1. © 2015 Pearson Education, Inc. Chapter 20 Electrochemistry James F. Kirby Quinnipiac University Hamden, CT Lecture Presentation

2. Electrochemistry © 2015 Pearson Education, Inc. Electrochemistry Electrochemistry is the study of the relationships between electricity and chemical reactions. It includes the study of both spontaneous and nonspontaneous processes.

3. Electrochemistry © 2015 Pearson Education, Inc. Synopsis of Assigning Oxidation Numbers (as a Reminder) 1.Elements = 0 2.Monatomic ion = charge 3.F: –1 4.O: –2 (unless peroxide = –1) 5.H: +1 (unless a metal hydride = –1) 6.The sum of the oxidation numbers equals the overall charge (0 in a compound).

4. Electrochemistry © 2015 Pearson Education, Inc. Oxidation Numbers To keep track of what loses electrons and what gains them, we assign oxidation numbers. If the oxidation number increases for an element, that element is oxidized. If the oxidation number decreases for an element, that element is reduced.

5. Electrochemistry © 2015 Pearson Education, Inc. Oxidation and Reduction A species is oxidized when it loses electrons. –Zinc loses two electrons, forming the Zn 2+ ion. A species is reduced when it gains electrons. –H + gains an electron, forming H 2. An oxidizing agent causes something else to be oxidized (H + ); a reducing agent causes something else to be reduced (Zn).

6. Electrochemistry © 2015 Pearson Education, Inc. Half-Reactions The oxidation and reduction are written and balanced separately. We will use them to balance a redox reaction. For example, when Sn 2+ and Fe 3+ react,

7. Electrochemistry © 2015 Pearson Education, Inc. Balancing Redox Equations 19.1 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+

8. Electrochemistry © 2015 Pearson Education, Inc. 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 19.1

9. Electrochemistry © 2015 Pearson Education, Inc. 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 19.1 9.For reactions in basic solutions, add OH - to both sides of the equation for every H + that appears in the final equation.

10. Electrochemistry © 2015 Pearson Education, Inc. Balancing Redox Equations: The Half-Reactions Method (a Synopsis) 1)Make two half-reactions (oxidation and reduction). 2)Balance atoms other than O and H. 3)Then, balance O and H using H 2 O/H +. 4) Add electrons to balance charges. 5) Multiply by common factor to make electrons in half- reactions equal. 6) Add the half-reactions. 7) Simplify by dividing by common factor or converting H + to OH – if basic. 8) Double-check atoms and charges balance!

11. Electrochemistry © 2015 Pearson Education, Inc. The Half-Reaction Method Consider the reaction between MnO 4 – and C 2 O 4 2– : MnO 4 – (aq) + C 2 O 4 2– (aq)  Mn 2+ (aq) + CO 2 (aq) Assigning oxidation numbers shows that Mn is reduced (+7  +2) and C is oxidized (+3  +4).

12. Electrochemistry © 2015 Pearson Education, Inc. Oxidation Half-Reaction C 2 O 4 2–  CO 2 To balance the carbon, we add a coefficient of 2: C 2 O 4 2–  2 CO 2

13. Electrochemistry © 2015 Pearson Education, Inc. Oxidation Half-Reaction C 2 O 4 2–  2 CO 2 The oxygen is now balanced as well. To balance the charge, we must add two electrons to the right side: C 2 O 4 2–  2 CO 2 + 2e –

14. Electrochemistry © 2015 Pearson Education, Inc. Reduction Half-Reaction MnO 4 –  Mn 2+ The manganese is balanced; to balance the oxygen, we must add four waters to the right side: MnO 4 –  Mn 2+ + 4 H 2 O

15. Electrochemistry © 2015 Pearson Education, Inc. Reduction Half-Reaction MnO 4 –  Mn 2+ + 4 H 2 O To balance the hydrogen, we add 8H + to the left side: 8 H + + MnO 4 –  Mn 2+ + 4 H 2 O

