1. Homework Problems Chapter 18 Homework Problems: 4, 7, 16, 40, 42, 46, 50, 53, 63, 64, 66, 68, 74, 99, 102, 118, 122

2. CHAPTER 18 Electrochemistry

3. Oxidation Number An oxidation number is a number assigned to an atom in a molecule or ion indicating whether that atom is electron rich (negative oxidation number), electron poor (positive oxidation number), or neutral (oxidation number of zero). Oxidation numbers do not indicate charges of atoms, but changes in oxidation number indicate whether atoms have gained or lost electron density. A reaction where some of the oxidation numbers change in going from reactants to products is called an oxidation-reduction (redox) reaction.

4. Rules For Assigning Oxidation Numbers 1) The sum of the oxidation numbers of a species is equal to the charge of the species. 2) Atoms in elemental forms have an oxidation number of 0. 3) Hydrogen has an oxidation number of +1 when bonded with nonmetals (molecular compounds) and -1 when combined with metals (ionic compounds). 4) Oxygen usually has an oxidation number of -2. Exceptions: Elemental forms, peroxides (H 2 O 2 ), OF 2. H 2 O 2 H +1, O -1OF 2 F -1, O +2 5) Halogens. F in compounds is always -1. Other halogens are often -1, but in compounds with oxygen or other halogens the oxidation number may take on different values. 6) For assigning oxidation numbers in ionic compounds, it is useful to break the compounds up into ions.

5. Examples: Assign oxidation numbers for each atom in the following substances. H 2 O Fe(NO 3 ) 2 SO 4 2- O 3 MnO 4 - I 3 -

6. H 2 OH +1; O -2 Fe(NO 3 ) 2 = Fe 2+ and NO 3 - so Fe +2, O -2, and so N +5 SO 4 2- O -2, and so S +6 O 3 elemental form, so O = 0 MnO 4 - O -2, and so Mn +7 I 3 - I = - 1/3 (unusual, but not impossible)

7. Oxidation-Reduction Reactions In an oxidation-reduction (redox) reaction the oxidation numbers of some of the atoms involved in the reaction change in going from reactants to products. 2 Fe 2 O 3 (s) + 3 C(s)  4 Fe(s) + 3 CO 2 (g) +3 -2 0 0 +4 -2 We use the following terms in reference to redox reactions. Oxidation - An increase in the value for the oxidation number. Reduction - A decrease in the value for the oxidation number. There will always be one oxidation and one reduction process per reaction. C0 to +4, and so is oxidized. Fe+3 to 0, and so is reduced.

8. Oxidizing and Reducing Agents We call the species that is oxidized in a redox reaction the reducing agent, and the species that is reduced in a redox reaction an oxidizing agent. This seems confusing, but is based on the idea that there must be both an oxidation and a reduction taking place in a redox reaction. SO 2 (g) + 2 Fe 3+ (aq) + 2 H 2 O( )  2 Fe 2+ (aq) + SO 4 2- (aq) + 4 H + (aq) +4 -2 +3 +1 -2 +2 +6 -2 +1 In the above reaction S is oxidized (from +4 to +6) and so SO 2 is a reducing agent, and Fe is reduced (from +3 to +2) and so Fe 3+ is an oxidizing agent.

9. Balancing Oxidation-Reduction Reactions (acid or base conditions) 1) Write the unbalanced net ionic equation. 2) Assign oxidation numbers; identify the substance being oxidized and the substance being reduced. 3) Write balanced half reactions for oxidation and reduction. Include electrons as products (oxidation) or reactants (reduction) to change oxidation numbers. You may use H 2 O and H + in balancing the half reactions. 4) If necessary, multiply one or both of the half reactions by whole numbers to get the electrons to cancel. 5) Combine the half reactions to get the net ionic equation. 6) If the reaction is carried out in base conditions, add OH - to both sides of the equation to cancel any H + ions present.

