1. Electrochemistry Interchange of electrical and chemical energy.

2. Electron Transfer Reactions Electron transfer reactions are oxidation-reduction or redox reactions. the energy released by a spontaneous reaction is converted to electricity or electrical energy is used to cause a nonspontaneous reaction to occur Therefore, this field of chemistry is often called ELECTROCHEMISTRY.

3. Electrochemistry Terminology #1  Oxidation  Oxidation – A process in which an element attains a more positive oxidation state Na(s)  Na + + e -  Reduction  Reduction – A process in which an element attains a more negative oxidation state Cl 2 + 2e -  2Cl -

4. Electrochemistry Terminology #2 G ain E lectrons = R eduction An old memory device for oxidation and reduction goes like this… LEO says GER L ose E lectrons = O xidation

5. Electrochemistry Terminology #3 Oxidizing agent  Oxidizing agent The substance that is reduced is the oxidizing agent Reducing agent  Reducing agent The substance that is oxidized is the reducing agent

6. 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-) 19.1 0 0 2+ 2-

7. Electrochemistry Terminology #4 Anode  Anode The electrode where oxidation occurs Cathode  Cathode The electrode where reduction occurs Memory device: Red Cat An ox

8. 8 Cr 2 O 7 2- (aq) + SO 3 2- (aq)  Cr 3+ (aq) + SO 4 2- (aq) How can we balance this equation? First Steps:  Separate into half-reactions.  Balance elements except H and O.

9. 9 Cr 2 O 7 2- (aq)  Cr 3+ (aq) SO 3 2- (aq)  SO 4 2- (aq) Now, balance all elements except O and H. Method of Half Reactions

10. Copyright © Cengage Learning. All rights reserved 10 Cr 2 O 7 2- (aq)  2Cr 3+ (aq) SO 3 2- (aq)  + SO 4 2- (aq) How can we balance the oxygen atoms? Method of Half Reactions (continued)

11. 11 Cr 2 O 7 2- (aq)  Cr 3+ (aq) + 7H 2 O H 2 O +SO 3 2- (aq)  + SO 4 2- (aq) How can we balance the hydrogen atoms? Method of Half Reactions (continued)

12. 12 This reaction occurs in an acidic solution. 14H + + Cr 2 O 7 2-  2Cr 3+ + 7H 2 O H 2 O +SO 3 2-  SO 4 2- + 2H + How can we write the electrons? Method of Half Reactions (continued)

13. 13 This reaction occurs in an acidic solution. 6e - + 14H + + Cr 2 O 7 2-  2Cr 3+ + 7H 2 O H 2 O +SO 3 2-  SO 4 2- + 2H + + 2e - How can we balance the electrons? Method of Half Reactions (continued)

14. 14 14H + + 6e - + Cr 2 O 7 2-  2Cr 3+ + 7H 2 O 3[H 2 O +SO 3 2-  SO 4 2- + 2e - + 2H + ] Final Balanced Equation: Cr 2 O 7 2- + 3SO 3 2- + 8H +  2Cr 3+ + 3SO 4 2- + 4H 2 O Method of Half Reactions (continued)

15. 15 Exercise Balance the following oxidation – reduction reaction that occurs in acidic solution. Br – (aq) + MnO 4 – (aq)  Br 2 (l)+ Mn 2+ (aq) 10Br – (aq) + 16H + (aq) + 2MnO 4 – (aq)  5Br 2 (l)+ 2Mn 2+ (aq) + 8H 2 O(l)

16. 16 Half–Reactions The overall reaction is split into two half – reactions, one involving oxidation and one reduction. 8H + + MnO 4 - + 5Fe 2+  Mn 2+ + 5Fe 3+ + 4H 2 O Reduction: 8H + + MnO 4 - + 5e -  Mn 2+ + 4H 2 O Oxidation: 5Fe 2+  5Fe 3+ + 5e -

17. 17 How do we harness the energy from the reaction? If they are in the same solution; energy is released as heat and can do no work. Separate the solution; MnO 4 - in one beaker and Fe 2+ in another requiring the e - to be transferred through a wire producing a current.

18. 18 Figure 17.1: Schematic of a method to separate the oxidizing and reducing agents of a redox reaction. (The solutions also contain counterions to balance the charge.) Current stop flowing due to separation of charge.

19. 19 Figure 17.2: Galvanic cells can contain a salt bridge as in (a) or a porous-disk connection as in (b). Keeps the net charge at zero; allows the ions to flow w/o mixing the solutions.

20. 20 Galvanic Cell A device in which chemical energy is changed to electrical energy.

21. 19.2 spontaneous redox reaction anode oxidation cathode reduction - + http://www.mhhe.com/physsci/chemistry/essentialchemistry/flash/galvan5.swf

22. 22 Cell Potential A galvanic cell consists of an oxidizing agent in one compartment that pulls electrons through a wire from a reducing agent in the other compartment. The “pull”, or driving force, on the electrons is called the cell potential (  º), or the electromotive force (emf) of the cell.  Unit of electrical potential is the volt (V).  1 joule of work per coulomb of charge transferred. (volt= Joule/coulomb)

23. Table of Reduction Potentials Measured against the Standard Hydrogen Electrode

24. 24 Figure 17.5: (a) A galvanic cell involving the reactions Zn → Zn 2+ + 2e- (at the anode/oxidation) and 2H + + 2e- → H 2 (at the cathode/reduction) has a potential of 0.76 V. (b) The standard hydrogen electrode where H 2 (g) at 1 atm is passed over a platinum electrode in contact with 1 M H+ ions. This electrode process (assuming ideal behavior) is arbitrarily assigned a value of exactly zero volts.

