1. Based on the standard reduction potentials in Table 2, which of the following metals could provide cathodic protection to iron: Al, Cu, Ni, Zn?
1. Based on the standard reduction potentials in Table 2, which of the following metals could provide cathodic protection to iron: Al, Cu, Ni, Zn?
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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Question
please answer #4
type the answer or write it clearly so i can understand please
![Half-Reaction
E°(V)
F(g) + 2e - 2F (aq)
O,(g) + 2H*(aq) + 2e
Co**(aq) + e
H,O,(aq) + 2H *(ag) + 2e 2H,0
PhO2(s) + 4H*(aq) + SO (aq) + 2e PhSO,(s) + 2H2O
Ce**(ag) + e Ce (aq)
MnO, (aq) + 8H*(aq) 5e
Aut
+2.87
→ 0;(g) + H;O
+2.07
Co (aq)
+1.82
+1.77
+1.70
+1.61
- Mn* (aq) + 4H,0
+1.51
(aq) + 3e
Cl,(g) + 2e 2CI (aq)
Cr,0 (aq) + 14H*(aq) + 6e
MnO,(s) + 4H (aq) + 2e
O,(g) + 4H (aq) + 4e
Br;(1) + 2e
NO, (aq) + 4H*(aq) + 3e-
2Hg (aq) + 2e
Hg3"(aq) + 2e
Ag*(aq) + e
Fe* (aq) + e
O,(g) + 2H*(aq) + 2e
MnO, (aq) + 2H,0 + 3e
L,(s) + 2e -→ 21 (aq)
0,(g) + 2H,0 + 4e
Cu2* (ag) + 2e
AgCl(s) + e
so (aq) + 4H (aq) + 2e
Cu²*(aq) + e
Sn** (aq) + 2e
2H*(aq) + 2e-
Pb2* (aq) + 2e
Sn2*(aq) + 2e
Ni²* (aq) + 2e
Co*(aq) + 2e
PbSO,(s) + 2e
Cd* (ag) + 2e
Fe*(aq) + 2e
Cr* (aq) + 3e
Zn*(aq) + 2e
2H,0 + 2e
Mn*(ag) + 2e
Al*(aq) + 3e
Be? (aq) + 2e
Mg2*(ag) + 2e
Na*(aq) + e
Ca* (aq) + 2e
Sr* (ag) + 2e
Ba*(ag) + 2e Ba(s)
K*(aq) + e
Li (aq) + e
Au(s)
+1.50
+1.36
2Cr*(aq) + 7H,0
Mn2 (ag) + 2H,O
+1.33
+1.23
2H,0
+1.23
2Br (aq)
+1.07
NO(g) + 2H,0
+0.96
Hg" (aq)
2Hg(l)
Ag(s)
Fe (aq)
+0.92
+0.85
+0.80
+0.77
→ H,O(aq)
MnO,(s) + 40H (aq)
+0.68
+0.59
+0.53
40H (aq)
→ Cu(s)
+ Ag(s) + Cl (aq)
+0.40
+0.34
+0.22
→ SO,(g) + 2H,0
+0.20
Cu*(aq)
Sn (aq)
+0.15
+0.13
H2(g)
0.00
Ph(s)
-0.13
→ Sn(s)
→ Ni(s)
→ Co(s)
→ Pb(s) + S0 (aq)
-0.14
-0.25
-0.28
-0.31
Cd(s)
-0.40
→ Fe(s)
- Cr(s)
-0.44
-0.74
Zn(s)
-0.76
H2(g) + 20H (aq)
- Mn(s)
-0.83
-1.18
Al(s)
-1.66
Be(s)
-1.85
→ Mg(s)
- 2.37
Na(s)
-2.71
→ Ca(s)
-2.87
Sr(s)
-2.89
-2.90
» K(s)
→ Li(s)
-2.93
