无机化学课件：14 Ch19 Electrochemistry

1、Chemistry: Atoms FirstJulia Burdge & Jason OverbyChapter 19ElectrochemistryElectrochemistry1919.2 Galvanic Cells19.3 Standard Reduction Potentials19.4 Spontaneity of Redox Reactions Under Standard-State Conditions19.5 Spontaneity of Redox Reactions Under Conditions Other Than Standard StateThe Nerns

2、t EquationConcentration Cells19.6 BatteriesDry Cells and Alkaline BatteriesLead Storage BatteriesLithium-ion BatteriesFuel Cells19.7 ElectrolysisElectrolysis of Molten Sodium ChlorideElectrolysis of WaterElectrolysis of an Aqueous Sodium Chloride SolutionQuantitative Applications of Electrolysis19.8

3、 CorrosionStatic Electricity Happens EverywhereLightningTriboelectricity 摩擦电Atoms bind electrons slightly differently.If you rub any two neutral different materials together, some electrons may transfer from one to the other resulting in oppositely charged objects.The amount of charge is generally s

4、mall - a nanocoulomb or so.Tribo means rubbingTriboelectricityTriboelectric SeriesZn(s) + Cu2+(aq) Zn2+(aq) + Cu(s)Galvanic Cells19.2When zinc metal is placed in a copper(II) solution, Zn is oxidized and Cu2+ ions are reduced.Galvanic CellsThe experimental apparatus for generating electricity throug

5、h the use of a spontaneous reaction is called a galvanic cell.Galvanic CellsElectric current flows from anode to cathode because there is a difference in electrical potential energy between the electrodes.The electrical potential is measured by a voltmeter and is called the cell potential (Ecell ).C

6、ell notation:Zn(s) Zn2+ (1 M) Cu2+ (1 M) Cu(s)anodecathodeNotation for a Voltaic Cellcomponents of anode compartment(oxidation half-cell)components of cathode compartment(reduction half-cell)phase of lower oxidation statephase of higher oxidation statephase of higher oxidation statephase of lower ox

7、idation statephase boundary between half-cellsExamples:Zn(s) | Zn2+(aq) | Cu2+(aq) | Cu (s)Zn(s) Zn2+(aq) + 2e-Cu2+(aq) + 2e- Cu(s) graphite | I-(aq) | I2(s) | H+(aq), MnO4-(aq), Mn2+(aq) | graphiteinert electrodeStandard Reduction PotentialsWhen the concentrations of Zn2+ and Cu2+ are 1 molar at 25

8、C, the cell voltage is 1.10 V.19.3Zn(s) + Cu2+(aq) Zn2+(aq) + Cu(s)Voltages of Some Voltaic CellsVoltaic CellVoltage (V)Common alkaline batteryLead-acid car battery (6 cells = 12V)Calculator battery (mercury)Electric eel (5000 cells in 6-ft eel = 750V)Nerve of giant squid (across cell membrane)2.01.

9、51.30.150.070Metal SurfaceSolutionElectrode Potential 电极电位Potential Difference here and forms a Bilayer 双电层Electrode potential depends on the properties of the metal, temperature, solution characteristics and concentration.Agar-agar 琼脂 + KNO3A voltaic cell based on the zinc-copper reaction.Oxidation

10、 half-reactionZn(s) Zn2+(aq) + 2e-Reduction half-reactionCu2+(aq) + 2e- Cu(s)Overall (cell) reactionZn(s) + Cu2+(aq) Zn2+(aq) + Cu(s)A voltaic cell using inactive electrodes.Reduction half-reactionMnO4-(aq) + 8H+(aq) + 5e- Mn2+(aq) + 4H2O(l)Oxidation half-reaction2I-(aq) I2(s) + 2e-Overall (cell) re

11、action2MnO4-(aq) + 16H+(aq) + 10I-(aq) 2Mn2+(aq) + 5I2(s) + 8H2O(l)Standard Reduction PotentialsThe hydrogen electrode is used as the arbitrary standard in measuring standard cell potentials.H2 2H+ + 2eE = 0 VStandard Reduction PotentialsZn(s) Zn2+ (1 M) H+ (1 M) H2(g) Pt(s)Zn(s) Zn2+ (1 M) + 2e2H+(

