2. 1 http:\\academicstaff.kmu.ac.ir\aliasadipour 76 slides940701

3. 2 2 Electrochemistry (Mark=2) http:\\academicstaff.kmu.ac.ir\aliasadipour 76 slides940701

4. 3 3 Electrochemistry All of Chemical reactins are related to ELECTRONS Redox reactions http:\\academicstaff.kmu.ac.ir\aliasadipour 76 slides940701

5. 4 Chemical Reactions Electric Power Power consumption Power generation Electrolytic cells Galvanic cells http:\\academicstaff.kmu.ac.ir\aliasadipour 76 slides940701

6. 5 5 Electrochemistry Conduction 1)Metalic 2)Electrolytic Temprature  Motion of ions  Resistance  -------------------------------- ----- http:\\academicstaff.kmu.ac.ir\aliasadipour 76 slides940701

7. 6 battery +- power source e-e- e-e- Ions Chemical change (-)(+) Aqueous NaCl Interionic attractions Ions Solvation Solvent viscosity Na + Cl - H2OH2O Electrolytic conduction Ion-Ion Attr. Ion- Solvent Attr. Solvent–Solvent Attr. Temprature  Attractions  & Kinetic energy  Conduction  http:\\academicstaff.kmu.ac.ir\aliasadipour 76 slides940701 Conduction ≈ Ions mobility

8. 7 battery +- inert electrodes power source vessel e-e- e-e- conductive medium Electrolytic Cell Construction http:\\academicstaff.kmu.ac.ir\aliasadipour 76 slides 940701

9. 8 +- battery Na (l) electrode half-cell Molten NaCl Na + Cl - Na + Na + + e -  Na2Cl -  Cl 2 + 2e - Cl 2 (g) escapes Observe the reactions at the electrodes NaCl (l) (-) Cl - (+) http:\\academicstaff.kmu.ac.ir\aliasadipour 76 slides940701

10. 9 +- battery e-e- e-e- NaCl (l) (-)(-)(+)(+) cathode anode Molten NaCl Na + Cl - Na + Na + + e -  Na 2Cl -  Cl 2 + 2e - cations migrate toward (-) electrode anions migrate toward (+) electrode At the microscopic level http:\\academicstaff.kmu.ac.ir\aliasadipour 76 slides940701

11. 10 Molten NaCl Electrolytic Cell cathode half-cell (-) REDUCTION Na + + e -  Na anode half-cell (+) OXIDATION2Cl -  Cl 2 + 2e - overall cell reaction 2Na + + 2Cl -  2Na + Cl 2 X 2 Non-spontaneous reaction! http:\\academicstaff.kmu.ac.ir\aliasadipour 76 slides940701

12. 11 What chemical species would be present in a vessel of aqueous sodium chloride, NaCl (aq)? Na + Cl - H2OH2O Will the half-cell reactions be the same or different? http:\\academicstaff.kmu.ac.ir\aliasadipour 76 slides940701

13. 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 (H 2 O) can be oxidized or reduced according to the reaction shown. ElectrodeIons...Anode RxnCathode Rxn E° Pt (inert)H 2 O H 2 O (l) + 2e-  H 2(g) + 2OH - (aq) -0.83 V H 2 O 2 H 2 O (l)  4e - + 4H + (g) + O 2(g) -1.23 V Net Rxn Occurring: 2 H 2 O  2 H 2(g) + O 2 (g) E° = - 2.06 V

14. 13 http:\\academicstaff.kmu.ac.ir\aliasadipour 76 slides940701

15. 14 battery +- power source e-e- e-e- NaCl (aq) (-)(+) cathode different half-cell Aqueous NaCl anode 2Cl -  Cl 2 + 2e - Na + Cl - H2OH2O What could be reduced at the cathode? http:\\academicstaff.kmu.ac.ir\aliasadipour 76 slides 2H 2 O + 2e -  H 2 + 2OH - 940701

16. 15 Aqueous NaCl Electrolysis possible cathode half-cells (-) REDUCTION Na + + e -  Na 2H 2 O + 2e -  H 2 + 2OH - possible anode half-cells (+) OXIDATION2Cl -  Cl 2 + 2e - 2H 2 O  O 2 + 4H + + 4e - overall cell reaction 2Cl - + 2H 2 O  H 2 + Cl 2 + 2OH - http:\\academicstaff.kmu.ac.ir\aliasadipour 76 slides940701

