1、 Design of electrochemical cell; Prediction of cell voltage and outcome; Protect of metal materials from corrosion; Fundamentals and Concepts Chemical energy Electrical energy Chapter 21 ElectrochemistryCu (s) + 2Ag+ (aq, 1M) 2 Ag(s) + Cu2+ (aq, 1M)Cu (s) = Cu2+(aq,1M) + 2e2Ag+(aq,1M) + 2e = 2Ag(s)H

2、alf-cell reactionHalf-cell reaction Cu (s) + 2Ag+(aq, 1M) 2 Ag(s) + Cu2+(aq,1M)Net reaction+- Voltaic cell or Daniel CellCell diagramCu (s) | Cu2+ (1 M) | Ag+ (1 M) | Ag(s)Ecell = 0.457 VCell Potential (Voltage) Chemical energyElectric energyLemon juiceZn-Cu stripsMn+-+ + + + + + + + +Model of Doubl

3、e LayerElectrode PotentialMThe more active the metal, the more the negative charges on the surface, and thus the larger the electrode potential.Mn+MStandard Electrode PotentialsZn (s) |Zn2+ (1 M) | H+ (1 M) |H2 (1 atm) |Pt (s) Zn (s) + 2H+ (1 M) Zn2+ + H2 (1 atm)E0 = 0.76 VcellE0 = EH+/H - EZn /Zn c

4、ell002+2E = E+ - E-cell0.76 V = 0 - EZn /Zn 02+EZn /Zn = -0.76 V02+2e- + 2H+ (1 M) H2(1 atm)Zn (s) Zn2+(1 M) + 2e-Standard Electrode PotentialsDefine:Standard hydrogen electrodeZn2+ (1 M) + 2e- ZnGas (1 atm), Solute (1M)Standard Electrode PotentialsPt (s) | H2 (1 atm) | H+ (1 M) | Cu2+ (1 M) | Cu (s

5、)2e- + Cu2+ (1 M) Cu (s)H2 (1 atm) 2H+ (1 M) + 2e-H2 (1 atm) + Cu2+ (1 M) Cu (s) + 2H+ (1 M)E0 = 0.34 VcellEcell = ECu /Cu EH /H 2+20000.34 = ECu /Cu - 002+ECu /Cu = 0.34 V2+0Cu2+(1M)+2e- CuAll E0 listed are in the form of Mn+ + ne = M The more positive E0, the greater the tendency for Mn+ to be red

6、uced. The more positive E0, the greater the tendency for M to be oxidized. Under standard conditions, the left table shows that F2(g) is the strongest oxidant, and Li(s) is the strongest reductant.E0 is pH-dependent. EM /M = ? Vn+02H+(aq) + 2e = H2(g) EA0 = 0.00 V 2H2O(aq) + 2e = H2(g) + 2OH-(aq) EB

7、0 = -0.83 V What is the standard cell potential of an electrochemical cell made of a Cd electrode in a 1.0 M Cd(NO3)2 solution and a Cr electrode in a 1.0 M Cr(NO3)3 solution?Cd2+ (aq) + 2e- Cd (s) E0 = -0.40 VCr3+ (aq) + 3e- Cr (s) E0 = -0.74 V2e- + Cd2+ (1 M) Cd (s)Cr (s) Cr3+ (1 M) + 3e-2Cr (s) +

8、 3Cd2+ (1 M) 3Cd (s) + 2Cr3+ (1 M)x 2x 3E0 = -0.40 (-0.74) cellE0 = 0.34 V cellCr (s) | Cr3+ (1 M) | Cd2+ (1 M) | Cd(s)Cell diagram ?Redox reaction described by thermodynamicsaA + bB cC + dDDG = DGo + RT ln Q DG0 = - RT lnKn = number of moles of electrons passed through the circuit.Faraday constant

9、F = 96,500 C/mol (C: Coulomb)DG = - n F EcellDG = - n F EcellDG0 = -n F E0cell- n F Ecell = - n F E 0cell + RTln QNernst equationAt 25 oC Ecell = E 0cell -The more positive Ecell, the greater the electricity produced by the reaction.The more negative DG, the greater the tendency for the reaction to

10、occur.Electric work gained by surroundings Ecell = E 0cell -lnK =Will the following reaction occur spontaneously at 250C if Fe2+ = 0.60 M and Cd2+ = 0.010 M? Fe2+ (aq) + Cd (s) Fe (s) + Cd2+ (aq)2e- + Fe2+ 2FeCd Cd2+ + 2e-E0 = -0.44 (-0.40) E0 = EFe /Fe ECd /Cd002+2+-0.05916nlog QE0E = -0.059162log

11、-0.04 VE = 0.0100.60SpontaneousAt standard conditions, the reaction can not happen spontaneously. 0Note that the initial voltage is 0.013 V.But it would decrease with time, until Ecell = 0 V (discharge).At this time, the equilibrium is reached.lnK =K=e-3.1=0.04= -0.04 V= 0.013 VAgCl (s) Ag+ (aq) + C

12、l- (aq)The constants Ka, Kb, Ksp, Kf measured via an electrochemical cell. Ksp = Ag+Cl-Ag (s) | Ag+(1M) | Cl- (1 M) | AgCl (s) | Ag (s) DG = DGo + RT ln Q DG0 = - RT ln Ksp Ecell = E 0cell -lnKsp =E0cell = E0AgCl/Ag - E0Ag+/Ag= 0.2223 -0.7996= -0.5773 VKsp = 1.75 10-10DGo = DHo T DSoAg(s) + AgCl(s)

13、= Ag(s) + Cl-(aq) + Ag+(aq)Cathod (reduction)Anode (oxidation)Batteries19.6Leclanch cellDry cellZn (s) Zn2+ (aq) + 2e-Anode: Cathode:2NH4 (aq) + 2MnO2 (s) + 2e- Mn2O3 (s) + 2NH3 (aq) + H2O (l)+Zn (s) + 2NH4 (aq) + 2MnO2 (s) Zn2+ (aq) + 2NH3 (aq) + H2O (l) + Mn2O3 (s)BatteriesZn(Hg) + 2OH- (aq) ZnO (

14、s) + H2O (l) + 2e-Anode: Cathode:HgO (s) + H2O (l) + 2e- Hg (l) + 2OH- (aq)Zn(Hg) + HgO (s) ZnO (s) + Hg (l)Mercury Battery19.6BatteriesAnode: Cathode:Lead storagebatteryPbO2 (s) + 4H+ (aq) + SO2- (aq) + 2e- PbSO4 (s) + 2H2O (l)4Pb (s) + SO2- (aq) PbSO4 (s) + 2e-4Pb (s) + PbO2 (s) + 4H+ (aq) + 2SO2-

15、 (aq) 2PbSO4 (s) + 2H2O (l)4Sulfonic acidInitial 38% (d=1.29)Final 26%(d=1.19)Batteries19.6Solid State Lithium BatteryA fuel cellAnode: Cathode:O2 (g) + 2H2O (l) + 4e- 4OH- (aq)2H2 (g) + 4OH- (aq) 4H2O (l) + 4e-2H2 (g) + O2 (g) 2H2O (l)EoH2/H2O = -0.83 V EoO2/OH- = 0.40V Eocell = 1.23 V Each year about 20 % of iron production in the world is lost due to corrosion.CorrosionCathodic Protection o