1 / 45

Please Sit with Your Group

Please Sit with Your Group. Please be sure each member of your team has a copy of Electrochemical Determination of Equilibrium Constants Lecture Notes Electrochemical Equilibrium Problem Set Today’s reporter is the person who began watching Super Bowl coverage the earliest.

samara
Télécharger la présentation

Please Sit with Your Group

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Please Sit with Your Group • Please be sure each member of your team has a copy of • Electrochemical Determination of Equilibrium Constants Lecture Notes • Electrochemical Equilibrium Problem Set • Today’s reporter is the person who began watching Super Bowl coverage the earliest. • Next reading assignment: • Zumdahl Chapter 11

  2. Cu(s) Cu2+(aq) + 2 e– Cu2+(aq) + 2 e– Cu(s) 0.030 V 0.000 V 0.020 V 0.010 V Cathode Electrode Anode Electrode e– Salt Bridge e– Cu Cu Cu Cu Cu Cu Cu Cu Cu Cu [Cu2+]o= 0.10 M [Cu2+]o= 1.00 M 0.10 M Cu2+ 0.55 M Cu2+ 1.00 M Cu2+ 0.55 M Cu2+ If the voltmeter is connected for a long period of time, what will be the final concentrationsif the volumes are the same?

  3. Cu(s) Cu2+(aq) + 2 e– Cu2+(aq) + 2 e– Cu(s) 0.030 V 0.010 V 0.020 V 0.000 V Cathode Electrode Anode Electrode e– Salt Bridge e– Cu Cu Cu Cu Cu Cu Cu Cu If the voltmeter is connected for a long period of time, what will happen to the masses of the anode and the cathode?

  4. Cu(s) Cu2+(aq) + 2 e– Cu2+(aq) + 2 e– Cu(s) [Cu2+]o= 0.10 M [Cu2+]o= 1.00 M 0.000 V Cathode Electrode Anode Electrode Salt Bridge Cu Cu 0.55 M Cu2+ 0.55 M Cu2+ The anode will lose mass The cathode will gain mass

  5. Electrochemical Determination of Equilibrium Constants Edward A. Mottel Integrated, First-Year Curriculum in Science, Engineering and Mathematics

  6. Electrochemical Determination of Equilibrium ConstantsAgenda • Equilibrium conditions • Determination of equilibrium constants • Solubility product • Molar solubility

  7. Concentration term RT o Ecell = Ecell - × ln Q nF Standard cell potential The Observed Cell Potential is a Function of Two Terms This term remains constant throughout the reaction, because the underlying reaction doesn't change. This term becomes more negative and its effect on the observed cell potential (Ecell) becomes greater as the reaction proceeds towards equilibrium

  8. Would you like to buy a battery at equilibrium?

  9. RT o Ecell = Ecell - × ln Q nF standard cell potential equals the concentration term At Equilibrium observed cell potential is zero reaction quotient equals the equilibrium constant the cell is “dead” Keq 0.00

  10. Procedure to Determine an Equilibrium Constant • Divide the target reaction into an oxidation and a reduction half-cell. • Determine the standard cell potential of the cell. • Use the Nernst equation to determine the equilibrium constant.

  11. Solubility product expression Ksp = [Ag+][Cl–] • Chemical equation that corresponds to this “target” process: AgCl(s) Ag+(aq) + Cl–(aq) Determine the Solubility Product of Silver Chloride

  12. AgCl(s) Ag+(aq) + Cl–(aq) Procedure • Divide the target reaction into an oxidation and a reduction half-cell. • Every reaction can be divided into at least one oxidation-reduction half-cell pair. • Even reactions not normally considered redox equations can be described as a redox half-cell pair.

  13. AgCl(s) +e– Ag(s) + Cl–(aq) Ag+(aq) + e– Ag(s) Ag(s) + AgCl(s) Ag+(aq) + Cl–(aq) + Ag(s) Divide the Target Reaction into Reduction and Oxidation Equations reduction oxidation target Find an equation that “looks like” the target equation Find the oxidation half-cell equation needed to give the target equation The oxidation and reduction reactions added together give the target equation

  14. AgCl(s) +e– Ag(s) + Cl–(aq) Ag+(aq) + e– Ag(s) AgCl(s) Ag+(aq) + Cl–(aq) Determine the Standard Cell Potential of the Target Equation reduction E°½= + 0.222 V oxidation E°½= - 0.800 V target E°cell= - 0.578 V

  15. 0.0592 0.0592 0.0592 products products =Eºcell - · log n n n reactants reactants · log · log Keq At Equilibrium Ecell = 0.00V Eºcell = Eºcell = Apply this to the silver chloride solubility product.

