Question 8 Based on the following data of electrode reduction potentials, formation constants and solubility product constants, answer the following questions: Reaction Cu2+ + 2 e Cuº Cu²+ + e = Cut Cut + e = Cuº Cut +4 CN = Cu(CN)4³- Cu2+ + 4 CN = Cu(CN)4²- CuCN= Cu + CN Ered 0.34 V 0.16 V 0.52 V Kea Kf = 2.0x1030 Kf = 1.0x1025 Ksp = 3.47x10-20 Calculate the Eºred potential for the corresponding reactions at 298 K. Give your answer to two decimal places. (0.0592) log(K) E = = a) Cu(CN)42 + e* = Cu(CN)4³-
Question 8 Based on the following data of electrode reduction potentials, formation constants and solubility product constants, answer the following questions: Reaction Cu2+ + 2 e Cuº Cu²+ + e = Cut Cut + e = Cuº Cut +4 CN = Cu(CN)4³- Cu2+ + 4 CN = Cu(CN)4²- CuCN= Cu + CN Ered 0.34 V 0.16 V 0.52 V Kea Kf = 2.0x1030 Kf = 1.0x1025 Ksp = 3.47x10-20 Calculate the Eºred potential for the corresponding reactions at 298 K. Give your answer to two decimal places. (0.0592) log(K) E = = a) Cu(CN)42 + e* = Cu(CN)4³-
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 8
Based on the following data of electrode reduction potentials, formation
constants and solubility product constants, answer the following
questions:
Reaction
Cu2+ + 2 e
Cuº
Cu²+ + e = Cut
Cut + e = Cuº
Cut + 4 CN = Cu(CN)4³-
Cu2+ + 4 CN = Cu(CN)4²-
CuCN= Cut + CN°
a) Cu(CN)42 + e = Cu(CN)4³-
Calculate the Eºred potential for the corresponding reactions at 298 K.
Give your answer to two decimal places.
E =
b) Cu(CN)4² +2 e² = Cu° + 4 CN
V
V
c) CuCN + 3 CN = Cu(CN)4² + e
V
Ered
0.34 V
0.16 V
0.52 V
Kea
(0.0592) log(K)
71
Kf = 2.0x1030
Kf = 1.0x1025
Ksp = 3.47x10-20](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F93cc6d0b-8497-479b-9715-842a08134eab%2Fa082b3ff-23a4-46a4-ac31-74332e2e5415%2Fr4rwpbj_processed.jpeg&w=3840&q=75)
Transcribed Image Text:Question 8
Based on the following data of electrode reduction potentials, formation
constants and solubility product constants, answer the following
questions:
Reaction
Cu2+ + 2 e
Cuº
Cu²+ + e = Cut
Cut + e = Cuº
Cut + 4 CN = Cu(CN)4³-
Cu2+ + 4 CN = Cu(CN)4²-
CuCN= Cut + CN°
a) Cu(CN)42 + e = Cu(CN)4³-
Calculate the Eºred potential for the corresponding reactions at 298 K.
Give your answer to two decimal places.
E =
b) Cu(CN)4² +2 e² = Cu° + 4 CN
V
V
c) CuCN + 3 CN = Cu(CN)4² + e
V
Ered
0.34 V
0.16 V
0.52 V
Kea
(0.0592) log(K)
71
Kf = 2.0x1030
Kf = 1.0x1025
Ksp = 3.47x10-20
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