1. CHAPTER SIX(19) Electrochemistry

2. Chapter 6 / Electrochemistry Chapter Six Contains: 6.1 Redox Reactions 6.2 Galvanic Cells 6.3 Standard Reduction Potentials 6.4 Thermodynamics of Redox Reactions 6.5 The Effect of Concentration on Cell Emf 6.6 Batteries 6.7 Corrosion 6.8 Electrolysis Chapter Six outline 2

3. 6.1 Redox Reactions Chapter 6 / Electrochemistry 3 2Mg (s) + O 2 (g) 2MgO (s) 2Mg 2Mg 2+ + 4e - O 2 + 4e - 2O 2- Oxidation half-reaction (lose e - ) Reduction half-reaction (gain e - ) Electrochemical processes are redox (oxidation-reduction) reactions in which: the energy released by a spontaneous reaction is converted to electricity or electrical energy is used to cause a nonspontaneous reaction to occur 002+2- Electrochemistry is the branch of chemistry that deals with the interconversion of electrical energy and chemical energy.

4. 6.1 Redox Reactions Chapter 6 / Electrochemistry 4 Oxidation number The charge the atom would have in a molecule (or an ionic compound) if electrons were completely transferred. 1.Free elements (uncombined state) have an oxidation number of zero. Na, Be, K, Pb, H 2, O 2, P 4 = 0 2.In monatomic ions, the oxidation number is equal to the charge on the ion. Li +, Li = +1; Fe 3+, Fe = +3; O 2-, O = -2 3.The oxidation number of oxygen is usually –2. In H 2 O 2 and O 2 2- it is –1.

5. 6.1 Redox Reactions Chapter 6 / Electrochemistry 5 4.The oxidation number of hydrogen is +1 except when it is bonded to metals in binary compounds. In these cases, its oxidation number is –1. 6. The sum of the oxidation numbers of all the atoms in a molecule or ion is equal to the charge on the molecule or ion. 5.Group IA metals are +1, IIA metals are +2 and fluorine is always –1. HCO 3 - O = -2H = +1 3x(-2) + 1 + ? = -1 C = +4 Oxidation numbers of all the atoms in HCO 3 - ?

6. 6.1 Redox Reactions Chapter 6 / Electrochemistry 6 Balancing Redox Equations 1.Write the unbalanced equation for the reaction ion ionic form. The oxidation of Fe 2+ to Fe 3+ by Cr 2 O 7 2- in acid solution? Fe 2+ + Cr 2 O 7 2- Fe 3+ + Cr 3+ 2.Separate the equation into two half-reactions. Oxidation: Cr 2 O 7 2- Cr 3+ +6+3 Reduction: Fe 2+ Fe 3+ +2+3 3.Balance the atoms other than O and H in each half-reaction. Cr 2 O 7 2- 2Cr 3+

7. 6.1 Redox Reactions Chapter 6 / Electrochemistry 7 Balancing Redox Equations 4.For reactions in acid, add H 2 O to balance O atoms and H + to balance H atoms. Cr 2 O 7 2- 2Cr 3+ + 7H 2 O 14H + + Cr 2 O 7 2- 2Cr 3+ + 7H 2 O 5.Add electrons to one side of each half-reaction to balance the charges on the half- reaction. Fe 2+ Fe 3+ + 1e - 6e - + 14H + + Cr 2 O 7 2- 2Cr 3+ + 7H 2 O 6.If necessary, equalize the number of electrons in the two half-reactions by multiplying the half-reactions by appropriate coefficients. 6Fe 2+ 6Fe 3+ + 6e - 6e - + 14H + + Cr 2 O 7 2- 2Cr 3+ + 7H 2 O

8. 6.1 Redox Reactions Chapter 6 / Electrochemistry 8 Balancing Redox Equations 7.Add the two half-reactions together and balance the final equation by inspection. The number of electrons on both sides must cancel. 6e - + 14H + + Cr 2 O 7 2- 2Cr 3+ + 7H 2 O 6Fe 2+ 6Fe 3+ + 6e - Oxidation: Reduction: 14H + + Cr 2 O 7 2- + 6Fe 2+ 6Fe 3+ + 2Cr 3+ + 7H 2 O 8.Verify that the number of atoms and the charges are balanced. 14x1 – 2 + 6x2 = 24 = 6x3 + 2x3 9.For reactions in basic solutions, add OH - to both sides of the equation for every H + that appears in the final equation.

