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Passage II
The standard half-cell reduction potential of
Fe3+(aq) | Fe is -0.036V and that of OH- | Fe(OH3)(s) | Feis -0.786V.
Q.
The value of loge Ksp for Fe(OH)3 at 298 K is
- a)-38.1
- b)+87.7
- c)-96.0
- d)-87.7
Correct answer is option 'D'. Can you explain this answer?
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Verified Answer
Passage IIThe standard half-cell reduction potentialof Fe3+(aq) | Fe i...
Above reaction is the net reaction in a cell for determination of Ksp.
Thus,
Anode
Fe → Fe3+ (aq) + 3e-, E0ox = 0.036V
Half - cell is Fe|Fe3+ (aq). OH-(aq)
Cathode
Fe(OH)3(s) + 3e- → Fe+ 3OH-, E0red = -0.786 V
Half-cell set up is Fe(OH)3(s)|Fe
complete cell is Fe|Fe3+(aq), OH- (aq)||Fe(OH)3 (s)|Fe
and E0cell = E0ox + E0red = 0.036 + (-0.786) = -0.750V
By net reaction, K = Ksp = [Fe3+][OH-]3
∴ 
Most Upvoted Answer
Passage IIThe standard half-cell reduction potentialof Fe3+(aq) | Fe i...
Calculation of Ksp for Fe(OH)3
Half-cell reactions:
Fe3+ (aq) + 3e- → Fe(s) E° = -0.036 V
Fe(OH)3 (s) + 3e- → Fe(s) + 3OH- (aq) E° = -0.786 V
Overall reaction: Fe3+ (aq) + 3OH- (aq) → Fe(OH)3 (s)
Standard emf of the cell, E°cell = E°cathode – E°anode
E°cell = 0.786 – (-0.036) = 0.822 V
The equilibrium constant expression for the above overall reaction is:
Ksp = [Fe3+] [OH-]^3 / [Fe(OH)3]
The standard free energy change for the above reaction is:
ΔG° = -nFE°cell = -3 x 96500 x 0.822 = -237310 J/mol
The relationship between ΔG° and Ksp is given by the equation:
ΔG° = -RT ln Ksp
-237310 = -8.314 x 298 x ln Ksp
ln Ksp = -87.7
Therefore, Ksp = e^-87.7 = 2.42 x 10^-38
Hence, the value of logeKsp for Fe(OH)3 at 298 K is -87.7.
Half-cell reactions:
Fe3+ (aq) + 3e- → Fe(s) E° = -0.036 V
Fe(OH)3 (s) + 3e- → Fe(s) + 3OH- (aq) E° = -0.786 V
Overall reaction: Fe3+ (aq) + 3OH- (aq) → Fe(OH)3 (s)
Standard emf of the cell, E°cell = E°cathode – E°anode
E°cell = 0.786 – (-0.036) = 0.822 V
The equilibrium constant expression for the above overall reaction is:
Ksp = [Fe3+] [OH-]^3 / [Fe(OH)3]
The standard free energy change for the above reaction is:
ΔG° = -nFE°cell = -3 x 96500 x 0.822 = -237310 J/mol
The relationship between ΔG° and Ksp is given by the equation:
ΔG° = -RT ln Ksp
-237310 = -8.314 x 298 x ln Ksp
ln Ksp = -87.7
Therefore, Ksp = e^-87.7 = 2.42 x 10^-38
Hence, the value of logeKsp for Fe(OH)3 at 298 K is -87.7.
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Passage IIThe standard half-cell reduction potentialof Fe3+(aq) | Fe is -0.036Vand that ofOH- | Fe(OH3)(s) | Feis -0.786V.Q. The value of logeKsp for Fe(OH)3 at 298 K isa)-38.1b)+87.7c)-96.0d)-87.7Correct answer is option 'D'. Can you explain this answer?
