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20 Electrochemistry Solutions to Exercises 20.91 Analyze/Plan. Follow the logic in Sample Exercise 20.14, paying close attention to units. Coulombs = amps-s; since this is a reduction, each mole of Cr(s) requires 3 Faradays. Solve. (a) 7.60 A 2.00 d X 24 1d hr X 60 min X 1 60 min S X 1 amp-s 1C X 96,485 1F C X 1 mol Cr 52.00 1 mol g Cr Cr = 236 g Cr(s) 3F (b) 0.250 mol Cr 1 mol 3F Cr X 96,485 F C 1 amp-s 1C 8.00 1 hr 60 1hr min 1 60 min = 2.51 A 20.92 Coulombs = amps-s; since this is a 2e⁻ reduction, each mole of Mg(s) requires 2 Faradays. (a) 4.55 A 4.50 d 24 1d h 60 1h min 1 60 min amp- 1C 96,485 C 1F 1 mol 2F Mg 1 mol Mg Mg = 223 g Mg (b) 25.00 g Mg 24.31 1 mol g Mg Mg 1 mol 2F Mg 96,485 F C 1 amp- C 1 60 min 3.50 1 A = 945 min 20.93 Analyze/Plan. Combine the ideas in Sample Exercises 20.14 and 20.15, paying close attention to units. Li+ is reduced at the anode; Cl⁻ is oxidized at the anode. Solve. (a) If the cell is 85% efficient, F 0.85 = 1.13512 10⁵ = 1.1 10⁵ C/mol Li required 7.5 X 10⁴ A 24 3600 1h 1amp-s 1C 1.13512 1mol Li 10⁵ C, 6.94 1 mol Li Li = 3.962 10⁵ = 4.0 10⁵ g Li (b) = (cathode) (anode) = -3.05 V (1.359 V) = -4.409 = -4.41 V The minimum voltage required to drive the reaction is the magnitude of 4.41 V. 20.94 (a) 7.5 X 10³ A 48 h 3600s 1h 1amp-s 1C 96,485 C 1 mol 2F Ca x0.68 40.0 1 mol Ca Ca = 1.830 = 1.8 10⁵ g Ca (b) = (cathode)- (anode) = -2.87 (1.359 V) = -4.229 = -4.23 V The minimum voltage required to drive the reaction is the magnitude of 4.23 V. 637