17. 16 Aqueous CuCl 2 Electrolysis possible cathode half-cells (-) REDUCTION Cu 2+ + 2e -  Cu 2H 2 O + 2e -  H 2 + 2OH - possible anode half-cells (+) OXIDATION2Cl -  Cl 2 + 2e - 2H 2 O  O 2 + 4H + + 4e - overall cell reaction Cu 2+ + 2Cl -  Cu (s) + Cl 2(g) http:\\academicstaff.kmu.ac.ir\aliasadipour 76 slides940701

18. 17 Aqueous Na 2 SO 4 Electrolysis possible cathode half-cells (-) REDUCTION Na + + e -  Na [2H 2 O + 2e -  H 2 + 2OH - ] possible anode half-cells (+) OXIDATION SO 4 2-  S 4 O 8 2_ + 2e - 2H 2 O  O 2 + 4H + + 4e - overall cell reaction 6H 2 O  2H 2 + O 2 +4H + + 4OH - http:\\academicstaff.kmu.ac.ir\aliasadipour 76 slides940701 2×

19. 18 Faraday’s Law The mass deposited or eroded from an electrode depends on the quantity of electricity. Quantity of electricity = coulomb (Q) Q = It coulomb current in amperes (amp) time in seconds http:\\academicstaff.kmu.ac.ir\aliasadipour 76 slides940701

20. 19 e-e- Ag + Ag 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 AgNO 3 (aq) under standard conditions, will deposit silver at the rate of 0.001118 g Ag/sec Experimentally: 1 amp = 0.001118 g Ag/sec http:\\academicstaff.kmu.ac.ir\aliasadipour 76 slides940701 1 coulomb = 1 amp-sec = 0.001118 g Ag Q = It

21. 20 Ag + + e -  Ag 1.00 mole Ag =1.00 mole e - = 107.87 g Ag 107.87 g Ag/mole e - 0.001118 g Ag/coul = 96,485 coul/mole e - 1 Faraday ( F ) mole e - = Q/ F http:\\academicstaff.kmu.ac.ir\aliasadipour 76 slides940701 1C=1AS /// 1J=1CV

22. 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. battery -+ +++--- 1.0 M Au +3 1.0 M Zn +2 1.0 M Ag + Au +3 + 3e -  AuZn +2 + 2e -  ZnAg + + e -  Ag e-e- e-e- e-e- e-e- http:\\academicstaff.kmu.ac.ir\aliasadipour 76 slides940701

23. 22 Examples using Faraday’s Law 1)How many grams of Cu will be deposited in 1L of A)0.1 M CuSO 4 B) 1 M CuSO 4 After 3.00 hours electrolysis by a current of 4.00 amps?(Cu=64) Cu +2 + 2e -  Cu 2)The charge on a single electron is 1.6021 x 10 -19 coulomb. Calculate Avogadro’s number from the fact that 1 F = 96,487 coulombs/mole e -. http:\\academicstaff.kmu.ac.ir\aliasadipour 76 slides940701

24. 23 http:\\academicstaff.kmu.ac.ir\aliasadipour 76 slides940701

25. 24 http:\\academicstaff.kmu.ac.ir\aliasadipour 76 slides940701

26. 25 Industrial Electrolysis Processes Slide 25 of 52 http:\\academicstaff.kmu.ac.ir\aliasadipour 76 slides940701

27. 26 http:\\academicstaff.kmu.ac.ir\aliasadipour 76 slides940701

28. 27 Volta’s battery (1800) Alessandro Volta 1745 - 1827 Paper moisturized with NaCl solution Cu Zn http:\\academicstaff.kmu.ac.ir\aliasadipour 76 slides940701

29. 28 Cu 1.0 M CuSO 4 Zn 1.0 M ZnSO 4 Salt bridge – KCl in agar Provides conduction between half-cells Galvanic Cell Construction Observe the electrodes to see what is occurring. http:\\academicstaff.kmu.ac.ir\aliasadipour 76 slides940701

