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2 2 Electrochemistry

3 3 Electrochemistry All of Chemical reactins are related to ELECTRONS Redox reactions Voltaic or Galvanic cells Electrochemical cells

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

5 5 Electrochemistry Conduction 1)Metalic 2)Electrolytic Temprature  Motion of ions  Resistance  1C=1AS /// 1J=1CV

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  Conduction ≈ Ions mobility

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

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 - (+)

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

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!

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?

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) V H 2 O 2 H 2 O (l)  4e - + 4H + (g) + O 2(g) V Net Rxn Occurring: 2 H 2 O  2 H 2(g) + O 2 (g) E° = V

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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? 2H 2 O + 2e -  H 2 + 2OH

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 -

16 Aqueous CuCl 2 Electrolysis possible cathode half-cells (-) REDUCTION Cu e -  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 Cl -  Cu (s) + Cl 2(g)

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 - 2×

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

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 g Ag/sec 1 amp = g Ag/sec 1 coulomb = 1 amp-sec = g Ag

20 Ag + + e -  Ag 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

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 M Au M Zn M Ag + Au e -  AuZn e -  ZnAg + + e -  Ag e-e- e-e- e-e- e-e-

22 Examples using Faraday’s Law 1)How many grams of Cu will be deposited in 3.00 hours 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 -.

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Industrial Electrolysis Processes Slide 25 of 52

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

28 Galvanic Cells 19.2 spontaneous redox reaction anode oxidation cathode reduction

29 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.

30 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 e -  Cu cathode half-cell Zn  Zn e - anode half-cell Anod - Cathod + What about half-cell reactions? What about the sign of the electrodes? What happened at each electrode? Why? Compare with Electrolytic cells

31 +- 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 -

32 Electrodes are passive (not involved in the reaction) Olmsted Williams

33 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 atm H 2 Half-cell 2H + + 2e -  H 2 E o SHE = 0.0 volts

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

35 Cell EMF Oxidizing and Reducing Agents

36 Measuring E 0 red Cu 2+ & Zn 2+ Slide 36 of 52 cathode anode Cu e -  Cu E=E 0 red Zn  Zn e - E=E 0 ox -E=E 0 red

37 Cu 1.0 M CuSO 4 Zn 1.0 M ZnSO 4 cathode half-cell Cu e -  Cu anode half-cell Zn  Zn e Measuring E 0 of a cell 1.1 volts

38 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 = V Cr 3+ (aq) + 3e -  Cr (s) E 0 = 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 = =0.34 cell E 0 = 0.34 V cell E 0 = V E 0 = 0.74 V

39 Calculating the cell potential, E o cell, at standard conditions Fe e -  Fe E o = v O 2 (g) + 2H 2 O + 4e -  4 OH - E o = 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 e - -E o = v2x 2Fe + O 2 (g) + 2H 2 O  2Fe(OH) 2 (s) E o cell = v reverse Fe + O 2 (g) + H 2 O  Fe(OH) 2 (s)

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41  G o = -n F E o cell Free Energy and the Cell Potential Cu  Cu e - E o = Ag + + e -  Ag E o = v 2x Cu + 2Ag +  Cu Ag E o cell = 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 Cu + 2Ag +  Cu Ag  G o = -2×96500×0.46= J

42 - E depends on: -Related half reaction - Concentration -kinetic e - +2H +  H 2 E 0 = Fe  3e - +Fe 3+ E 0 = Fe +H +  Fe 3+ +H 2 E 0 = Spontaneous redox reaction ????? !!!!!!!No =========================================================================================== V

V e - +Cu +  Cu E 0 = V Cu +  Cu 2+ +e - E 0 = V Cu +  Cu 2+ +Cu E 0 = 0.368V 2Cu +  Cu 2+ +Cu Auto redox=Dis proportionation

44http:\\aliasadipour.kmu.ac.ir V Auto redox=Dis proportionation ?????? 2e - +Fe 2+  Fe E 0 = V Fe 2+  Fe 3+ +e - E 0 = V × Fe 2+  2Fe 3+ +Fe E 0 = V NO