16. Electrochemistry © 2015 Pearson Education, Inc. Reduction Half-Reaction 8 H + + MnO 4 –  Mn 2+ + 4 H 2 O To balance the charge, we add 5e – to the left side: 5e – + 8 H + + MnO 4 –  Mn 2+ + 4 H 2 O

17. Electrochemistry © 2015 Pearson Education, Inc. Combining the Half-Reactions Now we combine the two half-reactions together: C 2 O 4 2–  2 CO 2 + 2e – 5e – + 8 H + + MnO 4 –  Mn 2+ + 4 H 2 O To make the number of electrons equal on each side, we will multiply the first reaction by 5 and the second by 2:

18. Electrochemistry © 2015 Pearson Education, Inc. Combining the Half-Reactions 5 C 2 O 4 2–  10 CO 2 + 10e – 10e – + 16 H + + 2 MnO 4 –  2 Mn 2+ + 8 H 2 O When we add these together, we get 10e – + 16 H + + 2 MnO 4 – + 5 C 2 O 4 2–  2 Mn 2+ + 8 H 2 O + 10 CO 2 +10e –

19. Electrochemistry © 2015 Pearson Education, Inc. Combining the Half-Reactions 10e – + 16 H + + 2 MnO 4 – + 5 C 2 O 4 2–  2 Mn 2+ + 8 H 2 O + 10 CO 2 +10e – The only thing that appears on both sides is the electrons. Subtracting them, we are left with 16 H + + 2 MnO 4 – + 5 C 2 O 4 2–  2 Mn 2+ + 8 H 2 O + 10 CO 2 (Verify that the equation is balanced by counting atoms and charges on each side of the equation.)

20. Electrochemistry © 2015 Pearson Education, Inc. Balancing in Basic Solution A reaction that occurs in basic solution can be balanced as if it occurred in acid. Once the equation is balanced, add OH – to each side to “neutralize” the H + in the equation and create water in its place. If this produces water on both sides, subtract water from each side so it appears on only one side of the equation.

21. Electrochemistry © 2015 Pearson Education, Inc. Galvanic Cells 19.2 spontaneous redox reaction anode oxidation cathode reduction

22. Electrochemistry © 2015 Pearson Education, Inc. Voltaic Cells In spontaneous redox reactions, electrons are transferred and energy is released. That energy can do work if the electrons flow through an external device. This is a voltaic cell.

23. Electrochemistry © 2015 Pearson Education, Inc. Voltaic Cells The oxidation occurs at the anode. The reduction occurs at the cathode. When electrons flow, charges aren’t balanced. So, a salt bridge, usually a U-shaped tube that contains a salt/agar solution, is used to keep the charges balanced.

24. Electrochemistry © 2015 Pearson Education, Inc. Voltaic Cells In the cell, electrons leave the anode and flow through the wire to the cathode. Cations are formed in the anode compartment. As the electrons reach the cathode, cations in solution are attracted to the now negative cathode. The cations gain electrons and are deposited as metal on the cathode.

25. Electrochemistry © 2015 Pearson Education, Inc. Electromotive Force (emf) Water flows spontaneously one way in a waterfall. Comparably, electrons flow spontaneously one way in a redox reaction, from high to low potential energy.

26. Electrochemistry © 2015 Pearson Education, Inc. Electromotive Force (emf) The potential difference between the anode and cathode in a cell is called the electromotive force (emf). It is also called the cell potential and is designated E cell. It is measured in volts (V). One volt is one joule per coulomb (1 V = 1 J/C).

27. Electrochemistry © 2015 Pearson Education, Inc. Standard Reduction Potentials Reduction potentials for many electrodes have been measured and tabulated. The values are compared to the reduction of hydrogen as a standard.

28. Electrochemistry © 2015 Pearson Education, Inc. Standard Hydrogen Electrode Their reference is called the standard hydrogen electrode (SHE). By definition as the standard, the reduction potential for hydrogen is 0 V: 2 H + (aq, 1M) + 2e –  H 2 (g, 1 atm)

29. Electrochemistry © 2015 Pearson Education, Inc. Standard Cell Potentials The cell potential at standard conditions can be found through this equation: E cell ° = E red (cathode) – E red (anode) °° Because cell potential is based on the potential energy per unit of charge, it is an intensive property.