10. Example: Balance the following reaction for acid conditions. Ag + (aq) + Cu(s)  Ag(s) + Cu 2+ (aq)

11. Example: Balance the following reaction for acid conditions. Ag + (aq) + Cu(s)  Ag(s) + Cu 2+ (aq) +1 0 0 +2 oxCu(s)  Cu 2+ + 2 e - redAg + (aq) + e -  Ag(s) Need to multiply the reduction reaction by 2 to get electrons to cancel. oxCu(s)  Cu 2+ + 2 e - red2 Ag + (aq) + 2 e -  2 Ag(s) netCu(s) + 2 Ag + (aq)  Cu 2+ (aq) + 2 Ag(s)

12. Example: Balance the following reaction for acid conditions. PbO 2 (s) + Mn 2+ (aq)  Pb 2+ (aq) + MnO 4 - (aq)

13. Example: Balance the following reaction for acid conditions. PbO 2 (s) + Mn 2+ (aq)  Pb 2+ (aq) + MnO 4 - (aq) +4 -2 +2 +2 +7 -2 oxMn 2+ (aq)  MnO 4 - (aq) + 5 e - Mn 2+ (aq) + 4 H 2 O( )  MnO 4 - (aq) + 5 e - + 8 H + (aq) redPbO 2 (s) + 2 e -  Pb 2+ (aq) PbO 2 (s) + 2 e - + 4 H + (aq)  Pb 2+ (aq) + 2 H 2 O( ) Now, must multiply the oxidation reaction by 2 and the reduction reaction by 5 to get electrons to cancel, giving us ox 2 Mn 2+ (aq) + 8 H 2 O( )  2 MnO 4 - (aq) + 10 e - + 16 H + (aq) red 5 PbO 2 (s) + 10 e - + 20 H + (aq)  5 Pb 2+ (aq) + 10 H 2 O( ) net 2 Mn 2+ (aq) + 5 PbO 2 (s) + 4 H + (aq)  2 MnO 4 - (aq) + 5 Pb 2+ (aq) + 2 H 2 O( )

14. Example: Balance the following reaction for base conditions. SeO 3 2- (aq) + Cl 2 (g)  SeO 4 2- (aq) + Cl - (aq)

15. Example: Balance the following reaction for base conditions. SeO 3 2- (aq) + Cl 2 (g)  SeO 4 2- (aq) + Cl - (aq) +4 -2 0 +6 -2 -1 oxSeO 3 2- (aq)  SeO 4 2- (aq) + 2 e - SeO 3 2- (aq) + H 2 O( )  SeO 4 2- (aq) + 2 e - + 2 H + (aq) redCl 2 (g) + 2 e -  2 Cl - (aq) ox SeO 3 2- (aq) + H 2 O( )  SeO 4 2- (aq) + 2 e - + 2 H + (aq) red Cl 2 (g) + 2 e -  2 Cl - (aq) net SeO 3 2- (aq) + Cl 2 (g) + H 2 O( )  SeO 4 2- (aq) + 2 Cl - (aq) + 2 H + (aq) 2 OH - (aq)  2 OH - (aq) base SeO 3 2- (aq) + Cl 2 (g) + 2 OH - (aq)  SeO 4 2- (aq) + 2 Cl - (aq) + H 2 O( ) Notice that to balance for base conditions we add OH - (aq) to convert all H + (aq) ions into H 2 O( ).

16. Electrochemical Cells An electrochemical cell is a device that can interconvert chemical and electrical energy. There are two general types of cells: Galvanic cell (battery) - In a galvanic cell a chemical reaction is used to generate a voltage. Electrolytic cell - In an electrolytic cell an external voltage is used to drive a chemical reaction in a particular direction.

17. Galvanic Cells In a galvanic cell a chemical reaction is used to generate a voltage. There will be a half-cell oxidation reaction and a half-cell reduction reaction, which combine to give the net cell reaction. The sites where the oxidation and reduction reactions take place are called electrodes. By convention anode - electrode where oxidation reaction occurs cathode - electrode where reduction reaction occurs In a galvanic cell the anode is negative and the cathode is positive. Current flow is from the anode to the cathode. Salt bridge - A gel containing ions that can transfer charge and therefore complete the electrical circuit.

18. Cell convention: Anode - Electrode where oxidation occurs. Cathode - Electrode where reduction occurs.

19. Cell Notation It is convenient to have a method for indicating the elements of a galvanic cell. This is done as follows: 1) List the elements of the cell, in order, from the anode (left) to the cathode (right). 2) Use a single vertical line to indicate a change in phase. 3) Use a double vertical line to indicate a salt bridge. We can also indicate the concentration or partial pressure of substances in the galvanic cell for cases where that information is useful.

