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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...
Calculation of Electrode Potential at 0.01 M concentration of Cu2+
- Given:
Standard electrode potential of Cu2+/Cu electrode (E°) = 0.34 V
Concentration of Cu2+ (Cu2+) = 0.01 M
Temperature (T) = 298 K
- Nernst Equation:
The Nernst equation relates the electrode potential of an electrochemical cell to the concentration of the reactants and products involved in the cell reaction.
The Nernst equation is given by:
E = E° - (0.0592/n) * log(Q)
Where:
E = Electrode potential at a given concentration
E° = Standard electrode potential
n = Number of electrons involved in the cell reaction
Q = Reaction quotient
- Calculation:
The cell reaction for Cu2+/Cu electrode is:
Cu2+ + 2e- -> Cu
Number of electrons (n) = 2
The reaction quotient (Q) can be calculated as:
Q = [Cu]/[Cu2+]
Q = 1 / 0.01
Q = 100
Now, substitute the values into the Nernst equation:
E = 0.34 - (0.0592/2) * log(100)
E = 0.34 - 0.0296 * 2
E = 0.34 - 0.0592
E = 0.281 V
Therefore, the electrode potential at 0.01 M concentration of Cu2+ is 0.281 V, which corresponds to option 'B'.
- Given:
Standard electrode potential of Cu2+/Cu electrode (E°) = 0.34 V
Concentration of Cu2+ (Cu2+) = 0.01 M
Temperature (T) = 298 K
- Nernst Equation:
The Nernst equation relates the electrode potential of an electrochemical cell to the concentration of the reactants and products involved in the cell reaction.
The Nernst equation is given by:
E = E° - (0.0592/n) * log(Q)
Where:
E = Electrode potential at a given concentration
E° = Standard electrode potential
n = Number of electrons involved in the cell reaction
Q = Reaction quotient
- Calculation:
The cell reaction for Cu2+/Cu electrode is:
Cu2+ + 2e- -> Cu
Number of electrons (n) = 2
The reaction quotient (Q) can be calculated as:
Q = [Cu]/[Cu2+]
Q = 1 / 0.01
Q = 100
Now, substitute the values into the Nernst equation:
E = 0.34 - (0.0592/2) * log(100)
E = 0.34 - 0.0296 * 2
E = 0.34 - 0.0592
E = 0.281 V
Therefore, the electrode potential at 0.01 M concentration of Cu2+ is 0.281 V, which corresponds to option 'B'.
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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?.
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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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?, a detailed solution 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? has been provided alongside types of 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? theory, EduRev gives you an
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