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Electric power conversion in electrochemistry

Electrolysis / Power consumption. Chemical Reactions. Electric Power. Electrochemical battery / Power generation. Electric power conversion in electrochemistry. Cell Construction. vessel. -. +. battery. power source. e -. e -. conductive medium. (-). (+). inert electrodes.

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Electric power conversion in electrochemistry

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  1. Electrolysis / Power consumption Chemical Reactions Electric Power Electrochemical battery / Power generation Electric power conversion in electrochemistry

  2. Cell Construction vessel - + battery power source e- e- conductive medium (-) (+) inert electrodes Sign or polarity of electrodes

  3. What chemical species would be present in a vessel of molten sodium chloride, NaCl (l)? Na+ Cl- Let’s examine the electrolytic cell for molten NaCl.

  4. 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-

  5. 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

  6. 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!

  7. Definitions: CATHODE REDUCTION occurs at this electrode ANODE OXIDATION occurs at this electrode

  8. 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?

  9. 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- g H2(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

  10. anode 2Cl- Cl2 + 2e- - + Aqueous NaCl battery power source e- e- NaCl (aq) What could be reduced at the cathode? Na+ Cl- (-) (+) H2O cathode different half-cell

  11. Aqueous NaCl Electrolytic Cell possible cathode half-cells (-) REDUCTION Na+ + e- Na 2H2O + 2e- H2 + 2OH- possible anode half-cells (+) OXIDATION 2Cl- Cl2 + 2e- 2H2O  O2 + 4H+ + 4e- overall cell reaction 2Cl- + 2H2O  H2 + Cl2 + 2OH-

  12. For every electron, an atom of silver is plated on the electrode. Ag+ + e- Ag e- 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 1 amp = 0.001118 g Ag/sec

  13. time in seconds coulomb current in amperes (amp) Faraday’s Law The mass deposited or eroded from an electrode depends on the quantity of electricity. Quantity of electricity – coulomb (Q) Q is the product of current in amps times time in seconds Q = It 1 coulomb = 1 amp-sec = 0.001118 g Ag

  14. 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 mass = molemetal x MM molemetal depends on the half-cell reaction

  15. Examples using Faraday’s Law • How many grams of Cu will be deposited in 3.00 hours by a current of 4.00 amps? Cu+2 + 2e- Cu • 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-.

  16. 21-8 Industrial Electrolysis Processes General Chemistry: Chapter 21

  17. 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

  18. Volta’s battery (1800) Alessandro Volta 1745 - 1827 Paper moisturized with NaCl solution Cu Zn

  19. Galvanic cells and electrodes • To sustain the cell reaction, the charge carried by the electrons through the external circuit must be accompanied by a compensating transport of ions between the two cells. • Relative amounts of charge can be carried by negative or positive ions (depends on their relative mobilities) through the solution. • Salt bridge, consists of an intermediate compartment filled with saturated salt solution and fitted with porous barriers at each end, is used for precise measurements. The purpose of salt bridge is to minimize the natural potential difference (junction potential). NEEP 423

  20. with Galvanic Cells anode oxidation cathode reduction spontaneous redox reaction 19.2

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

  22. 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

  23. What about half-cell reactions? What about the sign of the electrodes? - + Why? cathode half-cell Cu+2 + 2e- Cu anode half-cell Zn  Zn+2 + 2e- 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

  24. Now replace the light bulb with a volt meter. - + 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

  25. 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+

  26. How do we calculate Standard Redox Potentials? We must compare the half reactions to a standard What is that standard? 2 H3O+(aq) + 2e-  H2(g) + 2 H2O(l) E°= 0.00 V This is called the standard hydrogen electrode or SHE Now that we have a standard, we can calculate standard redox potential by using the table of standard redox potentials

  27. E0 is for the reaction as written • 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 19.3

  28. Cell EMF Oxidizing and Reducing Agents Chapter 20

  29. Now let’s combine the copper half-cell with the SHE Eo = + 0.34 v + 0.34 v cathode half-cell Cu+2 + 2e- Cu anode half-cell H2 2H+ + 2e- H2 1.00 atm KCl in agar Cu Pt 1.0 M CuSO4 1.0 M H+

  30. Now let’s combine the zinc half-cell with the SHE Eo = - 0.76 v - 0.76 v anode half-cell Zn  Zn+2 + 2e- cathode half-cell 2H+ + 2e- H2 H2 1.00 atm KCl in agar Pt Zn 1.0 M ZnSO4 1.0 M H+

  31. Increasing activity Assigning the Eo Write a reduction half-cell, assign the voltage measured, and the sign of the electrode to the voltage. Al+3 + 3e-  Al Eo = - 1.66 v Zn+2 + 2e-  Zn Eo = - 0.76 v 2H+ + 2e- H2 Eo = 0.00 v Cu+2 + 2e- Cu Eo = + 0.34 Ag+ + e-  Ag Eo = + 0.80 v