30. Electrochemistry © 2015 Pearson Education, Inc. Cell Potentials For the anode in this cell, E ° red = –0.76 V For the cathode, E ° red = +0.34 V So, for the cell, E ° cell = E ° red (anode) – E ° red (cathode) = +0.34 V – (–0.76 V) = +1.10 V

31. Electrochemistry © 2015 Pearson Education, Inc. Oxidizing and Reducing Agents The more positive the value of E ° red, the greater the tendency for reduction under standard conditions. The strongest oxidizers have the most positive reduction potentials. The strongest reducers have the most negative reduction potentials.

32. Electrochemistry © 2015 Pearson Education, Inc. Free Energy and Redox Spontaneous redox reactions produce a positive cell potential, or emf. E ° = E ° red (reduction) – E ° red (oxidation) Note that this is true for ALL redox reactions, not only for voltaic cells. Since Gibbs free energy is the measure of spontaneity, positive emf corresponds to negative ΔG. How do they relate? ΔG = –nFE (F is the Faraday constant, 96,485 C/mol.)

33. Electrochemistry © 2015 Pearson Education, Inc. Free Energy, Redox, and K How is everything related? ΔG ° = –nFE ° = –RT ln K

34. Electrochemistry © 2015 Pearson Education, Inc. 19.4 Spontaneity of Redox Reactions  G = -nFE cell  G 0 = -nFE cell 0 n = number of moles of electrons in reaction F = 96,500 J V mol = 96,500 C/mol  G 0 = -RT ln K = -nFE cell 0 E cell 0 = RT nF ln K (8.314 J/K mol)(298 K) n (96,500 J/V mol) ln K = = 0.0257 V n ln K E cell 0 = 0.0592 V n log K E cell 0 F (Faraday constant)= electrical charge contained in 1 mole of electrons 1 F = 96,485.3 coulombs or 96,500 coulombs or 96,500 J/V mol e - ∆G = change in free energy; max. amount of work that can be obtained K = equlibrium constant of a redox reaction

35. Electrochemistry © 2015 Pearson Education, Inc. Spontaneity of Redox Reactions 19.4

36. Electrochemistry © 2015 Pearson Education, Inc. 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) = 0.0257 V n ln K E cell 0 19.4 E 0 = -0.44 – (0.80) E 0 = -1.24 V 0.0257 V x nE0E0 cell ln K = n = 2 0.0257 V x 2x 2-1.24 V ln K = K = 1.23 x 10 -42 E 0 = E Fe /Fe – E Ag /Ag 00 2+ + K = e -96.5

37. Electrochemistry © 2015 Pearson Education, Inc. The Effect of Concentration on Cell Emf  G =  G 0 + RT ln Q  G = -nFE  G 0 = -nFE 0 -nFE = -nFE 0 + RT ln Q E = E 0 - ln Q RT nF Nernst equation At 298 19.5 - 0.0257 V n ln Q E 0 E = - 0.0592 V n log Q E 0 E = Q is the reaction quotient

38. Electrochemistry © 2015 Pearson Education, Inc. 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 Cd Cd 2+ + 2e - Oxidation: Reduction: n = 2 E 0 = -0.44 – (-0.40) E 0 = -0.04 V E 0 = E Fe /Fe – E Cd /Cd 00 2+ - 0.0257 V n ln Q E 0 E = - 0.0257 V 2 ln -0.04 VE = 0.010 0.60 E = 0.013 E > 0Spontaneous 19.5

39. Electrochemistry © 2015 Pearson Education, Inc. Nernst Equation Remember, ΔG = ΔG ° + RT ln Q So, –nFE = nFE ° + RT ln Q Dividing both sides by –nF, we get the Nernst equation: E = E ° – (RT/nF) ln Q OR E = E ° – (2.303 RT/nF) log Q Using standard thermodynamic temperature and the constants R and F, E = E ° – (0.0592/n) log Q

40. Electrochemistry © 2015 Pearson Education, Inc. Concentration Cells Notice that the Nernst equation implies that a cell could be created that has the same substance at both electrodes, called a concentration cell. For such a cell,would be 0, but Q would not. E cell ° Therefore, as long as the concentrations are different, E will not be 0.

