Presentation is loading. Please wait.

Presentation is loading. Please wait.

1http:\\asadipour.kmu.ac.ir 76 slides920311 2 2 Electrochemistry 76 slides920311.

Similar presentations


Presentation on theme: "1http:\\asadipour.kmu.ac.ir 76 slides920311 2 2 Electrochemistry 76 slides920311."— Presentation transcript:

1

2 1http:\\asadipour.kmu.ac.ir 76 slides920311

3 2 2 Electrochemistry http:\\asadipour.kmu.ac.ir 76 slides920311

4 3 3 Electrochemistry All of Chemical reactins are related to ELECTRONS Redox reactions http:\\asadipour.kmu.ac.ir 76 slides920311

5 4 Electric power conversion in electrochemistry Chemical Reactions Electric Power Power consumption Power generation Electrolysis Galvanic cells http:\\asadipour.kmu.ac.ir 76 slides920311

6 5 5 Electrochemistry Conduction 1)Metalic 2)Electrolytic Temprature  Motion of ions  Resistance  -------------------------------- ----- http:\\asadipour.kmu.ac.ir 76 slides920311

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:\\asadipour.kmu.ac.ir 76 slides920311 Conduction ≈ Ions mobility

8 7 battery +- inert electrodes power source vessel e-e- e-e- conductive medium Electrolytic Cell Construction http:\\asadipour.kmu.ac.ir 76 slides 920311

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:\\asadipour.kmu.ac.ir 76 slides920311

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:\\asadipour.kmu.ac.ir 76 slides920311

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:\\asadipour.kmu.ac.ir 76 slides920311

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:\\asadipour.kmu.ac.ir 76 slides920311

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 13http:\\asadipour.kmu.ac.ir 76 slides920311

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:\\asadipour.kmu.ac.ir 76 slides 2H 2 O + 2e -  H 2 + 2OH - 920311

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:\\asadipour.kmu.ac.ir 76 slides920311

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:\\asadipour.kmu.ac.ir 76 slides920311

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:\\asadipour.kmu.ac.ir 76 slides920311 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:\\asadipour.kmu.ac.ir 76 slides920311

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:\\asadipour.kmu.ac.ir 76 slides920311 1 coulomb = 1 amp-sec = 0.001118 g Ag

21 20 Ag + + e -  Ag 1.00 mole e - = 1.00 mole Ag = 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:\\asadipour.kmu.ac.ir 76 slides920311 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:\\asadipour.kmu.ac.ir 76 slides920311

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:\\asadipour.kmu.ac.ir 76 slides920311

24 23http:\\asadipour.kmu.ac.ir 76 slides920311

25 24http:\\asadipour.kmu.ac.ir 76 slides920311

26 25 21-8 Industrial Electrolysis Processes Slide 25 of 52 http:\\asadipour.kmu.ac.ir 76 slides920311

27 26http:\\asadipour.kmu.ac.ir 76 slides920311

28 27 Volta’s battery (1800) Alessandro Volta 1745 - 1827 Paper moisturized with NaCl solution Cu Zn http:\\asadipour.kmu.ac.ir 76 slides920311

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:\\asadipour.kmu.ac.ir 76 slides920311

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:\\asadipour.kmu.ac.ir 76 slides920311 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:\\asadipour.kmu.ac.ir 76 slides920311

32 31 Electrodes are passive (not involved in the reaction) Olmsted Williams http:\\asadipour.kmu.ac.ir 76 slides920311

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:\\asadipour.kmu.ac.ir 76 slides920311

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:\\asadipour.kmu.ac.ir 76 slides920311 Strongest oxidunt Strongest reductant

35 34 Measuring E 0 red Cu 2+ & Zn 2+ Slide 34 of 52 cathode anode http:\\asadipour.kmu.ac.ir 76 slides920311 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:\\asadipour.kmu.ac.ir 76 slides920311 ?

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:\\asadipour.kmu.ac.ir 76 slides E 0 = -0.40 V E 0 = 0.74 V 920311 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:\\asadipour.kmu.ac.ir 76 slides Fe + O 2 (g) + H 2 O  Fe(OH) 2 (s) 920311 Which one is oxidunt ?

