1. Please Pick Up  Electrochemical Equilibrium Problem Set

2. 7/15/2015 If the voltmeter is connected for a long period of time, Cu(s)Cu 2+ (aq) + 2 e – Cu(s) 0.030 V Anode Electrode Cathode Electrode Salt Bridge Cu e–e– 1.00 M Cu 2+ e–e– Cu 0.10 M Cu 2+ what will be the final concentrations if the volumes are the same? [Cu 2+ ] o = 0.10 M [Cu 2+ ] o = 1.00 M Cu 0.020 V Cu 0.010 V Cu 0.000 V 0.55 M Cu 2+ 0.55 M Cu 2+

3. 7/15/2015 If the voltmeter is connected for a long period of time, Cu(s)Cu 2+ (aq) + 2 e – Cu(s) 0.030 V Anode Electrode Cathode Electrode Salt Bridge Cu e–e– e–e– what will happen to the masses of the anode and the cathode? 0.020 V Cu 0.010 V Cu 0.000 V Cu

4. 7/15/2015 Anode Electrode Cathode Electrode Salt Bridge 0.000 V Cu(s)Cu 2+ (aq) + 2 e – Cu(s) [Cu 2+ ] o = 0.10 M [Cu 2+ ] o = 1.00 M 0.55 M Cu 2+ Cu 0.55 M Cu 2+ The anode will lose mass. The cathode will gain mass.

5. Electrochemical Determination of Equilibrium Constants Edward A. Mottel Department of Chemistry Rose-Hulman Institute of Technology

6. 7/15/2015 Electrochemical Determination of Equilibrium Constants  Reading assignment: Fine, Beall & Stuehr, Chapter 14.3-14.4  The Nernst equation can be used to determine equilibrium constants and related terms.

7. 7/15/2015 Nernst Equation E cell = E ° cell - RT nF · ln Q Standard cell potential Concentration term Under what conditions would the observed cell be zero?

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

9. 7/15/2015 At Equilibrium observed cell potential is zero reaction quotient equals the equilibrium constant the cell is “dead” standard cell potential equals the concentration term E cell = E ° cell - RT nF · ln Q 0.00 K eq

10. 7/15/2015 Equilibrium Constant Determination  Determine the standard cell potential.  Use the Nernst equation to determine the equilibrium constant. Solubility is an equilibrium process. Determine the solubility product of silver chloride. Procedure

11. 7/15/2015 Determine the Numeric Value of the Solubility Product of Silver Chloride K sp = u Solubility product expression u Chemical equation that corresponds to this “target” process: AgCl(s)Ag + (aq) + Cl – (aq) [Ag + ] [Cl – ]

12. 7/15/2015 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. AgCl(s)Ag + (aq) + Cl – (aq)

13. 7/15/2015 Divide the Target Reaction into Reduction and Oxidation Equations target Ag + (aq) + Cl – (aq)AgCl(s) Find an equation that “looks like” the target equation reduction oxidation

14. 7/15/2015 Divide the Target Reaction into Reduction and Oxidation Equations target Ag + (aq) + Cl – (aq)AgCl(s) Find an equation that “looks like” the target equation reduction AgCl(s) + e – Ag(s) + Cl – (aq) oxidation Find the oxidation half-cell equation needed to complete the target equation Ag + (aq) + e – Ag(s) The oxidation and reduction reactions added together give the target equation e – + Ag(s) ++ Ag(s) + e –

15. 7/15/2015 Determine the Standard Cell Potential of the Target Equation AgCl(s)Ag + (aq) + Cl – (aq) AgCl(s) + e – Ag(s) + Cl – (aq) E ° ½ = Ag + (aq) + e – Ag(s) + 0.222 V - 0.800 V - 0.578 V target reduction oxidation E ° ½ = E ° cell =

16. 7/15/2015 At Equilibrium E cell = 0.00V E ° cell = 0.0592 n products reactants · log E ° cell = 0.0592 n · log K eq 0.0592 n products reactants · log = E ° cell - Apply this to the silver chloride solubility product.

17. 7/15/2015 Solubility Product for AgCl AgCl(s)Ag + (aq) + Cl – (aq) E cell = E º cell - 0.0592 n products reactants · log E cell = -0.578 V - 0.0592 1 · log [Ag + ][Cl – ] 0.000 V = -0.578 V - 0.0592 1 · log K sp

18. 7/15/2015 AgCl(s)Ag + (aq) + Cl – (aq) log K sp = - 0.578 0.0592 = -9.76 K sp = 10 – 9.76 = 1.7 x 10 – 10 M 2 Solubility Product for AgCl

19. 7/15/2015 What is the molar solubility of silver chloride in pure water? Molar solubility is the maximum number of moles of solute which will dissolve to form a liter of solution.