12、1 M) + 2e H2 (1 atm)Zn(s) + 2H+(1 M) Zn2+ (1 M) H2 (1 atm)Anode (oxidation)Cathode (reduction)Overall:Standard Reduction PotentialsZn(s) Zn2+ (1 M) H+ (1 M) H2(g) Pt(s)Zn(s) Zn2+ (1 M) + 2e2H+(1 M) + 2e H2 (1 atm)Zn(s) + 2H+(1 M) Zn2+ (1 M) + H2 (1 atm)Anode (oxidation)Cathode (reduction)Overall:Ece

13、ll = Ecathode Eanode Ecell = EH+ /H2 EZn2+/Zn 0.76 V = 0 EZn2+/Zn EZn2+/Zn = 0.76 VStandard Reduction PotentialsPt(s) H2(g) H+ (1 M) Cu2+ (1 M) Cu(s)H2(1 atm) 2H+ (1 M) + 2eCu2+ (1 M) + 2e Cu (s)H2(1 atm) + Cu2+ (1 M) 2H+ (1 M) + Cu(s)Anode (oxidation)Cathode (reduction)Overall:Standard Reduction Po

14、tentialsPt(s) H2(g) H+ (1 M) Cu2+ (1 M) Cu(s)H2(1 atm) 2H+ (1 M) + 2eCu2+ (1 M) + 2e Cu (s)H2(1 atm) + Cu2+ (1 M) 2H+ (1 M) + Cu(s)Anode (oxidation)Cathode (reduction)Overall:Ecell = Ecathode Eanode Ecell = ECu2+/Cu EH+/H2 0.34 V = 0 EH+/H2 EH+/H2 = 0.34 VStandard Reduction PotentialsCu2+ (1 M) + 2e

15、 Cu (s)Zn(s) + Cu2+ (1 M) Zn2+ (1 M) + Cu(s)Anode (oxidation)Cathode (reduction)Overall:Ecell = Ecathode Eanode Ecell = ECu2+ /Cu EZn2+/Zn Ecell = 0.34 V (0.76 V) Ecell = 1.10 VZn(s) Zn2+ (1 M) Cu2+ (1 M) Cu(s)Zn(s) Zn2+ (1 M) + 2eStandard Reduction PotentialsStandard Reduction Potentials at 25C (Se

16、e Table 19.1)Half-ReactionE(V)Cl2 (l) + 2e 2Cl(aq)1.36Br2 (l) + 2e 2Br(aq)1.07Ag+ (aq) + e Ag(s)0.80Cu2+ (aq) + 2e Cu(s)0.342H+(aq) + 2e H2(g)0.00Pb2+ (aq) + 2e Pb(s)0.13Cd2+ (aq) + 2e Cd(s)0.40Zn2+ (aq) + 2e Zn(s)0.76Mn2+ (aq) + 2e Mn(s)1.66Increasing strength as reducing agentsIncreasing strength

17、as oxidizing agentsStandard Reduction Potentials2Ag+ (1 M) + 2e 2Ag(s)Zn(s) + 2Ag+ (1 M) Zn2+ (1 M) + 2Ag(s)Anode (oxidation)Cathode (reduction)Overall:Zn(s) Zn2+ (1 M) Ag+ (1 M) Ag(s)Zn(s) Zn2+ (1 M) + 2eHalf-ReactionE(V)Ag+ (aq) + e Ag(s)0.80Cu2+ (aq) + 2e Cu(s)0.342H+(aq) + 2e H2(g)0.00Zn2+ (aq)

18、+ 2e Zn(s)0.76Mn2+ (aq) + 2e Mn(s)1.66Ecell = EAg+/Ag EZn2+/Zn = 0.80 V (0.76 V) = 1.56 VDetermine the overall cell reaction and Ecell at 25C of a galvanic cell made of a Cd electrode in a 1.0 M Cd(NO3)2 solution and a Pb electrode in a 1.0 M Pb(NO3)2 solution.SolutionStep 1:Use a standard reduction

