Use the standard reduction potentials located in the 'Tables' linked above to calculate the equilibrium constant for the reaction: 2+ Sn²+ (aq) + Cd(s) → Sn(s) + Cd²+ (aq) Hint: Carry at least 5 significant figures during intermediate calculations to avoid round off error when taking the antilogarithm. K = AGⓇ for this reaction would be greater than zero.
Use the standard reduction potentials located in the 'Tables' linked above to calculate the equilibrium constant for the reaction: 2+ Sn²+ (aq) + Cd(s) → Sn(s) + Cd²+ (aq) Hint: Carry at least 5 significant figures during intermediate calculations to avoid round off error when taking the antilogarithm. K = AGⓇ for this reaction would be greater than zero.
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...
Related questions
Question
![Use the standard reduction potentials located in the 'Tables' linked
above to calculate the equilibrium constant for the reaction:
2+
Sn²+ (aq) + Cd(s) → Sn(s) + Cd²+ (aq)
Hint: Carry at least 5 significant figures during intermediate
calculations to avoid round off error when taking the antilogarithm.
K=
AGO for this reaction would be greater
than zero.
Use the standard reduction potentials located in the 'Tables' linked
above to calculate the equilibrium constant for the reaction:
2+
Sn²+ (aq) + 2Ag(s) → Sn(s) + 2Ag+ (aq)
Hint: Carry at least 5 significant figures during intermediate
calculations to avoid round off error when taking the antilogarithm.
K
=
AGO for this reaction would be less
than zero.
Use the standard reduction potentials located in the 'Tables' linked
above to calculate the equilibrium constant for the reaction:
Sn²
1²+ (aq) + Co(s) → Sn(s) + Co²+ (aq)
Hint: Carry at least 5 significant figures during intermediate
calculations to avoid round off error when taking the antilogarithm.
K =
AG for this reaction would be greater
than zero.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F11f65768-3637-42df-b106-adece588ddc5%2F75ae3498-16f2-4829-a2c6-889e139b5cf8%2F7a1jyiw_processed.png&w=3840&q=75)
Transcribed Image Text:Use the standard reduction potentials located in the 'Tables' linked
above to calculate the equilibrium constant for the reaction:
2+
Sn²+ (aq) + Cd(s) → Sn(s) + Cd²+ (aq)
Hint: Carry at least 5 significant figures during intermediate
calculations to avoid round off error when taking the antilogarithm.
K=
AGO for this reaction would be greater
than zero.
Use the standard reduction potentials located in the 'Tables' linked
above to calculate the equilibrium constant for the reaction:
2+
Sn²+ (aq) + 2Ag(s) → Sn(s) + 2Ag+ (aq)
Hint: Carry at least 5 significant figures during intermediate
calculations to avoid round off error when taking the antilogarithm.
K
=
AGO for this reaction would be less
than zero.
Use the standard reduction potentials located in the 'Tables' linked
above to calculate the equilibrium constant for the reaction:
Sn²
1²+ (aq) + Co(s) → Sn(s) + Co²+ (aq)
Hint: Carry at least 5 significant figures during intermediate
calculations to avoid round off error when taking the antilogarithm.
K =
AG for this reaction would be greater
than zero.
![Standard Reduction (Electrode) Potentials at 25 °C
Half-Cell Reaction
Eº (volts)
F₂(g) +2 e
2 F¯(aq)
2.87
Ce4+ (aq) + e
Ce³+ (aq)
1.61
1.51
MnO4 (aq) + 8 H*(aq) + 5 e
Cl₂(g) +2 e
2 Cl¯(aq)
1.36
Cr₂O7²-(aq) + 14 H+ (aq) + 6 e¯ -
1.33
O₂(g) + 4 H(aq) + 4 e
2 H₂O(1)
1.229
Br₂(1) +2 e¯ 2 Br¯(aq)
1.08
0.96
NO3(aq) + 4 H (aq) + 3 e¯ →→→ NO(g) + 2 H₂O(1)
2 Hg2+ (aq) + 2 eHg₂²+ (aq)
0.920
DOCE
Screen Shot 2022...