9. 6.8 Electrolysis Chapter 6 / Electrochemistry 9 Electrolysis is the process in which electrical energy is used to cause a nonspontaneous chemical reaction to occur. In electrolytic solution, the charge is carried by ions. Ions must be free to move. This is only occur at molten salts or aqueous solutions of electrolyte.

10. 6.8 Electrolysis Chapter 6 / Electrochemistry 10 Electrolysis of molten NaCl Battery pump electrons into the right hand electrode.(-ve)

11. 6.8 Electrolysis Chapter 6 / Electrochemistry 11 Electrolysis of molten NaCl - + anode cathode Cathode reaction Na + + e -  Na Anode reaction 2 Cl -  Cl 2 +2e - Always reduction occurs at the cathode Always oxidization occurs at the anode

12. 6.8 Electrolysis Chapter 6 / Electrochemistry 12 Electrolysis of Water Dose water conduct electricity?

13. 6.8 Electrolysis Chapter 6 / Electrochemistry 13 Electrolysis of Water 2H 2 O+2e -  2OH - (aq) + H 2 (g) (3) By summation eqs. 1 and 2,we get For reduction,we have ; 2 H + + 2e - → H 2 (g) (2) 2H 2 O ⇌ 2OH - + 2H +, (1) anode cathode - + For oxidizes 4H 2 O  4H + + 4OH -, (1) 4OH -  O 2 +2H 2 O + 4e -, (2) 2H 2 O  O 2 (g)+4H + +4e - (4) The summation of eqs. 1 and 2, we get

14. 6.8 Electrolysis Chapter 6 / Electrochemistry 14 Electrolysis of aqueous NaCl Anode reaction 2 Cl -  Cl 2 +2e - Cathode reaction Na + + e -  Na NaCl H2OH2O 2H 2 O  O 2 (g)+4H + +4e - 2H 2 O+2e -  2OH - (aq) + H 2 (g) √ X √ X 2H 2 O + 2Cl - + 2Na +  H 2 (g) +Cl 2 (g) +2 Na + OH -

15. 6.8 Electrolysis Chapter 6 / Electrochemistry 15 Electrolysis of aqueous CuSO 4 Anode reaction √ X Cathode reaction 2H 2 O+2e -  2OH - (aq) + H 2 (g) √ X 2H 2 O  O 2 (g) + 4H + + e - 2SO 4 --  S 2 O 8 -- + 2e - Cu ++ + 2e -  Cu (s) 2Cu ++ + 2H 2 O  O 2 (g) + 2Cu(s) + 4 H +

16. 6.8 Electrolysis Chapter 6 / Electrochemistry 16 Electrolysis of aqueous CuCl 2 Anode reaction 2 Cl -  Cl 2 +2e - 2H 2 O  O 2 (g)+4H + +4e - √ X Cathode reaction 2H 2 O+2e -  2OH - (aq) + H 2 (g) √ X Cu ++ + 2e -  Cu (s) Cu ++ + 2Cl -  Cl 2 (g) + Cu (s)

17. 6.8 Electrolysis Chapter 6 / Electrochemistry 17 Electrolysis of aqueous CuSO 4 with Cu electrode Cu electrode Cathode reaction 2H 2 O+2e -  2OH - (aq) + H 2 (g) √ X Cu ++ + 2e -  Cu (s) Anode reaction √ X 2H 2 O  O 2 (g) + 4H + + e - 2SO 4 --  S 2 O 8 -- + 2e - Cu  Cu ++ + 2e - X Cu ++ + Cu  Cu + Cu ++