41. Electrochemistry © 2015 Pearson Education, Inc. Some Applications of Cells Electrochemistry can be applied as follows:  Batteries: a portable, self-contained electrochemical power source that consists of one or more voltaic cells.  Batteries can be primary cells (cannot be recharged when “dead”—the reaction is complete) or secondary cells (can be recharged).  Prevention of corrosion (“rust-proofing”)  Electrolysis

42. Electrochemistry © 2015 Pearson Education, Inc. Some Examples of Batteries Lead–acid battery: reactants and products are solids, so Q is 1 and the potential is independent of concentrations; however, made with lead and sulfuric acid (hazards). Alkaline battery: most common primary battery. Ni–Cd and Ni–metal hydride batteries: lightweight, rechargeable; Cd is toxic and heavy, so hydrides are replacing it. Lithium-ion batteries: rechargeable, light; produce more voltage than Ni-based batteries.

43. Electrochemistry © 2015 Pearson Education, Inc. Some Batteries Lead–Acid BatteryAlkaline Battery

44. Electrochemistry © 2015 Pearson Education, Inc. Lithium-Ion Battery

45. Electrochemistry © 2015 Pearson Education, Inc. Fuel Cells When a fuel is burned, the energy created can be converted to electrical energy. Usually, this conversion is only 40% efficient, with the remainder lost as heat. The direct conversion of chemical to electrical energy is expected to be more efficient and is the basis for fuel cells. Fuel cells are NOT batteries; the source of energy must be continuously provided.

46. Electrochemistry © 2015 Pearson Education, Inc. Hydrogen Fuel Cells In this cell, hydrogen and oxygen form water. The cells are twice as efficient as combustion. The cells use hydrogen gas as the fuel and oxygen from the air.

47. Electrochemistry © 2015 Pearson Education, Inc. Corrosion Corrosion is oxidation. Its common name is rusting.

48. Electrochemistry © 2015 Pearson Education, Inc. Preventing Corrosion Corrosion is prevented by coating iron with a metal that is more readily oxidized. Cathodic protection occurs when zinc is more easily oxidized, so that metal is sacrificed to keep the iron from rusting.

49. Electrochemistry © 2015 Pearson Education, Inc. Preventing Corrosion Another method to prevent corrosion is used for underground pipes. A sacrificial anode is attached to the pipe. The anode is oxidized before the pipe.

50. Electrochemistry © 2015 Pearson Education, Inc. Electrolysis Nonspontaneous reactions can occur in electrochemistry IF outside electricity is used to drive the reaction. Use of electrical energy to create chemical reactions is called electrolysis.

51. Electrochemistry © 2015 Pearson Education, Inc. Electrolysis of Water 19.8

52. Electrochemistry © 2015 Pearson Education, Inc. Electrolysis and “Stoichiometry” 1 coulomb = 1 ampere × 1 second Q = It = nF Q = charge (C) I = current (A) t = time (s) n = moles of electrons that travel through the wire in the given time F = Faraday’s constant NOTE: n is different than that for the Nernst equation!

53. Electrochemistry © 2015 Pearson Education, Inc. Electrolysis and Mass Changes charge (C) = current (A) x time (s) 1 mole e - = 96,500 C 19.8

54. Electrochemistry © 2015 Pearson Education, Inc. 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) 2 mole e - = 1 mole Ca mol Ca = 0.452 C s x 1.5 hr x 3600 s hr96,500 C 1 mol e - x 2 mol e - 1 mol Ca x = 0.0126 mol Ca = 0.50 g Ca 19.8