20. Cu(s) | Cu 2+ (aq) || Ag + (aq) | Ag(s) oxCu(s)  Cu +2 (aq) + 2 e - red2 ( Ag + (aq) + e -  Ag(s) ) cellCu(s) + 2 Ag + (aq)  Cu 2+ (aq) + 2 Ag(s)

21. Example: Give the half-cell oxidation reaction, the half cell reduction reaction, and the net cell reaction for the galvanic cells with the following shorthand notation. Zn(s) | Zn 2+ (aq) || Cu 2+ (aq) | Cu Pt(s) | H 2 (g) | H + (aq) || Cl - (aq) | AgCl(s) | Ag(s)

22. Zn(s) | Zn 2+ (aq) || Cu 2+ (aq) | Cu anode cathode oxZn(s)  Zn 2+ (aq) + 2 e - redCu 2+ (aq) + 2 e -  Cu(s) cellZn(s) + Cu 2+ (aq)  Zn 2+ (aq) + Cu(s)

23. Pt(s) | H 2 (g) | H + (aq) || Cl - (aq) | AgCl(s) | Ag(s) anode cathode oxH 2 (g)  2 H + (aq) + 2 e - red2 ( AgCl(s) + e -  Ag(s) + Cl - (aq) ) cellH 2 (g) + 2 AgCl(s)  2 Ag(s) + 2 H + (aq) + 2 Cl - (aq) Standard hydrogen electrode (SHE) - Electrode where the following process takes place (either as an oxidation or as a reduction half reaction). oxH 2 (g)  2 H + (aq) + 2 e - red 2 H + (aq) + 2 e -  H 2 (g) Platinum is used as the electrode since it is chemically inert.

24. Standard Cell Potential The standard half-cell potential (E  ) is the voltage generated by a half-cell reaction. We can talk about either a standard half-cell reduction potential (E  red ) or a standard half-cell oxidation potential (E  ox ) which correspond to the potential generated by these processes when all reactants and products are present at standard concentration. Since galvanic cells must have both an oxidation and a reduction half-cell reaction, the standard half-cell reduction potential is defined relative to the potential generated by the standard hydrogen electrode (SHE). 2 H + (aq) + 2 e -  H 2 (g)E  red = 0.00 v (by definition) H 2 (g)  2 H + (aq) + 2 e - E  ox = 0.00 v Based on this definition experimental values for other half-cell reduction potentials can be found.

25. Example: Consider the following galvanic cell. Pt(s) | H 2 (g) | H + (aq) || Cu 2+ (aq) | Cu(s) oxH 2 (g)  2 H + (aq) + 2 e - E  ox redCu 2+ (aq) + 2 e -  Cu(s)E  red cellH 2 (g) + Cu 2+ (aq)  2 H + (aq) + Cu(s) E  cell The experimental value for the cell potential is E  cell = 0.34 v. Since E  cell = E  ox + E  red E  red = E  cell - E  ox = 0.34 v - 0.00 v = 0.34 v All half-cell reduction potentials can be found by the use of this method. A table of half-cell reduction potentials is found in Table 18.1, or Appendix II-D of the text.

26. Properties of Half-Cell Potentials 1) Reversing the direction of the reaction (changing the reaction from a reduction to an oxidation) changes the sign of the potential. Cu 2+ (aq) + 2 e -  Cu(s) E  red = 0.34 v Cu(s)  Cu 2+ (aq) + 2 e - E  ox = - 0.34 v 2) Multiplying a reaction by a positive constant has no effect on the value of the half-cell potential. This is because the potential is defined as that found for standard conditions. Cu 2+ (aq) + 2 e -  Cu(s) E  red = 0.34 v 2 Cu 2+ (aq) + 4 e -  2 Cu(s) E  red = 0.34 v

27. Finding Cell Potentials For Standard Conditions The standard cell potential for a galvanic cell can be found by adding together the half-cell oxidation potential and the half-cell reduction potential. Example: Zn(s) | Zn 2+ (aq) || Cu 2+ (aq) | Cu oxZn(s)  Zn 2+ (aq) + 2 e - E  ox = 0.76 v redCu 2+ (aq) + 2 e -  Cu(s)E  red = 0.34 v cellZn(s) + Cu 2+ (aq)  Zn 2+ (aq) + Cu(s)E  cell = 1.10 v Because half-cell oxidation potentials can be found from half-cell reduction potentials (by reversing the direction of the reaction and the sign of the potential) we typically only keep tables of the half-cell reduction potentials.

