1. 1 17.7-8 Electrolysis & Applications Since chemical oxidation-reduction involves the transfer of electrons from one substance to another, it should be possible to harness the flow of electrons to produce electricity. We do this with voltaic cells. Since chemical oxidation-reduction involves the transfer of electrons from one substance to another, it should be possible to harness the flow of electrons to produce electricity. We do this with voltaic cells. Electricity can also be used to cause non- spontaneous chemical reactions (i.e. recharging batteries). This process is called electrolysis (carried out in electrolytic cells) Electricity can also be used to cause non- spontaneous chemical reactions (i.e. recharging batteries). This process is called electrolysis (carried out in electrolytic cells)

2. 2 Opening Demo Electrolyis of KI(aq) Electrolyis of KI(aq) Discuss which ions attract to which electrodes. Discuss which ions attract to which electrodes. Names / charges of electrodes Names / charges of electrodes Negative cell potential and meaning. Negative cell potential and meaning.

3. 3 17.7 Electrolysis Electrolysis is used for isolating active elements, purifying metals, and electroplating. Electrolysis is used for isolating active elements, purifying metals, and electroplating. Pure compounds: H 2 O, molten salts Pure compounds: H 2 O, molten salts Use inert electrodes in the liquid and pass electricity through the system Use inert electrodes in the liquid and pass electricity through the system

4. 4 17.7 Electrolysis The negative electrode (cathode) attracts cations; reduction occurs. The negative electrode (cathode) attracts cations; reduction occurs. The positive electrode (anode) attracts anions; oxidation occurs. The positive electrode (anode) attracts anions; oxidation occurs.

5. 5 Electrolysis of Molten NaCl

6. 6 Electrolysis of NaCl Cathode: Cathode: Na + (l) + e -  Na(l)

7. 7 Electrolysis of NaCl Anode: Anode: 2Cl - (l)  Cl 2 (g)+ 2e -

8. 8 Electrolysis of NaCl 2Na + (l) + 2Cl - (l)  2Na(l) + Cl 2 (g) E o = -4.07 V 2Na + (l) + 2Cl - (l)  2Na(l) + Cl 2 (g) E o = -4.07 V Must supply at least 4.07 V to electrolyze molten sodium chloride. Must supply at least 4.07 V to electrolyze molten sodium chloride. NaCl melts at 804 o C, where Na vaporizes and burns. NaCl melts at 804 o C, where Na vaporizes and burns. Lower the temperature by adding CaCl 2. (Why does this work?) Lower the temperature by adding CaCl 2. (Why does this work?)

9. 9 Electrolysis of NaCl Other problem, Na reacts with Cl 2, even at room temperature. Other problem, Na reacts with Cl 2, even at room temperature. Commercial operations use a Downs Cell. (described in 17.8, pg 859) Commercial operations use a Downs Cell. (described in 17.8, pg 859)

10. 10 Downs Cell How does the Downs Cell solve the problem of reaction between Na and Cl 2 ? How does the Downs Cell solve the problem of reaction between Na and Cl 2 ?

11. 11 Applications of Electrolysis Electrolysis can be used in a variety of applications: Electrolysis can be used in a variety of applications: Chemical recovery of elements in mixturesChemical recovery of elements in mixtures Industrial recovery of elements, miningIndustrial recovery of elements, mining Plating out of metals, electroplating.Plating out of metals, electroplating. And many more!And many more!

12. 12 Industrial Processes Purification of Copper: Recovered from its ores by chemical reduction. Purification of Copper: Recovered from its ores by chemical reduction. Purified by electrolysis. Purified by electrolysis. Recover impurities: Recover impurities: Mo (25%)Mo (25%) Se (93%)Se (93%) Te (96%)Te (96%) Au (32%)Au (32%) Ag (28%)Ag (28%)