  32. Measuring Standard Reduction Potential anode cathode cathode anode General Chemistry: Chapter 21

  33. Cd2+(aq) + 2e- Cd (s)E0 = -0.40 V Cr3+(aq) + 3e- Cr (s)E0 = -0.74 V Cr (s) Cr3+ (1 M) + 3e- E0 = 0.34 V E0 = -0.40 – (-0.74) E0 =Ecathode- Eanode cell cell cell 2Cr (s) + 3Cd2+ (1 M)  3Cd (s) + 2Cr3+ (1 M) 0 0 2e- + Cd2+ (1 M) Cd (s) What is the standard emf of an electrochemical cell made of a Cd electrode in a 1.0 M Cd(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 2 Anode (oxidation): Cathode (reduction): x 3 19.3

  34. H2O with O2 Consider a drop of oxygenated water on an iron object Calculating the cell potential, Eocell, at standard conditions Fe 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 corrosion or the oxidation of a metal.

  35. E=?-Concentration-Related half reactionkinetic-------------------------------------------------------3e- +2Fe3+  Fe E0 =-0.0362e- +2H+  H2 E0=0.000Fe +H+ Fe3+ !!!!!!!!!!!!!!!!!!!!!!!!===================================================================================================e- +Fe 3+  Fe 2+ E0=0.7712e +Fe2+  Fe E0=-0.44Fe3+ ---------Fe2+-----------Fe0.771 -0.440-0.036=================================================Cu2+---------Cu+-----------Cu+0.153 +0.5210.337

  36. Equilibrium Constants in Redox Reactions • Whereas potential and free energy are related, and free energy and equilibrium are related, equilibrium and potential must be related to one another. • DG° = –nFE°cell and DG° = –RTln Keq therefore –RTln Keq = –nFEocell RTln KeqRT E°cell = ––––––––– = –––– ln Keq nF nF R and F are constant, therefore at 298 K: 0.025693 V E°cell = –––––––– ln Keq n

  37. 1) 2e +Fe2+  Fe E0=-0.4402) e +Fe3+  Fe2+ E0=-0.7713e +Fe3+  Fe E0=??? ===============

  38. 1)G0=-2(-0.440)F=+0.880F2)G0=-1(+0.771)F=-0.771F--------------------------------+0.109F G0 =-nfE0=+0.109F=3FE0E)=-0.036

  39. 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 • The Nernst equation relates emf to concentration using • and noting that

  40. at 25oC: Eocell = 0.0591 log K n from thermodynamics: DGo = -2.303RT log K and the previous relationship: DGo = -nFEocell -nFEocell = -2.303RT log K where n is the number of electrons for the balanced reaction

  41. What happens to the electrode potential if conditions are not at standard conditions? The Nernst equation adjusts for non-standard conditions For a reduction potential: ox + ne  red in general: E = Eo – RT ln (red) nF (ox) at 25oC: E = Eo - 0.0591 log (red) n (ox) Calculate the E for the hydrogen electrode where 0.50 M H+ and 0.95 atm H2.

  42. 1) An example: Ni(s) | Ni2+(0.600M)|| Sn2+(0.300M) | Sn(s) According to the reduction potentials: 2 e- + Ni2+34 Ni(s) -0.230 V 2 e- + Sn2+34 Sn(s) -0.140V One of these needs to be reversed to get a positive voltage, and a spontaneous reaction.

  43. Ni(s)34 2 e- + Ni2+ +0.230 V 2 e- + Sn2+34 Sn(s) -0.140V Ni(s) + Sn2+34 Ni2+ + Sn(s) 0.090 V = Eo

  44. E) Calculation of the equilibrium constant 1) at equilibrium E = 0 ; Q = ____ From the Nernst Equation: For the cell: Ni(s) | Ni2+(0.600M)|| Sn2+(0.300M) | Sn(s)

  45. 2) An example: Sn(s) | Sn2+(1.0M)|| Pb2+(0.0010M) | Pb(s) According to the reduction potentials: 2 e- + Pb2+34 Pb(s) E0=-0.126 V 2 e- + Sn2+34 Sn(s) E0=-0.136V One of these needs to be reversed to get a positive voltage, and a spontaneous reaction.!!!!!!!!!!!!!!!!!!!!!!!!!

  46. E=E0-0.059/2log[Sn2+]/[Pb2+]E=-0.079

  47. Free Energy and the Cell Potential 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 where n is the number of electrons for the balanced reaction What is the free energy for the cell? 1F= 96,500 J/v

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