30. 29 Cu 1.0 M CuSO 4 Zn 1.0 M ZnSO 4 Cu plates out or deposits on electrode Zn electrode erodes or dissolves Cu +2 + 2e -  Cu cathode half-cell Zn  Zn +2 + 2e - anode half-cell Anod - Cathod + What about half-cell reactions? What about the sign of the electrodes? What happened at each electrode? Why? http:\\academicstaff.kmu.ac.ir\aliasadipour 76 slides940701 Compare with Electrolytic cells

31. 30 +- battery e-e- e-e- NaCl (l) (-)(-)(+)(+) Cathode - Anode + Electrolytic cells sign of the electrodes? Na + Cl - Na + Na + + e -  Na 2Cl -  Cl 2 + 2e - http:\\academicstaff.kmu.ac.ir\aliasadipour 76 slides940701

32. 31 Electrodes are passive (not involved in the reaction) Olmsted Williams http:\\academicstaff.kmu.ac.ir\aliasadipour 76 slides940701

33. 32 H 2 input 1.00 atm inert metal How do we calculate Standard Redox Potentials? We need a standard electrode to make measurements against! The Standard Hydrogen Electrode (SHE) Pt 1.00 M H + 25 o C 1.00 M H + 1.00 atm H 2 Half-cell 2H + + 2e -  H 2 E o SHE = 0.0 volts http:\\academicstaff.kmu.ac.ir\aliasadipour 76 slides940701

34. 33 19.3 E 0 is for the reaction as written E 0 red // E 0 ox The more positive E 0 the greater the tendency for the substance to be reduced The half-cell reactions are reversible The sign of E 0 changes when the reaction is reversed Changing the stoichiometric coefficients of a half-cell reaction does not change the value of E 0 http:\\academicstaff.kmu.ac.ir\aliasadipour 76 slides940701 Strongest oxidunt Strongest reductant

35. 34 Measuring E 0 red Cu 2+ & Zn 2+ Slide 34 of 52 cathode anode http:\\academicstaff.kmu.ac.ir\aliasadipour 76 slides940701 Cu +2 + 2e -  Cu E=E 0 red Zn  Zn +2 + 2e - E=E 0 ox -E=E 0 red

36. 35 Cu 1.0 M CuSO 4 Zn 1.0 M ZnSO 4 cathode half-cell Cu +2 + 2e -  Cu anode half-cell Zn  Zn +2 + 2e - - + Measuring E 0 of a cell 1.1 volts http:\\academicstaff.kmu.ac.ir\aliasadipour 76 slides940701 ?

37. 36 What is the standard emf of an electrochemical cell made of a Cd electrode in a 1.0 M Cd(NO 3 ) 2 solution and a Cr electrode in a 1.0 M Cr(NO 3 ) 3 solution? Cd 2+ (aq) + 2e -  Cd (s) E 0 = -0.40 V Cr 3+ (aq) + 3e -  Cr (s) E 0 = -0.74 V Cd is the stronger oxidizer Cd will oxidize Cr 2e - + Cd 2+ (1 M) Cd (s) Cr (s) Cr 3+ (1 M) + 3e - Anode (oxidation): Cathode (reduction): 2Cr (s) + 3Cd 2+ (1 M)  3Cd (s) + 2Cr 3+ (1 M) x 2 x 3 E 0 cell = -0.40 +0.74=0.34 cell E 0 = 0.34 V cell 19.3 http:\\academicstaff.kmu.ac.ir\aliasadipour 76 slides E 0 = -0.40 V E 0 = 0.74 V 940701 E 0 cell = ? !!

38. 37 Calculating the cell potential, E o cell, at standard conditions Fe +2 + 2e -  Fe E o = -0.44 v O 2 (g) + 2H 2 O + 4e -  4 OH - E o = +0.40 v This is spontaneoues corrosion or the oxidation of a metal. Consider a drop of oxygenated water on an iron object Fe H 2 O with O 2 Fe  Fe +2 + 2e - -E o = +0.44 v2x 2Fe + O 2 (g) + 2H 2 O  2Fe(OH) 2 (s) E o cell = +0.84 v reverse http:\\academicstaff.kmu.ac.ir\aliasadipour 76 slides Fe + O 2 (g) + H 2 O  Fe(OH) 2 (s) 940701 Which one is oxidunt ?

