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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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