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MSJ2e_Ch19_ISM2_June26_p833

# MSJ2e_Ch19_ISM2_June26_p833 - Chapter 19 Electrochemistry...

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Chapter 19: Electrochemistry and Its Applications G° = – (2 mol) × (96485 C/mol) × (3.45 V) × 1 ϑ 1 Χ 1 ς 1 κϑ 1000 = –666 kJ (b) When the reaction is written with all the coefficients doubled: double = original + original = –666 kJ + (–666 kJ) = –1332 kJ E ° cell is independent of scale, so E ° cell = 3.45 V. 38. Adapt the method described in the answer to Question 36 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. G° = – (2 mol) × (96485 C/mol) × (2.12 V) × 1 ϑ 1 Χ 1 ς 1 κϑ 1000 = –409 kJ 39. Adapt the method described in the answer to Question 37 and Problem-Solving Example 19.8. Cd(s) Cd 2+ (aq) + 2 e E ° anode = – 0.40 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) – (– 0.40 V) = 0.74 V K° = 10 νΕ χελλ ο 0.0592 ς = 10 (2 μολ 29 (0.74 29 0.0592 ς = 10 25 G° = – (2 mol) × (96485 C/mol) × (0.74 V) × 1 ϑ 1 Χ 1 ς 1 κϑ 1000 = –1.4 × 10 2 kJ 40 . Adapt the method described in the answer to Question 36 and Problem-Solving Example 19.8. NOTE: Table 19.1 has a different value for the reduction potential of liquid bromine than Appendix I. Here we are using the value from Table 19.1. 2 Br (aq) Br 2 ( ) + 2 e E ° anode = + 1.08 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 ( ) E ° cell = (+0.535 V) – (+1.08 V) = –0.55 V K° = 10 νΕ χελλ ο 0.0592 ς = 10 (2 μολ 29 (-0.55 ς 29 0.0592 ς = 10 –19 G° = – (2 mol) × (96485 C/mol) × (–0.55 V) × 1 ϑ 1 Χ 1 ς 1 κϑ 1000 = 1.1 × 10 2 kJ 41. Adapt the method described in the answer to Question 37 and Problem-Solving Example 19.8. 2 Ag(s) 2 Ag + (aq) + 2 e oxidation half-reaction E ° anode = + 0.80 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) – (+ 0.80 V) = –1.56 V 49

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Chapter 19: Electrochemistry and Its Applications K° = 10 νΕ χελλ ο 0.0592 ς = 10 (2 μολ 29 (-1.56 ς 29 0.0592 ς = 10 –52.8 = 2 × 10 –53 G° = – (2 mol) × (96485 C/mol) × (–1.56 V) × 1 ϑ 1 Χ 1 ς 1 κϑ 1000 = 301 kJ 42 . Adapt the method described in the answer to Question 36 and Problem-Solving Example 19.8. 2 Br (aq) Br 2 ( ) + 2 e oxidation half-reaction E ° anode = + 1.08 V Cl 2 (g) + 2 e 2 Cl (aq) reduction half-reaction E ° cathode = +1.360 V Cl 2 (g) + 2 Br (aq) Br 2 ( ) + 2 Cl (aq) E ° cell = (+1.360 V) – (+1.08 V) = 0.28 V K° = 10 νΕ χελλ ο 0.0592 ς = 10 (2 μολ 29 (0.28 ς 29 0.0592 ς = 10 9.5 = 3 × 10 9 G° = – (2 mol) × (96485 C/mol) × (0.28 V) × 1 ϑ 1 Χ 1 ς 1 κϑ 1000 = –54 kJ Effect of Concentration on Cell Potential 43. Adapt the method described in the answer to Question 29 and Problem-Solving Example 19.7, then use the Nernst Equation at T = 298 K, by adapting the method described in the Problem-Solving Example 19.10.
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