39. 38 http:\\academicstaff.kmu.ac.ir\aliasadipour 76 slides940701

40. 39  G o = -n F E o cell Free Energy and the Cell Potential Cu  Cu +2 + 2e - E o = - 0.34 Ag + + e -  Ag E o = + 0.80 v 2x Cu + 2Ag +  Cu +2 + 2Ag E o cell = +0.46 v where n is the number of electrons for the balanced reaction What is the free energy for the cell? 1 F = 96,500 J/v http:\\academicstaff.kmu.ac.ir\aliasadipour 76 slides Cu + 2Ag +  Cu +2 + 2Ag  G o = -2×96500×0.46=-88780 J 940701

41. 40 - E depends on: -Related half reaction - Concentration -kinetic ------------------------------------------------------ 2e - +2H +  H 2 E 0 = 0.000 Fe  3e - +Fe 3+ E 0 = 0.036 ------------------------------------------ Fe +H +  Fe 3+ +H 2 E 0 = 0.036 Spontaneous redox reaction ????? !!!!!!!No =========================================================================================== http:\\academicstaff.kmu.ac.ir\aliasadipour 76 slides 0.036 V 940701 - 0.337 V

42. 41 - 0.337 V http:\\academicstaff.kmu.ac.ir\aliasadipour 76 slides e - +Cu +  Cu E 0 = 0.521 V Cu +  Cu 2+ +e - E 0 = -0.153 V ------------------------------------------- 2Cu +  Cu 2+ +Cu E 0 = 0.368V 2Cu +  Cu 2+ +Cu Auto redox=Dis proportionation 940701

43. 42 http:\\academicstaff.kmu.ac.ir\aliasadipour 76 slides 0.036 V Auto redox=Dis proportionation ?????? 2e - +Fe 2+  Fe E 0 = -0.440 V Fe 2+  Fe 3+ +e - E 0 = -0.771 V 940701 2 × ------------------------------------------- 3Fe 2+  2Fe 3+ +Fe E 0 = -1.221V NO

44. 43 http:\\academicstaff.kmu.ac.ir\aliasadipour 76 slides -0.036 V 940701 ------------------------------------------------------- 3e +Fe 3+  Fe E0=-0.331 ? No e isn’t a function state 1) e +Fe 3+  Fe 2+ E0= 0.771 2) 2e +Fe 2+  Fe E0=-0.440

45. 44  G 0 =-nE 0 f= -3E 0 f http:\\academicstaff.kmu.ac.ir\aliasadipour 76 slides940701  G 0 =-nE 0 f 2) 2e +Fe 2+  Fe E 0 =-0.440 1) e +Fe 3+  Fe 2+ E 0 = 0.771  G 0 =-1(+0.771) F=-0.771f  G 0 =-2(-0.440) F=+0.880f 3e +Fe 3+  Fe  G 0 =+0.109f ------------------------------------------------------ =+0.109f 3E 0 =-0.109 E 0 =-0.036 v

46. Free Energy and Chemical Reactions 45 W W q q ΔGΔG ΔGΔG ΔHΔH ΔHΔH TΔSTΔS TΔSTΔS Spontaneous reaction Ideal reverse cell Operating cell 940701 http:\\academicstaff.kmu.ac.ir\aliasadipour 76 slides ΔG = ΔH - T·ΔS W = ΔH - q

47. 46 http:\\academicstaff.kmu.ac.ir\aliasadipour 76 slides Ni (s) | Ni 2+ (XM) || Sn 2+ (YM) | Sn (s) A cell 2 e - + Sn 2+ → Sn (s) Ni (s) → 2 e - + Ni 2+ Ni (s) + Sn 2+ → Ni 2+ + Sn (s) Redox reaction Cathode Anode Representation of a cell 940701

48. 47 http:\\academicstaff.kmu.ac.ir\aliasadipour 76 slides Ni (s) | Ni 2+ (1M) || Sn 2+ (1M) | Sn (s) Ni (s) → 2 e - + Ni 2+ Eº =0.230 V Ni (s) + Sn 2+ (1M) → Ni 2+ (1M) + Sn (s) CathodeAnode Emf of a standard cell Eº =0.230 -0.140 =0.090V 2 e - + Sn 2+ → Sn (s) Eº=-0.140V 940701 ------------------------------------

49. 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 http:\\academicstaff.kmu.ac.ir\aliasadipour 76 slides940701 E = E o – RT ln Q nf /-nf E = E o - 0.0591 log Q n

