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1 920311 76 slides

2 Electrochemistry 2 920311 76 slides

3 All of Chemical reactins are related to
Electrochemistry All of Chemical reactins are related to ELECTRONS Redox reactions 3 920311 76 slides

4 Electric power conversion in electrochemistry
Power consumption Electrolysis Electric Power Chemical Reactions Galvanic cells Power generation 920311 76 slides

5 Electrochemistry Conduction 1)Metalic 2)Electrolytic
Temprature Motion of ions  Resistance  ----- ----- 5 920311 76 slides

6 Electrolytic conduction
- + battery power source Electrolytic conduction e- Ions Chemical change e- Aqueous NaCl Conduction ≈ Ions mobility Interionic attractions Ions Solvation …………………………………………. Solvent viscosity …………………………………….. Ion-Ion Attr. Ion- Solvent Attr. Solvent–Solvent Attr. Na+ Cl- Temprature Attractions & Kinetic energy Conduction (-) (+) H2O 920311 76 slides

7 - + Electrolytic Cell Construction power battery source e- e-
conductive medium vessel inert electrodes 920311 76 slides

8 Observe the reactions at the electrodes
Molten NaCl Observe the reactions at the electrodes - + battery Cl2 (g) escapes Na (l) NaCl (l) Na+ Cl- Na+ Cl- (-) (+) electrode half-cell electrode half-cell Cl- Na+ Na+ + e-  Na 2Cl-  Cl e- 920311 76 slides

9 At the microscopic level
Molten NaCl At the microscopic level - + battery e- NaCl (l) cations migrate toward (-) electrode anions migrate toward (+) electrode Na+ Cl- Na+ e- Cl- (-) (+) anode cathode Cl- Na+ 2Cl-  Cl e- Na+ + e-  Na 920311 76 slides

10 Molten NaCl Electrolytic Cell
cathode half-cell (-) REDUCTION Na+ + e-  Na anode half-cell (+) OXIDATION 2Cl-  Cl e- overall cell reaction 2Na Cl-  2Na Cl2 X 2 Non-spontaneous reaction! 920311 76 slides

11 Will the half-cell reactions be the same or different?
What chemical species would be present in a vessel of aqueous sodium chloride, NaCl (aq)? Na+ Cl- H2O Will the half-cell reactions be the same or different? 920311 76 slides

12 Water Complications in Electrolysis
In an electrolysis, the most easily oxidized and most easily reduced reaction occurs. When water is present in an electrolysis reaction, then water (H2O) can be oxidized or reduced according to the reaction shown. Electrode Ions ... Anode Rxn Cathode Rxn E° Pt (inert) H2O H2O(l)+ 2e- g H2(g)+ 2OH-(aq) V H2O H2O(l) g 4e H+(g) + O2(g) V Net Rxn Occurring: 2 H2O g 2 H2(g)+ O2 (g) E° = V

13 CHM 102 Sinex 920311 76 slides

14 What could be reduced at the cathode?
- + Aqueous NaCl battery power source e- e- 2H2O + 2e-  H OH- NaCl (aq) What could be reduced at the cathode? Na+ Cl- (-) (+) H2O cathode different half-cell anode 2Cl-  Cl e- 920311 76 slides

15 Aqueous NaCl Electrolysis
possible cathode half-cells (-) REDUCTION Na+ + e-  Na 2H2O + 2e-  H OH- possible anode half-cells (+) OXIDATION2Cl-  Cl e- 2H2O  O H e- overall cell reaction 2Cl H2O  H2 + Cl OH- 920311 76 slides

16 Aqueous CuCl2 Electrolysis
possible cathode half-cells (-) REDUCTION Cu e-  Cu 2H2O + 2e-  H OH- possible anode half-cells (+) OXIDATION2Cl-  Cl e- 2H2O  O H e- overall cell reaction Cu Cl-  Cu(s) + Cl2(g) 920311 76 slides

17 Aqueous Na2SO4 Electrolysis
possible cathode half-cells (-) REDUCTION Na+ + e-  Na [2H2O + 2e-  H OH- ] possible anode half-cells (+) OXIDATION SO42-  S4O82_ + 2e- 2H2O  O H e- overall cell reaction 6H2O  2H2 + O2 +4H+ + 4OH- 920311 76 slides

18 Faraday’s Law Q = It Quantity of electricity = coulomb (Q)
The mass deposited or eroded from an electrode depends on the quantity of electricity. Q = It time in seconds coulomb current in amperes (amp) 920311 76 slides