18. 6.8 Electrolysis Chapter 6 / Electrochemistry 18 charge (C) = current (A) x time (s) 1 mole e - = 96,500 C = 1 Faraday Electric current is measured in amperes (A). Quantity of electricity is measured in coulombs (C). C = A x time ( in second) = A x t(s) Electromotive force is measured in Volts,V. V = J/C V x C = J

19. 6.8 Electrolysis Chapter 6 / Electrochemistry 19 Na + + e -  Na From this half cell reaction we have 1 mole of e -  1 mol of Na 1 mole e -  mwt of Na 1 mole e -  23g of Na ≡ 96500C ½ mole of e - will ppt. ½ mole of Na,23/2 g Na.

20. 6.8 Electrolysis Chapter 6 / Electrochemistry 20 Example: How much Ca will be produced in an electrolytic cell of molten CaCl 2 if a current of 0.452 A is passed through the cell for 1.5 hours? Anode: Cathode: Ca 2+ (l) + 2e - Ca (s) 2Cl - (l) Cl 2 (g) + 2e - Ca 2+ (l) + 2Cl - (l) Ca (s) + Cl 2 (g) C = A x t s = 0.452 x 1.5 x 60x60 = 2440.8C ? g of Ca = 2440.8 x 40g / 2 x 96500 = 0.5g Ca From the equation we have: 2440.8 C =====  ?g of Ca Ca 2+ (l) + 2e - Ca (s) 2 F = 2 x 96500C ≡ 1 mol Ca = 40g Ca

21. 6.8 Electrolysis Chapter 6 / Electrochemistry 21 Example: a) In the electrolysis of CuSO 4, How much copper is plated out on the cathode by a current of 0.75 A in 10 min ? b) What the volume of O 2 (g) at STP is librated ? a) C = A x t s = 0.75 x 10 x 60 = 450C Cu ++ + 2 e -  Cu (s) 2 F = 2 x 96500C ≡ 1 mol Cu = 63.5g Cu ? g of Cu = 450 x 63.5g / 2 x 96500 = 0.148g Cu From the equation we have: 450 C =====  ?g of Cu

22. 6.8 Electrolysis Chapter 6 / Electrochemistry 22 b) 2H 2 O  O 2 (g) + 4H + + 4 e - 4 F ≡ 4x96500 C= 1 mole O 2 ≡ 24.5 L 450C → ? L ?L = 450 x 24.5 / 4 x 96500 = 2.83 x 10 -2 L O 2 (g) PV=nRT

23. 6.8 Electrolysis Chapter 6 / Electrochemistry 23 Number of F in CuSO 4 cell = Number of F passed in AgNo 3

24. 6.8 Electrolysis Chapter 6 / Electrochemistry 24 Example: a) What mass of copper is plated in the electrolysis of CuSO 4 in the same time 1.0g of Ag is plated in a silver coulometer that arranged in series with CuSO 4 cell ? b) I f 1.0 A is passed, how many minutes are required to plate this quantity ? a) Ag + + e -  Ag(s) 1 mole of e =1 F ≡ 96500 C ≡107.868 g Ag ?C ≡ 1.0 g C= 1 x 96500 / 107.868 = 894.61 C Cu ++ + 2e -  Cu 2 mole of e = 2 F = 2 x 96500 C = 63.5 g 894.61 C = ? g ? G of Cu = 894.61 x 63.5 / 2x 96600 = 0.2948 g

25. 6.8 Electrolysis Chapter 6 / Electrochemistry 25 b) C = A x t s 894.61 = 1.0 x t s t = 894.615 s = 894.615 /60 = 14.9 min

26. 6.2 Galvanic Cells Chapter 6 / Electrochemistry 26 Galvanic cell (Voltaic cell) – device in which transfer of electrons takes place through an external circuit rather than directly between reactants. A spontaneous redox reaction generates an electric current when a piece of zinc metal is placed in a CuSO 4 solution, Zn is oxidized to Zn 2+ ions while Cu 2+ ions are reduced to metallic copper Cu 2+ ions Zn Electrons are transferred from Zn to Cu ++, but there is no useful electric current. The electrons are transferred directly from the reducing agent (Zn) to the oxidizing agent (Cu 2+ ) in solution. However, if we physically separate the oxidizing agent from the reducing agent, the transfer of electrons can take place via an external conducting medium (a metal wire).

