1. Redox Reactions & Electrochemistry Chapter 19 Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

2. 2Mg (s) + O 2 (g) 2MgO (s) 2Mg 2Mg 2+ + 4e - O 2 + 4e - 2O 2- _______ half-reaction (____ e - ) 19.1 ______________________ processes are 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

3. Review Oxidation – a species is oxidized when it loses one or more electrons, and it is called a reducing agent Reduction – a species is reduced when it gains one or more electrons, and it is called an oxidizing agent Oxidation and reduction always occur together, never in isolation. If something gains electrons, something else had to lose them.

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 ______. Na, Be, K, Pb, H 2, O 2, P 4 = 0 2.In monatomic ions, the oxidation number is equal to the __________________. Li +, Li = +1; Fe 3+, Fe = +3; O 2-, O = -2 3.The oxidation number of oxygen is usually _________. In H 2 O 2 and O 2 2- it is __________. 19.4

5. 4.The oxidation number of hydrogen is ___ except when it is bonded to metals in binary compounds. In these cases, its oxidation number is ___. (LiAlH 4 ) 6. The sum of the oxidation numbers of all the atoms in a molecule or ion is equal to ________________________. 5.Group IA metals are ___, IIA metals are ___ and fluorine is always ___. 19.4 Oxidation Number

6. Identify all of the oxidation numbers of the atoms in HCO 3 — 19.4 Oxidation Number H = C = O =

7. Balancing Redox Equations 19.1 1.Write the unbalanced equation for the reaction in ionic form. Balance 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+ p.606f

8. 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. 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. Electrochemical Cells 19.2 spontaneous redox reaction _______ __________ _______ __________ p.610

11. Electrochemical Cells 19.2 The difference in electrical potential between the anode and cathode is called: ____________________ 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 p.611

12. Standard Electrode Potentials 19.3 Zn (s) | Zn 2+ (1 M) || H + (1 M) | H 2 (1 atm) | Pt (s) 2e - + 2H + (1 M) 2H 2 (1 atm) Zn (s) Zn 2+ (1 M) + 2e - Anode (oxidation): Cathode (reduction): p.612

13. Standard Electrode Potentials 19.3 ____________ ____________ ____________ (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) 2e - + 2H + (1 M) 2H 2 (1 atm) Reduction Reaction Any time you see º, think “standard state conditions” p.611

14. 19.3 E 0 = 0.76 V cell Standard emf (E 0 ) cell 0.76 V = 0 - E Zn /Zn 0 2+ E Zn /Zn = -0.76 V 0 2+ Zn 2+ (1 M) + 2e - Zn E 0 = ____________ E 0 = E H /H - E Zn /Zn cell 00 + 2+ 2 Standard Electrode Potentials E 0 = E cathode - E anode cell 00 Zn (s) | Zn 2+ (1 M) || H + (1 M) | H 2 (1 atm) | Pt (s) p.612 E 0 = E reduct’n - E oxid cell 00

15. Standard Electrode Potentials 19.3 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): E 0 = E cathode - E anode cell 00 E 0 = 0.34 V cell E cell = E Cu /Cu – E H /H 2++ 2 000 0.34 = E Cu /Cu - 0 0 2+ E Cu /Cu = 0.34 V 2+ 0

16. 19.3 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 negative E 0 the greater the tendency for the substance to be oxidized Under standard-state conditions, any species on the left of a given half-reaction will react spontaneously with a species that appears on the right of any half-reaction located below it in the table (the diagonal rule) p.614f

17. 19.3 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

18. Let’s look at p.616 and work through Example 19.2 Then, let’s work Practice Exercise 19.2.

19. Can Sn reduce Zn 2+ under standard-state conditions? Practice Exercise 19.2, p.616 How do we find the answer? Look up the Eº values in Table 19.1, p.616. Zn 2+ (aq) + 2e - —> Zn(s) (Is this oxidation or reduction?) Which reactions in the table will reduce Zn 2+ (aq)?

20. 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): x 2 x 3 E 0 = E cathode - E anode cell 00 E 0 = -0.40 – (-0.74) cell E 0 = _________ cell 19.3 Practice Exercise 19.3, p.617

21. 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 Follow along on pp.617ff!

