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Using Table 1, rank the following species from the strongest to the weakest agent: I' (aq), Fe(s), Al(s). Explain why.

Using Table 1, rank the following species from the strongest to the weakest agent: I' (aq), Fe(s), Al(s). Explain why.

Chemistry
Chemistry
10th Edition
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Chapter1: Chemical Foundations
Section: Chapter Questions
Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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please answer #2 type the answer or write it clearly so i can understand please
1. Using data in Table 1 (see next page), calculate the standard emf for a cell that employs the following overall cell reaction: 2AI(s) + 31(s) → 2 Al3+ (ag) + 6I° (ag) 2. Using Table 1, rank the following species from the strongest to the weakest reducing agent: I (ag), Fe(s), Al(s). Explain why. 3. Using the standard reduction potentials in Table 2, calculate the equilibrium constant at 25°C for the reaction Br2(1) + 2C1(ag) Cl2(g) + 2 Br(aq) 4. Based on the standard reduction potentials in Table 2, which of the following metals could provide çathodic protection to iron: Al, Cu, Ni, Zn? 5. The electrolysis of CuCI(ag) produces Cu(s) and Cl2(g). What is the minimum external emf needed to drive this electrolysis under standard conditions?
Transcribed Image Text:1. Using data in Table 1 (see next page), calculate the standard emf for a cell that employs the following overall cell reaction: 2AI(s) + 31(s) → 2 Al3+ (ag) + 6I° (ag) 2. Using Table 1, rank the following species from the strongest to the weakest reducing agent: I (ag), Fe(s), Al(s). Explain why. 3. Using the standard reduction potentials in Table 2, calculate the equilibrium constant at 25°C for the reaction Br2(1) + 2C1(ag) Cl2(g) + 2 Br(aq) 4. Based on the standard reduction potentials in Table 2, which of the following metals could provide çathodic protection to iron: Al, Cu, Ni, Zn? 5. The electrolysis of CuCI(ag) produces Cu(s) and Cl2(g). What is the minimum external emf needed to drive this electrolysis under standard conditions?
Potential (V) Reduction Half-Reaction F2(8) + 2 e -→ 2F (aq) MnO4 (aq) + 8 H (aq) + 5 e Cl2(8) + 2 e Cr20,2 (ag) + 14 H*(aq) + 6 e → O2(g) + 4 H*(aq) + 4 e 2 H20(1) Br2(1) + 2 e → 2 Br (ag) NO3 (ag) + 4 H*(aq) + 3 e Ag"(aq) + e Fe* (aq) + e O2lg) + 2 H*(aq) + 2 e → MnO, (ag) + 2 H2O(!) + 3 e I2 (s) + 2 e → O2(8) + 2 H20(I) + 4 e Cu*(aq) + 2 e 2 H*(aq) + 2 e Ni2 (aq) + 2 e Fe2 (aq) + 2 e Zn2* (aq) + 2 e 2 H20(1) + 2 e Al3* (aq) + 3 e Na*(aq) + e¯→ Li*(aq) + e +2.87 +1.51 Mn²+(aq) + 4 H20(1) +1.36 →2 CI (aq) +1.33 2 Cr" (aq) + 7 H2O(1) +1.23 +1.06 +0.96 NO(g) + 2 H2O(1) +0.80 Ag(s) Fe2"(aq) +0.77 - H2O2(aq) MnO2(s) + 4 OH (aq) +0.68 +0.59 21 (aq) → 4 OH (aq) → Cu(s) → H2(8) Ni(s) Fe(s) Zn(s) +0.54 +0.40 +0.34 0 [defined] -0.28 -0.44 -0.76 > H2(g) + 2 OH (ag) Al(s) Na(s) → Li(s) -0.83 > -1.66 -2.71 -3.05
Transcribed Image Text:Potential (V) Reduction Half-Reaction F2(8) + 2 e -→ 2F (aq) MnO4 (aq) + 8 H (aq) + 5 e Cl2(8) + 2 e Cr20,2 (ag) + 14 H*(aq) + 6 e → O2(g) + 4 H*(aq) + 4 e 2 H20(1) Br2(1) + 2 e → 2 Br (ag) NO3 (ag) + 4 H*(aq) + 3 e Ag"(aq) + e Fe* (aq) + e O2lg) + 2 H*(aq) + 2 e → MnO, (ag) + 2 H2O(!) + 3 e I2 (s) + 2 e → O2(8) + 2 H20(I) + 4 e Cu*(aq) + 2 e 2 H*(aq) + 2 e Ni2 (aq) + 2 e Fe2 (aq) + 2 e Zn2* (aq) + 2 e 2 H20(1) + 2 e Al3* (aq) + 3 e Na*(aq) + e¯→ Li*(aq) + e +2.87 +1.51 Mn²+(aq) + 4 H20(1) +1.36 →2 CI (aq) +1.33 2 Cr" (aq) + 7 H2O(1) +1.23 +1.06 +0.96 NO(g) + 2 H2O(1) +0.80 Ag(s) Fe2"(aq) +0.77 - H2O2(aq) MnO2(s) + 4 OH (aq) +0.68 +0.59 21 (aq) → 4 OH (aq) → Cu(s) → H2(8) Ni(s) Fe(s) Zn(s) +0.54 +0.40 +0.34 0 [defined] -0.28 -0.44 -0.76 > H2(g) + 2 OH (ag) Al(s) Na(s) → Li(s) -0.83 > -1.66 -2.71 -3.05
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