27. 6.2 Galvanic Cells Chapter 6 / Electrochemistry 27

28. 6.2 Galvanic Cells Chapter 6 / Electrochemistry 28 ZnSO 4 solution Zn 2+ SO 4 2–

29. 6.2 Galvanic Cells Chapter 6 / Electrochemistry 29 ZnSO 4 solution CuSO 4 solution Zn 2+ Cu 2+ SO 4 2– SO 4 2–

30. 6.2 Galvanic Cells Chapter 6 / Electrochemistry 30 ZnSO 4 solution CuSO 4 solution Zn 2+ Cu 2+ Salt bridge Cotton plugs SO 4 2– SO 4 2–

31. 6.2 Galvanic Cells Chapter 6 / Electrochemistry 31 Zinc anode Copper cathode ZnSO 4 solution CuSO 4 solution Zn 2+ Cu 2+ Salt bridge e–e– e–e– Cl – K+K+ Voltmeter Cotton plugs SO 4 2– SO 4 2–

32. 6.2 Galvanic Cells Chapter 6 / Electrochemistry 32 Zinc anode Copper cathode ZnSO 4 solution CuSO 4 solution Zn 2+ Cu 2+ Salt bridge Zn e–e– e–e– Cl – K+K+ Voltmeter Cotton plugs SO 4 2– SO 4 2–

33. 6.2 Galvanic Cells Chapter 6 / Electrochemistry 33 Zinc anode Copper cathode ZnSO 4 solution CuSO 4 solution Zn 2+ Cu 2+ Salt bridge Zn Zn 2+ 2e–2e– e–e– e–e– Cl – K+K+ Voltmeter Cotton plugs SO 4 2– SO 4 2–

34. 6.2 Galvanic Cells Chapter 6 / Electrochemistry 34 Zinc anode Copper cathode ZnSO 4 solution CuSO 4 solution Zn 2+ Cu 2+ Salt bridge Zn is oxidized to Zn 2+ at anode. Zn(s) Zn 2+ (aq) + 2e – Zn Zn 2+ 2e–2e– e–e– e–e– Cl – K+K+ Voltmeter Cotton plugs SO 4 2– SO 4 2–

35. 6.2 Galvanic Cells Chapter 6 / Electrochemistry 35 Zinc anode Copper cathode ZnSO 4 solution CuSO 4 solution Zn 2+ Cu 2+ Salt bridge Cu Zn is oxidized to Zn 2+ at anode. Zn(s) Zn 2+ (aq) + 2e – Zn Zn 2+ 2e–2e– e–e– e–e– Cl – K+K+ Voltmeter Cotton plugs SO 4 2– SO 4 2–

36. 6.2 Galvanic Cells Chapter 6 / Electrochemistry 36 Zinc anode Copper cathode ZnSO 4 solution CuSO 4 solution Zn 2+ Cu 2+ Salt bridge 2e–2e– Cu 2+ Cu Zn is oxidized to Zn 2+ at anode. Zn(s) Zn 2+ (aq) + 2e – Zn Zn 2+ 2e–2e– e–e– e–e– Cl – K+K+ Voltmeter Cotton plugs SO 4 2– SO 4 2–

37. 6.2 Galvanic Cells Chapter 6 / Electrochemistry 37 Zinc anode Copper cathode ZnSO 4 solution CuSO 4 solution Zn 2+ Cu 2+ Salt bridge 2e–2e– Cu 2+ Cu 2e – + Cu 2+ (aq)Cu(s) Cu 2+ is reduced to Cu at cathode. Zn is oxidized to Zn 2+ at anode. Zn(s) Zn 2+ (aq) + 2e – Zn Zn 2+ 2e–2e– e–e– e–e– Cl – K+K+ Voltmeter Cotton plugs SO 4 2– SO 4 2–

