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-

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Chapter19: Transition Metals And Coordination Chemistry
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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
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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