22. 19.4 Spontaneity of Redox Reactions If you know one, you can calculate the other… If you know K, you can calculate  Eº and  Gº If you know  Eº, you can calculate  Gº

23. Spontaneity of Redox Reactions 19.4 p.619 Relationships among  G º, K, and Eº cell

24. Let’s look at p.618 and work through Example 19.4 Then, let’s work Practice Exercise 19.4.

25. 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 = ___________ 0.0257 V x nE0E0 cell exp K = n = ___ 0.0257 V x 2x 2-1.24 V = exp K = ________________ E 0 = E Fe /Fe – E Ag /Ag 00 2+ + Practice Exercise 19.4, p.619

26. Let’s look at p.619 and work through Example 19.5 Then, let’s work Practice Exercise 19.5.

27. 6e - + 3Al 3+ 3Al 2Mg 2Mg 2+ + 6e - Oxidation: Reduction: Calculate  G 0 for the following reaction at 25 0 C. 2Al 3+ (aq) + 3Mg (s) 2Al (s) + 3Mg +2 (aq) 19.4 n = ? Practice Exercise 19.5, p.620  G 0 = -nFE cell 0 E 0 = E cathode - E anode cell 00

28. 6e - + 3Al 3+ 3Al 2Mg 2Mg 2+ + 6e - Oxidation: Reduction: Calculate  G 0 for the following reaction at 25 0 C. 2Al 3+ (aq) + 3Mg (s) 2Al (s) + 3Mg +2 (aq) 19.4 n = __ Practice Exercise 19.5, p.620  G 0 = -nFE cell 0 E 0 = __ V E 0 = E cathode - E anode cell 00 = ___ V = ___ X (96,500 J/V mol) X ___ V  G 0 = _______ kJ/mol

29. 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 _____________ equation At 298K 19.5 - 0.0257 V n ln Q E 0 E = - 0.0592 V n log Q E 0 E = Work through derivation on pp.620ff

30. The Nernst equation enables us to calculate E as a function of [reactants] and [products] in a redox reaction.

31. Let’s look at p.621 and work through Example 19.6 Then, let’s work Practice Exercise 19.6, p.621.

32. 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 = ___ 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 = ____________ E ___ 0________________ 19.5 Practice Exercise 19.6, p.621

33. Batteries 19.6 Leclanché cell Dry cell Zn (s) Zn 2+ (aq) + 2e - Anode: Cathode: 2NH 4 (aq) + 2MnO 2 (s) + 2e - Mn 2 O 3 (s) + 2NH 3 (aq) + H 2 O (l) + Zn (s) + 2NH 4 (aq) + 2MnO 2 (s) Zn 2+ (aq) + 2NH 3 (aq) + H 2 O (l) + Mn 2 O 3 (s)

34. Batteries Zn(Hg) + 2OH - (aq) ZnO (s) + H 2 O (l) + 2e - Anode: Cathode: HgO (s) + H 2 O (l) + 2e - Hg (l) + 2OH - (aq) Zn(Hg) + HgO (s) ZnO (s) + Hg (l) Mercury Battery 19.6

35. Batteries 19.6 Anode: Cathode: Lead storage battery PbO 2 (s) + 4H + (aq) + SO 2- (aq) + 2e - PbSO 4 (s) + 2H 2 O (l) 4 Pb (s) + SO 2- (aq) PbSO 4 (s) + 2e - 4 Pb (s) + PbO 2 (s) + 4H + (aq) + 2SO 2- (aq) 2PbSO 4 (s) + 2H 2 O (l) 4

36. Batteries 19.6 Solid State Lithium Battery

37. Batteries 19.6 A ______ ______ is an electrochemical cell that requires a continuous supply of reactants to keep functioning Anode: Cathode: O 2 (g) + 2H 2 O (l) + 4e - 4OH - (aq) 2H 2 (g) + 4OH - (aq) 4H 2 O (l) + 4e - 2H 2 (g) + O 2 (g) 2H 2 O (l)

38. Corrosion 19.7

39. Cathodic Protection of an Iron Storage Tank 19.7 p.631

40. 19.8 _______________ is the process in which electrical energy is used to cause a nonspontaneous chemical reaction to occur.

41. Electrolysis of Water 19.8

42. Electrolysis and Mass Changes charge (C) = current (A) x time (s) 1 mole e - = 96,500 C 19.8 So what is the charge on a single electron?