38. 6.2 Galvanic Cells Chapter 6 / Electrochemistry 38 Zinc anode Copper cathode ZnSO 4 solution CuSO 4 solution Zn 2+ Cu 2+ Salt bridge Zn(s) + Cu 2+ (aq) Zn 2+ (aq) + Cu(s) Net reaction 2e–2e– Cu 2+ Cu 2e – + Cu 2+ (aq)Cu(s) Cu 2+ is reduced to Cu at cathode. Zn is oxidized to Zn 2+ at anode. Zn(s) Zn 2+ (aq) + 2e – Zn Zn 2+ 2e–2e– e–e– e–e– Cl – K+K+ Voltmeter Cotton plugs SO 4 2– SO 4 2– Daniell cell

39. 6.2 Galvanic Cells Chapter 6 / Electrochemistry 39 Anode – electrode at which oxidation occurs Cathode – electrode at which reduction occurs Electrons always flow from anode to cathode Salt bridge – tube that contains an electrolytes, maintains charge neutrality for a voltaic cell

40. 6.2 Galvanic Cells Chapter 6 / Electrochemistry 40 Cell Diagram Zn (s) + Cu 2+ (aq) Cu (s) + Zn 2+ (aq) [Cu 2+ ] = 1 M & [Zn 2+ ] = 1 M Zn (s) | Zn 2+ (1 M) || Cu 2+ (1 M) | Cu (s) anodecathode Cl 2 (g) + 2 I – (aq)  2Cl - (aq)+I 2 (s) Cell notation I - (aq)/ I(s) /Pt // Cl 2 (g)/ Cl - (aq)/ Pt anodecathode Cell Diagram

41. 6.3 Standard Reduction Potentials Chapter 6 / Electrochemistry 41 = electromotive force (EMF) = E cell = Cell potential = Cell voltage ( V = J/C) = driving force that moves electrons from anode to cathode The difference in electrical potential between the anode and cathode is called: cell voltage electromotive force (emf) cell potential An electric current flows from the anode to the cathode because there is a difference in electrical potential energy between the electrodes.

42. Chapter 6 / Electrochemistry 42 E cell = E rd. + E ox. (potential difference) E° cell = standard cell potential E° cell = E° re. + E° ox. Zn (s) + Cu 2+ (aq)  Zn 2+ (aq) + Cu (s), E° cell = 1.10 V Note: E˚ cell must be +ve value for cell to operate, (spontaneous) It is impossible to measure the potential of just a single electrode, but if we arbitrarily set the potential value of a particular electrode at zero, we can use it to determine the relative potentials of other electrodes. 6.3 Standard Reduction Potentials

43. Chapter 6 / Electrochemistry 43 Standard reduction potential (E 0 ) is the voltage associated with a reduction reaction at an electrode when all solutes are 1 M and all gases are at 1 atm. E 0 = 0 V Standard hydrogen electrode (SHE) Reduction Reaction 2e - + 2H + (1 M) H 2 (1 atm) 6.3 Standard Reduction Potentials

44. Chapter 6 / Electrochemistry 44 The standard hydrogen electrode (SHE) is used as reference electrode. 6.3 Standard Reduction Potentials

45. Chapter 6 / Electrochemistry 45 E 0 = 0.76 V cell Zn (s) | Zn 2+ (1 M) || H + (1 M) | H 2 (1 atm) | Pt (s) 2e - + H + (1 M) H 2 (s) Zn (1 atm) Zn +2 (1 M) + 2e - Anode (oxidation): Cathode (reduction): Zn (1 atm) + 2H + (1 M) H 2 (s) + Zn +2 (1 M) 6.3 Standard Reduction Potentials