20. 7/15/2015 Molar Solubility of Silver Chloride AgCl(s)Ag + (aq) + Cl – (aq) -x+x 1.3 x 10 – 5 moles of AgCl will dissolve to form a liter of solution. K sp = [Ag + ] [Cl – ] = 1.7 x 10 – 10 M 2 K sp = [ x ] [ x ] = 1.7 x 10 – 10 M 2 [ x ] = 1.3 x 10 – 5 M

21. 7/15/2015 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.

22. 7/15/2015 Determine the Standard Cell Potential of the Target Equation E ° ½ = +1.662 V E ° cell = +2.002 V E ° ½ = +0.340 V 2 Al(s) + 3 Cu 2+ (aq)2 Al 3+ (aq) + 3 Cu(s) target reduction Cu 2+ (aq) + 2 e – Cu(s) oxidation Al 3+ (aq) + 3 e – Al(s) 3  ( ) 2  ( )

23. 7/15/2015 0.0592 n products reactants · log E ° cell = At Equilibrium [Al 3+ ] 2 [Cu 2+ ] 3 2.002 6 2.002 = 0.0592 6 · log K eq 2 Al(s) + 3 Cu 2+ (aq)2 Al 3+ (aq) + 3 Cu(s) = E ° cell - 0.0592 n products reactants · log E cell = 0.00 V

24. 7/15/2015 2.002 = 0.0592 6 · log K eq log K eq = 6 x 2.002 0.0592 = 202.9 K eq = At Equilibrium 2 Al(s) + 3 Cu 2+ (aq)2 Al 3+ (aq) + 3 Cu(s) What exactly does 8 × 10 202 mean? 10 202.9 = 8 x 10 202 M – 1

25. 7/15/2015 Example  The solubility product of lead(II) chloride is 1.7 × 10 –5 M 3.  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?

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

27. 7/15/2015 Use the Solubility Product Equation as the Target Equation PbCl 2 (s)Pb 2+ (aq) + 2 Cl – (aq) target

28. 7/15/2015 Break the Equation into a Reduction and an Oxidation Half-cell reduction PbCl 2 (s) + 2 e – Pb(s) + 2 Cl – (aq) oxidation Pb 2+ (aq) + 2 e – Pb(s) PbCl 2 (s)Pb 2+ (aq) + 2 Cl – (aq) target The reduction potential isn’t listed in the table. Now what?

29. 7/15/2015 Use the Equilibrium Constant to Determine the Standard Cell Potential E cell = 0.00 V = E ° cell - 0.0592 n · log K eq E ° cell = 0.0592 n · log K eq (1.7 × 10 –5 M 3 ) E ° cell = -0.141 V 2

30. 7/15/2015 Work the Problem Backwards PbCl 2 (s) + 2 e – Pb(s) + 2 Cl – (aq) Pb 2+ (aq) + 2 e – Pb(s) PbCl 2 (s)Pb 2+ (aq) + 2 Cl – (aq) E ° ½ = E ° cell = - 0.141 V E ° ½ = reduction oxidation target - 0.267 V +0.126 V

31. 7/15/2015 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.

32. 7/15/2015 Reverse the Equations PbCl 2 (s) + 2 e – Pb(s) + 2 Cl – (aq) Pb 2+ (aq) + 2 e – Pb(s) PbCl 2 (s)Pb 2+ (aq) + 2 Cl – (aq) E ° ½ = + 0.126 V E ° cell = - 0.141 V E ° ½ = - 0.267 V reduction oxidation target

33. 7/15/2015 Reverse the Equations Pb(s) + 2 Cl – (aq) PbCl 2 (s) + 2 e – Pb 2+ (aq) + 2 e – Pb(s) PbCl 2 (s)Pb 2+ (aq) + 2 Cl – (aq) oxidation target E ° ½ = + 0.267 V E ° ½ = + 0.126 V E ° cell = - 0.141 V

