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Electrochemistry

Electrochemistry. 2. Electrochemistry. All of Chemical reactins are related to ELECTRONS Redox reactions Voltaic or Galvanic cells Electrochemical cells. 3. Power consumption. Chemical Reactions. Electric Power. Power generation. Electric power conversion in electrochemistry.

sawyer-thompson
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Electrochemistry

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

  2. Electrochemistry 2 http:\\aliasadipour.kmu.ac.ir

  3. Electrochemistry • All of Chemical reactins are related to ELECTRONS • Redox reactions • Voltaic or Galvanic cells • Electrochemical cells 3 http:\\aliasadipour.kmu.ac.ir

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

  5. Electrochemistry • Conduction • 1)Metalic • 2)Electrolytic • TempratureMotion of ions Resistance  • 1C=1AS /// 1J=1CV -------------------------------- ----- ----- 5 http:\\aliasadipour.kmu.ac.ir

  6. - + 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 http:\\aliasadipour.kmu.ac.ir

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

  8. 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- Cl2 + 2e- http:\\aliasadipour.kmu.ac.ir

  9. 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- Cl2 + 2e- Na+ + e- Na http:\\aliasadipour.kmu.ac.ir

  10. Molten NaCl Electrolytic Cell cathode half-cell (-) REDUCTION Na+ + e- Na anode half-cell (+) OXIDATION 2Cl- Cl2 + 2e- overall cell reaction 2Na+ + 2Cl- 2Na + Cl2 X 2 Non-spontaneous reaction! http:\\aliasadipour.kmu.ac.ir

  11. 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? http:\\aliasadipour.kmu.ac.ir

  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- gH2(g)+ 2OH-(aq) -0.83 V H2O 2 H2O(l)g 4e- + 4H+(g) + O2(g) -1.23 V Net Rxn Occurring: 2 H2O g 2 H2(g)+ O2 (g) E°= - 2.06 V

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

  14. anode 2Cl- Cl2 + 2e- - + Aqueous NaCl battery power source e- e- 2H2O + 2e- H2 + 2OH- NaCl (aq) What could be reduced at the cathode? Na+ Cl- (-) (+) H2O cathode different half-cell http:\\aliasadipour.kmu.ac.ir

  15. Aqueous NaCl Electrolysis possible cathode half-cells (-) REDUCTION Na+ + e- Na 2H2O + 2e- H2 + 2OH- possible anode half-cells (+) OXIDATION2Cl- Cl2 + 2e- 2H2O  O2 + 4H+ + 4e- overall cell reaction 2Cl- + 2H2O  H2 + Cl2 + 2OH- http:\\aliasadipour.kmu.ac.ir

  16. Aqueous CuCl2 Electrolysis possible cathode half-cells (-) REDUCTION Cu2+ + 2e- Cu 2H2O + 2e- H2 + 2OH- possible anode half-cells (+) OXIDATION2Cl- Cl2 + 2e- 2H2O  O2 + 4H+ + 4e- overall cell reaction Cu2+ + 2Cl- Cu(s) + Cl2(g) http:\\aliasadipour.kmu.ac.ir

  17. Aqueous Na2SO4 Electrolysis possible cathode half-cells (-) REDUCTION Na+ + e- Na [2H2O + 2e- H2 + 2OH- ] possible anode half-cells (+) OXIDATION SO42- S4O82_ + 2e- 2H2O  O2 + 4H+ + 4e- overall cell reaction 6H2O  2H2 + O2 +4H+ + 4OH- 2× http:\\aliasadipour.kmu.ac.ir

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

  19. 1 coulomb = 1 amp-sec = 0.001118 g Ag e- 1 amp = 0.001118 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 0.001118 g Ag/sec Ag+ Ag http:\\aliasadipour.kmu.ac.ir

  20. 107.87 g Ag/mole e- 0.001118 g Ag/coul 1 Faraday (F ) Ag+ + e- Ag 1.00 mole e- = 1.00 mole Ag = 107.87 g Ag =96,485 coul/mole e- mole e- = Q/F http:\\aliasadipour.kmu.ac.ir

  21. battery • 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- - + - + - + - + e- e- e- 1.0 M Au+3 1.0 M Zn+2 1.0 M Ag+ Au+3 + 3e- Au Zn+2 + 2e- Zn Ag+ + e- Ag http:\\aliasadipour.kmu.ac.ir

  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+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:\\aliasadipour.kmu.ac.ir

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

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

  25. 21-8 Industrial Electrolysis Processes http:\\aliasadipour.kmu.ac.ir Slide 25 of 52

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

  28. Galvanic Cells anode oxidation cathode reduction spontaneous redox reaction http:\\aliasadipour.kmu.ac.ir 19.2

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

  30. What about half-cell reactions? What about the sign of the electrodes? Anod - Cathod + Why? Compare with Electrolytic cells Cu+2+ 2e- Cu cathode half-cell Zn  Zn+2 + 2e- 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 http:\\aliasadipour.kmu.ac.ir

  31. Electrolytic cells sign of the electrodes? - + battery e- NaCl (l) Na+ Cl- Na+ e- Cl- (-) (+) Anode + Cathode - Cl- Na+ 2Cl- Cl2 + 2e- Na+ + e- Na http:\\aliasadipour.kmu.ac.ir

