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Given : 2H2O → O2 + 4H+ + 4e–
Eº = –1.23 V Calculate electrode potential at pH = 5.
Eº = –1.23 V Calculate electrode potential at pH = 5.
Correct answer is '–00.93'. Can you explain this answer?
Most Upvoted Answer
Given : 2H2O → O2 + 4H+ + 4e–Eº = –1.23 V Calcul...
Understanding the Reaction
The reaction given is:
2H2O → O2 + 4H+ + 4e
The standard electrode potential (E°) for this reaction is -1.23 V.
Effect of pH on Electrode Potential
The potential of the electrode can change with pH, primarily due to the concentration of H+ ions. The Nernst equation is used to calculate this change:
E = E° + (RT/nF) * ln(Q)
Where:
- E = Electrode potential under non-standard conditions
- E° = Standard electrode potential (-1.23 V)
- R = Universal gas constant (8.314 J/(mol K))
- T = Temperature in Kelvin (assume 298 K for standard conditions)
- n = Number of moles of electrons transferred (4 in this case)
- F = Faraday's constant (96485 C/mol)
- Q = Reaction quotient
Calculating Q
At pH 5:
- [H+] = 10^(-5) M
- Q = [H+]^4 = (10^(-5))^4 = 10^(-20)
Substituting Values in Nernst Equation
E = -1.23 V + (0.0592 V/4) * log(10^20)
Calculating:
- E = -1.23 V + 0.0148 V * 20
- E = -1.23 V + 0.296 V
- E = -0.934 V
Rounding off gives:
E ≈ -0.93 V
Conclusion
Thus, the electrode potential at pH 5 is approximately -0.93 V, confirming the answer.
The reaction given is:
2H2O → O2 + 4H+ + 4e
The standard electrode potential (E°) for this reaction is -1.23 V.
Effect of pH on Electrode Potential
The potential of the electrode can change with pH, primarily due to the concentration of H+ ions. The Nernst equation is used to calculate this change:
E = E° + (RT/nF) * ln(Q)
Where:
- E = Electrode potential under non-standard conditions
- E° = Standard electrode potential (-1.23 V)
- R = Universal gas constant (8.314 J/(mol K))
- T = Temperature in Kelvin (assume 298 K for standard conditions)
- n = Number of moles of electrons transferred (4 in this case)
- F = Faraday's constant (96485 C/mol)
- Q = Reaction quotient
Calculating Q
At pH 5:
- [H+] = 10^(-5) M
- Q = [H+]^4 = (10^(-5))^4 = 10^(-20)
Substituting Values in Nernst Equation
E = -1.23 V + (0.0592 V/4) * log(10^20)
Calculating:
- E = -1.23 V + 0.0148 V * 20
- E = -1.23 V + 0.296 V
- E = -0.934 V
Rounding off gives:
E ≈ -0.93 V
Conclusion
Thus, the electrode potential at pH 5 is approximately -0.93 V, confirming the answer.
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Given : 2H2O → O2 + 4H+ + 4e–Eº = –1.23 V Calculate electrode potential at pH = 5.Correct answer is '–00.93'. Can you explain this answer?
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Given : 2H2O → O2 + 4H+ + 4e–Eº = –1.23 V Calculate electrode potential at pH = 5.Correct answer is '–00.93'. Can you explain this answer? for JEE 2025 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about Given : 2H2O → O2 + 4H+ + 4e–Eº = –1.23 V Calculate electrode potential at pH = 5.Correct answer is '–00.93'. Can you explain this answer? covers all topics & solutions for JEE 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Given : 2H2O → O2 + 4H+ + 4e–Eº = –1.23 V Calculate electrode potential at pH = 5.Correct answer is '–00.93'. Can you explain this answer?.
Given : 2H2O → O2 + 4H+ + 4e–Eº = –1.23 V Calculate electrode potential at pH = 5.Correct answer is '–00.93'. Can you explain this answer? for JEE 2025 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about Given : 2H2O → O2 + 4H+ + 4e–Eº = –1.23 V Calculate electrode potential at pH = 5.Correct answer is '–00.93'. Can you explain this answer? covers all topics & solutions for JEE 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Given : 2H2O → O2 + 4H+ + 4e–Eº = –1.23 V Calculate electrode potential at pH = 5.Correct answer is '–00.93'. Can you explain this answer?.
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