25. 25 Standard Reduction Potentials 2H + (aq) + Zn(s) → Zn 2+ (aq) + H 2 (g) The anode compartment? The cathode compartment? Cathode consists of a platinum electrode (inert conduction) in contact w/1M H + and bathed by H gas at 1atm. Standard Hydrogen Electrode. Potential of half-reactions are summed to obtain cell potential. Zn → Zn 2+ + 2e- oxidation 2H+ + 2e- → H 2 reduction

26. 26 Total cell potential = 0.76V; must be divided to get the half- potentials Where: [H + ]= 1M and P H2 =1 atm A potential of exactly 0 V, then the reaction will have a potential of 0.76 V because Assign this reaction 2H + + 2e- → H 2 reduction Zn → Zn 2+ + 2e- oxidation

27. 27 Example: Fe 3+ (aq) + Cu(s) → Cu 2+ (aq) + Fe 2+ (aq) Half-Reactions: Fe 3+ + e – → Fe 2+ E ° = 0.77 V Cu 2+ + 2e – → Cu E ° = 0.34 V To calculate the cell potential, we must reverse reaction 2. Cu → Cu 2+ + 2e – – E ° = – 0.34 V

28. 28 Overall Balanced Cell Reaction 2Fe 3+ + 2e – → 2Fe 2+ E ° = 0.77 V (cathode) Cu → Cu 2+ + 2e – – E ° = – 0.34 V (anode) Cell Potential: E ° cell = E °(cathode) – E °(anode) **not on formula sheet! *** the – in the equation automatically flips the anode E ° cell = 0.77 V – 0.34 V = 0.43 V

29. 29 Manipulations to obtain a balanced oxidation-reduction reaction 1.One half-reaction must be reversed, which means the sign must be reversed. 2.Since the # of e- gained must equal the # of e- lost, the half-reactions must be multiplied by integers. But the value of  º is not changed since it is an intensive property. 3.A galvanic cell runs spontaneously in the direction that gives a positive value for E ° cell.

30. 30 Line Notation Anode components are listed on the left and the cathode on the right. Mg(s) Mg 2+ (aq) Al 3+ (aq) Al(s)  Mg → Mg 2+ + 2e – (anode)  Al 3+ + 3e – → Al(cathode) Phase difference- single line Salt bridge- double line

31. The relationship between thermodynamics and electrochemistry

32. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 32 Cell Potential, Electrical Work, and Free Energy Work that can be done by a cell depends on the “push” driving force. (emf) Defined in terms of potential difference (V) between two points in a circuit.

33. 33 Volt – J of work/C of charge transferred emf = potential difference(V) = work (J) charge(C) 1J work is produced or needed, when 1C of charge is transferred between two points in the circuit that differ by a potential of 1 volt.

34. 34 Work is from the point of view of the system. Work flowing out is (-) Cell produces current, ξ is (+), current is used to do work (-). Thus, ξ and w have opposite signs ξ= -w/q (charge) -w= ξq Max work is obtained at max ξ (cell potential) -w max =q ξ max w max = -q ξ max

35. 35 Problem? W max will never be reached Why?? *in any real spontaneous process some E is always lost. Actual work is less that max work. Entropy??? What is the only type of process in which you can reach w max ?

36. 36 Although we may never achieve w max, we can calculate max potential.We use w max to calculate efficiency. Suppose we have a galvanic cell w max = 2.50V And 1.33 mol of e- were passed through the cell and the actual potential of 2.10V is registered. The actual work done is: w= - ξq ξ = actual potential difference at which the current flowed (2.10V or 2.10J/C) q = quantity of charge in coulombs transferred

37. 37 Faraday (F) The charge on 1mol of e- = faraday(F) 96,485 C of charge/mol e- q=nF= (1.33mol)(96,485C/mol) w= -q ξ = -(1.33mol)(96,485C/mol)(2.10J/C) = -2.96x10 5 J w max = -(1.33mol)(96,485C)(2.50J/C) = -3.21x10 5 J Efficiency = w/w max x 100% = 83.8%

38. 38 Relate potential of cell to free energy at constant T and P : w max =ΔG w max = -qξ max = ΔG q=nF ΔG=-nF ξ Assume at this point any potential is max.

39. 39 For standard conditions: ΔG  =  nFE  The max cell potential is directly related to the free energy difference between the reactants and the products in the cell. (experimental means!) A positive cell potential yeilds a -ΔG

40. 40 Calculate ΔG° for the reaction Cu 2+ (aq) + Fe(s) → Cu(s) + Fe 2+ (aq) Is this reaction spontaneous? Half reaction: Cu 2+ + 2e- → Cu ξ° = 0.34 V Fe(s) → Fe 2+ + 2e- -ξ° = 0.44 V Cu 2+ + Fe → Fe 2+ + Cu ξ° = 0.78 V

41. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 41 Predict whether 1M HNO 3 will dissolve gold metal to form a 1 M Au 3+ solution.