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

  33. 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+ + 2e- H2 inert metal EoSHE = 0.0 volts 1.00 M H+ http:\\aliasadipour.kmu.ac.ir

  34. E0 is for the reaction as writtenE0red // E0ox • The more positive E0 the greater the tendency for the substance to be reduced • The half-cell reactions are reversible • The sign of E0changes when the reaction is reversed • Changing the stoichiometric coefficients of a half-cell reaction does not change the value of E0 http:\\aliasadipour.kmu.ac.ir 19.3

  35. Cell EMF Oxidizing and Reducing Agents http:\\aliasadipour.kmu.ac.ir 35

  36. -E=E0red MeasuringE0red Cu2+& Zn2+ anode cathode cathode anode Cu+2+ 2e- Cu E=E0red Zn  Zn+2 + 2e- E=E0ox http:\\aliasadipour.kmu.ac.ir Slide 36 of 52

  37. Measuring E0of a cell - + 1.1 volts cathode half-cell Cu+2 + 2e- Cu anode half-cell Zn  Zn+2 + 2e- Cu Zn 1.0 M CuSO4 1.0 M ZnSO4 http:\\aliasadipour.kmu.ac.ir

  38. Cd2+(aq) + 2e-Cd(s)E0 = -0.40 V Cr3+(aq) + 3e-Cr (s)E0 = -0.74 V Cr (s) Cr3+ (1 M) + 3e- E0cell = -0.40 +0.74=0.34 E0 = 0.34 V cell cell 2Cr (s) + 3Cd2+ (1 M)  3Cd (s) + 2Cr3+ (1 M) 2e- + Cd2+ (1 M) Cd (s) What is the standard emf of an electrochemical cell made of a Cd electrode in a 1.0 MCd(NO3)2 solution and a Cr electrode in a 1.0 M Cr(NO3)3 solution? Cd is the stronger oxidizer Cd will oxidize Cr x 3 Cathode (reduction): E0 = -0.40 V x 2 E0 = 0.74 V Anode (oxidation): http:\\aliasadipour.kmu.ac.ir 19.3

  39. H2O with O2 Consider a drop of oxygenated water on an iron object Calculating the cell potential, Eocell, at standard conditions Fe Fe + O2 (g) + H2O Fe(OH)2(s) Fe+2 + 2e- Fe Eo = -0.44 v reverse 2x Fe Fe+2 + 2e- -Eo = +0.44 v O2 (g) + 2H2O + 4e- 4 OH-Eo = +0.40 v 2Fe + O2 (g) + 2H2O  2Fe(OH)2 (s) Eocell= +0.84 v This is spontaneoues corrosion or the oxidation of a metal. http:\\aliasadipour.kmu.ac.ir

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  41. Free Energy and the Cell Potential Cu + 2Ag+ Cu+2 + 2Ag Cu Cu+2 + 2e-Eo= - 0.34 Ag+ + e-  Ag Eo = + 0.80 v 2x Eocell= +0.46 v Cu + 2Ag+ Cu+2 + 2Ag 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=-88780 J http:\\aliasadipour.kmu.ac.ir

  42. -Edepends on: -Related half reaction -Concentration -kinetic------------------------------------------------------2e- +2H+  H2 E0 = 0.000 Fe  3e- +Fe3+E0 = 0.036 ------------------------------------------ Fe +H+ Fe3+ +H2E0 = 0.036 Spontaneous redox reaction ?????!!!!!!!No=========================================================================================== 0.036 V http:\\aliasadipour.kmu.ac.ir

  43. 0.337 V 2Cu+ Cu2++Cu Auto redox=Disproportionation e- +Cu+ Cu E0 = 0.521 V Cu+ Cu2++e- E0 = -0.153 V ------------------------------------------- 2Cu+ Cu2++Cu E0 = 0.368V http:\\aliasadipour.kmu.ac.ir

  44. 0.036 V Auto redox=Disproportionation?????? NO 2e- +Fe2+ Fe E0 = -0.440 V Fe2+ Fe3++e- E0 = -0.771 V 2 × ------------------------------------------- 3Fe2+ 2Fe3++Fe E0 = -1.221V http:\\aliasadipour.kmu.ac.ir

  45. 1) e +Fe3+  Fe2+ E0= 0.771 2) 2e +Fe2+  Fe E0=-0.440 ------------------------------------------------------- 3e +Fe3+  Fe E0=+0.331 ? No e isn’t a function state 2e- +Fe2+ Fe E0 = -0.440 V Fe2+ Fe3++e- E0 = -0.771 V ------------------------------------------- 3Fe2+ 2Fe3++Fe E0 = -1.221V http:\\aliasadipour.kmu.ac.ir -0.036 V

  46. G0 =-nE0f G0 =-nE0f= -3E0f 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 =+0.109f 3E0=-0.109E0=-0.036 v http:\\aliasadipour.kmu.ac.ir

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

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

  49. 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 =0.230 -0.140 =0.090V http:\\aliasadipour.kmu.ac.ir

  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 /-nf E = Eo – RTln Q n E = Eo - 0.0591 log Q n 50 http:\\aliasadipour.kmu.ac.ir

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