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If Eo = 0.34 V then reduction potential of Cu2 /Cu electrode at pH = 14 is (Ksp of Cu(OH)2 = 1 x 10-19) A)-0.92 V A)- 0.22 V C)-0.65 V D)-0.76 V Solutions please?
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If Eo = 0.34 V then reduction potential of Cu2 /Cu electrode at pH = ...
Solution:
Given: Eo = 0.34 V, pH = 14, Ksp of Cu(OH)2 = 1 x 10-19
Reduction potential of Cu2+/Cu electrode can be calculated using the Nernst equation:
E = Eo - (0.0591/n) log(Q)
where, E is the reduction potential, Eo is the standard reduction potential, n is the number of electrons involved, and Q is the reaction quotient.
1. Write the half-reaction for Cu2+/Cu electrode
Cu2+ + 2e- → Cu
2. Calculate the reaction quotient (Q)
Q = [Cu]/[Cu2+]
At pH 14, the concentration of Cu2+ can be calculated using the solubility product (Ksp) of Cu(OH)2
Ksp = [Cu2+][OH-]2
[Cu2+] = Ksp/[OH-]2
[Cu2+] = 1 x 10-19/10-28
[Cu2+] = 106 M
[OH-] = 10-14/106
[OH-] = 9.43 x 10-13 M
Q = [Cu]/[Cu2+]
Q = 1/[Cu2+]
Q = 1/106
Q = 9.43 x 10-12
3. Calculate the reduction potential (E)
E = Eo - (0.0591/n) log(Q)
n = 2 (as 2 electrons are involved in the half-reaction)
E = 0.34 - (0.0591/2) log(9.43 x 10-12)
E = -0.76 V
Therefore, the reduction potential of Cu2+/Cu electrode at pH 14 is -0.76 V.
Answer: D) -0.76 V
Given: Eo = 0.34 V, pH = 14, Ksp of Cu(OH)2 = 1 x 10-19
Reduction potential of Cu2+/Cu electrode can be calculated using the Nernst equation:
E = Eo - (0.0591/n) log(Q)
where, E is the reduction potential, Eo is the standard reduction potential, n is the number of electrons involved, and Q is the reaction quotient.
1. Write the half-reaction for Cu2+/Cu electrode
Cu2+ + 2e- → Cu
2. Calculate the reaction quotient (Q)
Q = [Cu]/[Cu2+]
At pH 14, the concentration of Cu2+ can be calculated using the solubility product (Ksp) of Cu(OH)2
Ksp = [Cu2+][OH-]2
[Cu2+] = Ksp/[OH-]2
[Cu2+] = 1 x 10-19/10-28
[Cu2+] = 106 M
[OH-] = 10-14/106
[OH-] = 9.43 x 10-13 M
Q = [Cu]/[Cu2+]
Q = 1/[Cu2+]
Q = 1/106
Q = 9.43 x 10-12
3. Calculate the reduction potential (E)
E = Eo - (0.0591/n) log(Q)
n = 2 (as 2 electrons are involved in the half-reaction)
E = 0.34 - (0.0591/2) log(9.43 x 10-12)
E = -0.76 V
Therefore, the reduction potential of Cu2+/Cu electrode at pH 14 is -0.76 V.
Answer: D) -0.76 V
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If Eo = 0.34 V then reduction potential of Cu2 /Cu electrode at pH = ...
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If Eo = 0.34 V then reduction potential of Cu2 /Cu electrode at pH = 14 is (Ksp of Cu(OH)2 = 1 x 10-19) A)-0.92 V A)- 0.22 V C)-0.65 V D)-0.76 V Solutions please?
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If Eo = 0.34 V then reduction potential of Cu2 /Cu electrode at pH = 14 is (Ksp of Cu(OH)2 = 1 x 10-19) A)-0.92 V A)- 0.22 V C)-0.65 V D)-0.76 V Solutions please? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about If Eo = 0.34 V then reduction potential of Cu2 /Cu electrode at pH = 14 is (Ksp of Cu(OH)2 = 1 x 10-19) A)-0.92 V A)- 0.22 V C)-0.65 V D)-0.76 V Solutions please? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for If Eo = 0.34 V then reduction potential of Cu2 /Cu electrode at pH = 14 is (Ksp of Cu(OH)2 = 1 x 10-19) A)-0.92 V A)- 0.22 V C)-0.65 V D)-0.76 V Solutions please?.
If Eo = 0.34 V then reduction potential of Cu2 /Cu electrode at pH = 14 is (Ksp of Cu(OH)2 = 1 x 10-19) A)-0.92 V A)- 0.22 V C)-0.65 V D)-0.76 V Solutions please? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about If Eo = 0.34 V then reduction potential of Cu2 /Cu electrode at pH = 14 is (Ksp of Cu(OH)2 = 1 x 10-19) A)-0.92 V A)- 0.22 V C)-0.65 V D)-0.76 V Solutions please? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for If Eo = 0.34 V then reduction potential of Cu2 /Cu electrode at pH = 14 is (Ksp of Cu(OH)2 = 1 x 10-19) A)-0.92 V A)- 0.22 V C)-0.65 V D)-0.76 V Solutions please?.
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