19 1 coulomb = 1 amp-sec = 0.001118 g Ag e-
Experimentally: 1 amp = g Ag/sec For every electron, an atom of silver is plated on the electrode. Ag e-  Ag Electrical current is expressed in terms of the ampere, which is defined as that strength of current which, when passed thru a solution of AgNO3 (aq) under standard conditions, will deposit silver at the rate of g Ag/sec Ag+ Ag 920311 76 slides

20 Ag+ + e-  Ag 1 Faraday (F ) mole e- = Q/F
1.00 mole e- = 1.00 mole Ag = g Ag g Ag/mole e- g Ag/coul = 96,485 coul/mole e- 1 Faraday (F ) mole e- = Q/F 1C=1AS /// 1J=1CV 920311 76 slides

21 A series of solutions have 50,000 coulombs passed thru them, if the solutions were Au+3, Zn+2, and Ag+, and Au, Zn, and Ag were plated out respectively, calculate the amount of metal deposited at each anode. e- - + battery - + - + - + e- e- e- 1.0 M Au+3 1.0 M Zn+2 1.0 M Ag+ Au e-  Au Zn e-  Zn Ag+ + e-  Ag 920311 76 slides

22 Examples using Faraday’s Law
1)How many grams of Cu will be deposited in 1L of A)0.1 M CuSO4 B) 1 M CuSO4 After 3.00 hours electrolysis by a current of 4.00 amps?(Cu=64) Cu e-  Cu 2)The charge on a single electron is x coulomb. Calculate Avogadro’s number from the fact that 1 F = 96,487 coulombs/mole e-. 920311 76 slides

23 920311 76 slides

24 920311 76 slides

25 21-8 Industrial Electrolysis Processes
920311 76 slides Slide 25 of 52

26 CHM 102 Sinex 920311 76 slides

27 Paper moisturized with NaCl solution
Volta’s battery (1800) Alessandro Volta Paper moisturized with NaCl solution Cu Zn 920311 76 slides

28 Observe the electrodes to see what is occurring.
Galvanic Cell Construction Salt bridge – KCl in agar Provides conduction between half-cells Observe the electrodes to see what is occurring. Cu Zn 1.0 M CuSO4 1.0 M ZnSO4 920311 76 slides

29 Anod - Cathod + What about half-cell reactions?
What about the sign of the electrodes? Anod - Cathod + Why? Compare with Electrolytic cells Cu e-  Cu cathode half-cell Zn  Zn e- anode half-cell Cu plates out or deposits on electrode Zn electrode erodes or dissolves What happened at each electrode? Cu Zn 1.0 M CuSO4 1.0 M ZnSO4 920311 76 slides

30 Electrolytic cells sign of the electrodes?
- + battery e- NaCl (l) Na+ Cl- Na+ e- Cl- (-) (+) Anode + Cathode - Cl- Na+ 2Cl-  Cl e- Na+ + e-  Na 920311 76 slides

31 Electrodes are passive (not involved in the reaction)
Olmsted Williams Electrodes are passive (not involved in the reaction) 920311 76 slides

32 How do we calculate Standard Redox Potentials?
We need a standard electrode to make measurements against! The Standard Hydrogen Electrode (SHE) H2 input 1.00 atm 25oC 1.00 M H+ 1.00 atm H2 Pt Half-cell 2H e-  H2 inert metal EoSHE = 0.0 volts 1.00 M H+ 920311 76 slides

33 The half-cell reactions are reversible
E0 is for the reaction as written E0red // E0ox The more positive E0 the greater the tendency for the substance to be reduced The half-cell reactions are reversible The sign of E0 changes when the reaction is reversed Changing the stoichiometric coefficients of a half-cell reaction does not change the value of E0 Strongest oxidunt Strongest reductant 920311 76 slides 19.3

34 Measuring E0red Cu2+& Zn2+
-E=E0red anode cathode cathode anode Cu e-  Cu E=E0red Zn  Zn e- E=E0ox 920311 76 slides Slide 34 of 52

35 - + Measuring E0 of a cell ? 1.1 volts cathode half-cell
Cu e-  Cu anode half-cell Zn  Zn e- Cu Zn 1.0 M CuSO4 1.0 M ZnSO4 920311 76 slides

36 Cd will oxidize Cr What is the standard emf 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 = V Cr3+ (aq) + 3e-  Cr (s) E0 = V 2e- + Cd2+ (1 M) Cd (s) x 3 Cathode (reduction): E0 = V Cd is the stronger oxidizer Cd will oxidize Cr x 2 E0 = V Cr (s) Cr3+ (1 M) + 3e- Anode (oxidation): 2Cr (s) + 3Cd2+ (1 M)  Cd (s) + 2Cr3+ (1 M) E0 cell = ? !! E0 cell = =0.34 cell E0 = 0.34 V cell 920311 76 slides 19.3

