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)
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)
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
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![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](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F637f7bdc-ad18-43dd-b09d-1aaef64771cb%2F37582562-804f-4c80-9f3a-92e3310637f4%2F45h000f_processed.png&w=3840&q=75)
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

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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