19、 potential table to determine which electrode is the cathode and which is the anode.Pb/Pb2+ is the cathode; Cd/Cd2+ is the anode Standard Reduction PotentialsHalf-ReactionE(V)Ag+ (aq) + e Ag(s)0.80Cu2+ (aq) + 2e Cu(s)0.342H+(aq) + 2e H2(g)0.00Pb2+ (aq) + 2e Pb(s)0.13Cd2+ (aq) + 2e Cd(s)0.40Standard

20、Reduction PotentialsDetermine the overall cell reaction and Ecell at 25C of a galvanic cell made of a Cd electrode in a 1.0 M Cd(NO3)2 solution and a Pb electrode in a 1.0 M Pb(NO3)2 solution.SolutionStep 2:Use the equation below to calculate Ecell.Half-ReactionE(V)Pb2+ (aq) + 2e Pb(s)0.13Cd2+ (aq)

21、+ 2e Cd(s)0.40Ecell = Ecathode Eanode Ecell = EPb2+/Pb ECd2+/Cd Ecell = 0.13 V (0.40 V) = 0.27 V Worked Example 19.2Strategy Use the tabulated values of E to determine which electrode is the cathode and which is the anode, combine cathode and anode half-cell reactions to get the overall cell reactio

22、n, and use Ecell = Ecathode Eanode to calculate Ecell.A galvanic cell consists of an Mg electrode in a 1.0 M Mg(NO3)2 solution and a Cd electrode in a 1.0 M Cd(NO3)2 solution. Determine the overall cell reaction, and calculate the standard cell potential at 25C.Solution The half-cell reactions and t

23、heir standard reduction potentials areMg2+ + 2e- Mg E = 2.37 VCd2+ + 2e- Cd E = 0.40 VBecause the Cd half-cell reaction has the greater (less negative) standard reduction potential, it will occur as the reduction. The Mg half-cell reaction will occur as the oxidation. Therefore, Ecathode = 0.40 V an

24、d Eanode = 2.37 V. Worked Example 19.2 (cont.)Solution Adding the two half-cell reactions together gives the overall cell reaction:The standard cell potential isMg Mg2+ + 2e-Cd2+ + 2e- CdOverall:Mg + Cd2+ Mg2+ + CdEcell = Ecathode Eanode = ECd2+/Cd EMg2+/Mg = (0.40 V) (2.37 V)= 1.97 VThink About It

25、If you ever calculate a negative voltage for a galvanic cell potential, you have done something wrongcheck your work. Under standard-state conditions, the overall cell reaction will proceed in the direction that gives a positive Ecell. Worked Example 19.3Strategy In each case, write the equation fro

26、m the redox reaction that might take place and use E values to determine whether or not the proposed reaction will actually occur.Predict what redox reaction will take place, if any, when molecular bromine (Br2) is added to (a) a 1-M solution of NaI and (b) a 1-M solution of NaCl. (Assume a temperat

27、ure of 25C.)Solution From Table 19.1:Br2(l) + 2e- 2Br-(aq) E = 1.07 VI2(l) + 2e- 2I-(aq) E = 0.53 VCl2(l) + 2e- 2Cl-(aq) E = 1.36 V Worked Example 19.3 (cont.)Solution (a) If a redox reaction is to occur, it will be oxidation of I- ions by Br2:Br2(l) + 2I-(aq) 2Br-(aq) + I2(s)Because the reduction p

28、otential of Br2 is greater than that of I2, Br2 will be reduced to Br- and I- will be oxidized to I2. Thus, the preceding reaction will occur.(b) In this case, the proposed reaction is the reduction of Br2 by Cl- ions:Br2(l) + 2Cl-(aq) 2Br-(aq) + Cl2(g)Because the reduction potential of Br2 is great

29、er than that of I2, Br2 will be reduced to Br- and I- will be oxidized to I2. Thus, the preceding reaction will occur.Think About It We can use Ecell = Ecathode Eanode and treat problems of this type like galvanic cell problems. Write the proposed redox reaction, and identify the “cathode” and the “

30、anode.” If the calculated Ecell is positive, the reaction will occur. If the calculated Ecell is negative, the reaction will not occur.Spontaneity of Redox Reactions Under Standard-State ConditionsEcell is related to the thermodynamic quantities G and K.n is the number of moles electronsF is the Far