  16. 0.0592 0.0592 products AgCl(s) Ag+(aq) + Cl–(aq) n 1 reactants 0.0592 Ecell = Eºcell - · log 0.000 V = -0.578 V- · log Ksp Ecell = -0.578 V- · log [Ag+][Cl–] 1 Solubility Product for AgCl

  17. AgCl(s) Ag+(aq) + Cl–(aq) 0.578 log Ksp = - = -9.76 0.0592 Solubility Product for AgCl Ksp = 10–9.76 = 1.7 x 10–10M2

  18. What is Molar Solubility? • Molar solubility is the maximum number of moles of solute which will dissolve to form a liter of solution. What is the molar solubility of silver chloride in pure water?

  19. AgCl(s) Ag+(aq) + Cl–(aq) Molar Solubility of Silver Chloride -x +x +x Ksp = [Ag+][Cl–] = 1.7 x 10–10M2 Ksp = [ x ][ x ] = 1.7 x 10–10M2 [ x ]= 1.3 x 10–5M 1.3 x 10–5 moles of AgCl will dissolve to form a liter of solution.

  20. Determine the Equilibrium Constant for the Reaction of Aluminum Metal and Copper(II) Ion • Write the target equation for the reaction. • Break the equation into a reduction and an oxidation half-cell. • Determine the standard cell potential. • Use the Nernst equation to solve for the equilibrium constant.

  21. 3´ ( 2´ ( Cu2+(aq) + 2 e– ) ) Cu(s) Al3+(aq) + 3 e– Al(s) 2 Al(s) + 3 Cu2+(aq) 2 Al3+(aq) + 3 Cu(s) Determine the Standard Cell Potential of the Target Equation reduction E°½=+0.340 V oxidation E°½=+1.662 V target E°cell=+2.002 V

  22. 0.0592 0.0592 0.0592 products products · log =Eºcell - n n 6 reactants reactants · log 2 Al(s) + 3 Cu2+(aq) 2 Al3+(aq) + 3 Cu(s) · log Keq 2.002= At Equilibrium Ecell = 0.00 V [Al3+]2 Eºcell = 2.002 6 [Cu2+]3

  23. 0.0592 6 6 x 2.002 0.0592 2 Al(s) + 3 Cu2+(aq) 2 Al3+(aq) + 3 Cu(s) · log Keq 2.002= At Equilibrium = 202.9 log Keq = 10202.9 = 8 x 10202M–1 Keq = What exactly does 8 × 10202 mean?

  24. Example • The solubility product of lead(II) chloride is 1.7 × 10–5M3. • Using this equilibrium constant, determine a half-cell potential involving lead(II) chloride. • Design an electrochemical cell to determine this value. How are the half-cells and the cell equation related to other terms?

  25. PbCl2(s) Pb2+(aq) + 2 Cl–(aq) Relationship of the Cell Equations reduction Usually listed in a table of half-cell potentials oxidation target Equals sum of oxidation and reduction half-cells. Related to the equilibrium constant Ksp = [Pb2+] [Cl–]2

  26. PbCl2(s) Pb2+(aq) + 2 Cl–(aq) Use the Solubility Product Equation as the Target Equation target

  27. PbCl2(s) + 2 e– Pb(s) + 2 Cl–(aq) Pb2+(aq) + 2 e– Pb(s) PbCl2(s) Pb2+(aq) + 2 Cl–(aq) Break the Equation into a Reduction and an Oxidation Half-cell reduction oxidation target The reduction potential isn’t listed in the table. Now what?

  28. 0.0592 0.0592 · log Keq Ecell = 0.00 V =Eºcell - n n · log Keq Eºcell = Use the Equilibrium Constant to Determine the Standard Cell Potential (1.7 × 10–5M3 ) 2 Eºcell = -0.141 V

  29. PbCl2(s) + 2 e– Pb(s) + 2 Cl–(aq) Pb2+(aq) + 2 e– Pb(s) PbCl2(s) Pb2+(aq) + 2 Cl–(aq) Work the Problem Backwards reduction - 0.267 V E°½= oxidation +0.126 V E°½= target E°cell = - 0.141 V

  30. Design an Electrochemical Cell to Determine this Value Can the potential of a non-spontaneous reaction be measured? No. It would need to be powered by the voltmeter. Only spontaneous reactions can be measured. What can you do? Set up the reverse reaction, which is spontaneous.

