = 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 = ΔG / nf = ( x 10 4 J) / (1 ml e ) (96485 C / ml e )[V / (J / C)] = = 0.31 V 1-17

18 1.66 Use lg K = n and ΔG = nf. T = ( )K = 98 K = lg K = lg 65 = = V n ΔG = RT ln K = (8.314 J/ml K) (98 K) ln (65) = x 10 4 = 1.0 x 10 4 J n 1.67 Use lg K = and ΔG = RT ln K. T = (73 + 5)K = 98 K = (0.059 / n) lg K = (0.059 / ) lg = = V ΔG = RT ln K = (8.314 J/ml K) (98 K) ln (0.065) = x 10 3 = 6.8 x 10 3 J 1.68 Since this is a vltaic, a spntaneus reactin is ccurring. Fr a spntaneus reactin between H /H + and Cu/Cu, Cu must be reduced and H must be xidized: + Oxidatin: H (g) Cu(s) + H (aq) + e = 0.00 V Reductin: Cu (aq) + e Cu(s) = 0.34 V + Cu (aq) + H (g) Cu(s) + H (aq) = cathde ande = 0.34 V 0.00 V = 0.34 V = lg Q n + H = lg n Cu [ H ] Fr a standard hydrgen electrde [H + ] = 1.0 M and [H ] = 1.0 atm. 0. V = 0.34 V lg Cu V 0.34 V = lg Cu V = lg Cu = lg Raise each side t 10 x. Cu x 10 = [Cu ] 5 5 [Cu ] = x 10 = 8.8 x 10 M 1.69 The reactin is: Pb (aq) + Mn(s) Pb(s) + Mn (aq) = Pb Mn = 0.13 V (1.18 V) = 1.05 V Mn = lg n Pb [ ] 0.44 V = 1.05 V lg Pb 1-18

19 (0.44 V 1.05 V) ( / 0.059) = lg [ 1. 4] = Pb + [Pb ] = x 10 1 = 3.5 x 10 1 M [ 1.4] Pb + = (unrunded) 1.70 The spntaneus reactin (vltaic ) invlves the xidatin f C and the reductin f Ni. Oxidatin: C(s) C (aq) + e = 0.8 V Reductin: Ni (aq) + e Ni(s) = 0.5 V Ni (aq) + C(s) Ni(s) + C (aq) = = 0.5 V (0.8 V) = 0.03 V cathde ande C a) Use the Nernst equatin: = lg n = e n Ni [ 0.0 ] = 0.03 V lg [ 0.80] = 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 (There is a 1:1 mle rati) [ x] V = 0.03 V lg [ x] 0 = [ x ] lg [ x] [ x ] 0 = lg Raise each side t 10 [ x] x. [ x ] 1 = [ x] x = 0.80 x x = 0.30 M [Ni ] = = 0.50 M c) At equilibrium = 0.00, t decrease the ptential t 0.00, [C ] increases and [Ni ] decreases [ x ] 0.00 V = V lg [ x] [ x ] 0.03 V = lg [ x] [ x] Raise each side t 10 x x = lg [ ] = [ x] [ x] x = (unrunded) [C ] = = = 0.91 M [Ni ] = = = 0.09 M 1-19

20 1.71 The spntaneus reactin (vltaic ) is Cd (aq) + Mn(s) Cd(s) + Mn (aq) with = 0.40 V (1.18 V) = 0.78 V. a) Use the Nernst equatin: = n = lg Mn Cd [ 0.090] lg [ ] n = e = V = 0.77 V b) Fr the [Cd ] t decrease frm M t M, a change f M, the [Mn ] must increase by the same amunt, frm M t M [ 0.10] = 0.78 lg [ ] = V = V c) Increase the manganese an d decrease the cadmiu m by equal amunts. Ttal = M Mn = 0.78 lg Cd + ([Mn ] / [Cd ]) = x 10 4 (unrunded) [Mn ] + [Cd ] = M 6 [Cd ] = x 10 M (unrunded) [Mn ] = M [Cd ] = M d) At equilibrium = = Mn lg 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. Oxidatin: H : 0.10 M) + e (g:0.95 atm) H + (aq = 0.00 V Reductin: H + (aq:.0 M) + e H (g: 0.60 atm) = 0.00 V H + (aq:.0 M) + H (g:0.95 ) H + atm ( aq: 0.10 M) + H (g: 0.60 atm) = 0.00 V n = e Q fr the equals = 0.00 V H H ande + cathde P H(cathde) P H(ande) = ( 0.10) ( 0.60) (.0) ( 0.95) lg ( ) = = V = (unrunded) 1-0