45http:\\aliasadipour.kmu.ac.ir V e +Fe 3+  Fe E0= ? No e isn’t a function state 1) e +Fe 3+  Fe 2+ E0= ) 2e +Fe 2+  Fe E0= e - +Fe 2+  Fe E 0 = V Fe 2+  Fe 3+ +e - E 0 = V Fe 2+  2Fe 3+ +Fe E 0 = V

46  G 0 =-nE 0 f= -3E 0 f  G 0 =-nE 0 f 2) 2e +Fe 2+  Fe E 0 = ) e +Fe 3+  Fe 2+ E 0 =  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 = E 0 = v

Free Energy and Chemical Reactions 47 W W q q ΔGΔG ΔGΔG ΔHΔH ΔHΔH TΔSTΔS TΔSTΔS Spontaneous reaction Ideal reverse cell Operating cell http:\\aliasadipour.kmu.ac.ir ΔG = ΔH - T·ΔS W = ΔH - q

48http:\\aliasadipour.kmu.ac.ir 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

49http:\\aliasadipour.kmu.ac.ir 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 = =0.090V 2 e - + Sn 2+ → Sn (s) Eº=-0.140V

50 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 E = E o – RT ln Q n /-nf E = E o log Q n

51 Effect of Concentration on Cell EMF at 25 o C: E = E o log Ni 2+ / Sn 2+ n Calculate the E red for the hydrogen electrode where 0.50 M H + and 0.95 atm H 2. Ni (s) | Ni 2+ (XM) || Sn 2+ (YM) | Sn (s) Ni (s) + Sn 2+ (YM) → Ni 2+ (XM) + Sn (s) Eº= V Q= Ni 2+ / Sn 2+ E= /2×logx/y E= /2×logpH2/[H + ] 2 2H + +2e →H 2 Q=X/Y

52 Ni (s) + Sn 2+ → Ni 2+ + Sn (s) Eº= V Ni (s) | Ni 2+ (0.600M) || Sn 2+ (0.300M) | Sn (s) Emf of a cell

53 Emf of a cell Sn (s) | Sn 2+ (1.0M) || Pb 2+ (0.0010M) | Pb (s) 2 e - + Pb 2+ → Pb (s) Eº= V Sn (s) → 2 e - + Sn 2+ Eº=0.136V E=E º /2log[Sn 2+ ]/[Pb 2+ ] E= !!!= 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) E= (Electrolytic cell) (Galvanic cell)

54 equilibrium constant of a cell at equilibrium E = 0 Nernst Equation: E = E o log B n A A BA B

55http:\\aliasadipour.kmu.ac.ir Ni (s) | Ni 2+ (0.600M) || Sn 2+ (0.300M) | Sn (s) Ni (s) + Sn 2+ → Ni 2+ + Sn (s) Eº= V equilibrium constant of a cell

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57 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

58 Electrod potential and electrolysis E° = V) 2H 2 O + 2e -  H 2 + 2OH - (E° = V) In aqueous salts electrolysis [OH - ] =1× M E = E° - log Q n V E = E° - log [OH - ] 2 pH V E = log [1*10 -7 ] 2 *1=

59 Electrod potential and electrolysis E° = V) 2H 2 O  O 2 + 4H + + 4e - (E° = V) In aqueous salts electrolysis [H + ] =1× M E = E° - log Q n V E = E° - log [H + ] 4 pO V E = log [1*10 -7 ] 4 *1=

60 Effect of concentration in aqueous Na 2 SO 4 electrolysis possible cathode half-cells (-) E° = V) REDUCTION Na + + e -  Na (E° = V) E° = V) [2H 2 O + 2e -  H 2 + 2OH - ] (E° = V) E = V) (E = V) possible anode half-cells (+) E° = V) OXIDATION SO 4 2-  S 4 O 8 2_ + 2e - (E° = V) E° = V) 2H 2 O  O 2 + 4H + + 4e - (E° = V) E = V) (E = V) overall cell reaction 6H 2 O  2H 2 + O 2 + 4H + + 4OH - E = = ×

61 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 )