13. 13 Electroplating of Nickel

14. 14 Electrolytic Processes with Metals A variety of metals can be prepared by electrolysis, if a cheap source of electricity is available. In addition, some metals* are purified by electrolysis. A variety of metals can be prepared by electrolysis, if a cheap source of electricity is available. In addition, some metals* are purified by electrolysis. aluminumcadmiumaluminumcadmium calciumcopper*calciumcopper* gold*lead*gold*lead* magnesiumsodiummagnesiumsodium zinczinc

15. 15 Faraday’s Law Recall… Recall… F = charge on 1 mol e - = 96500 coul/molF = charge on 1 mol e - = 96500 coul/mol and Electrical Current = charge / time and Electrical Current = charge / time 1 ampere = 1 coulomb of charge / second1 ampere = 1 coulomb of charge / second 1 A = 1 coul / s1 A = 1 coul / s Use these relationships to analyze electrolytic processes Use these relationships to analyze electrolytic processes 77 a, 79 a, 81 77 a, 79 a, 81

16. 16 17.7 Faraday’s Law Faraday’s Law: the mass of product produced by a given amount of current is proportional to the number of electrons transferred. Faraday’s Law: the mass of product produced by a given amount of current is proportional to the number of electrons transferred.

17. 17 Faraday’s Law Moles deposited: Moles deposited: Ag +, 1 mole e -  1 mol AgAg +, 1 mole e -  1 mol Ag Cu 2 +, 1 mole e -  1/2 mol CuCu 2 +, 1 mole e -  1/2 mol Cu Au 3 +, 1 mole e -  1/3 mol AuAu 3 +, 1 mole e -  1/3 mol Au H +, 1 mole e -  1/2 mol H 2H +, 1 mole e -  1/2 mol H 2 F = charge on 1 mol e - = 96500 coul/mol F = charge on 1 mol e - = 96500 coul/mol charge = current x time charge = current x time 1 coul = 1 A s 1 coul = 1 A s moles e - = charge (coul) x 1 mol/96500 coul moles e - = charge (coul) x 1 mol/96500 coul

18. 18 Faraday’s Law If we electrolyze molten NaCl with a current of 50.0 A for 30. min (or 1800 s), what mass of Na is produced? If we electrolyze molten NaCl with a current of 50.0 A for 30. min (or 1800 s), what mass of Na is produced? Na + + e -  Na Na + + e -  Na 50.0 C x 1800 s x 1 mol e - 50.0 C x 1800 s x 1 mol e - s96500 C s96500 C = 0.9326 mol e - = 0.9326 mol e -

19. 19 Faraday’s Law moles Na = 0.9326 mol e - x 1 mol Na/1 mol e - moles Na = 0.9326 mol e - x 1 mol Na/1 mol e - = 0.9326 mol = 0.9326 mol mass Na = 0.9326 mol x 22.99 g/mol = 21.44 g mass Na = 0.9326 mol x 22.99 g/mol = 21.44 g We can also calculate how much electrical energy it will take for an electrolysis. We will not pursue these calculations. We can also calculate how much electrical energy it will take for an electrolysis. We will not pursue these calculations.

20. 20 “Stoich Map”…

21. 21 Revisit Electrolysis of KI(aq) Introduction

22. 22 Electrolysis of Ion Mixtures Ce 4 + + e -  Ce 3 + E o = 1.70 V Ce 4 + + e -  Ce 3 + E o = 1.70 V VO 2 + + 2H + + e -  VO 2 + + H 2 O E o = 1.00 V VO 2 + + 2H + + e -  VO 2 + + H 2 O E o = 1.00 V Fe 3 + + e -  Fe 2 + E o = 0.77 V Fe 3 + + e -  Fe 2 + E o = 0.77 V Rank the above in terms of strength as an oxidizing agent.Rank the above in terms of strength as an oxidizing agent. Which species would be reduced at the cathode in an electrolytic cell at the lowest voltage?Which species would be reduced at the cathode in an electrolytic cell at the lowest voltage?

23. 23 Electrolysis of H 2 O We need to be careful using this principle with aqueous solutions as water is a major species along with the ions. We need to be careful using this principle with aqueous solutions as water is a major species along with the ions.

