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17.7-8 Electrolysis & Applications

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.

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17.7-8 Electrolysis & Applications

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  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. • 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. 17.7 Electrolysis • Electrolysis is used for isolating active elements, purifying metals, and electroplating. • Pure compounds: H2O, molten salts • Use inert electrodes in the liquid and pass electricity through the system

  3. 17.7 Electrolysis • The negative electrode (cathode) attracts cations; reduction occurs. • The positive electrode (anode) attracts anions; oxidation occurs.

  4. Electrolysis of Molten NaCl

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

  6. Electrolysis of NaCl • Anode: 2Cl-(l) Cl2(g)+ 2e-

  7. Electrolysis of NaCl • 2Na+(l) + 2Cl-(l) 2Na(l) + Cl2(g) Eo = -4.07 V • Must supply at least 4.07 V to electrolyze molten sodium chloride. • NaCl melts at 804oC, where Na vaporizes and burns. • Lower the temperature by adding CaCl2. (Why does this work?)

  8. Electrolysis of NaCl • Other problem, Na reacts with Cl2, even at room temperature. • Commercial operations use a Downs Cell. (described in 17.8, pg 859)

  9. Downs Cell • How does the Downs Cell solve the problem of reaction between Na and Cl2?

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

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

  12. Electroplating of Nickel

  13. 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. • aluminum cadmium • calcium copper* • gold* lead* • magnesium sodium • zinc

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

  15. 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.

  16. 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? • Na+ + e- Na • 50.0 C x 1800 s x 1 mol e- s 96500 C = 0.9326 mol e-

  17. Faraday’s Law • moles Na = 0.9326 mol e- x 1 mol Na/1 mol e- = 0.9326 mol • 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.

  18. “Stoich Map”…

  19. Revisit Electrolysis of KI(aq) Introduction

  20. Electrolysis of H2O • Anode (oxidation): 2H2O O2(g) + 4H+ + 4e- Eoxo = -1.23 V • Cathode (reduction): 2H2O + 2e-H2(g) + 2OH- Eredo = -0.83 V 2H2O  2H2(g) + O2(g) Ecello = -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) • 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

  21. Electrolysis of Aqueous Solutions • Products depend on whether it is easier to oxidize or reduce the dissolved ions or water. • Sample Problem: Consider a solution of NiCl2 under standard conditions.

  22. Electrolysis of Aqueous Solutions • Possible Anode Oxidations: 2Cl- Cl2 + 2e- Eo = -1.36 V 2H2O  O2 + 4H+ + 4e- Eo = -1.23 V Because of more positive voltage, we would predict H2O will oxidize before Cl-.* • Possible Cathode Reductions: Ni2+ + 2e- Ni Eo = -0.25 V 2H2O + 2e- H2 + 2OH- Eo = -0.83 V Because of more positive voltage, Ni2+ will reduce before H2O. • Products are O2* and Ni.

  23. Group Work • What are the products of electrolysis of an aqueous NiBr2 solution? • What are the products of electrolysis of an aqueous CuF2 solution? • What are the products of electrolysis of a mixture of aqueous CuBr2 and NiF2?

  24. Group Work 1 • What are the products of electrolysis of a NiBr2 solution? • Anode reactions: 2Br- Br2 + 2e- Eo = -1.07 V 2H2O  O2 + 4H+ + 4e-Eo = -1.23 V • Cathode reactions: Ni2+ + 2e- Ni Eo = -0.25 V 2H2O + 2e- H2 + 2OH- Eo = -0.83 V • Ni and Br2

  25. Group Work 2 • What are the products of electrolysis of a CuF2 aqueous solution? • Anode: 2F- F2 + 2e- Eo = -2.87 V 2H2O  O2 + 4H+ + 4e- Eo = -1.23 V • Cathode: Cu2+ + 2e- Cu Eo = 0.34 V 2H2O + 2e- H2 + 2OH- Eo = -0.83 V • Cu and O2

  26. Group Work 3 • What are the products of electrolysis of a mixture of aqueous CuCl2 and NiCl2? • Anode: 2Br- Br2 + 2e- Eo = -1.07 V 2F- F2 + 2e- Eo = -2.87 V 2H2O  O2 + 4H+ + 4e- Eo = -1.23 V • Cathode: Ni2+ + 2e- Ni Eo = -0.25 V Cu2+ + 2e- Cu Eo = 0.34 V 2H2O + 2e- H2 + 2OH- Eo = -0.83 V • Cu and Br2

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