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If the standard electrode potential of Cu2+/Cu electrode is 0.34 V, what is the electrode potential at 0.01 M concentration of Cu2+? (T = 298K)
- a)0.399 V
- b)0.281 V
- c)0.222 V
- d)0.176 V
Correct answer is option 'B'. Can you explain this answer?
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If the standard electrode potential of Cu2+/Cu electrode is 0.34 V, wh...
Given:
Standard electrode potential of Cu2+/Cu electrode = 0.34 V
Concentration of Cu2+ = 0.01 M
Temperature, T = 298 K
To find: Electrode potential at 0.01 M concentration of Cu2+
Solution:
The Nernst equation relates the electrode potential of a half-cell to the activities (or concentrations) of the species involved in the redox reaction.
Nernst equation: E = E° - (RT/nF) ln(Q)
where,
E = Electrode potential
E° = Standard electrode potential
R = Gas constant = 8.314 J mol-1 K-1
T = Temperature in Kelvin
n = number of electrons transferred in the redox reaction
F = Faraday constant = 96,485 C mol-1
Q = reaction quotient
The reaction taking place at the Cu2+/Cu electrode is:
Cu2+ + 2e- → Cu
The standard electrode potential for this reaction is given as 0.34 V, which means that under standard conditions, the Cu2+/Cu half-cell has a potential of 0.34 V.
Now, let's calculate the reaction quotient, Q, for the given concentration of Cu2+.
Q = [Cu2+]/[Cu]2
Substituting the given values, we get:
Q = (0.01)/(1x10-4)2 = 100
Substituting the values in the Nernst equation, we get:
E = 0.34 - (8.314 x 298 / (2 x 96,485)) ln(100) = 0.281 V
Therefore, the electrode potential at 0.01 M concentration of Cu2+ is 0.281 V.
Hence, option (b) is the correct answer.
Standard electrode potential of Cu2+/Cu electrode = 0.34 V
Concentration of Cu2+ = 0.01 M
Temperature, T = 298 K
To find: Electrode potential at 0.01 M concentration of Cu2+
Solution:
The Nernst equation relates the electrode potential of a half-cell to the activities (or concentrations) of the species involved in the redox reaction.
Nernst equation: E = E° - (RT/nF) ln(Q)
where,
E = Electrode potential
E° = Standard electrode potential
R = Gas constant = 8.314 J mol-1 K-1
T = Temperature in Kelvin
n = number of electrons transferred in the redox reaction
F = Faraday constant = 96,485 C mol-1
Q = reaction quotient
The reaction taking place at the Cu2+/Cu electrode is:
Cu2+ + 2e- → Cu
The standard electrode potential for this reaction is given as 0.34 V, which means that under standard conditions, the Cu2+/Cu half-cell has a potential of 0.34 V.
Now, let's calculate the reaction quotient, Q, for the given concentration of Cu2+.
Q = [Cu2+]/[Cu]2
Substituting the given values, we get:
Q = (0.01)/(1x10-4)2 = 100
Substituting the values in the Nernst equation, we get:
E = 0.34 - (8.314 x 298 / (2 x 96,485)) ln(100) = 0.281 V
Therefore, the electrode potential at 0.01 M concentration of Cu2+ is 0.281 V.
Hence, option (b) is the correct answer.
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If the standard electrode potential of Cu2+/Cu electrode is 0.34 V, what is the electrode potential at 0.01 M concentration of Cu2+? (T = 298K)a)0.399 Vb)0.281 Vc)0.222 Vd)0.176 VCorrect answer is option 'B'. Can you explain this answer?
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If the standard electrode potential of Cu2+/Cu electrode is 0.34 V, what is the electrode potential at 0.01 M concentration of Cu2+? (T = 298K)a)0.399 Vb)0.281 Vc)0.222 Vd)0.176 VCorrect answer is option 'B'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about If the standard electrode potential of Cu2+/Cu electrode is 0.34 V, what is the electrode potential at 0.01 M concentration of Cu2+? (T = 298K)a)0.399 Vb)0.281 Vc)0.222 Vd)0.176 VCorrect answer is option 'B'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for If the standard electrode potential of Cu2+/Cu electrode is 0.34 V, what is the electrode potential at 0.01 M concentration of Cu2+? (T = 298K)a)0.399 Vb)0.281 Vc)0.222 Vd)0.176 VCorrect answer is option 'B'. Can you explain this answer?.
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