Slide 1“minimal” galvanic cellsElectrolytic CellsSlide 4Slide 5Slide 6Electrolysis of waterSlide 8Slide 9Slide 10ElectroplatingSlide 12Slide 13Slide 14Slide 15Slide 16Slide 17Hour Exam IIWednesday, March 157:00 – 9:00 pm103 MumfordHall AQG, AQIAllen AQJBlair AQF150 Animal ScienceFisher AQB, AQCKoys AGDPearson AQAconflict exam4:30 – 6:30162 NoyesReview session7:00 – 9:00, Monday124 Burrill Hall“minimal” galvanic cellsonly need reactants ZnH+o = 0.34 VZn2+(aq) + 2e-  Zn(s) o = -0.76 V2H+(aq) + 2e-  H2(g) o = 0.00 Vreductionb) oxidationZn  Zn2+ + 2e-  = 0.76 V2H+ (aq)0 = .76 V  =very largeQ =a) oxidationCu Cu2+(aq) + 2e-  Cu(s)+ Zn (s) Zn2+ (aq) + H2(g)0Electrolytic Cellscell > 0cell < 0G < 0spontaneousgalvanic cellG > 0non-spontaneous electrolytic cell2 H2(g)G = -474 kJredox reaction:OspontaneousH 00+ O2(g) 2 H2O(l) 1+ 2-Electrolytic Cells2 H2O(l)G = 474 kJoxygen half-cell: H2OH2O  O2a) oxidationb) reductiona) anodeb) cathode2 + 4 H++ 4 e- O2 reaction 2 H2(g)+ O2(g)Electrolytic Cells2 H2O(l)  2 H2(g) + O2(g) G = 474 kJhydrogen half-cell:H2O  H2 H2Oreduction reaction cathode+ 2e- H2+ OH-22Electrolytic Cells2 H2O2 H2Ooxidation:anodereduction:cathode2( )____________________________________6 H2O 2 H2O  O2+ 2 H2+ 4H++ 4 OH-O2+ 2 H2 O2+ 4 H++ 4 e-+ 2e- H2+ 2 OH-Electrolysis of wateroxidationreduction2H2O  O2+ 4H++ 4e-4H2O + 4e- 2H2 + 4OH-Pt electrodesbattery+-e-e-anodecathodeacidbase1 mol gas2 mol gasElectrolysis of water2.5 amp Power source current =charge mol e-molproductgramproductcurrent and timeA(C/s) xsx 1mol e- 96,500 Cx mol product mol e- x g product mol product3.2 g O2amperes (A)= coulombs/sec (C/s)Electrolysis of watercharge mol e-molproductgramproductcurrent and time2 H2O  O2 + 4 H+ + 4 e-2.5 A, 3.2 g O2 (C/s) x s2.5 A1432.0 g/mol3.2x mol e- Cx mol O2mol e-mol O2x g O2= g O21 mol e-96500 CElectrolysis of water2 H2O  O2 + 4 H+ + 4 e-2.5 A, 3.2 g O23.2 g O2 x2.5 C x s15440 s x1 mol O2 x32 g O24 mol e- x1 mol O296500 C =1 mol e-38600 C = 38600 C s = 15440 s1 min x 60 s 1 hr60 min= 4.3 hrsElectroplatingCu2+ + 2e-  Cuanode cathodeoxidation reductionCu(s) Cu2+(aq) Cu2+(aq)+ 2e-+ 2e- Cu(s)o = 0.34 VElectroplatinganode cathodeoxidationreductionCu(s)  Cu2+(aq) + 2e-Cu2+(aq) + 2e- Cu(s)0.75 Aatomic mass of Cu =for 25 min. deposits 0.37 g CuElectroplatingCu2+(aq) + 2e- Cu(s)0.75 A for 25 min deposits 0.37 g CuA (C) sx s x 1 mol e- 96500 Cx 1 mol Cu 2 mol e-x g Cu mol CuElectroplating0.75(C) sx 25 min 0.37 g Cu= 1.125 x 103 C 1.125 x 103 Cx 1 mol e- 96500 C= 1.17 x 10-2 mol e-1.17 x 10-2 mol e-x 1mol Cu 2mol e-=5.83 x 10-3mol Cu= 63.5 g mol5.8x10-3 mol Cuatomic mass Cux 60s minElectrolysis of salt solutionsCuBr2Cu2+Br-+2e-  Cu2 Br2 + 2e-0.153-1.083-.934H2O+2e-  H2+2OH--0.832H2O  O2+2H++4e--1.23-1.91-1.08-2.06Electrolysis of salt solutionsCaI2Ca2+I-+2e-  Ca2 I2 + 2e--2.86-0.535-3.40H2O+2e-  H2+2OH--0.832H2O  O2+2H++4e--1.23-1.36-4.09-2.06Electrolysis of salt solutionsNaClNa+Cl-+ e-  Na2 Cl2 + 2e--2.71-1.36-4.07H2O+2e-  H2+2OH--0.832H2O 