37 Calculating the cell potential, Eocell, at standard conditions
H2O with O2 Consider a drop of oxygenated water on an iron object Fe Fe + O2 (g) + H2O  Fe(OH)2 (s) Fe e-  Fe Eo = v reverse 2x Which one is oxidunt? Fe  Fe e- -Eo = v O2 (g) H2O e-  4 OH- Eo = v 2Fe + O2 (g) H2O  2Fe(OH)2 (s) Eocell= v This is spontaneoues corrosion or the oxidation of a metal. 920311 76 slides

38 920311 76 slides

39 Free Energy and the Cell Potential
Cu Ag+  Cu Ag Cu  Cu e- Eo = 2x Ag+ + e-  Ag Eo = v Cu Ag+  Cu Ag Eocell= v DGo = -nFEocell 1F = 96,500 J/v where n is the number of electrons for the balanced reaction What is the free energy for the cell? DGo = -2×96500×0.46= J 76 slides 920311

40 -E depends on: Related half reaction -Concentration kinetic e- +2H+  H E0 = Fe  3e- +Fe E0 = Fe +H+ Fe3+ +H2 E0 = Spontaneous redox reaction ????? !!!!!!!No =========================================================================================== V 0.036 V 920311 76 slides

41 Auto redox=Dis proportionation
V 2Cu+  Cu2++Cu Auto redox=Dis proportionation e- +Cu+  Cu E0 = V Cu+  Cu2++e E0 = V 2Cu+  Cu2++Cu E0 = 0.368V 920311 76 slides

42 Auto redox=Dis proportionation ?????? NO
0.036 V Auto redox=Dis proportionation ?????? NO 2e- +Fe2+  Fe E0 = V Fe2+  Fe3++e E0 = V 2 × 3Fe2+  2Fe3++Fe E0 = V 920311 76 slides

43 -------------------------------------------
1) e +Fe3+  Fe2+ E0= 0.771 2) 2e +Fe2+  Fe E0=-0.440 3e +Fe3+  Fe E0= ? No e isn’t a function state 2e- +Fe2+  Fe E0 = V Fe2+  Fe3++e E0 = V 3Fe2+  2Fe3++Fe E0 = V 920311 76 slides V

44 ------------------------------------------------------
G0 =-nE0f 1) e +Fe3+  Fe2+ E0= 0.771 G0=-1(+0.771) F=-0.771f 2) 2e +Fe2+  Fe E0=-0.440 G0=-2(-0.440) F=+0.880f 3e +Fe3+  Fe G0=+0.109f G0 =-nE0f= -3E0f =+0.109f 3E0= E0= v 920311 76 slides

45 Free Energy and Chemical Reactions
W = ΔH - q ΔG = ΔH - T·ΔS q ΔH TΔS ΔG W Ideal reverse cell Operating cell Spontaneous reaction 920311 76 slides

46 Representation of a cell
Ni(s) + Sn2+→ Ni2+ + Sn(s) Redox reaction 2 e- + Sn2+ → Sn(s) Ni(s) →2 e- + Ni2+ Ni(s) | Ni2+(XM) || Sn2+(YM) | Sn(s) A cell Cathode Anode 920311 76 slides

47 Emf of a standard cell Ni(s) + Sn2+(1M)→ Ni2+(1M) + Sn(s)
Ni(s) | Ni2+(1M)|| Sn2+(1M) | Sn(s) Anode Cathode Ni(s) →2 e- + Ni2+ Eº =0.230 V 2 e- + Sn2+ → Sn(s) Eº=-0.140V Eº = =0.090V 920311 76 slides

48 Effect of Concentration on Cell EMF
A voltaic cell is functional until E = 0 at which point equilibrium has been reached. The point at which E = 0 is determined by the concentrations of the species involved in the redox reaction. The Nernst Equation /-nf E = Eo – RT ln Q n E = Eo log Q n 48 920311 76 slides

49 Effect of Concentration on Cell EMF
Ni(s) | Ni2+ (XM) || Sn2+ (YM) | Sn(s) Q= Ni2+/ Sn2+ Ni(s) + Sn2+ (YM) → Ni2+ (XM) + Sn(s) Eº= V at 25oC: E = Eo log Ni2+/ Sn n Q=X/Y E= /2×logx/y Calculate the Ered for the hydrogen electrode where 0.50 M H+ and 0.95 atm H2. 2H++2e →H2 E= /2×logpH2/[H+]2 920311 76 slides

