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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.
For the determination of solubility product of Fe(OH)3 (Ksp) the appropriate cell representation and its EMF respectively are
- a)Fe | Fe(OH)3(s) || OH-. (aq) Fe3+ (aq) | Fe -0.750 V
- b)Fe | Fe3+ (aq), OH- (aq) || Fe(OH)3(s) | | Fe -0.750 V
- c)Fe|Fe(OH3)(s) || OH-(aq), Fe3+(aq) | Fe +0.750V
- d)Fe|Fe3+(aq), OH-(aq)| Fe(OH3)(s)|Fe -0.822V
Correct answer is option 'B'. Can you explain this answer?
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Passage IIThe standard half-cell reduction potentialof Fe3+(aq) | Fe i...
Above reaction is the net reaction in a cell for determination of Ksp.
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Passage IIThe standard half-cell reduction potentialof Fe3+(aq) | Fe i...
Cell Representation for Determination of Ksp of Fe(OH)3
The appropriate cell representation for the determination of solubility product of Fe(OH)3 is:
Fe | Fe3+ (aq), OH- (aq) || Fe(OH)3 (s) | | Fe
Explanation:
- The cell consists of two half-cells: Fe | Fe3+ (aq) and Fe(OH)3 (s) | | Fe.
- The vertical line (||) represents the boundary between the two half-cells.
- The half-cell on the left contains Fe in contact with Fe3+ (aq) and OH- (aq) ions. This half-cell acts as the anode and undergoes oxidation.
- The half-cell on the right contains Fe(OH)3 (s) in contact with Fe2+ (aq) ions. This half-cell acts as the cathode and undergoes reduction.
- The overall cell reaction is Fe(OH)3 (s) + 3e- → Fe(s) + 3OH- (aq)
- The standard reduction potentials of Fe3+ (aq) | Fe and Fe(OH)3 (s) | Fe are given as -0.036 V and -0.786 V, respectively.
EMF of the Cell
The EMF (electromotive force) of the cell can be calculated using the Nernst equation:
Ecell = E°cell - (RT/nF) ln Q
where, E°cell is the standard cell potential, R is the gas constant, T is the temperature in Kelvin, n is the number of electrons transferred in the cell reaction, F is the Faraday constant, and Q is the reaction quotient.
For the given cell representation, the standard cell potential (E°cell) can be calculated as follows:
E°cell = E°cathode - E°anode
= (-0.036) - (-0.786)
= 0.750 V
Since the reaction involves the transfer of three electrons, n = 3.
The reaction quotient (Q) for the cell reaction can be expressed as follows:
Q = [Fe2+] / [OH-]^3
At equilibrium, Q = Ksp, where Ksp is the solubility product of Fe(OH)3.
Therefore, the EMF of the cell can be expressed as:
Ecell = 0.750 - (0.0257/3) log Ksp
Solving for Ksp, we get:
Ksp = 2.8 x 10^-39
Therefore, the appropriate cell representation and its EMF for the determination of solubility product of Fe(OH)3 is option B: Fe | Fe3+ (aq), OH- (aq) || Fe(OH)3 (s) | | Fe with an EMF of -0.750 V.
The appropriate cell representation for the determination of solubility product of Fe(OH)3 is:
Fe | Fe3+ (aq), OH- (aq) || Fe(OH)3 (s) | | Fe
Explanation:
- The cell consists of two half-cells: Fe | Fe3+ (aq) and Fe(OH)3 (s) | | Fe.
- The vertical line (||) represents the boundary between the two half-cells.
- The half-cell on the left contains Fe in contact with Fe3+ (aq) and OH- (aq) ions. This half-cell acts as the anode and undergoes oxidation.
- The half-cell on the right contains Fe(OH)3 (s) in contact with Fe2+ (aq) ions. This half-cell acts as the cathode and undergoes reduction.
- The overall cell reaction is Fe(OH)3 (s) + 3e- → Fe(s) + 3OH- (aq)
- The standard reduction potentials of Fe3+ (aq) | Fe and Fe(OH)3 (s) | Fe are given as -0.036 V and -0.786 V, respectively.
EMF of the Cell
The EMF (electromotive force) of the cell can be calculated using the Nernst equation:
Ecell = E°cell - (RT/nF) ln Q
where, E°cell is the standard cell potential, R is the gas constant, T is the temperature in Kelvin, n is the number of electrons transferred in the cell reaction, F is the Faraday constant, and Q is the reaction quotient.
For the given cell representation, the standard cell potential (E°cell) can be calculated as follows:
E°cell = E°cathode - E°anode
= (-0.036) - (-0.786)
= 0.750 V
Since the reaction involves the transfer of three electrons, n = 3.
The reaction quotient (Q) for the cell reaction can be expressed as follows:
Q = [Fe2+] / [OH-]^3
At equilibrium, Q = Ksp, where Ksp is the solubility product of Fe(OH)3.
Therefore, the EMF of the cell can be expressed as:
Ecell = 0.750 - (0.0257/3) log Ksp
Solving for Ksp, we get:
Ksp = 2.8 x 10^-39
Therefore, the appropriate cell representation and its EMF for the determination of solubility product of Fe(OH)3 is option B: Fe | Fe3+ (aq), OH- (aq) || Fe(OH)3 (s) | | Fe with an EMF of -0.750 V.
