NEET Exam > NEET Questions > If Eo = 0.34 V then reduction potential of C...
Start Learning for Free
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?
Most Upvoted Answer
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
Community Answer
If Eo = 0.34 V then reduction potential of Cu2 /Cu electrode at pH = ...
Attention NEET Students!
To make sure you are not studying endlessly, EduRev has designed NEET study material, with Structured Courses, Videos, & Test Series. Plus get personalized analysis, doubt solving and improvement plans to achieve a great score in NEET.
|
Explore Courses for NEET exam
|
|
Similar NEET Doubts
Top Courses for NEETView all
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?
Question Description
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?.
Solutions 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? in English & in Hindi are available as part of our courses for NEET.
Download more important topics, notes, lectures and mock test series for NEET Exam by signing up for free.
Here you can find the meaning of 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? defined & explained in the simplest way possible. Besides giving the explanation of
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?, a detailed solution 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? has been provided alongside types of 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? theory, EduRev gives you an
ample number of questions to practice 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? tests, examples and also practice NEET tests.
|
|
Explore Courses for NEET exam
|
|
Suggested Free Tests
Signup for Free!
Signup to see your scores go up within 7 days! Learn & Practice with 1000+ FREE Notes, Videos & Tests.
|
© EduRev
|
Education Revolution
|
Follow Us
|
Signup to see your scores
go up within 7 days!
Access 1000+ FREE Docs, Videos and Tests
Takes less than 10 seconds to signup