50. 49 Effect of Concentration on Cell EMF at 25 o C: E = E o - 0.0591 log Ni 2+ / Sn 2+ n Calculate the E red for the hydrogen electrode where 0.1 M H + and 0.2 atm H 2. http:\\academicstaff.kmu.ac.ir\aliasadipour 76 slides Ni (s) | Ni 2+ (XM) || Sn 2+ (YM) | Sn (s) Ni (s) + Sn 2+ (YM) → Ni 2+ (XM) + Sn (s) Eº= 0.090 V 940701 Q= Ni 2+ / Sn 2+ E=0.090-0.059/2×logx/y E=0.000-0.059/2×logpH2/[H + ] 2 2H + +2e →H 2 Q=X/Y -------------------------------------------------------

51. 50 Ni (s) + Sn 2+ → Ni 2+ + Sn (s) Eº= 0.090 V http:\\academicstaff.kmu.ac.ir\aliasadipour 76 slides Ni (s) | Ni 2+ (0.600M) || Sn 2+ (0.300M) | Sn (s) Emf of a cell 940701

52. 51 Emf of a cell Sn (s) | Sn 2+ (1.0M) || Pb 2+ (0.0010M) | Pb (s) 2 e - + Pb 2+ → Pb (s) Eº=-0.126 V Sn (s) → 2 e - + Sn 2+ Eº=0.136V http:\\academicstaff.kmu.ac.ir\aliasadipour 76 slides E=E º -0.059/2log[Sn 2+ ]/[Pb 2+ ] E=-0.079 !!!= Reversed cell Sn (s) + Pb 2+ (0.0010M) → Sn 2+ (1.0M) + Pb (s) Eº cell =0.010 V pb (s) | pb 2+ (1.0M) || sn 2+ (0.0010M) | sn (s) 940701 E=+0.079 (Electrolytic cell) (Galvanic cell)

53. 52 equilibrium constant of a cell at equilibrium E = 0 Nernst Equation: http:\\academicstaff.kmu.ac.ir\aliasadipour 76 slides E = E o - 0.0591 log B n A 940701 A BA B

54. 53 http:\\academicstaff.kmu.ac.ir\aliasadipour 76 slides Ni (s) | Ni 2+ (0.600M) || Sn 2+ (0.300M) | Sn (s) Ni (s) + Sn 2+ → Ni 2+ + Sn (s) Eº= 0.090 V equilibrium constant of a cell 940701

55. 54 http:\\academicstaff.kmu.ac.ir\aliasadipour 76 slides940701

56. 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 http:\\academicstaff.kmu.ac.ir\aliasadipour 76 slides 940701

57. 56 Electrod potential and electrolysis E° = - 0.83 V) 2H 2 O + 2e -  H 2 + 2OH - (E° = - 0.83 V) In aqueous salts electrolysis [OH - ] =1× 10 -7 M http:\\academicstaff.kmu.ac.ir\aliasadipour 76 slides E = E° - log Q n 0.059 V E = E° - log [OH - ] 2 pH 2 2 0.059 V E = -0.83 -0.0295 log [1*10 -7 ] 2 *1=-0.417 940701

58. 57 Electrod potential and electrolysis E° = - 1.23V) 2H 2 O  O 2 + 4H + + 4e - (E° = - 1.23V) In aqueous salts electrolysis [H + ] =1× 10 -7 M http:\\academicstaff.kmu.ac.ir\aliasadipour 76 slides E = E° - log Q n 0.059 V E = E° - log [H + ] 4 pO 2 4 0.059 V E = -1.23 -0.01475 log [1*10 -7 ] 4 *1=-0.817 940701

59. 58 Effect of concentration in aqueous Na 2 SO 4 electrolysis possible cathode half-cells (-) E° = - 2.71 V) REDUCTION Na + + e -  Na (E° = - 2.71 V) E° = - 0.83 V) [2H 2 O + 2e -  H 2 + 2OH - ] (E° = - 0.83 V) E = - 0.417 V) (E = - 0.417 V) possible anode half-cells (+) E° = - 2.01 V) OXIDATION SO 4 2-  S 4 O 8 2_ + 2e - (E° = - 2.01 V) E° = - 1.23V) 2H 2 O  O 2 + 4H + + 4e - (E° = - 1.23V) E = - 0.817 V) (E = - 0.817 V) overall cell reaction 6H 2 O  2H 2 + O 2 + 4H + + 4OH - E =-0.417-0.817=-1.234 http:\\academicstaff.kmu.ac.ir\aliasadipour 76 slides940701 2 ×