  31. PbCl2(s) + 2 e– Pb(s) + 2 Cl–(aq) Pb2+(aq) + 2 e– Pb(s) PbCl2(s) Pb2+(aq) + 2 Cl–(aq) Reverse the Equations reduction E°½= - 0.267 V oxidation E°½=+ 0.126 V target E°cell = - 0.141 V

  32. PbCl2(s) + 2 e– Pb(s) + 2 Cl–(aq) Pb2+(aq) + 2 e– Pb(s) PbCl2(s) Pb2+(aq) + 2 Cl–(aq) Reverse the Equations oxidation E°½= + 0.267 V oxidation E°½=+ 0.126 V target E°cell= - 0.141 V

  33. PbCl2(s) + 2 e– Pb(s) + 2 Cl–(aq) Pb2+(aq) + 2 e– Pb(s) PbCl2(s) Pb2+(aq) + 2 Cl–(aq) Reverse the Equations oxidation E°½= +0.267 V reduction E°½= -0.126 V target E°cell= -0.141 V

  34. Reverse the Equations oxidation PbCl2(s) + 2 e– E°½= + 0.267 V Pb(s) + 2 Cl–(aq) reduction E°½= - 0.126 V Pb2+(aq) + 2 e– Pb(s) target E°cell= + 0.141 V Pb2+(aq) + 2 Cl–(aq) PbCl2(s) Write the shorthand notation for the cell.

  35. A Precipitation Cell is the Reverse of the Solubility Product Cell Pb(s) | Cl–(aq)(1.00 M) | | Pb2+ (1.00 M) | Pb(s) oxidation PbCl2(s) + 2 e– E°½= + 0.267 V Pb(s) + 2 Cl–(aq) reduction E°½= - 0.126 V Pb2+(aq) + 2 e– Pb(s) target E°cell = + 0.141 V Pb2+(aq) + 2 Cl–(aq) PbCl2(s)

  36. PbCl2(s) + 2 e– Pb2+(aq) + 2 e– Pb(s) Pb(s) + 2 Cl–(aq) 0.141 V Cathode Electrode Anode Electrode e– Salt Bridge e– Pb Pb 1 M NaCl 1 M Pb(NO3)2 Pb(s) | Cl–(aq)(1.00 M) | | Pb2+ (1.00 M) | Pb(s)

  37. Consider the Following Electrochemical Cell Involving an Inert Platinum Electrode Cr(s) | Cr2+(0.400 M) | | Sn2+(0.018 M), Sn4+(0.680 M) | Pt(s) • Identify the reaction occurring at each electrode. • Determine the observed cell potential at 25 °C.

  38. Cr2+(aq) + 2 e– Cr2+(aq) + 2 e– E°½= +0.86 V E°½= +0.86 V Cr(s) Cr(s) Cr3+(aq) + 3 e– E°½= +0.74 V Cr(s) Cr3+(aq) +e– E°½= +0.41 V Cr2+(aq) Possible Anode ReactionsCr(s) | Cr2+ Which reaction occurs at the anode? What reactions might occur at the cathode?

  39. E°½= -0.136 V Sn2+(aq) + 2 e– Sn(s) E°½= +0.150 V E°½= +0.150 V Sn4+(aq) + 2 e– Sn4+(aq) + 2 e– Sn2+(aq) Sn2+(aq) Possible Cathode ReactionsSn2+, Sn4+ | Pt(s) Which reaction occurs at the cathode?

  40. Analysis • Chromium metal is oxidized to chromium(II) ion at the anode. • Tin(IV) ion is reduced to tin(II) ion, but no precipitate is formed. Reduction occurs at the inert electrode.

  41. oxidation Cr2+(aq) + 2 e– Cr(s) reduction Sn4+(aq) + 2 e– Sn2+(aq) net reaction Cr(s) + Sn4+(aq) Sn2+(aq) + Cr2+(aq) Overall Reaction E°½= +0.86 V E°½= +0.150 V E°cell= +1.01 V

  42. 0.0592 products Cr(s) + Sn4+(aq) Sn2+(aq) + Cr2+(aq) n reactants [Sn2+][Cr2+] · log Ecell = Eºcell - [Sn4+] [0.018][0.400] [0.680] Cr(s) | Cr2+(0.400 M) | | Sn2+(0.018 M), Sn4+(0.680 M) | Pt(s) 1.01 2 Ecell = 1.07 V

  43. The End

More Related