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Passage IIThe standard half-cell reduction potentialof Fe3+(aq) | Fe is -0.036Vand that ofOH- | Fe(OH3)(s) | Feis -0.786V.Q. The value of logeKsp for Fe(OH)3 at 298 K isa)-38.1b)+87.7c)-96.0d)-87.7Correct answer is option 'D'. Can you explain this answer? for Class 12 2024 is part of Class 12 preparation. The Question and answers have been prepared according to the Class 12 exam syllabus. Information about Passage IIThe standard half-cell reduction potentialof Fe3+(aq) | Fe is -0.036Vand that ofOH- | Fe(OH3)(s) | Feis -0.786V.Q. The value of logeKsp for Fe(OH)3 at 298 K isa)-38.1b)+87.7c)-96.0d)-87.7Correct answer is option 'D'. Can you explain this answer? covers all topics & solutions for Class 12 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Passage IIThe standard half-cell reduction potentialof Fe3+(aq) | Fe is -0.036Vand that ofOH- | Fe(OH3)(s) | Feis -0.786V.Q. The value of logeKsp for Fe(OH)3 at 298 K isa)-38.1b)+87.7c)-96.0d)-87.7Correct answer is option 'D'. Can you explain this answer?.
Passage IIThe standard half-cell reduction potentialof Fe3+(aq) | Fe is -0.036Vand that ofOH- | Fe(OH3)(s) | Feis -0.786V.Q. The value of logeKsp for Fe(OH)3 at 298 K isa)-38.1b)+87.7c)-96.0d)-87.7Correct answer is option 'D'. Can you explain this answer? for Class 12 2024 is part of Class 12 preparation. The Question and answers have been prepared according to the Class 12 exam syllabus. Information about Passage IIThe standard half-cell reduction potentialof Fe3+(aq) | Fe is -0.036Vand that ofOH- | Fe(OH3)(s) | Feis -0.786V.Q. The value of logeKsp for Fe(OH)3 at 298 K isa)-38.1b)+87.7c)-96.0d)-87.7Correct answer is option 'D'. Can you explain this answer? covers all topics & solutions for Class 12 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Passage IIThe standard half-cell reduction potentialof Fe3+(aq) | Fe is -0.036Vand that ofOH- | Fe(OH3)(s) | Feis -0.786V.Q. The value of logeKsp for Fe(OH)3 at 298 K isa)-38.1b)+87.7c)-96.0d)-87.7Correct answer is option 'D'. Can you explain this answer?.
Solutions for Passage IIThe standard half-cell reduction potentialof Fe3+(aq) | Fe is -0.036Vand that ofOH- | Fe(OH3)(s) | Feis -0.786V.Q. The value of logeKsp for Fe(OH)3 at 298 K isa)-38.1b)+87.7c)-96.0d)-87.7Correct answer is option 'D'. Can you explain this answer? in English & in Hindi are available as part of our courses for Class 12.
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Passage IIThe standard half-cell reduction potentialof Fe3+(aq) | Fe is -0.036Vand that ofOH- | Fe(OH3)(s) | Feis -0.786V.Q. The value of logeKsp for Fe(OH)3 at 298 K isa)-38.1b)+87.7c)-96.0d)-87.7Correct answer is option 'D'. Can you explain this answer?, a detailed solution for Passage IIThe standard half-cell reduction potentialof Fe3+(aq) | Fe is -0.036Vand that ofOH- | Fe(OH3)(s) | Feis -0.786V.Q. The value of logeKsp for Fe(OH)3 at 298 K isa)-38.1b)+87.7c)-96.0d)-87.7Correct answer is option 'D'. Can you explain this answer? has been provided alongside types of Passage IIThe standard half-cell reduction potentialof Fe3+(aq) | Fe is -0.036Vand that ofOH- | Fe(OH3)(s) | Feis -0.786V.Q. The value of logeKsp for Fe(OH)3 at 298 K isa)-38.1b)+87.7c)-96.0d)-87.7Correct answer is option 'D'. Can you explain this answer? theory, EduRev gives you an
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