-3.05](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F637f7bdc-ad18-43dd-b09d-1aaef64771cb%2F170cd344-f2f1-421b-8037-4e2ddacfbe85%2Fprb5pd_processed.jpeg&w=3840&q=75)
Transcribed Image Text:Half-Reaction
E°(V)
F(g) + 2e - 2F (aq)
O,(g) + 2H*(aq) + 2e
Co**(aq) + e
H,O,(aq) + 2H *(ag) + 2e 2H,0
PhO2(s) + 4H*(aq) + SO (aq) + 2e PhSO,(s) + 2H2O
Ce**(ag) + e Ce (aq)
MnO, (aq) + 8H*(aq) 5e
Aut
+2.87
→ 0;(g) + H;O
+2.07
Co (aq)
+1.82
+1.77
+1.70
+1.61
- Mn* (aq) + 4H,0
+1.51
(aq) + 3e
Cl,(g) + 2e 2CI (aq)
Cr,0 (aq) + 14H*(aq) + 6e
MnO,(s) + 4H (aq) + 2e
O,(g) + 4H (aq) + 4e
Br;(1) + 2e
NO, (aq) + 4H*(aq) + 3e-
2Hg (aq) + 2e
Hg3"(aq) + 2e
Ag*(aq) + e
Fe* (aq) + e
O,(g) + 2H*(aq) + 2e
MnO, (aq) + 2H,0 + 3e
L,(s) + 2e -→ 21 (aq)
0,(g) + 2H,0 + 4e
Cu2* (ag) + 2e
AgCl(s) + e
so (aq) + 4H (aq) + 2e
Cu²*(aq) + e
Sn** (aq) + 2e
2H*(aq) + 2e-
Pb2* (aq) + 2e
Sn2*(aq) + 2e
Ni²* (aq) + 2e
Co*(aq) + 2e
PbSO,(s) + 2e
Cd* (ag) + 2e
Fe*(aq) + 2e
Cr* (aq) + 3e
Zn*(aq) + 2e
2H,0 + 2e
Mn*(ag) + 2e
Al*(aq) + 3e
Be? (aq) + 2e
Mg2*(ag) + 2e
Na*(aq) + e
Ca* (aq) + 2e
Sr* (ag) + 2e
Ba*(ag) + 2e Ba(s)
K*(aq) + e
Li (aq) + e
Au(s)
+1.50
+1.36
2Cr*(aq) + 7H,0
Mn2 (ag) + 2H,O
+1.33
+1.23
2H,0
+1.23
2Br (aq)
+1.07
NO(g) + 2H,0
+0.96
Hg" (aq)
2Hg(l)
Ag(s)
Fe (aq)
+0.92
+0.85
+0.80
+0.77
→ H,O(aq)
MnO,(s) + 40H (aq)
+0.68
+0.59
+0.53
40H (aq)
→ Cu(s)
+ Ag(s) + Cl (aq)
+0.40
+0.34
+0.22
→ SO,(g) + 2H,0
+0.20
Cu*(aq)
Sn (aq)
+0.15
+0.13
H2(g)
0.00
Ph(s)
-0.13
→ Sn(s)
→ Ni(s)
→ Co(s)
→ Pb(s) + S0 (aq)
-0.14
-0.25
-0.28
-0.31
Cd(s)
-0.40
→ Fe(s)
- Cr(s)
-0.44
-0.74
Zn(s)
-0.76
H2(g) + 20H (aq)
- Mn(s)
-0.83
-1.18
Al(s)
-1.66
Be(s)
-1.85
→ Mg(s)
- 2.37
Na(s)
-2.71
→ Ca(s)
-2.87
Sr(s)
-2.89
-2.90
» K(s)
→ Li(s)
-2.93
-3.05
![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?](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F637f7bdc-ad18-43dd-b09d-1aaef64771cb%2F170cd344-f2f1-421b-8037-4e2ddacfbe85%2Fsjfx1mb_processed.png&w=3840&q=75)
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?
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