Use the standard reduction potentials located in the 'Tables' linked above to calculate the equilibrium constant for the reaction: Fe2+(aq) + 2Cu*(aq) → Fe(s) + 2Cu²+(aq) Hint: Carry at least 5 significant figures during intermediate calculations to avoid round off error when taking the antilogarithm. You may use the OWL references to find the values you may need in this question. Equilibrium constant: AG° for this reaction would bị than zero. greater less Standard Reduction (Electrode) Potentials at 25 °C Half-Cell Reaction E° (volts) F2(g) + 2 e→2 F(aq) 2.87 Ce4*(aq) + e –→ Ce"(aq) 3+ 1.61 2+ MnO4 (aq) + 8 H"(aq) + 5 e¯ –→ Mn-T(aq) + 4 H2O(1) 1.51 Cl2(g) + 2 e' —2 СГ (aq) 1.36 Cr2072"(aq) + 14 H*(aq) + 6 e¯ →2 Cr"(aq) + 7 H2O(1) 1.33 O2(g) + 4 H*(aq) + 4 e→2 H2O(1) 1.229 Br2(1) + 2 e¯ –→2 Br¯(aq) 1.08 NO3 (aq) + 4 H"(aq) +3 e¯ –→ NO(g) + 2 H2O(1) 0.96 2 Hg-*(aq) + 2 e¯ → Hg22 (aq) 0.920 Hg-"(aq) + 2 e →Hg(1) 0.855 Ag"(aq) + e¯ – Ag(s) 0.799 2+, Hg2"(aq) + 2 e¯ →2 Hg(1) 0.789 Fe*(aq) + e → Fe2*(aq) 0.771 I2(s) + 2 e → 2 1 (aq) 0.535 Fe(CN)6° (aq) + e→ Fe(CN)64 (aq) 0.48 Cu2*(aq) + 2 e → Cu(s) 0.337 Cu2*(aq) + e" – Cu"(aq) 0.153 S(s) + 2 H*(aq) + 2 e → H2S(aq) 0.14 2 H*(aq) + 2 e → H2(g) 0.0000 Pb2*(aq) + 2 e → Pb(s) -0.126 Sn-"(aq) + 2 e¯ → Sn(s) „2+ -0.14 Ni2+(ag) + 2 e →Ni(s) -0.25 Co2+(ag) + 2 e →Co(s) -0.28 Cd2*(aq) + 2 e →Cd(s) -0.403 Cr3+ C* (aq) (aq) + e -0.41 Fe2*(aq) + 2 e →Fe(s) -0.44 Cr3+ (aq) + 3 e → Cr(s) -0.74 Zn2*(aq) + 2 e →Zn(s) -0.763 2 H20(1) + 2 e → H2(g) + 2 OH (aq) -0.83 Mn-"(aq) + 2 e¯ → Mn(s) 2+ -1.18 Als*(aq) + 3 e → Al(s) -1.66 2+ Mg"(aq) + 2 e →Mg(s) -2.37 Na"(aq) + e –→ Na(s) -2.714 K(aq) + e→ K(s) -2.925 Li*(aq) + e Li(s) -3.045
Use the standard reduction potentials located in the 'Tables' linked above to calculate the equilibrium constant for the reaction: Fe2+(aq) + 2Cu*(aq) → Fe(s) + 2Cu²+(aq) Hint: Carry at least 5 significant figures during intermediate calculations to avoid round off error when taking the antilogarithm. You may use the OWL references to find the values you may need in this question. Equilibrium constant: AG° for this reaction would bị than zero. greater less Standard Reduction (Electrode) Potentials at 25 °C Half-Cell Reaction E° (volts) F2(g) + 2 e→2 F(aq) 2.87 Ce4*(aq) + e –→ Ce"(aq) 3+ 1.61 2+ MnO4 (aq) + 8 H"(aq) + 5 e¯ –→ Mn-T(aq) + 4 H2O(1) 1.51 Cl2(g) + 2 e' —2 СГ (aq) 1.36 Cr2072"(aq) + 14 H*(aq) + 6 e¯ →2 Cr"(aq) + 7 H2O(1) 1.33 O2(g) + 4 H*(aq) + 4 e→2 H2O(1) 1.229 Br2(1) + 2 e¯ –→2 Br¯(aq) 1.08 NO3 (aq) + 4 H"(aq) +3 e¯ –→ NO(g) + 2 H2O(1) 0.96 2 Hg-*(aq) + 2 e¯ → Hg22 (aq) 0.920 Hg-"(aq) + 2 e →Hg(1) 0.855 Ag"(aq) + e¯ – Ag(s) 0.799 2+, Hg2"(aq) + 2 e¯ →2 Hg(1) 0.789 Fe*(aq) + e → Fe2*(aq) 0.771 I2(s) + 2 e → 2 1 (aq) 0.535 Fe(CN)6° (aq) + e→ Fe(CN)64 (aq) 0.48 Cu2*(aq) + 2 e → Cu(s) 0.337 Cu2*(aq) + e" – Cu"(aq) 0.153 S(s) + 2 H*(aq) + 2 e → H2S(aq) 0.14 2 H*(aq) + 2 e → H2(g) 0.0000 Pb2*(aq) + 2 e → Pb(s) -0.126 Sn-"(aq) + 2 e¯ → Sn(s) „2+ -0.14 Ni2+(ag) + 2 e →Ni(s) -0.25 Co2+(ag) + 2 e →Co(s) -0.28 Cd2*(aq) + 2 e →Cd(s) -0.403 Cr3+ C* (aq) (aq) + e -0.41 Fe2*(aq) + 2 e →Fe(s) -0.44 Cr3+ (aq) + 3 e → Cr(s) -0.74 Zn2*(aq) + 2 e →Zn(s) -0.763 2 H20(1) + 2 e → H2(g) + 2 OH (aq) -0.83 Mn-"(aq) + 2 e¯ → Mn(s) 2+ -1.18 Als*(aq) + 3 e → Al(s) -1.66 2+ Mg"(aq) + 2 e →Mg(s) -2.37 Na"(aq) + e –→ Na(s) -2.714 K(aq) + e→ K(s) -2.925 Li*(aq) + e Li(s) -3.045
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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Transcribed Image Text:Use the standard reduction potentials located in the 'Tables' linked above to calculate the equilibrium constant for the reaction:
Fe2+(aq) + 2Cu*(aq)
→ Fe(s) + 2Cu²+(aq)
Hint: Carry at least 5 significant figures during intermediate calculations to avoid round off error when taking the antilogarithm. You may use the OWL
references to find the values you may need in this question.
Equilibrium constant:
AG° for this reaction would bị
than zero.
greater
less
Transcribed Image Text:Standard Reduction (Electrode) Potentials at 25 °C
Half-Cell Reaction
E° (volts)
F2(g) + 2 e→2 F(aq)
2.87
Ce4*(aq) + e –→ Ce"(aq)
3+
1.61
2+
MnO4 (aq) + 8 H"(aq) + 5 e¯ –→ Mn-T(aq) + 4 H2O(1)
1.51
Cl2(g) + 2 e' —2 СГ (aq)
1.36
Cr2072"(aq) + 14 H*(aq) + 6 e¯
→2 Cr"(aq) + 7 H2O(1)
1.33
O2(g) + 4 H*(aq) + 4 e→2 H2O(1)
1.229
Br2(1) + 2 e¯ –→2 Br¯(aq)
1.08
NO3 (aq) + 4 H"(aq) +3 e¯ –→ NO(g) + 2 H2O(1)
0.96
2 Hg-*(aq) + 2 e¯ →
Hg22 (aq)
0.920
Hg-"(aq) + 2 e →Hg(1)
0.855
Ag"(aq) + e¯ – Ag(s)
0.799
2+,
Hg2"(aq) + 2 e¯ →2 Hg(1)
0.789
Fe*(aq) + e →
Fe2*(aq)
0.771
I2(s) + 2 e → 2 1 (aq)
0.535
Fe(CN)6° (aq) + e→ Fe(CN)64 (aq)
0.48
Cu2*(aq) + 2 e → Cu(s)
0.337
Cu2*(aq) + e" – Cu"(aq)
0.153
S(s) + 2 H*(aq) + 2 e → H2S(aq)
0.14
2 H*(aq) + 2 e → H2(g)
0.0000
Pb2*(aq) + 2 e → Pb(s)
-0.126
Sn-"(aq) + 2 e¯ → Sn(s)
„2+
-0.14
Ni2+(ag) + 2 e →Ni(s)
-0.25
Co2+(ag) + 2 e →Co(s)
-0.28
Cd2*(aq) + 2 e →Cd(s)
-0.403
Cr3+
C* (aq)
(aq) + e
-0.41
Fe2*(aq) + 2 e →Fe(s)
-0.44
Cr3+
(aq) + 3 e → Cr(s)
-0.74
Zn2*(aq) + 2 e →Zn(s)
-0.763
2 H20(1) + 2 e → H2(g) + 2 OH (aq)
-0.83
Mn-"(aq) + 2 e¯ → Mn(s)
2+
-1.18
Als*(aq) + 3 e → Al(s)
-1.66
2+
Mg"(aq) + 2 e →Mg(s)
-2.37
Na"(aq) + e –→ Na(s)
-2.714
K(aq) + e→ K(s)
-2.925
Li*(aq) + e Li(s)
-3.045
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