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Electrochemistry

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

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Electrochemistry

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  1. http:\\asadipour.kmu.ac.ir 76 slides

  2. Electrochemistry 2 http:\\asadipour.kmu.ac.ir 76 slides

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

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

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

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

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

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

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

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

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

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

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

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

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

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

  19. 1 coulomb = 1 amp-sec = 0.001118 g Ag e- Experimentally: 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:\\asadipour.kmu.ac.ir 76 slides

  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 • 1C=1AS /// 1J=1CV http:\\asadipour.kmu.ac.ir 76 slides

  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:\\asadipour.kmu.ac.ir 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+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 slides

  23. http:\\asadipour.kmu.ac.ir 76 slides

  24. http:\\asadipour.kmu.ac.ir 76 slides

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

  26. http:\\asadipour.kmu.ac.ir 76 slides

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

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

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

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

  31. Olmsted Williams Electrodes are passive (not involved in the reaction) http:\\asadipour.kmu.ac.ir 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+ + 2e- H2 inert metal EoSHE = 0.0 volts 1.00 M H+ http:\\asadipour.kmu.ac.ir 76 slides

  33. 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 Strongest oxidunt Strongest reductant http:\\asadipour.kmu.ac.ir 76 slides 19.3

  34. -E=E0red MeasuringE0red Cu2+& Zn2+ anode cathode cathode anode Cu+2+ 2e- Cu E=E0red Zn  Zn+2 + 2e- E=E0ox http:\\asadipour.kmu.ac.ir 76 slides Slide 34 of 52

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

  36. 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? x 3 Cathode (reduction): E0 = -0.40 V Cd is the stronger oxidizer Cd will oxidize Cr x 2 E0 = 0.74 V Anode (oxidation): E0cell = ? !! http:\\asadipour.kmu.ac.ir 76 slides 19.3

  37. 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 Which one is oxidunt? 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:\\asadipour.kmu.ac.ir 76 slides

  38. http:\\asadipour.kmu.ac.ir 76 slides

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

  40. -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=========================================================================================== - 0.337 V 0.036 V http:\\asadipour.kmu.ac.ir 76 slides

  41. - 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:\\asadipour.kmu.ac.ir 76 slides

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

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

  44. 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:\\asadipour.kmu.ac.ir 76 slides

  45. Free Energy and Chemical Reactions • ΔG = ΔH - T·ΔS W = ΔH - q q ΔH ΔG TΔS W Ideal reverse cell Operating cell Spontaneous reaction http:\\asadipour.kmu.ac.ir 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 http:\\asadipour.kmu.ac.ir 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 =0.230 -0.140 =0.090V http:\\asadipour.kmu.ac.ir 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 – RTln Q n E = Eo - 0.0591 log Q n 48 http:\\asadipour.kmu.ac.ir 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= 0.090 V at 25oC: E = Eo - 0.0591logNi2+/ Sn2+ n Q=X/Y E=0.090-0.059/2×logx/y ------------------------------------------------------- Calculate the Eredfor the hydrogen electrode where 0.50 M H+ and 0.95 atm H2. 2H++2e →H2 E=0.000-0.059/2×logpH2/[H+]2 http:\\asadipour.kmu.ac.ir 76 slides

  50. Emf of a cell Ni(s) | Ni2+(0.600M)|| Sn2+(0.300M) | Sn(s) Ni(s) + Sn2+→ Ni2+ + Sn(s)Eº= 0.090 V http:\\asadipour.kmu.ac.ir 76 slides

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