34. 7/15/2015 Reverse the Equations Pb(s) + 2 Cl – (aq) PbCl 2 (s) + 2 e – Pb 2+ (aq) + 2 e – Pb(s) PbCl 2 (s)Pb 2+ (aq) + 2 Cl – (aq) oxidation reduction target E ° ½ = + 0.267 V E ° ½ = - 0.126 V E ° cell = - 0.141 V

35. 7/15/2015 Pb 2+ (aq) + 2 Cl – (aq) Pb 2+ (aq) + 2 e – Pb(s) + 2 Cl – (aq) Reverse the Equations PbCl 2 (s) + 2 e – Pb(s) PbCl 2 (s) oxidation reduction target Write the shorthand notation for the cell. E ° ½ = + 0.267 V E ° ½ = - 0.126 V E ° cell = + 0.141 V

36. 7/15/2015 Pb 2+ (aq) + 2 Cl – (aq) A Precipitation Cell is the Reverse of the Solubility Product Cell PbCl 2 (s) Pb 2+ (aq) + 2 e – Pb(s) Pb(s) + 2 Cl – (aq) PbCl 2 (s) + 2 e – Pb(s) | Cl – (aq) (1.00 M) | | Pb 2+ (1.00 M) | Pb(s) oxidation reduction target E ° ½ = + 0.267 V E ° ½ = - 0.126 V E ° cell = + 0.141 V

37. 7/15/2015 Pb(s) | Cl – (aq) (1.00 M) | | Pb 2+ (1.00 M) | Pb(s) 0.141 V e–e– Salt Bridge 1 M NaCl Anode Electrode Pb 1 M Pb(NO 3 ) 2 Cathode Electrode e–e– Pb Pb(s) + 2 Cl – (aq) PbCl 2 (s) + 2 e – Pb 2+ (aq) + 2 e – Pb(s)

38. 7/15/2015 Pb(s) | Cl – (aq) (1.00 M) | | Pb 2+ (1.00 M) | Pb(s) 0.141 V e–e– Salt Bridge 1 M NaCl Anode Electrode Pb 1 M Pb(NO 3 ) 2 Cathode Electrode e–e– Pb Pb(s) + 2 Cl – (aq) PbCl 2 (s) + 2 e – Pb 2+ (aq) + 2 e – Pb(s)

39. 7/15/2015 Consider the Following Electrochemical Cell Involving an Inert Platinum Electrode Cr(s) | Cr 2+ (0.400 M) | | Sn 2+ (0.018 M), Sn 4+ (0.680 M) | Pt(s)  Identify the reaction occurring at each electrode.  Determine the observed cell potential at 25 °C.

40. 7/15/2015 Possible Anode Reactions Cr(s) | Cr 2+ Cr(s) Cr 2+ (aq) + 2 e – E ° ½ = +0.86 V Cr(s) Cr 2+ (aq) + 2 e – E ° ½ = +0.86 V Cr(s) Cr 3+ (aq) + 3 e – E ° ½ = +0.74 V Cr 2+ (aq) Cr 3+ (aq) + e – E ° ½ = +0.41 V Which reaction occurs at the anode? What reactions might occur at the cathode?

41. 7/15/2015 Possible Cathode Reactions Sn 2+, Sn 4+ | Pt(s) Sn 2+ (aq) + 2 e – Sn(s) E ° ½ = -0.136 V Sn 4+ (aq) + 2 e – Sn 2+ (aq) E ° ½ = +0.150 V Sn 4+ (aq) + 2 e – Sn 2+ (aq) E ° ½ = +0.150 V Which reaction occurs at the cathode?

42. 7/15/2015 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.

43. 7/15/2015 E ° ½ = +0.150 V E ° ½ = +0.86 V E ° cell = +1.01 V oxidation reduction net reaction Overall Reaction Cr(s) + Sn 4+ (aq) Sn 2+ (aq) + Cr 2+ (aq) Sn 4+ (aq) + 2 e – Sn 2+ (aq) Cr(s) Cr 2+ (aq) + 2 e –

44. 7/15/2015 Cr(s) | Cr 2+ (0.400 M) | | Sn 2+ (0.018 M), Sn 4+ (0.680 M) | Pt(s) E cell = E ° cell - 0.0592 n products reactants · log [Sn 2+ ] [Cr 2+ ] [Sn 4+ ] 2 E cell = 1.07 V [0.018] [0.400] [0.680] 1.01 Cr(s) + Sn 4+ (aq) Sn 2+ (aq) + Cr 2+ (aq)

45. 7/15/2015 The End

46. 7/15/2015