28. Free Energy and Cell Potential The connection between galvanic cells and thermodynamics is through the following relationship (which can be derived).  G rxn = - n F E cell  G  rxn = - n F E  cell In the above expressions n is the number of electrons transferred per mole of reaction (which can be found from either the oxidation or the reduction half-cell reaction) and F is the Faraday constant, a conversion factor between Coulombs (MKS unit of charge) and moles of charge. F = 96485 C/molNote (1 volt) (1 Coulomb) = 1 Joule In the previous galvanic cell n = 2 and E  cell = 1.10 v, so  G  rxn = - (2) (96485 C/mol) (1.10 v) = - 212.3 kJ/mol (spontaneous)

29. Spontaneous Cell Reactions Based on the above results we can determine when a cell reaction will occur spontaneously. Since  G rxn = - n F E cell we have the following possibilities E cell > 0 means  G rxn < 0 and the reaction is spontaneous E cell = 0 means  G rxn = 0 and the system is at equilibrium E cell 0 and the reaction is not spontaneous Note that in the last case the reverse reaction will be spontaneous (a consequence of free energy being a state function). Since the above is true in general it is also true for cells where standard conditions (so true if we add “  ” to  G rxn and E cell in the above expressions.

30. Electrochemical Series The electrochemical series is a list of half-cell reduction potentials, in order, from most positive to most negative voltage.

31. Properties of the Electrochemical Series Since reactions where E  cell is positive are spontaneous, we may say the more positive a half-cell potential the more likely it will lead to a spontaneous cell reaction. That means that reactions at the top of the series are most likely to occur as reduction reactions, and reactions at the bottom of the series are least likely to occur as reduction reactions. Because changing from a reduction to an oxidation reaction changes the sign of the half-cell potential, it also means that reactions at the bottom of the series are most likely to occur as oxidation reactions, and reactions at the top of the series are least likely to occur as oxidation reactions.

32. Oxidizing and Reducing Agents Oxidizing agent – Is reduced in a chemical reaction, and so oxidizes another substance. Reducing agent – Is oxidized in a chemical reaction, and so reduces another substance. In the electrochemical series substances at the top left of the series are the best oxidizing agents (since they are most likely to be reduced) and substances at the lower right of the series are the best reducing agents (since they are most likely to be oxidized).

33. Example: Which of the following reactions will occur spontaneously for standard conditions? Ni 2+ (aq) + Zn(s)  Ni(s) + Zn 2+ (aq) Zn 2+ (aq) + Ni(s)  Zn(s) + Ni 2+ (aq) In the first reaction Ni 2+ is an oxidizing agent, while in the second reaction Zn 2+ is an oxidizing agent. Based on the above portion of the electrochemical series Ni 2+ is more likely to be reduced than Zn 2+, and so it is the first reaction that will be spontaneous (in fact E  cell = + 0.50 v for the first process and – 0.50 v for the second process).

34. The Nernst Equation The central equation for galvanic cells is the Nernst equation. It may be derived in the following manner. Recall the following thermochemical relationship  G rxn =  G  rxn + RT ln Q However, we have a connection between free energy and cell potential  G rxn = - n F E cell If we use the above to substitute, we get - n F E cell = - n F E  cell + RT ln Q E cell = E  cell – RT ln Qthe Nernst equation. nF

35. Uses of the Nernst Equation The Nernst equation can be used to predict the value for the cell potential for cases where reactants and products are not at their standard concentrations. Example: Consider the following galvanic cell Zn(s) | Zn 2+ (aq, 0.010 M) || Cu 2+ (aq, 1.00 M) | Cu What is E cell for the above galvanic cell?

36. Zn(s) | Zn +2 (aq, 0.010 M) || Cu 2+ (aq, 1.00 M) | Cu What is E cell for the above galvanic cell? From the Nernst equation E cell = E  cell – RT ln Q nF oxZn(s)  Zn 2+ (aq) + 2 e - E  ox = 0.76 v redCu 2+ (aq) + 2 e -  Cu(s)E  red = 0.34 v cellZn(s) + Cu 2+ (aq)  Zn 2+ (aq) + Cu(s)E  cell = 1.10 v n = 2 Q = [Zn 2+ ] = (0.010) = 0.010 [Cu 2+ ] (1.00) So E cell = 1.10 v – (8.314 J/mol. K) (298.2 K) ln(0.010) (2) (96485 C/mol) = 1.10 v + 0.059 v = 1.16 v

37. The Nernst Equation and Equilibrium The Nernst equation says E cell = E  cell – RT ln Q nF Consider a system at equilibrium. In this case  G rxn = 0, which means E cell = 0, and Q = K, the equilibrium constant. If we substitute this into the Nernst equation we get 0 = E  cell – RT ln K nF E  cell = RT ln K nF ln K = n F E  cell RT Using this expression we can obtain values for the equilibrium constant for any reaction that can be represented by a galvanic cell.