31、aday constant (96,500 C/mol e)R is the gas constant (8.314 J/mol K)19.4G = nFEcell Spontaneity of Redox Reactions Under Standard-State ConditionsBy converting to the base-10 logarithm of K, we getEcell =0.0592Vnlog K (at 25C)Spontaneity of Redox Reactions Under Standard-State ConditionsCalculate G a

32、nd K for the following reaction at 25C:3Mg(s) + 2Al3+(aq) 3Mg2+(aq) + 2Al(s)SolutionStep 1:Use E to calculate Ecell.Half-ReactionE(V)Al3+ (aq) + 3e Al(s)1.66Mg2+ (aq) + 2e Mg(s)2.37Ecell = EAl3+/Al EMg2+/Mg Ecell = 1.66 V (2.37 V) = 0.71 VSpontaneity of Redox Reactions Under Standard-State Condition

33、sCalculate G and K for the following reaction at 25C:3Mg(s) + 2Al3+(aq) 3Mg2+(aq) + 2Al(s)SolutionStep 2:Use the equation below to calculate G :G = (6 e)(96500 J/Vmol e )(0.71 V)G = 411090 J or 411 kJG = nFEcellSpontaneity of Redox Reactions Under Standard-State ConditionsCalculate G and K for the f

34、ollowing reaction at 25C:3Mg(s) + 2Al3+(aq) 3Mg2+(aq) + 2Al(s)SolutionStep 3:Use the equation below to calculate K :K = 8 x 10165 Worked Example 19.4Strategy Use E values from Table 19.1 to calculate E for the reaction, and then use G = nFEcell to calculate the standard free-energy change.Calculate

35、the standard free-energy change for the following reaction at 25C:2Au(s) + 3Ca2+(1.0 M) 2Au3+(1.0 M) + 3Ca(s)Solution The half-cell reactions areFrom Table 19.1, ECa2+/Ca = 2.87 V and EAu3+/Au = 1.50 V.Cathode (reduction):3Ca2+(aq) + 6e- Ca(s)Anode (oxidation): 2Au(s) 2Au3+(aq) + 6e-Ecell = Ecathode

36、 Eanode = ECa2+/Ca EAu3+/Au= 2.87 V 1.50 V= 4.37 V Worked Example 19.4 (cont.)Solution Next, substitute this value of E into G = nFEcell to obtain G. The overall reaction show that n = 6, soThink About It The large positive value of G indicates that reactants are favored at equilibrium, which is con

37、sistent with the fact that E for the reaction is negative.G = nFEcell = (6 mol e-)(96,500 J/Vmol e-)(4.37 V)= 2.53106 J/mol= 2.53103 kJ/mol Worked Example 19.5Strategy Use E values from Table 19.1 to calculate E for the reaction, and then calculate the equilibrium constant using Ecell = (0.0592 V/n)

38、log K (rearranged to solve for K.)Calculate the equilibrium constant for the following reaction at 25C:Sn(s) + 2Cu+(aq) Sn2+(aq) + 2Cu+(aq)Solution The half-cell reactions areFrom Table 19.1, ECu2+/Cu+ = 0.15 V and ESn2+/Sn = 0.14 V.Cathode (reduction):2Cu2+(aq) + 2e- 2Cu+(aq)Anode (oxidation): Sn(s

39、) Sn2+(aq) + 2e-Ecell = Ecathode Eanode = ECu2+/Cu+ ESn2+/Sn= 0.15 V (0.14 V)= 0.29 V Worked Example 19.5 (cont.)Solution Solving Ecell = (0.0592 V/n)log K for K givesThink About It A positive standard cell potential corresponds to a large equilibrium constant.K = 10= 10= 6109nE/0.0592 V(2)(0.29 V)/

40、0.0592 VSpontaneity of Redox Reactions Under Conditions Other Than Standard StateUnder non-standard conditions, cell potential is calculated with the Nernst equation:When Q 1, E EWhen Q E19.5Spontaneity of Redox Reactions Under Conditions Other Than Standard StateWill the following reaction occur sp

