1. Electrochemistry Chapter 3

2. 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 Electrochemical 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 002+2-

3. - Oxidation is the loss of electrons and reduction is the gain of electrons. One can’t occur without the other, but we frequently break them up into the two half reactions. Example: Zn o + Cu +2 -----) Zn +2 + Cu o Oxidn: Zn o -----) Zn +2 + 2 e - Redn: Cu +2 + 2 e - -----) Cu o a) Oxidizing Agent causes oxidn. & is reduced (Cu 2+ ) b) Reducing Agent causes redn. & is oxidized (Zn o ) (Know Terms & Definitions) - Classes of Redox Reactions: Combination, Decomposition, Displacement, Combustion

4. A Summary of an Oxidation-Reduction Process

5. Chemistry in Action: Breath Analyzer 4.4 3CH 3 COOH + 2Cr 2 (SO 4 ) 3 + 2K 2 SO 4 + 11H 2 O 3CH 3 CH 2 OH + 2K 2 Cr 2 O 7 + 8H 2 SO 4 +6 +3

6. 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. 4.4

7. 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 (H -1 when combined with Ia, IIa, IIIa). 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 - ? 4.4

8. Examples: Determine the ON of single N in each: N 2 N 2 O NO NF 3 N 2 O 4 NO 3 -1

9. - In any redox reaction you should be able to determine: The ON for each element Which reagent has been oxidized Which reagent has been reduced Which reagent is the oxidizing agent Which reagent is the reducing agent What are the two half reactions for each How many moles of e - are transferred in each half rxn

10. 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+

11. 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

12. 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

13. * Add OH - equal to # H + to both sides. ** Convert H + & OH - on same side to H 2 O & simplify by canceling H 2 O if needed. *** Check result by noting if following are both balanced: elements & charge. 9.For reactions in basic solutions : Do all previous steps from 1 to 6 and:

14. Examples (Acidic) Zn + NO 3 - + H + -----) Zn +2 + NH 4 + + H 2 O 1)Zn -----) Zn 2+ 0 ---) +2 on Zn (Oxidn.) NO 3 - -----) NH 4 + +5 ---) -3 on N (Redn.) 2)Zn -----) Zn 2+ + 2 e - (balance all but H&O) NO 3 - -----) NH 4 + + 3 H 2 O (balance O with H 2 O) 10 H + + NO 3 - -----) NH 4 + + 3 H 2 O (balance H with H + ) 10 H + + NO 3 - + 8 e - -----) NH 4 + + 3 H 2 O (bal. Charge w e - ) 3)4 [Zn -----) Zn 2+ + 2 e - ] (8e - lost) 1 [10 H + + NO 3 - + 8 e - -----) NH 4 + + 3 H 2 O] (8e - gained) 4 Zn + 10 H + + NO 3 - + 8 e - ---) 4 Zn 2+ + 8 e - + NH 4 + + 3 H 2 O 4 Zn + 10 H + + NO 3 - -----) 4 Zn 2+ + NH 4 + + 3 H 2 O

15. Example MnO 4 - + S 2 O 3 2- -----) Mn 2+ + SO 4 2- (acidic) 1) a) MnO 4 - ----) Mn 2+ [+7 ---) +2 Redn] b) S 2 O 3 2- ----) SO 4 2- [+2 ---) +6 per S Oxn] 2) a) 5e - + 8H + + MnO 4 - ----) Mn 2+ + 4H 2 O (x 8) b) 1S 2 O 3 2- -----) 2SO 4 2- 5H 2 O + S 2 O 3 2- -----) 2SO 4 2- + 10H + 5H 2 O + S 2 O 3 2- -----) 2SO 4 2- + 10H + + 8e - (x 5) 3) 40e - + 64H + + 8MnO 4 - + 25H 2 O + 5S 2 O 3 2- ---) 8Mn 2+ + 32H 2 O + 10SO 4 2- + 50H + + 40e - 14 H + + 8 MnO 4 - + 5 S 2 O 3 2- ----) 8 Mn 2+ + 7 H 2 O + 10 SO 4 2-