Standard Reduction (Electrode) Potentials at 25 °C
Half-Cell Reaction
Eº (volts)
Fe(CN)6³(aq) + e
Fe(CN)6+ (aq)
Cu²+ (aq) + 2 e¯→→→→→→ Cu(s)
Cu²+ (aq) + e
Cu (aq)
S(s) + 2 H(aq) + 2 e² →→→ H₂S(aq)
2 H(aq) + 2 e→→→→→ H₂(g)
Pb²+ (aq) + 2 e→→→→→→ Pb(s)
Sn²+ (aq) + 2e →→→→Sn(s)
Ni²+ (aq) + 2 e→→→→→→Ni(s)
Co²+ (aq) + 2 e→→→→→→Co(s)
Cd²+ (aq) +2 e
Cd(s)
Cr³+ (aq) + e
Cr²+ (aq)
Fe2+ (aq) + 2 e
Fe(s)
Cr³+ (aq) + 3 e→→→→→Cr(s)
Zn²+ (aq) + 2 e→→→→→→ Zn(s)
2 H₂O(1) + 2 e →→→ H₂(g) + 2 OH¯(aq)
Mn²+ (aq) + 4H₂O(1)
2 Cr³+ (aq) + 7 H₂O(1)
0.48
0.337
0.153
0.14
0.0000
-0.126
-0.14
-0.25
-0.28
-0.403
-0.41
-0.44
-0.74
-0.763
-0.83
Standard Reduction (Electrode) Potentials at 25 ºC
Half-Cell Reaction
Eº (volts)
Pb(aq) + 2 e →→→→→→ Pb(s)
2+
Sn²(aq) + 2 e→→→→→→→→Sn(s)
Ni²+ (aq) + 2 e¯→→→→→→Ni(s)
2+
Co²+ (aq) + 2 e
Co(s)
Cd²+ (aq) + 2 e¯ →→→→→→ Cd(s)
Cr³+ (aq) + e
> Cr²+ (aq)
Fe²+ (aq) + 2 e
→→→ Fe(s)
→→→→Cr(s)
Cr³+ (aq) + 3 e
2+
Zn²+ (aq) + 2e
→→→ Zn(s)
2 H₂O(1) + 2 e →→→ H₂(g) + 2 OH (aq)
Mn²+ (aq) + 2 e →→→→Mn(s)
Al³+ (aq) + 3 e¯→→→→→→ Al(s)
Mg2+ (aq) + 2 e→→→→→→ Mg(s)
Na+ (aq) + e→→→→→→→ Na(s)
K+ (aq) + e→→→→→→K(s)
Lit (aq) +eLi(s)
-0.126
-0.14
-0.25
-0.28
-0.403
-0.41
-0.44
-0.74
-0.763
-0.83
-1.18
-1.66
-2.37
-2.714
-2.925
-3.045](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F11f65768-3637-42df-b106-adece588ddc5%2F75ae3498-16f2-4829-a2c6-889e139b5cf8%2Ftj26sl_processed.png&w=3840&q=75)
Transcribed Image Text:Standard Reduction (Electrode) Potentials at 25 °C
Half-Cell Reaction
Eº (volts)
F₂(g) +2 e
2 F¯(aq)
2.87
Ce4+ (aq) + e
Ce³+ (aq)
1.61
1.51
MnO4 (aq) + 8 H*(aq) + 5 e
Cl₂(g) +2 e
2 Cl¯(aq)
1.36
Cr₂O7²-(aq) + 14 H+ (aq) + 6 e¯ -
1.33
O₂(g) + 4 H(aq) + 4 e
2 H₂O(1)
1.229
Br₂(1) +2 e¯ 2 Br¯(aq)
1.08
0.96
NO3(aq) + 4 H (aq) + 3 e¯ →→→ NO(g) + 2 H₂O(1)
2 Hg2+ (aq) + 2 eHg₂²+ (aq)
0.920
DOCE
Screen Shot 2022...
Standard Reduction (Electrode) Potentials at 25 °C
Half-Cell Reaction
Eº (volts)
Fe(CN)6³(aq) + e
Fe(CN)6+ (aq)
Cu²+ (aq) + 2 e¯→→→→→→ Cu(s)
Cu²+ (aq) + e
Cu (aq)
S(s) + 2 H(aq) + 2 e² →→→ H₂S(aq)
2 H(aq) + 2 e→→→→→ H₂(g)
Pb²+ (aq) + 2 e→→→→→→ Pb(s)
Sn²+ (aq) + 2e →→→→Sn(s)
Ni²+ (aq) + 2 e→→→→→→Ni(s)
Co²+ (aq) + 2 e→→→→→→Co(s)
Cd²+ (aq) +2 e
Cd(s)
Cr³+ (aq) + e
Cr²+ (aq)
Fe2+ (aq) + 2 e
Fe(s)
Cr³+ (aq) + 3 e→→→→→Cr(s)
Zn²+ (aq) + 2 e→→→→→→ Zn(s)
2 H₂O(1) + 2 e →→→ H₂(g) + 2 OH¯(aq)
Mn²+ (aq) + 4H₂O(1)
2 Cr³+ (aq) + 7 H₂O(1)
0.48
0.337
0.153
0.14
0.0000
-0.126
-0.14
-0.25
-0.28
-0.403
-0.41
-0.44
-0.74
-0.763
-0.83
Standard Reduction (Electrode) Potentials at 25 ºC
Half-Cell Reaction
Eº (volts)
Pb(aq) + 2 e →→→→→→ Pb(s)
2+
Sn²(aq) + 2 e→→→→→→→→Sn(s)
Ni²+ (aq) + 2 e¯→→→→→→Ni(s)
2+
Co²+ (aq) + 2 e
Co(s)
Cd²+ (aq) + 2 e¯ →→→→→→ Cd(s)
Cr³+ (aq) + e
> Cr²+ (aq)
Fe²+ (aq) + 2 e
→→→ Fe(s)
→→→→Cr(s)
Cr³+ (aq) + 3 e
2+
Zn²+ (aq) + 2e
→→→ Zn(s)
2 H₂O(1) + 2 e →→→ H₂(g) + 2 OH (aq)
Mn²+ (aq) + 2 e →→→→Mn(s)
Al³+ (aq) + 3 e¯→→→→→→ Al(s)
Mg2+ (aq) + 2 e→→→→→→ Mg(s)
Na+ (aq) + e→→→→→→→ Na(s)
K+ (aq) + e→→→→→→K(s)
Lit (aq) +eLi(s)
-0.126
-0.14
-0.25
-0.28
-0.403
-0.41
-0.44
-0.74
-0.763
-0.83
-1.18
-1.66
-2.37
-2.714
-2.925
-3.045
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