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20 Electrochemistry Solutions to Exercises 20.109 (a) The standard reduction potential for is much greater than that of V -2.37 V). In aqueous solution, would be preferentially reduced and no Mg(s) would be obtained. (b) 97,000 A 24 hr 3600 1 hr X 1A 1C 96,485 1F C 1 mol 2F Mg 24.31 1mol Mg Mg 0.96 = 1.0 10⁶ g Mg = 1.0 10³ kg Mg 20.110 Analyze. Given mass of aluminum desired, applied voltage and electrolysis efficiency, calculate kWh of electricity required. Plan. Beginning with mass Al and paying attention to units, calculate coulombs required if the process is 100% efficient. Then, take efficiency into account and then use V, C and the relationship between J and kWh to calculate kWh required. Solve. 1.0 10³ kg Al X 1000 1kg 26.98 1 mol g Al Al 1 mol 3F Al X 96,485 F C = 1.073 10¹⁰ = 1.1 10¹⁰ C If the cell is 45 efficient, (1.073 X 10¹⁰ /0.45) = 2.384 X 10¹⁰ = 2.4 10¹⁰ C are required to plate the bumper. 4.50 V X 2.384 X 10¹⁰ C 1C-V 1J X 3.6 1 kWh 10⁶ J = 29,801 = kWh 20.111 (a) 7 X 10⁸ mol H₂ X 1 mol 2F H₂ X 96,485 1F C = 1.35 10¹⁴ = 10¹⁴ C (b) O₂ + (aq) = 1.23 V + 2e⁻ H₂(g)] O₂ (g) Pₜ = 300 atm = Since H₂(g) and O₂ (g) are generated in a 2:1 mole ratio, PH₂ = 200 atm and = 100 atm. E = E° 0.0592 X = -1.23 V 0.0592 4 [100 (c) Energy = nFE = X 10⁸ mol) V) = 1.80 X 10¹⁴ = 2 10¹⁴ J (d) 1.80 X 10¹⁴ J 3.6 1 kWh 10⁶ J $0.85 kWh = $4.24 X 10⁷ = $4 X 10⁷ It would cost more than $40 million for the electricity alone. Integrative Exercises 20.112 2NH₃(g) (a) The oxidation number of H₂(g) and N₂(g) is 0. The oxidation number of N in NH₃ is -3, H in NH₃ is +1. H₂ is being oxidized and N₂ is being reduced. 643