chapter19_part2

chapter19_part2 - 886 Chapter 19: Electrochemistry and Its...

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Chapter 19: Electrochemistry and Its Applications 886 E ° cell = "# G o nF = " ( " 598 kJ) # 1000 J 1 kJ $ % ( ) # 1 C # 1 V 1 J $ % ( ) (4 mol)(96485 C/ mol) = 1.55 V (b) When the reaction is written with all the coefficients doubled: Δ double = 2 × Δ original = 2 × (–598 kJ) = –1196 kJ E ° cell is independent of scale, so E ° cell = 1.55 V. 39. Answer: (a) –666 kJ (b) –1332 kJ, 3.45 V Strategy and Explanation: (a) Adapt the method described in Problem-Solving Example 19.8: Δ G° = –nF E ° cell One Mg atom is going from Ox. # = 0 to Ox. # = +2. So, n = 2 mol. Δ G° = – (2 mol) × (96485 C/mol) × (3.45 V) × 1 J 1 C " 1 V " 1 kJ 1000 = –666 kJ (b) When the reaction is written with all the coefficients doubled: Δ double = Δ original + Δ original = –666 kJ + (–666 kJ) = –1332 kJ E ° cell is independent of scale, so E ° cell = 3.45 V. 40. Answer: –409 kJ Strategy and Explanation: Adapt the method described in the solution to Question 38 and Problem-Solving Example 19.8. Zn(s) + Cl 2 (g) Zn 2+ (aq) + 2 Cl (aq) Zn is going from Ox. # = 0 to Ox. # = +2. So, n = 2 mol. Δ G° = – (2 mol) × (96485 C/mol) × (2.12 V) × 1 J 1 C " 1 V " 1 kJ 1000 = –409 kJ 41. Answer: K° = 10 25 , Δ G° = –142 kJ Strategy and Explanation: Adapt the method described in the solution to Question 38 and Problem-Solving Example 19.9. Cd(s) Cd 2+ (aq) + 2 e E ° anode = – 0.403 V Cu 2+ (aq) + 2 e Cu(s) E ° cathode = 0.337 V Cd(s) + Cu 2+ (aq) Cd 2+ (aq) + Cu(s) E ° cell = (0.337 V) – (– 0.403 V) = 0.740 V log K° = nE cell o 0.0592 V = (2 mol) " (0.74) 0.0592 V = 25 K° = 10 25 Δ G° = – (2 mol) × (96485 C/mol) × (0.740 V) × 1 J 1 C " 1 V " 1 kJ 1000 = –142 kJ 42 . Answer: K° = 1 × 1 0 –18 , Δ G° = 102 kJ Strategy and Explanation: Adapt the method described in the solution to Question 38 and Problem-Solving Example 19.9. 2 Br (aq) Br 2 ( l ) + 2 e E ° anode = +1.066 V I 2 (s) + 2 e 2 I (aq) E ° cathode = +0.535 V I 2 (s) + 2 Br (aq) 2 I (aq) + Br 2 ( l ) E ° cell = (+0.535 V) – (+1.066 V) = –0.531 V
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Chapter 19: Electrochemistry and Its Applications 887 log K° = nE cell o 0.0592 V = (2 mol) " ( # 0.531 V) 0.0592 V = # 17.9 K° = 1 × 10 –18 Δ G° = – (2 mol) × (96485 C/mol) × (–0.531 V) × 1 J 1 C " 1 V " 1 kJ 1000 = 102 kJ 43. Answer: K° = 2 × 10 –53 , Δ G° = 301.5 kJ Strategy and Explanation: Adapt the method described in the solution to Question 42. 2 Ag(s) 2 Ag + (aq) + 2 e oxidation half-reaction E ° anode = + 0.7994 V Zn 2+ (aq) + 2 e Zn(s) reduction half-reaction E ° cathode = –0.763 V Zn 2+ (aq) + 2 Ag + (aq) Sn 4+ (aq) + 2 Ag(s) E ° cell = + (–0.763 V) – (+ 0.7994 V) = –1.562 V log K° = nE cell o 0.0592 V = (2 mol) " ( # 1.562 V) 0.0592 V = # 52.8 K° = 10 –52.8 = 2 × 10 –53 Δ G° = – (2 mol) × (96485 C/mol) × (–1.562 V) × 1 J 1 C " 1 V
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This note was uploaded on 07/15/2009 for the course CHE 131 taught by Professor Kerber during the Fall '08 term at SUNY Stony Brook.

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chapter19_part2 - 886 Chapter 19: Electrochemistry and Its...

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