41、ontaneously at 298 K if Fe2+ = 0.60 M and Cd2+ = 0.010 M?Cd(s) + Fe2+(aq) Cd2+(aq) + Fe(s)SolutionStep 1:Calculate E using data from a standard reduction potential table.Half-ReactionE(V)Cd2+ (aq) + 2e Cd(s)0.40Fe2+ (aq) + 2e Fe(s)0.44Ecell = EFe2+/Fe ECd2+/Cd Ecell = 0.44 V (0.40 V) = 0.04 VSpontan

42、eity of Redox Reactions Under Conditions Other Than Standard StateWill the following reaction occur spontaneously at 298 K if Fe2+ = 0.60 M and Cd2+ = 0.010 M?Cd(s) + Fe2+(aq) Cd2+(aq) + Fe(s)SolutionStep 2:Use the equation below to calculate E. If E is positive, the reaction remains spontaneous. Wo

43、rked Example 19.6Strategy Use E values from Table 19.1 to determine E for the reaction, and use E = E (0.0592 V/n)log Q to calculate E. If E is positive, the reaction will occur spontaneously.Predict whether the following reaction will occur spontaneously as written at 298 K:Co(s) + Fe2+(aq) Co2+(aq

44、) + Fe(s)assuming Co2+ = 0.15 M and Fe2+ = 0.68 M.Solution The half-cell reactions areCathode (reduction):Fe2+(aq) + 2e- Fe(s)Anode (oxidation): Co(s) Co2+(aq) + 2e- Worked Example 19.6 (cont.)Solution The reaction quotient, Q, for the reaction is Co2+Fe2+. Therefore, Q = (0.15/0.68) = 0.22.The nega

45、tive E value indicates that the reaction is not spontaneous as written under the conditions describe.Ecell = Ecathode Eanode = EFe2+/Fe ECo2+/Co= 0.44 V (0.28 V)= 0.16 VE = E 0.0592 Vnlog Q= 0.16 V 0.0592 V2log 0.22= 0.14 V Worked Example 19.6 (cont.)Think About It For this reaction to be spontaneou

46、s as written, the ratio of Fe2+ to Co2+ would have to be enormous. We can determine the required ratio by first setting E equal to zero:For E to be positive, therefore, the ratio of Fe2+ to Co2+, the reciprocal of Q, would have to be greater than 3105 to 1.0 V = 0.16 V 0.0592 Vnlog Q = log Q(0.16 V)

47、(2)0.0592 Vlog Q = 5.4Q = 105.4 = = 410-6Co2+Fe2+Spontaneity of Redox Reactions Under Conditions Other Than Standard StateA concentration cell is one with two half-cells containing the same components, but differing ion concentrations.Zn(s) Zn2+ (0.10 M) Zn2+ (1.0 M) Zn(s)Spontaneity of Redox Reacti

48、ons Under Conditions Other Than Standard StateA concentration cell is one with two half-cells containing the same components, but differing ion concentrations.Zn(s) Zn2+ (0.10 M) Zn2+ (1.0 M) Zn(s)Zn2+ (1.0 M) + 2e Zn (s)Zn2+ (1.0 M) Zn2+ (0.10 M)Anode (oxidation)Cathode (reduction)Overall:Zn(s) Zn2

49、+ (0.10 M) + 2e Worked Example 19.7Strategy Use E = E (0.0592 V/n)log Q to solve for the unknown concentration of silver ion. The half-cell with the higher Ag+ (1.0 M AgNO3) concentration will be the cathode; the half-cell with the lower, unknown Ag+ concentration (saturated AgCN solution) will be t

50、he anode. The overall reaction is Ag+(0.10 M) Ag+(x M).An electrochemical cell is constructed for the purpose of determining Ksp of silver cyanide (AgCN) at 25C. One half-cell consists of a silver electrode in a 1.00-M solution of silver nitrate. The other half-cell consists of a silver electrode in

51、 a saturated solution of silver cyanide. The cell potential is measured and found to be 0.470 V. Determine the concentration of silver ion in the saturated silver cyanide solution and the value of Ksp for AgCN. Worked Example 19.7 (cont.)Solution Because this is a concentration cell, Ecell = 0 V. Th