38. Example: The solubility product for silver bromide (AgBr) is K sp = [Ag + ][Br - ]. Use data from Appendix II-D for half-cell reduction potentials to find a numerical value for K sp.

39. The solubility product reaction is AgBr(s)  Ag + (aq) + Br - (aq) ox Ag(s)  Ag + (aq) + e - E  ox = - 0.80 v red AgBr(s) + e -  Ag(s) + Br - E  red = + 0.07 v cell AgBr(s)  Ag + (aq) + Br - (aq) E  cell = - 0.73 v So ln K = n F E  cell = (1) (96485 C/mol) (- 0.73 v) = - 28.41 RT (8.314 J/mol. K) (298.2 K) K sp = e -28.41 = 4.6 x 10 -13

40. oxH 2 (g)  2 H + (aq) + 2 e - E  ox = 0.00 v red“reference electrode”E  red cellE  cell = E  red E cell = E  cell - RT ln Q n = 2 ; Q = [H + ] 2 /(p H2 ) nF = E  red - (8.314 J/mol.K) (298.2 K) ln{([H + ] 2 /(p H2 )} (2) (96485 C/mol) For p H2 = 1.00 atm, this becomes E cell = E  red - (0.0256 v) ln[H + ] = E  red + (0.0256 v) { - ln[H + ]} = E  red + (0.0592 v) { - log 10 [H + ]} = E  red + (0.0592 v) pH

41. Electrolysis In electrolysis an external voltage is used to force a chemical reaction to take place in a particular direction. Anode - oxidation reaction (+) Cathode - reduction reaction (-)

42. As in galvanic cells, we can write down the half-cell oxidation and reduction reactions and the net cell reactions for an electrolytic cell. ox2 Cl - ( )  Cl 2 (g) + 2 e - red2 ( Na + ( ) + e -  Na( ) ) cell2 Na + ( ) + 2 Cl - ( )  2 Na( ) + Cl 2 (g) Notice that in electrolysis the reactions that take place are opposite of those that would normally occur. The Nernst equation does not apply for electrolysis, since we are not generating a voltage in an electrolytic cell. All that is required is that the external voltage being used is large enough to drive the reaction in the desired direction.

43. Calculations Involving Electrolytic Cells The main type of calculation done in connection with electrolytic cells is finding the mass of a particular substance that can be produced in a particular cell, given the operating conditions. To do these calculations requires: 1) The appropriate half cell reaction. 2) The number of moles of electrical charge transferred. 3) The molecular mass of the product.

44. Example: Electrolysis is carried out for molten sodium chloride. The cell is operated at a current i = 20.0 amp (1 amp = 1 C/s) for a period of 1.00 hour. How many grams of Cl 2 (MW = 70.90 g/mol) will be produced?

45. Electrolysis is carried out for molten sodium chloride. The cell is operated at a current i = 20.0 amp (1 amp = 1 C/s) for a period of 1.00 hour. How many grams of Cl 2 will be produced? 1) The appropriate half cell reaction. 2 Cl - ( )  Cl 2 (g) + 2 e - 2) The number of moles of electrical charge transferred. mol charge = 1.00 hr 3600 s 20.0 C 1 mol = 0.746 mol 1 hr s 96485 C 3) The molecular mass of the product. M(Cl 2 ) = 70.90 g/mol. g Cl 2 = 0.746 mol charge 1 mol Cl 2 70.90 g Cl 2 = 26.5 g 2 mol charge mol Cl 2

46. End of Chapter 18 “Anode comes from the Greek  (upward) and  (a way), and therefore suggests the rising of the sun in the east. Cathode comes from  (downward), and is related to the setting of the sun, in the west.” - Keith Laidler, in The World of Physical Chemistry, explaining how Michael Faraday came up with the terms “anode” and “cathode”. “…the total annual estimated direct cost of corrosion in the U.S. (in 1998) is a staggering $276 billion - approximately 3.1% of the nation’s Gross Domestic Product (GDP).” - Corrosion Costs and Preventive Strategies in the United States “Rust never sleeps.” - Neil Young (borrowed from Devo)