60. 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 (O 2 & H 2 >Cl 2 ) http:\\academicstaff.kmu.ac.ir\aliasadipour 76 slides 940701

61. 60 Effect of overvoltage & concentration in aqueous NaCl Electrolysis possible cathode half-cells (-) E° = - 2.71 V) REDUCTION Na + + e -  Na (E° = - 2.71 V) E° = - 0.83 V) [2H 2 O + 2e -  H 2 + 2OH - ] (E° = - 0.83 V) E = - 0.417 V) (E = - 0.417 V) possible anode half-cells (+) E° = - 1.36 V) OXIDATION2Cl -  Cl 2 + 2e - (E° = - 1.36 V) E° = - 1.23V) 2H 2 O  O 2 + 4H + + 4e - (E° = - 1.23V) E = - 0.817 V) (E = - 0.817 V) OVERVOLTAGE H2 & O2 > Cl 2 overall cell reaction 2Cl - + 2H 2 O  H 2 + Cl 2 + 2OH - http:\\academicstaff.kmu.ac.ir\aliasadipour 76 slides940701

62. 61 Effect of overvoltage & concentration in aqueous CuCl 2 Electrolysis possible cathode half-cells (-) E° = +0.337V) REDUCTION Cu 2+ + 2e -  Cu (E° = +0.337V) E° = - 0.83 V) 2H 2 O + 2e -  H 2 + 2OH - (E° = - 0.83 V) E = - 0.417 V) (E = - 0.417 V) possible anode half-cells (+) E° = - 1.36 V) OXIDATION2Cl -  Cl 2 + 2e - (E° = - 1.36 V) E° = - 1.23V) 2H 2 O  O 2 + 4H + + 4e - (E° = - 1.23V) E = -0.817V) (E = -0.817V) OVERVOLTAGE H2 & O2 > Cl 2 overall cell reaction Cu 2+ + 2Cl -  Cu (s) + Cl 2(g) http:\\academicstaff.kmu.ac.ir\aliasadipour 76 slides 940701

63. 62 Cu CuSO 4 Cu Cu +2 + 2e -  CuCu  Cu +2 + 2e - What happened at each electrode? http:\\academicstaff.kmu.ac.ir\aliasadipour 76 slides battery Impure Cu pure Cu Anode + Cathode - Pure Cu deposit on cathode = (Pure cathodic Cu) What happens if aqueous CuSO 4 electrolyze between 2 Cu electrodes ?=purification of Cu 940701

64. 63 What happens if aqueous CuSO 4 electrolyze between 2 Cu electrodes ?=purification of Cu possible anode half-cells (+) (Impure Cu) E° = -0.337 V) OXIDATION Cu  Cu 2+ + 2e - (E° = -0.337 V) E° = - 1.23V) 2H 2 O  O 2 + 4H + + 4e - (E° = - 1.23V) E = - 0.817 V) (E = - 0.817 V) 2SO 4 2- E° = -2.01V ) 2SO 4 2-  S 2 O 8 2- + 2e - (E° = -2.01V ) possible cathode half-cells (-) (Purified Cu) E° = +0.337V) REDUCTION Cu 2+ + 2e -  Cu (E° = +0.337V) E° = - 0.83 V) 2H 2 O + 2e -  H 2 + 2OH - (E° = - 0.83 V) E = - 0.417 V) (E = - 0.417 V) http:\\academicstaff.kmu.ac.ir\aliasadipour 76 slides (((Purified cathodic Cu))) overall cell reaction Cu 2+ + Cu (s) Anod  Cu 2+ + Cu (s) Cathode 940701

65. 64 Cu 1.0 M CuSO 4 Cu 1.0 M CuSO 4 A galvanic cell with the similar electrods and electrolytes 0.0 http:\\academicstaff.kmu.ac.ir\aliasadipour 76 slides volts 940701