50 Emf of a cell Ni(s) | Ni2+(0.600M)|| Sn2+(0.300M) | Sn(s)
Ni(s) + Sn2+→ Ni2+ + Sn(s) Eº= V 920311 76 slides

51 E=Eº-0.059/2log[Sn2+]/[Pb2+] E=-0.079 !!!= Reversed cell E=+0.079
Emf of a cell Sn(s) | Sn2+(1.0M)|| Pb2+(0.0010M) | Pb(s) Sn(s) + Pb2+(0.0010M) → Sn2+(1.0M) + Pb(s) Sn(s) → 2 e- + Sn2+ Eº=0.136V 2 e- + Pb2+ → Pb(s) Eº= V Eºcell=0.010 V E=Eº-0.059/2log[Sn2+]/[Pb2+] E= !!!= Reversed cell E=+0.079 (Electrolytic cell) pb(s) | pb2+(1.0M)|| sn2+(0.0010M) | sn(s) (Galvanic cell) 920311 76 slides

52 equilibrium constant of a cell
A B Nernst Equation: E = Eo log B n A at equilibrium E = 0 920311 76 slides

53 equilibrium constant of a cell
Ni(s) | Ni2+(0.600M)|| Sn2+(0.300M) | Sn(s) Ni(s) + Sn2+→ Ni2+ + Sn(s) Eº= V 920311 76 slides

54 920311 76 slides

55 Electrod potential and electrolysis
Theoritical emf of a Voltaic cell is maximum voltage. (Practically is less) Theoritical emf of an electrolysis cell is minimum voltage. (Practically is more) Emf is related to: Resistance Concentration Overvoltage 76 slides 920311

56 Electrod potential and electrolysis
2H2O + 2e-  H OH- (E° = V) In aqueous salts electrolysis [OH-] =1× 10-7M E = E° log Q n 0.059 V E = E° log [OH-] 2pH2 2 0.059 V E = log [1*10-7] 2*1=-0.417 76 slides 920311

57 Electrod potential and electrolysis
2H2O  O H e- (E° = V) In aqueous salts electrolysis [H+] =1× 10-7M E = E° log Q n 0.059 V E = E° log [H+] 4pO2 4 0.059 V E = log [1*10-7] 4*1=-0.817 76 slides 920311

58 Effect of concentration in aqueous Na2SO4 electrolysis
possible cathode half-cells (-) REDUCTION Na+ + e-  Na (E° = V) [2H2O + 2e-  H OH- ] (E° = V) (E = V) possible anode half-cells (+) OXIDATION SO42-  S4O82_ + 2e- (E° = V) 2H2O  O H e (E° = V) (E = V) overall cell reaction 6H2O  2H2 + O H+ + 4OH- E= =-1.234 2 × 920311 76 slides

59 Electrod potential and electrolysis
Overvoltage(OV): (Because of slow rate of reaction) OV of deposition of metals are low OV of liberation of gases are appreciable (O2 & H2 >Cl2) 76 slides 920311

60 Effect of overvoltage & concentration in aqueous NaCl Electrolysis
possible cathode half-cells (-) REDUCTION Na+ + e-  Na (E° = V) [2H2O + 2e-  H OH- ] (E° = V) (E = V) possible anode half-cells (+) OXIDATION2Cl-  Cl e (E° = V) 2H2O  O H e (E° = V) (E = V) OVERVOLTAGE H2 & O2 > Cl2 overall cell reaction 2Cl H2O  H2 + Cl OH- 920311 76 slides

61 Effect of overvoltage & concentration in aqueous CuCl2 Electrolysis
possible cathode half-cells (-) REDUCTION Cu e-  Cu (E° = V) 2H2O + 2e-  H OH- (E° = V) (E = V) possible anode half-cells (+) OXIDATION2Cl-  Cl e (E° = V) 2H2O  O H e (E° = V) (E = V) OVERVOLTAGE H2 & O2 > Cl2 overall cell reaction Cu Cl-  Cu(s) + Cl2(g) 920311 76 slides

62 What happens if aqueous CuSO4 electrolyze between 2 Cu electrodes
What happens if aqueous CuSO4 electrolyze between 2 Cu electrodes ?=purification of Cu Cathode - Anode + battery Cu What happened at each electrode? Cu e-  Cu Cu  Cu e- CuSO4 Cu Cu Impure Cu pure Cu Pure Cu deposit on cathode =(Pure cathodic Cu) 920311 76 slides