62 Effect of overvoltage & concentration in aqueous NaCl Electrolysis possible cathode half-cells (-) E° = V) REDUCTION Na + + e -  Na (E° = V) E° = V) [2H 2 O + 2e -  H 2 + 2OH - ] (E° = V) E = V) (E = V) possible anode half-cells (+) E° = V) OXIDATION2Cl -  Cl 2 + 2e - (E° = V) E° = V) 2H 2 O  O 2 + 4H + + 4e - (E° = V) E = V) (E = V) OVERVOLTAGE H2 & O2 > Cl 2 overall cell reaction 2Cl - + 2H 2 O  H 2 + Cl 2 + 2OH -

63 Effect of overvoltage & concentration in aqueous CuCl 2 Electrolysis possible cathode half-cells (-) E° = V) REDUCTION Cu e -  Cu (E° = V) E° = V) 2H 2 O + 2e -  H 2 + 2OH - (E° = V) E = V) (E = V) possible anode half-cells (+) E° = V) OXIDATION2Cl -  Cl 2 + 2e - (E° = V) E° = V) 2H 2 O  O 2 + 4H + + 4e - (E° = V) E = V) (E = V) OVERVOLTAGE H2 & O2 > Cl 2 overall cell reaction Cu Cl -  Cu (s) + Cl 2(g)

64 Cu CuSO 4 Cu Cu e -  CuCu  Cu e - What happened at each electrode? 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

65 What happens if aqueous CuSO 4 electrolyze between 2 Cu electrodes ?=purification of Cu possible anode half-cells (+) (Impure Cu) E° = V) OXIDATION Cu  Cu e - (E° = V) E° = V) 2H 2 O  O 2 + 4H + + 4e - (E° = V) E = V) (E = V) 2SO 4 2- E° = -2.01V ) 2SO 4 2-  S 2 O e - (E° = -2.01V ) possible cathode half-cells (-) (Purified Cu) E° = V) REDUCTION Cu e -  Cu (E° = V) E° = V) 2H 2 O + 2e -  H 2 + 2OH - (E° = V) E = V) (E = V) (((Purified cathodic Cu))) overall cell reaction Cu 2+ + Cu (s) Anod  Cu 2+ + Cu (s) Cathode

66 Cu 1.0 M CuSO 4 Cu 1.0 M CuSO 4 A cell with the similar electrods and electrolytes volts

67 Cu 1.0 M CuSO 4 Cu 1.0 M CuSO 4 A cell with the similar electrods but different concentration electrolytes ؟؟ volts

68 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.

Cu │ Cu 2+ (0.1M) ‖ Cu 2+ (1.0 M) │Cu Anod cathod E=E /2Log(0.1/1.0) = Concentration Cells Cu+Cu 2+ (1.0 M)  Cu 2+ (0.1M)+Cu

70 pH meter, A concentration Cell http:\\aliasadipour.kmu.ac.ir Slide 70 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)

71 Slide 71 of 52 E cell = E cell ° - log n V x2x E cell = 0 - log V x2x2 1 E cell = V log x E cell = (0.059 V) pH 2 H + (1 M) → 2 H + (x M) E cell = E cell ° - log Q n V pH = E cell /(0.059)

72 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.

73 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

74 Corrosion of Fe: Unwanted Voltaic Cells O2+2H 2 O+4e - →4OH - Rust formation: Fe 2+ →Fe 3+ +e E 0 = V O 2 (g) + 4 H + (aq) + 4 e - → 4 H 2 O (aq) E 0 = V Fe 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) E0=0.440 VE0=1.229 V E 0 =0.401 V

75 Prevention of Corrosion Cover the Fe surface with a protective coating Paint Tin (Tin plate) Zn (Galvanized iron)

76 Corrosion Protection Slide 76 of 52 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

77 Corrrosion Protection (cathode) (electrolyte) (anode) Fe →Fe 2+ +2e E 0 =0.440 V Mg →Mg 2+ +2e E 0 =2.363 V Steel pipe don’t rust Fe →Fe 2+ +2e E 0 =0.440 V

78 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.