66. 65 Cu 1.0M CuSO 4 Cu 0.1 M CuSO 4 A galvanic cell with the similar electrods but different concentration electrolytes ؟؟ http:\\academicstaff.kmu.ac.ir\alias adipour 76 slides volts 940701

67. 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. http:\\academicstaff.kmu.ac.ir\aliasadipour 76 slides940701

68. http:\\academicstaff.kmu.ac.ir\aliasadipour 76 slides67 Cu │ Cu 2+ (0.1M) ‖ Cu 2+ (1.0 M) │Cu Anod cathod E=E 0 -0.059/2Log(0.1/1.0) =+0.0296 Concentration Cells Cu+Cu 2+ (1.0 M)  Cu 2+ (0.1M)+Cu 940701

69. 68 pH meter, A concentration Cell 940701 http:\\academicstaff.kmu.ac.ir\aliasadipour 76 slides Slide 68 of 52 2 H + (1 M) → 2 H + (x M) Pt | H 2 (1 atm)|H + (x M) ||H + (1.0 M) |H 2 (1 atm) | Pt(s) 2 H + (1 M) + 2 e - → H 2 (g, 1 atm) H 2 (g, 1 atm) → 2 H + (x M) + 2 e - H 2 (g, 1 atm) +2 H + (1 M) → 2 H + (x M) + H 2 (g, 1 atm)

70. 69 Slide 69 of 52 E cell = E cell ° - log n 0.059 V x2x2 1212 E cell = 0 - log 2 0.059 V x2x2 1 E cell = - 0.059 V log x E cell = (0.059 V) pH 2 H + (1 M) → 2 H + (x M) E cell = E cell ° - log Q n 0.059 V http:\\academicstaff.kmu.ac.ir\aliasadipour 76 slides pH = E cell /(0.059) 940701

71. 70 The pH Meter In practice, a special pH electrode is much more convenient than using platinum electrodes and a tank of hydrogen gas! A stable reference electrode and a glass-membrane electrode are contained within a combination pH electrode. The electrode is merely dipped into a solution, and the potential difference between the electrodes is displayed as pH. http:\\academicstaff.kmu.ac.ir\aliasadipour 76 slides940701

72. 71 Corrosion of Fe: Unwanted Voltaic Cells O2+2H 2 O+4e - →4OH - Rust formation: Fe 2+ →Fe 3+ +e E 0 =-0.771 V O 2 (g) + 4 H + (aq) + 4 e - → 2 H 2 O (aq) E 0 = 1.229 V ---------------------------------------------------------------- 4Fe 2+ (aq) + O 2 (g) + 4H + (aq)  4Fe 3+ (aq) + 2H 2 O(l) E 0 =0.458V 2Fe 3+ (aq) + 4H 2 O(l)  Fe 2 O 3  H 2 O(s) + 6H + (aq) http:\\academicstaff.kmu.ac.ir\aliasadipour 76 slides E0=0.440 VE0=1.229 V E 0 =0.401 V 940701

73. 72 Prevention of Corrosion Cover the Fe surface with a protective coating Paint Tin (Tin plate) Zn (Galvanized iron) http:\\academicstaff.kmu.ac.ir\aliasadipour 76 slides940701

74. 73 Corrosion Protection Slide 73 of 52 http:\\academicstaff.kmu.ac.ir\aliasadipour 76 slides Fe →Fe 2+ +2e E 0 =0.440 V Cu →Cu 2+ +2e E 0 =0.337 V Fe →Fe 2+ +2e E 0 =0.440 V Zn →Zn 2+ +2e E 0 =0.763 V 940701

75. 74 Corrrosion Protection (cathode) (electrolyte) (anode) http:\\academicstaff.kmu.ac.ir\aliasadipour 76 slides Fe →Fe 2+ +2e E 0 =0.440 V Mg →Mg 2+ +2e E 0 =2.363 V Steel pipe don’t rust 940701 Fe →Fe 2+ +2e E 0 =0.440 V E 0 =0.401 V

76. 75 galvanicelectrolytic need power source two electrodes produces electrical current anode (-) cathode (+) anode (+) cathode (-) salt bridge vessel conductive medium Comparison of Electrochemical Cells  G < 0  G > 0 http:\\academicstaff.kmu.ac.ir\aliasadipour 76 slides940701

77. 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. http:\\academicstaff.kmu.ac.ir\aliasadipour 76 slides940701