39 38http:\\asadipour.kmu.ac.ir 76 slides920311

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:\\asadipour.kmu.ac.ir 76 slides Cu + 2Ag +  Cu +2 + 2Ag  G o = -2×96500×0.46=-88780 J 920311

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:\\asadipour.kmu.ac.ir 76 slides 0.036 V 920311 - 0.337 V

42 41 - 0.337 V http:\\asadipour.kmu.ac.ir 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 920311

43 42http:\\asadipour.kmu.ac.ir 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 920311 2 × ------------------------------------------- 3Fe 2+  2Fe 3+ +Fe E 0 = -1.221V NO

44 43http:\\asadipour.kmu.ac.ir 76 slides -0.036 V 920311 ------------------------------------------------------- 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 2e - +Fe 2+  Fe E 0 = -0.440 V Fe 2+  Fe 3+ +e - E 0 = -0.771 V ------------------------------------------- 3Fe 2+  2Fe 3+ +Fe E 0 = -1.221V

45 44  G 0 =-nE 0 f= -3E 0 f http:\\asadipour.kmu.ac.ir 76 slides920311  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 920311http:\\asadipour.kmu.ac.ir 76 slides ΔG = ΔH - T·ΔS W = ΔH - q

47 46http:\\asadipour.kmu.ac.ir 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 920311

48 47http:\\asadipour.kmu.ac.ir 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 920311 ------------------------------------

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:\\asadipour.kmu.ac.ir 76 slides920311 E = E o – RT ln Q n /-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 0.50 M H + and 0.95 atm H 2. http:\\asadipour.kmu.ac.ir 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 920311 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:\\asadipour.kmu.ac.ir 76 slides Ni (s) | Ni 2+ (0.600M) || Sn 2+ (0.300M) | Sn (s) Emf of a cell 920311

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:\\asadipour.kmu.ac.ir 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) 920311 E=+0.079 (Electrolytic cell) (Galvanic cell)

53 52 equilibrium constant of a cell at equilibrium E = 0 Nernst Equation: http:\\asadipour.kmu.ac.ir 76 slides E = E o - 0.0591 log B n A 920311 A BA B

54 53http:\\asadipour.kmu.ac.ir 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 920311

55 54http:\\asadipour.kmu.ac.ir 76 slides920311

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:\\asadipour.kmu.ac.ir 76 slides 920311

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:\\asadipour.kmu.ac.ir 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 920311

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:\\asadipour.kmu.ac.ir 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 920311

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:\\asadipour.kmu.ac.ir 76 slides920311 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:\\asadipour.kmu.ac.ir 76 slides 920311

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:\\asadipour.kmu.ac.ir 76 slides920311

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:\\asadipour.kmu.ac.ir 76 slides 920311

63 62 Cu CuSO 4 Cu Cu +2 + 2e -  CuCu  Cu +2 + 2e - What happened at each electrode? http:\\asadipour.kmu.ac.ir 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 920311

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:\\asadipour.kmu.ac.ir 76 slides (((Purified cathodic Cu))) overall cell reaction Cu 2+ + Cu (s) Anod  Cu 2+ + Cu (s) Cathode 920311

65 64 Cu 1.0 M CuSO 4 Cu 1.0 M CuSO 4 A cell with the similar electrods and electrolytes 0.0 http:\\asadipour.kmu.ac.ir 76 slides volts 920311

66 65 Cu 1.0M CuSO 4 Cu 0.1 M CuSO 4 A cell with the similar electrods but different concentration electrolytes ؟؟ http:\\asadipour.kmu.ac.ir 76 slides volts 920311

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:\\asadipour.kmu.ac.ir 76 slides920311

68 http:\\asadipour.kmu.ac.ir 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 920311

69 68 pH meter, A concentration Cell 920311http:\\asadipour.kmu.ac.ir 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:\\asadipour.kmu.ac.ir 76 slides pH = E cell /(0.059) 920311

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:\\asadipour.kmu.ac.ir 76 slides920311

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 - → 4 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:\\asadipour.kmu.ac.ir 76 slides E0=0.440 VE0=1.229 V E 0 =0.401 V 920311

73 72 Prevention of Corrosion Cover the Fe surface with a protective coating Paint Tin (Tin plate) Zn (Galvanized iron) http:\\asadipour.kmu.ac.ir 76 slides920311

74 73 Corrosion Protection Slide 73 of 52 http:\\asadipour.kmu.ac.ir 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 920311

75 74 Corrrosion Protection (cathode) (electrolyte) (anode) http:\\asadipour.kmu.ac.ir 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 920311 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:\\asadipour.kmu.ac.ir 76 slides920311

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:\\asadipour.kmu.ac.ir 76 slides920311


Download ppt "1http:\\asadipour.kmu.ac.ir 76 slides920311 2 2 Electrochemistry 76 slides920311."

Similar presentations


Ads by Google