21 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-size d battery i s 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 A g 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 Bridge supprts rust mre rapidly at the water line H O. due t the presence f large cncentratins f bth O and 1.80 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, 0.44 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 aluminu m 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) = 0.13 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) = 0.76 V. Yes, zinc is apprpriate t act as a sacrificial ande. g) (chrmium) = 0.74 V. Yes, chrmium is apprpriate t act as a sacrificial ande. 1.8 a) Oxidatin ccurs at the left electrde (ande). b) lemental M f rms 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) = 0.40 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. 1-1

22 1.85 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 ). Br (l) Br (l) + e b) At the cathde, sdium ins are reduced t frm sdium metal (Na). Na + (l) + e Na(s) 1.88 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 ptassiu m 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 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 Pssible reductins: Cu (aq) + e Cu(s) = V Ba (aq) + e Ba(s) =.90 V Al 3+ (aq) + 3 e Al(s) = 1.66 V HO(l) + e H (g) + OH (aq) = 1 V with vervltage Cpper can be prepared by electrlysis f its aqueus salt since its reductin half- ptential is mre psitive than the ptential fr the reductin f water. The reductin f cpper is mre spntaneus than the reductin f water. Since the reductin ptentials f Ba and Al 3+ are mre negative and therefre less spntaneus than the reductin f water, these ins cannt be reduced in the presence f water since the water is reduced instead. Pssible xidatins: Br (aq) Br (l) + e HO(l) O (g) + 4 H + (aq) + 4 e = V = 1. 4 V with vervltage Brmine can be prepared by electrlysis f its aqueus salt because its reductin half- ptential is mre negative than the ptential fr the xidatin f water with vervltage. The mre negative reductin ptential fr Br indicates that its xidatin is mre spntaneus than 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. 1-

23 1.95 Pssible reductins: Li + (aq) + 1 e Li(s) = 3.05 V Zn (aq) + e Zn(s) = 0.76 V Ag + (aq) + e Ag(s) = 0.80 V H O(l) + e H (g) + OH (aq) = 1 V with vervltage Zinc and silver can be prepared by electrlysis f their aqueus salt slutins since their reductin half- ptentials are mre psitive than the ptential fr the reductin f water. The reductin f zinc and silver is mre spntaneus than the reductin f water. Since the reductin ptential f Li + is mre negative and therefre less spntaneus than the reductin f water, this in cannt be reduced in the presence f water since the water is reduced instead. Pssible xidatins: I (aq) I (l) + e = V + H O(l) O (g) + 4 H (aq) + 4 e = 1.4 V with vervltage Idine can be prepared by electrlysis f its aqueus salt because its reductin half- ptential is mre negative than the ptential fr the xidatin f water with vervltage. The mre negative reductin ptential fr I indicates that its xidatin is mre spntaneus than the xidatin f water lectrlysis will separate bth irn and cadmium frm an 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 Since the reductin ptential f water is mre negative than the reductin ptential fr F, the xidatin f water is mre spntaneus than that f F. The xidatin f water prduces xygen gas (O ), and hydrnium ins (H 3 O + ) at the ande. Pssible reductins: HO(l) + e H (g) + OH (aq) = 1 V with vervltage Li + (aq) + e Li(s) = 3.05 V Since the reductin ptential f water is mre psitive than that f Li +, the reductin f water is mre spntaneus than the reductin f Li +. 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 ns (H 3 O + ), and hydrnium i ) at the ande. The SO 4 in cannt xidize as S is already in its highest xidatin state in SO 4. Pssible reductins: H O(l) + e H (g) + OH (aq) = 1 V with vervltage Sn (aq) + e Sn(s) = 0.14 V SO 4 (aq) + 4 H + (aq) + e SO (g) + H O(l) = 0.63 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 / ) l g [1 / (1 x 10 7 ) 4 ] = = 0.63 V The mst easily reduced in is Sn with the mst psitive reductin ptential, 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 hydrgen ins (H + ) and aqueus ) 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. NO 3 cannt xidize since N is in its highest xidatin state in NO 3. Pssible reductins: H O(l) + e H (g) + OH (aq) = 1 V with vervltage Cr 3+ (aq) + 3 e Cr(s) = 0.74 V NO 3 (aq) + 4 H + (aq) + 3 e NO(g) + H O(l) = V (apprximate) 1-3

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