16. Example (Basic) MnO 4 - + I - -----) MnO 2 + IO 3 - 1) MnO 4 - -----) MnO 2 +7 ---) +4 Mn I - -----) IO 3 - -1 ----) +5 I 2) 4H + + MnO 4 - + 3e - -----) MnO 2 + 2H 2 O 3H 2 O + I - -----) IO 3 - + 6H + + 6e - 3) 2 [4H + + MnO 4 - + 3e - -----) MnO 2 + 2H 2 O] 1 [3H 2 O + I - -----) IO 3 - + 6H + + 6e - ] 8H + + 2MnO 4 - + 6e - + 3H 2 O + I - ----) 2MnO 2 + 4H 2 O + IO 3 - + 6H + + 6e - 2H + + 2MnO 4 - + I - ----) 2MnO 2 + H 2 O + IO 3 - 4) 2OH - + 2H + + 2MnO 4 - + I - ----) 2MnO 2 + H 2 O + IO 3 - + 2OH - 5) 2H 2 O + 2MnO 4 - + I - ----) 2MnO 2 + H 2 O + IO 3 - + 2OH - H 2 O + 2MnO 4 - + I - ----) 2MnO 2 + IO 3 - + 2OH - 6) Double check balancing of atoms & charge.

17. Voltaic Cells A. Introduction - A voltaic cell (battery) is an electrochemical cell in which a spontaneous redox reaction generates an electric current. - It consists of two half cells which are connected: Oxidation half cell consists of electrode (anode = - ) & electrolyte. Reduction half cell consists of electrode (cathode = + ) & electrolyte. External connection of anode to the cathode. Internal connection of a salt bridge that allows the flow of ions but prevents mixing of electrolytes from two half cells. Electrons travel from anode to cathode.

18. Galvanic Cells 19.2 spontaneous redox reaction anode oxidation cathode reduction

19. Note the following Conventions: Zn + Cu 2+ -----) Zn 2+ + Cu Cell Diagram Zn | Zn 2+ (1.0 M) || Cu 2+ (1.0 M) | Cu Anode (-) (oxidation) Zn ---) Zn 2+ + 2e - Cathode (+) (reduction) Cu 2+ + 2e - ---) Cu Anode written on left Cathode written on right | = phase boundary || = salt bridge Zn = anode electrode Cu = cathode electrode, = separates ions in the same solution M = Molar concentration of ions

20. Galvanic Cells 19.2 The difference in electrical potential between the anode and cathode is called: cell voltage electromotive force (emf) cell potential 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

21. Standard Reduction Potentials 19.3 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 H + /H 2 E 0 = 0 V Standard hydrogen electrode (SHE) 2e - + 2H + (1 M) H 2 (1 atm) Reduction Reaction

22. Standard Reduction Potentials 19.3 Zn (s) | Zn 2+ (1 M) || H + (1 M) | H 2 (1 atm) | Pt (s) 2e - + 2H + (1 M) H 2 (1 atm) Zn (s) Zn 2+ (1 M) + 2e - Anode (oxidation): Cathode (reduction): Zn (s) + 2H + (1 M) Zn 2+ + H 2 (1 atm)

23. 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 = -0.76 V E 0 = E H /H - E Zn /Zn cell 00 + 2+ 2 Standard Reduction Potentials E 0 = E cathode - E anode cell 00 Zn (s) | Zn 2+ (1 M) || H + (1 M) | H 2 (1 atm) | Pt (s)

24. Standard Reduction 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): H 2 (1 atm) + Cu 2+ (1 M) Cu (s) + 2H + (1 M) 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

25. 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 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

26. 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 = E cathode - E anode cell 00 E 0 = -0.40 – (-0.74) cell E 0 = 0.34 V cell 19.3

27. 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

28. Spontaneity of Redox Reactions 19.4  G 0 = -RT ln K = -nFE cell 0

29. 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 exp K = n = 2 0.0257 V x 2x 2-1.24 V = exp K = 1.23 x 10 -42 E 0 = E Fe /Fe – E Ag /Ag 00 2+ +

30. 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 =

31. 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

32. 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)

33. 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

34. 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

35. Batteries 19.6 Solid State Lithium Battery

36. Batteries 19.6 A fuel cell 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)

37. Chemistry In Action: Bacteria Power CH 3 COO - + 2O 2 + H + 2CO 2 + 2H 2 O

38. Corrosion 19.7

39. Cathodic Protection of an Iron Storage Tank 19.7

40. 19.8 Electrolysis 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 The quantity of electricity carried by 1 mole of electrons is called a faraday, which is equal to 96,500 C Electroplating Plating means depositing the neutral metal on the electrode by reducing the metal ions in solution

43. 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

44. Chemistry In Action: Dental Filling Discomfort Hg 2 /Ag 2 Hg 3 0.85 V 2+ Sn /Ag 3 Sn -0.05 V 2+ Sn /Ag 3 Sn -0.05 V 2+