63 Cu2+ + Cu(s) Anod  Cu2+ + Cu(s) Cathode (((Purified cathodic Cu)))
What happens if aqueous CuSO4 electrolyze between 2 Cu electrodes ?=purification of Cu possible anode half-cells (+) (Impure Cu) OXIDATION Cu  Cu e (E° = V) 2H2O  O H e (E° = V) (E = V) 2SO42-  S2O e (E° = -2.01V) possible cathode half-cells (-) (Purified Cu) REDUCTION Cu e-  Cu (E° = V) 2H2O + 2e-  H OH- (E° = V) (E = V) overall cell reaction Cu2+ + Cu(s) Anod  Cu2+ + Cu(s) Cathode (((Purified cathodic Cu))) 920311 76 slides

64 electrods and electrolytes
A cell with the similar electrods and electrolytes 0.0 volts Cu Cu 1.0 M CuSO4 1.0 M CuSO4 920311 76 slides

65 A cell with the similar electrods but different concentration electrolytes
؟؟ volts Cu Cu 1.0M CuSO4 0.1 M CuSO4 920311 76 slides

66 Electrolysis of Copper
Concentration Cells A concentration cell based on the Cu/Cu2+ half-reaction. A, Even though the half-reactions involve the same components, the cell operates because the half-cell concentrations are different. B, The cell operates spontaneously until the half-cell concentrations are equal. Note the change in electrodes (exaggerated here for clarity) and the equal color of solutions. 920311 76 slides

67 Cu│Cu2+ (0.1M)‖ Cu2+ (1.0 M)│Cu Anod cathod
Concentration Cells Cu│Cu2+ (0.1M)‖ Cu2+ (1.0 M)│Cu Anod cathod E=E /2Log(0.1/1.0) = Cu+Cu2+ (1.0 M)Cu2+ (0.1M)+Cu 920311 76 slides

68 pH meter, A concentration Cell
Chemistry 140 Fall 2002 pH meter, A concentration Cell Pt | H2 (1 atm)|H+(x M) ||H+(1.0 M) |H2(1 atm) | Pt(s) H2(g, 1 atm) → 2 H+(x M) + 2 e- 2 H+(1 M) + 2 e- → H2(g, 1 atm) H2(g, 1 atm) +2 H+(1 M) → 2 H+(x M) + H2(g, 1 atm) 2 H+(1 M) → 2 H+(x M) 920311 76 slides Slide 68 of 52

69 pH = Ecell /(0.059) Ecell = Ecell° - log Q n 0.059 V
2 H+(1 M) → 2 H+(x M) Ecell = Ecell° log n 0.059 V x2 12 Ecell = log 2 0.059 V x2 1 Ecell = V log x Ecell = (0.059 V) pH pH = Ecell /(0.059) 920311 76 slides Slide 69 of 52

70 The pH Meter In practice, a special pH electrode is much more convenient than using platinum electrodes and a tank of hydrogen gas! The electrode is merely dipped into a solution, and the potential difference between the electrodes is displayed as pH. A stable reference electrode and a glass-membrane electrode are contained within a combination pH electrode. 920311 76 slides

71 Corrosion of Fe: Unwanted Voltaic Cells
E0=0.440 V E0=1.229 V O2+2H2O+4e-→4OH- E0=0.401 V Rust formation: Fe2+→Fe3+ +e E0= V O2(g) + 4 H+(aq) + 4 e- → 4 H2O (aq) E0 = V 4Fe2+(aq) + O2(g) + 4H+(aq)  4Fe3+(aq) + 2H2O(l) E0 =0.458V 2Fe3+(aq) + 4H2O(l)  Fe2O3·H2O(s) + 6H+(aq) 920311 76 slides

72 Prevention of Corrosion
Cover the Fe surface with a protective coating Paint Tin (Tin plate) Zn (Galvanized iron) 920311 76 slides

73 Corrosion Protection Fe→Fe2+ +2e E0=0.440 V Cu →Cu2+ +2e E0=0.337 V
Fe →Fe2+ +2e E0=0.440 V Zn →Zn2+ +2e E0=0.763 V 920311 76 slides Slide 73 of 52

74 Corrrosion Protection
(cathode) Steel pipe don’t rust E0=0.401 V (electrolyte) Fe →Fe2+ +2e E0=0.440 V Fe →Fe2+ +2e E0=0.440 V Mg→Mg2+ +2e E0=2.363 V (anode) 920311 76 slides

75 Comparison of Electrochemical Cells
galvanic electrolytic need power source produces electrical current two electrodes anode (-) cathode (+) anode (+) cathode (-) conductive medium DG < 0 DG > 0 vessel salt bridge 920311 76 slides

76 Cathodic Protection In cathodic protection, an iron object to be protected is connected to a chunk of an active metal. The iron serves as the reduction electrode and remains metallic. The active metal is oxidized. Water heaters often employ a magnesium anode for cathodic protection. 920311 76 slides


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