46. Chapter 6 / Electrochemistry 46 Pt (s) | H 2 (1 atm) | H + (1 M) || Cu 2+ (1 M) | Cu (s) 2e - + Cu 2+ (1 M) Cu (s) H 2 (1 atm) 2H + (1 M) + 2e - Anode (oxidation): Cathode (reduction): H 2 (1 atm) + Cu 2+ (1 M) Cu (s) + 2H + (1 M) E 0 = 0.34 V cell 6.3 Standard Reduction Potentials

47. Chapter 6 / Electrochemistry 47 E 0 is for the reaction as written The more positive E 0 the greater the tendency for the substance to be reduced (The more  E  is,the greater the driving force of the RXN) The half-cell reactions are reversible The sign of E 0 changes when the reaction is reversed Changing the stoichiometric coefficients of a half-cell reaction does not change the value of E 0 Zn 2+ + 2 e -  Zn E° red = -0.76V Zn  Zn 2+ +2e - E° oxid = +0.76V Zn 2+ +2e -  Zn E° red = -0.76V 2Zn 2+ +4e -  2 Zn E° red = -0.76V 6.3 Standard Reduction Potentials

48. Chapter 6 / Electrochemistry 48 Example: Use electrode potential to determine whether the following proposed reactions are spontaneous with all substances present at unit activity: a) Cl 2 (g)+ 2I - (aq)  2Cl - (aq)+I 2 (s) b)2Ag(s)+2H + (aq)  2Ag + (aq)+H 2 (g) a) Fist we have to see who is reduced and who is oxidized. Cl 2 is reducedI - is oxidized Anode reactor 2I -  I 2 (s) + 2e - E  ox = -0.536V Cathode Cl 2 +2e-  2Cl - E  =+1.36V Cl 2 +2I -  I 2 (s)+2Cl - (aq), E  cell = +0.824V Since E  cell is +ve the reaction is spontaneous Cell reaction 6.3 Standard Reduction Potentials

49. Chapter 6 / Electrochemistry 49 Ag is oxidized b: 2Ag (s) + 2H + (aq) → Ag + (aq) +H 2 (g) H + is reduced 2Ag(s)  2Ag + (aq) + 2e - E  ox. = -0.799V 2H + +2e-  H 2 (g)E  = 0.00V 2Ag + 2H + (aq) →2Ag + (aq) + H 2 (g), E  cell = -0.799V The result is nonspontaneous, E  cell = -ve Anode reaction: Cathode reaction: 6.3 Standard Reduction Potentials

50. Chapter 6 / Electrochemistry 50 What is the standard emf of an electrochemical cell made of a Cd electrode in a 1.0 M Cd(NO 3 ) 2 solution and a Cr electrode in a 1.0 M Cr(NO 3 ) 3 solution? Cd 2+ (aq) + 2e - Cd (s) E 0 = -0.40 V Cr 3+ (aq) + 3e - Cr (s) E 0 = -0.74 V Cd is the stronger oxidizer Cd will oxidize Cr 2e - + Cd 2+ (1 M) Cd (s) Cr (s) Cr 3+ (1 M) + 3e - Anode (oxidation): Cathode (reduction): 2Cr (s) + 3Cd 2+ (1 M) 3Cd (s) + 2Cr 3+ (1 M) x 2 x 3 E 0 = -0.40 V E 0 = 0.74 V E° cell = E° re. + E° ox. E° cell = -0.40 + 0.74 = 0.34 V. 6.3 Standard Reduction Potentials

51. Chapter 6 / Electrochemistry 51 6.3 Thermodynamics of Redox Reactions Spontaneity of Redox RXNs E  0, spontaneous  G = - n F E mol electrons cell potential Faraday’s Constant (96500 C/mol)  G  = - n F E° Eº and equilibrium constant  G  = - RT lnK……...1  G  = - n F E  ………2 From eqns. (1) and (2) we get E˚ = ( RT/n F ) lnK