52、e reaction quotient, Q, is (x M)/(1.00 M); and the value of n is 1.Therefore, Ag+ = 1.1510-8 M and Ksp for AgCN = x2 = 1.310-16.Ecell = Ecell 0.0592 Vnlog Q0.470 V = 0 V 0.0592 V1logx1.007.939 =logx1.00107.939 = 1.1510-8 =x1.00 x = 1.1510-8 Think About It Remember that in a saturated solution of a s

53、alt that dissociates into two ions, the ion concentrations are equal to each other and each ion concentration is equal to the square root of Ksp.BatteriesDry Cell Batteries No fluid component Zinc container (anode) Graphite cathode 1.5 V19.62 NH4+(aq) + 2MnO2(s) + 2e Mn2O3(s) + 2NH3(aq) + H2O(l)Zn(s

54、) + 2NH4+(aq) + 2MnO2(s) Zn2+(aq)+ Mn2O3(s) + 2NH3(aq) + H2O(l)Anode Cathode Overall:Zn(s) Zn2+ (aq) + 2eBatteriesAlkaline Batteries Basic medium2MnO2(s) + 2H2O(l) + 2e 2MnO(OH)(s) + 2OH(aq)Zn(s) + 2H2O(l) + 2MnO2(s) Zn(OH)2(s)+ 2MnO(OH)(s)Anode Cathode Overall:Zn(s) + 2OH(aq) Zn(OH)2(s) + 2eBatteri

55、esLead Storage Batteries 6 cells 2 V per cell 12 V total RechargeablePbO2(s) + 4H+(aq) + SO42(aq) + 2e PbSO4(s) + 2H2O(l)Pb(s) + PbO2(s) + 4H+(aq) + 2SO42(aq) 2PbSO4(s) + 2H2O(l)Anode Cathode Overall:Pb(s) + 2SO42(aq) PbSO4(s) + 2eBatteriesLithium ion batteries Cell potential: 3.4 V Rechargeable (ma

56、ny times)Li+(aq) + CoO2 + e LiCoO2(s)Li+(s) + CoO2 LiCoO2(s) Anode Cathode Overall:Li(s) Li+ + eBatteriesFuel Cells Cell potential: 1.23 V 70% efficientO2 (aq) + 2H2O(l) + 4e 4OH(aq)2H2(g) + O2(g) 2H2O(l) Anode Cathode Overall:2H2(g) + 4 OH(aq) 4H2O(l) + 4eElectrolysisThe use of electric energy to d

57、rive a non-spontaneous chemical reaction is called electrolysis.An electrolytic cell is used.19.7Cell TypeChemical ReactionElectric EnergyGalvanicSpontaneousProducedElectrolyticNonspontaneousConsumedEnergy is absorbed to drive a nonspontaneous redox reactionGeneral characteristics of voltaic and ele

58、ctrolytic cells.VOLTAIC CELLELECTROLYTIC CELLEnergy is released from spontaneous redox reactionReduction half-reactionY+ e- YOxidation half-reactionX X+ + e-System does work on its surroundingsReduction half-reactionB+ e- BOxidation half-reactionA- A + e-Surroundings(power supply)do work on system(c

59、ell)Overall (cell) reactionX + Y+ X+ + Y; DG 0Comparison of Voltaic and Electrolytic CellsCell TypeDGEcellElectrodeNameProcessSignVoltaicVoltaicElectrolyticElectrolytic 0 0 0 0 0 0 0AnodeAnodeCathodeCathodeOxidationOxidationReductionReduction-+ElectrolysisElectrolysis of molten sodium chloride; Ecel

60、l = 4 V2Na+(l) + 2e 2Na(l)2Na+(l) + 2Cl(l) 2Na(l) + Cl2(g) Anode Cathode Overall:2Cl(l) Cl2(g) + 2eElectrolysisElectrolysis of water; G = 474.4 kJ/mol4H+(aq) + 4e 2H2(g)2H2O(l) O2(g) + 2H2(g) Anode Cathode Overall:2H2O(l) O2(g) + 4H+(aq) + 4eElectrolysisFaraday developed the quantitative treatment o