Chemistry
Chemistry
10th Edition
ISBN: 9781305957404
Author: Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher: Cengage Learning
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Question

193.) The standard free energy change = blank 

 

2H (ag) +2e = (g) 0.00 Sn (aq) + 2e Sn+ (aq) 0.15 (bo) Cu (ag) + e = Cu* (aq) 0.16 Cu+ (ag) + 2e – Cu(s) 0.34 IO (aq) + H20(1) + 2e I (aq) + 20H (aq) I (aq) + 20H (ag) 0.49 Cu* (ag) +e Cu(s) 0.52 L(s) + 2e 21 (ag) 0.54 Fe (aq) +e Fe²+ (aq) 0.77 Hg 2+ (ag) + 2e 2Hg(1) 0.80 Ag (aq) +e = Ag(s) 0.80 Hg²+ (aq) + 2e = Hg(!) 0.85 CIO (ag) + H,0(1) + 2e Cl (aq) + 20H (ag) 0.90 2Hg (aq) + 2e Hg, (aq) 2+ 0.90 NO; (ag) + 4H* (ag) + 3e NO(g) + 2H20(1) 0.96 Br2 (1) + 2e 2Br (ag) 1.07 02 (g) + 4H (ag) + 4e 2H,0(1) 1.23 Cr2 O, (ag) + 14H (ag) + 6e H 2Cr** (aq) + 7H2O(1) 1.33 Cl2(g) + 2e 2C1 (aq) 1.36 MnO4 (ag) + 8H (aq) + 5e Mn2+ (aq) + 4H20(1) 1.49 H2O2 (aq) + 2H* (ag) + 2e - 2H20(1) 1.78 S20g (aq) + 2e = 2S0,2 (ag) 2.01 F2(g) +2e 2F (ag) 2.87 Standard free-energy change kJ
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Transcribed Image Text:2H (ag) +2e = (g) 0.00 Sn (aq) + 2e Sn+ (aq) 0.15 (bo) Cu (ag) + e = Cu* (aq) 0.16 Cu+ (ag) + 2e – Cu(s) 0.34 IO (aq) + H20(1) + 2e I (aq) + 20H (aq) I (aq) + 20H (ag) 0.49 Cu* (ag) +e Cu(s) 0.52 L(s) + 2e 21 (ag) 0.54 Fe (aq) +e Fe²+ (aq) 0.77 Hg 2+ (ag) + 2e 2Hg(1) 0.80 Ag (aq) +e = Ag(s) 0.80 Hg²+ (aq) + 2e = Hg(!) 0.85 CIO (ag) + H,0(1) + 2e Cl (aq) + 20H (ag) 0.90 2Hg (aq) + 2e Hg, (aq) 2+ 0.90 NO; (ag) + 4H* (ag) + 3e NO(g) + 2H20(1) 0.96 Br2 (1) + 2e 2Br (ag) 1.07 02 (g) + 4H (ag) + 4e 2H,0(1) 1.23 Cr2 O, (ag) + 14H (ag) + 6e H 2Cr** (aq) + 7H2O(1) 1.33 Cl2(g) + 2e 2C1 (aq) 1.36 MnO4 (ag) + 8H (aq) + 5e Mn2+ (aq) + 4H20(1) 1.49 H2O2 (aq) + 2H* (ag) + 2e - 2H20(1) 1.78 S20g (aq) + 2e = 2S0,2 (ag) 2.01 F2(g) +2e 2F (ag) 2.87 Standard free-energy change kJ
Calculate the standard free-energy change at 25°C for the following reaction: 4Fe(s) + 302(9) + 12H* (aq) →4Fe (aq) + 6H,0(1) 6H,O(1) Use standard electrode potentials. Standard Electrode (Reduction) Potentials in Aqueous Solution at 25°C Cathode (Reduction) Half-Reaction Standard Potential, E" (V) Lit (ag) + e Li(s) -3.04 Na (ag) + e Na(s) 2.71 - 2+ Mg (ag) + 2e - Mg(s) -2.38 Al* (aq) + 3e Al(s) -1.66 2H20(1) + 2e = H2(g) + 20H (aq) -0.83 Zn (aq) + 2e - Zn(s) -0.76 Cr+ (ag) + 3e - Cr(s) -0.74 Fe?+ (ag) + 2e = Fe(s) -0.41 Cd²+ (aq) + 2e = Cd(s) -0.40 Ni+ (ag) + 2e - Ni(s) -0.23 Sn²+ (aq) +2e Sn(s) -0.14 Pb (aq) + 2e Pb(s) -0.13 Fe (ag) + 3e Fe(s) -0.04 2H (aq) + 2e - H2(g) 0.00 Sn+ (ag) + 2e Sn2+ (ag) 0.15 Cu2+ (ag) + e Cu (aq) 0.16 Cu+ (ag) + 2e Cu(s) 0.34 I0 (aq) + H20(1)+2e I (aq) + 20H (ag) 0.49 Cu" (ag) +e = Cu(s) 0.52 I2 (8) + 2e 21 (ag) 0.54 0 77 Cengage Learning | Cengage Technical Sup
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Transcribed Image Text:Calculate the standard free-energy change at 25°C for the following reaction: 4Fe(s) + 302(9) + 12H* (aq) →4Fe (aq) + 6H,0(1) 6H,O(1) Use standard electrode potentials. Standard Electrode (Reduction) Potentials in Aqueous Solution at 25°C Cathode (Reduction) Half-Reaction Standard Potential, E" (V) Lit (ag) + e Li(s) -3.04 Na (ag) + e Na(s) 2.71 - 2+ Mg (ag) + 2e - Mg(s) -2.38 Al* (aq) + 3e Al(s) -1.66 2H20(1) + 2e = H2(g) + 20H (aq) -0.83 Zn (aq) + 2e - Zn(s) -0.76 Cr+ (ag) + 3e - Cr(s) -0.74 Fe?+ (ag) + 2e = Fe(s) -0.41 Cd²+ (aq) + 2e = Cd(s) -0.40 Ni+ (ag) + 2e - Ni(s) -0.23 Sn²+ (aq) +2e Sn(s) -0.14 Pb (aq) + 2e Pb(s) -0.13 Fe (ag) + 3e Fe(s) -0.04 2H (aq) + 2e - H2(g) 0.00 Sn+ (ag) + 2e Sn2+ (ag) 0.15 Cu2+ (ag) + e Cu (aq) 0.16 Cu+ (ag) + 2e Cu(s) 0.34 I0 (aq) + H20(1)+2e I (aq) + 20H (ag) 0.49 Cu" (ag) +e = Cu(s) 0.52 I2 (8) + 2e 21 (ag) 0.54 0 77 Cengage Learning | Cengage Technical Sup
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