52. Chapter 6 / Electrochemistry 52 6.3 Thermodynamics of Redox Reactions  G 0 = -RT ln K = -nFE cell 0

53. Chapter 6 / Electrochemistry 53 6.3 Thermodynamics of Redox Reactions 2e - + Fe 2+ Fe 2Ag 2Ag + + 2e - Oxidation: Reduction: What is the equilibrium constant for the following reaction at 25 0 C? Fe 2+ (aq) + 2Ag (s) Fe (s) + 2Ag + (aq) E 0 = -0.44 + (-0.80) E 0 = -1.24 V n = 2, T=25 °C= 298 K, F=96500 E˚ = ( RT/n F ) lnK E° cell = E° re. + E° ox. E 0 = -0.44 V E 0 = -0.80 V E˚ = ( RT/n F ) lnK lnK = E˚/ ( RT/n F ) = -1.24 /(8.314 X 298/2X96500) = -96.6 K =e -96.6 = 1.12x10 -42

54. Chapter 6 / Electrochemistry 54 6.3 Thermodynamics of Redox Reactions Example: a) Use electrochemical data to calculate the value of  G  for the reaction b) If  H  = - 254.0 KS, calculate  S  a) We have to calculate E  cell from the two half cell reactions. 2Ag (s) + Cl 2 (g) → 2AgCl (s) Anode reaction: 2Ag (s) + Cl – (aq)  2Ag Cl(s) + 2e - E  ox = -0.222V 2e - + Cl 2 (s)  2 Cl – (aq), E  = 1.369V 2Ag (s) + Cl 2 (g)  2AqCl (s), E  cell = 1.137V Cathode reaction:

55. Chapter 6 / Electrochemistry 55 6.3 Thermodynamics of Redox Reactions  G  = -nf E  cell = -2x96500 x 1.137 J = -219.4 kJ b)  G  =  H   T  S  - 219.4 x 1000J = - 254.0 x 1000 J - 298  SJ/K  S  = - ( 254.0  219.4 ) 1000J 298 K = - 116 J/K

56. Chapter 6 / Electrochemistry 56 6.3 Thermodynamics of Redox Reactions Example: Use electrochemical data to calculate the equilibrium constant K for the following reaction at 25ºC Fe ++ (aq) + Ag + (aq) ⇌ Fe +++ (aq) + Ag (s) Fe ++ (aq)  Fe +++ (aq) + e  E  = -0.771V n = 1, T=25 °C= 298 K, F=96500 e  + Ag + (aq)  Ag(s) E  = 0.799V E  cell = - (RT /nF ) ln K 0.028 = - ( 8.314x298 /1x96500 ) ln K  ln K = 1.091  K = 2.98 E 0 = -0.771 + (0.799) E 0 = 0.028 V E° cell = E° re. + E° ox.

57. Chapter 6 / Electrochemistry 57 6.3 Thermodynamics of Redox Reactions  G =  G  + RT ln Q -nFE = -nFE  + RTlQ E = E  - (RT / n F) ln Q Nonstandard standard At equilibrium, E=0, K=Q E  = (RT /n F) ln K Nernst equation

58. Chapter 6 / Electrochemistry 58 6.3 Thermodynamics of Redox Reactions Example: What is the electrode potential of Z n ++ / Z n electrode in which the concentration of Zn ++ = 0.1M. Z n ++ +2e -  Z n, E˚ = -0.76 V E = E  - (RT / n F) ln Q n = 2, T=25 °C= 298 K, F=96500, E˚ = -0.76 V, Q = ( Z n)/ ( Z n++) = 1 /0.1 E= - 0.76 –( 8.314 x 298 / 2x96500 ln (1/0.1) = -0.76 – 0.0128 (2.303) = - 0.76 – 0.0257 = - 0.79V