24. 24 Electrolysis of H 2 O Anode (oxidation): Anode (oxidation): 2H 2 O  O 2 (g) + 4H + + 4e - E ox o = -1.23 V Cathode (reduction): Cathode (reduction): 2H 2 O + 2e -  H 2 (g) + 2OH - E red o = -0.83 V 2H 2 O  2H 2 (g) + O 2 (g) E cell o = -2.06 V Must supply at least 2.06 V to electrolyze water (if anode [H + ] = 1.0 M and cathode [OH - ] = 1.0 M) Must supply at least 2.06 V to electrolyze water (if anode [H + ] = 1.0 M and cathode [OH - ] = 1.0 M) In pure water, [H + ] = [OH - ] = 10 -7 M and the overall potential is –1.23 V In pure water, [H + ] = [OH - ] = 10 -7 M and the overall potential is –1.23 V An electrolyte is usually added to increase electrical conductivity An electrolyte is usually added to increase electrical conductivity

25. 25 Electrolysis of Aqueous Solutions Products depend on whether it is easier to oxidize or reduce the dissolved ions or water. Products depend on whether it is easier to oxidize or reduce the dissolved ions or water. Sample Problem: Consider a solution of NiCl 2 under standard conditions. Sample Problem: Consider a solution of NiCl 2 under standard conditions.

26. 26 Electrolysis of Aqueous Solutions Possible Anode Oxidations: Possible Anode Oxidations: 2Cl -  Cl 2 + 2e - E o = -1.36 V 2H 2 O  O 2 + 4H + + 4e - E o = -1.23 V Because of more positive voltage, we would predict H 2 O will oxidize before Cl -.* Possible Cathode Reductions: Possible Cathode Reductions: Ni 2 + + 2e -  Ni E o = -0.25 V 2H 2 O + 2e -  H 2 + 2OH - E o = -0.83 V Because of more positive voltage, Ni 2+ will reduce before H 2 O. Products are O 2 * and Ni. Products are O 2 * and Ni.

27. 27 Group Work 1. What are the products of electrolysis of an aqueous NiBr 2 solution? 2. What are the products of electrolysis of an aqueous CuF 2 solution? 3. What are the products of electrolysis of a mixture of aqueous CuBr 2 and NiF 2 ?

28. 28 Group Work 1 What are the products of electrolysis of a NiBr 2 solution? What are the products of electrolysis of a NiBr 2 solution? Anode reactions:Anode reactions: 2Br -  Br 2 + 2e - E o = -1.07 V 2H 2 O  O 2 + 4H + + 4e - E o = -1.23 V Cathode reactions:Cathode reactions: Ni 2 + + 2e -  Ni E o = -0.25 V 2H 2 O + 2e -  H 2 + 2OH - E o = -0.83 V Ni and Br 2 Ni and Br 2

29. 29 Group Work 2 What are the products of electrolysis of a CuF 2 aqueous solution? What are the products of electrolysis of a CuF 2 aqueous solution? Anode: Anode: 2F -  F 2 + 2e - E o = -2.87 V 2H 2 O  O 2 + 4H + + 4e - E o = -1.23 V Cathode: Cathode: Cu 2 + + 2e -  Cu E o = 0.34 V 2H 2 O + 2e -  H 2 + 2OH - E o = -0.83 V Cu and O 2 Cu and O 2

30. 30 Group Work 3 What are the products of electrolysis of a mixture of aqueous CuCl 2 and NiCl 2 ? What are the products of electrolysis of a mixture of aqueous CuCl 2 and NiCl 2 ? Anode: Anode: 2Br -  Br 2 + 2e - E o = -1.07 V 2F -  F 2 + 2e - E o = -2.87 V 2H 2 O  O 2 + 4H + + 4e - E o = -1.23 V Cathode: Cathode: Ni 2 + + 2e -  Ni E o = -0.25 V Cu 2 + + 2e -  Cu E o = 0.34 V 2H 2 O + 2e -  H 2 + 2OH - E o = -0.83 V Cu and Br 2 Cu and Br 2