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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 • 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)
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
17.7 Electrolysis • The negative electrode (cathode) attracts cations; reduction occurs. • The positive electrode (anode) attracts anions; oxidation occurs.
Electrolysis of NaCl • Cathode: Na+(l) + e-Na(l)
Electrolysis of NaCl • Anode: 2Cl-(l) Cl2(g)+ 2e-
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?)
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)
Downs Cell • How does the Downs Cell solve the problem of reaction between Na and Cl2?
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!
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%)
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
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
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 • 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-
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.
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
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.
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.
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?
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
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
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