59. Chapter 6 / Electrochemistry 59 6.3 Thermodynamics of Redox Reactions Example: What is the potential for the cell : Ni / Ni ++ (0.01M) // Cl – (0.2 M) / Cl 2 M) / Pt Ni  N i ++ + 2e - E  = + 0.25V 2e - + Cl 2 (g)  2Cl - E  = 1.36V 2Ni + Cl 2 (g) → 2Cl - (aq) + Ni ++ (aq), E˚ cell = 1.61V n = 2, T=25 °C= 298 K, F=96500, E˚ = 1.61 V, Q =  Cl -  2  Ni -++  / P Cl2 = ((0.2) 2 (0.01)/1=0.0004 E = E  - (RT / n F) ln Q E= 1.61 –( 8.314 x 298 / 2x96500 ln (0.0004) = 1.71V

60. Chapter 6 / Electrochemistry 60 6.3 Thermodynamics of Redox Reactions Example: What is the E for the cell : Sn / Sn ++ (1.0M) // Pb ++ (0.001M) / Pb. Sn  Sn ++ + 2e - E . = 0.136 V Pb ++ + 2e -  Pb (s) E  = - 0.126 V Sn (s) + Pb ++  Sn ++ (aq) + Pb (s)E  cell = 0.010V n = 2, T=25 °C= 298 K, F=96500, E˚ = 0.01 V, Q = ( Sn +2 )/ ( Pb +2 ) = 1 /0.001 =- 0.079 V E = 0.01- ( 8.314x298/ 2x96500) ln (1/0.001) E = E  - (RT / n F) ln Q

61. Chapter 6 / Electrochemistry 61 6.3 Thermodynamics of Redox Reactions Will the following reaction occur spontaneously at 25 0 C if [Fe 2+ ] = 0.60 M and [Cd 2+ ] = 0.010 M? Fe 2+ (aq) + Cd (s) Fe (s) + Cd 2+ (aq) 2e - + Fe 2+ 2Fe E . = -0.44 V Cd Cd 2+ + 2e - E . = 0.40 V n = 2, T=25 °C= 298 K, F=96500, E˚ = -.04 V, Q = ( Cd +2 )/ ( Fe +2 ) = 0.01 /0.6 =0.013 V E = -0.04- ( 8.314x298/ 2x96500) ln (0.01/0.6) E = E  - (RT / n F) ln Q

62. Chapter 6 / Electrochemistry 62 6.3 Thermodynamics of Redox Reactions Example: Consider a cell reaction Mg(s) + 2H + (aq)  Mg ++ (aq)+ H 2 (g), E  cell =2.363V What is the concentration of H + (aq) in a cell in which  Mg ++  = 1.00M and P H2 =1.0atm, if the Ecell = 2.099 V? n = 2, T=25 °C= 298 K, F=96500, E˚ = 2.363 V, E= 2.099V Q = ( Mg +2 )P H2 / ( H + ) 2 = (1)(1) / ( H + ) 2 2.099 = 2.363- ( 8.314x298/ 2x96500) ln (1/ ( H + ) 2 ) E = E  - (RT / n F) ln Q 2.099 – 2.363 = -0.0128 ln [ H + ] -2 ln [ H + ] -2 = 20.625 [ H + ] -2 = 906407915.01 [H + ]= 3.3X10 -5 M

63. Chapter 6 / Electrochemistry 63 6.3 Thermodynamics of Redox Reactions Example: What is the E for the cell : Cu / Cu ++ (0.01M) // Cu ++ (0.1M) / Cu. Cu  Cu ++ + 2e - E . = 0.34 V Cu ++ + 2e -  Cu (s) E  = - 0.34 V Cu (s) + Cu ++  Cu ++ (aq) + Cu (s) E  cell = 0.0V n = 2, T=25 °C= 298 K, F=96500, E˚ = 0.0 V, Q = ( Cu +2 )/ ( Cu +2 ) = 0.01 /0.1 =- 0.0296 V E = 0.0- ( 8.314x298/ 2x96500) ln (0.01/0.1) E = E  - (RT / n F) ln Q Concentration cell : two half-cells composed of the same material but differing in ion concentrations High concentration = reduction, low concentration = oxidation

64. 64 Chapter 6 / Electrochemistry