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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. For the determination of solubility product of Fe(OH)3 (Ksp) the appropriate cell representation and its EMF respectively are a)Fe | Fe(OH)3(s) || OH-. (aq) Fe3+(aq) | Fe -0.750 Vb)Fe | Fe3+(aq), OH-(aq) || Fe(OH)3(s) | | Fe -0.750 Vc)Fe|Fe(OH3)(s) || OH-(aq), Fe3+(aq) | Fe +0.750Vd)Fe|Fe3+(aq), OH-(aq)| Fe(OH3)(s)|Fe -0.822VCorrect answer is option 'B'. 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. For the determination of solubility product of Fe(OH)3 (Ksp) the appropriate cell representation and its EMF respectively are a)Fe | Fe(OH)3(s) || OH-. (aq) Fe3+(aq) | Fe -0.750 Vb)Fe | Fe3+(aq), OH-(aq) || Fe(OH)3(s) | | Fe -0.750 Vc)Fe|Fe(OH3)(s) || OH-(aq), Fe3+(aq) | Fe +0.750Vd)Fe|Fe3+(aq), OH-(aq)| Fe(OH3)(s)|Fe -0.822VCorrect answer is option 'B'. 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. For the determination of solubility product of Fe(OH)3 (Ksp) the appropriate cell representation and its EMF respectively are a)Fe | Fe(OH)3(s) || OH-. (aq) Fe3+(aq) | Fe -0.750 Vb)Fe | Fe3+(aq), OH-(aq) || Fe(OH)3(s) | | Fe -0.750 Vc)Fe|Fe(OH3)(s) || OH-(aq), Fe3+(aq) | Fe +0.750Vd)Fe|Fe3+(aq), OH-(aq)| Fe(OH3)(s)|Fe -0.822VCorrect answer is option 'B'. 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. For the determination of solubility product of Fe(OH)3 (Ksp) the appropriate cell representation and its EMF respectively are a)Fe | Fe(OH)3(s) || OH-. (aq) Fe3+(aq) | Fe -0.750 Vb)Fe | Fe3+(aq), OH-(aq) || Fe(OH)3(s) | | Fe -0.750 Vc)Fe|Fe(OH3)(s) || OH-(aq), Fe3+(aq) | Fe +0.750Vd)Fe|Fe3+(aq), OH-(aq)| Fe(OH3)(s)|Fe -0.822VCorrect answer is option 'B'. 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. For the determination of solubility product of Fe(OH)3 (Ksp) the appropriate cell representation and its EMF respectively are a)Fe | Fe(OH)3(s) || OH-. (aq) Fe3+(aq) | Fe -0.750 Vb)Fe | Fe3+(aq), OH-(aq) || Fe(OH)3(s) | | Fe -0.750 Vc)Fe|Fe(OH3)(s) || OH-(aq), Fe3+(aq) | Fe +0.750Vd)Fe|Fe3+(aq), OH-(aq)| Fe(OH3)(s)|Fe -0.822VCorrect answer is option 'B'. 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. For the determination of solubility product of Fe(OH)3 (Ksp) the appropriate cell representation and its EMF respectively are a)Fe | Fe(OH)3(s) || OH-. (aq) Fe3+(aq) | Fe -0.750 Vb)Fe | Fe3+(aq), OH-(aq) || Fe(OH)3(s) | | Fe -0.750 Vc)Fe|Fe(OH3)(s) || OH-(aq), Fe3+(aq) | Fe +0.750Vd)Fe|Fe3+(aq), OH-(aq)| Fe(OH3)(s)|Fe -0.822VCorrect answer is option 'B'. 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. For the determination of solubility product of Fe(OH)3 (Ksp) the appropriate cell representation and its EMF respectively are a)Fe | Fe(OH)3(s) || OH-. (aq) Fe3+(aq) | Fe -0.750 Vb)Fe | Fe3+(aq), OH-(aq) || Fe(OH)3(s) | | Fe -0.750 Vc)Fe|Fe(OH3)(s) || OH-(aq), Fe3+(aq) | Fe +0.750Vd)Fe|Fe3+(aq), OH-(aq)| Fe(OH3)(s)|Fe -0.822VCorrect answer is option 'B'. 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. For the determination of solubility product of Fe(OH)3 (Ksp) the appropriate cell representation and its EMF respectively are a)Fe | Fe(OH)3(s) || OH-. (aq) Fe3+(aq) | Fe -0.750 Vb)Fe | Fe3+(aq), OH-(aq) || Fe(OH)3(s) | | Fe -0.750 Vc)Fe|Fe(OH3)(s) || OH-(aq), Fe3+(aq) | Fe +0.750Vd)Fe|Fe3+(aq), OH-(aq)| Fe(OH3)(s)|Fe -0.822VCorrect answer is option 'B'. Can you explain this answer? in English & in Hindi are available as part of our courses for Class 12.
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Here you can find the meaning of Passage IIThe standard half-cell reduction potentialof Fe3+(aq) | Fe is -0.036Vand that ofOH- | Fe(OH3)(s) | Feis -0.786V.Q. For the determination of solubility product of Fe(OH)3 (Ksp) the appropriate cell representation and its EMF respectively are a)Fe | Fe(OH)3(s) || OH-. (aq) Fe3+(aq) | Fe -0.750 Vb)Fe | Fe3+(aq), OH-(aq) || Fe(OH)3(s) | | Fe -0.750 Vc)Fe|Fe(OH3)(s) || OH-(aq), Fe3+(aq) | Fe +0.750Vd)Fe|Fe3+(aq), OH-(aq)| Fe(OH3)(s)|Fe -0.822VCorrect answer is option 'B'. Can you explain this answer? defined & explained in the simplest way possible. Besides giving the explanation of
Passage IIThe standard half-cell reduction potentialof Fe3+(aq) | Fe is -0.036Vand that ofOH- | Fe(OH3)(s) | Feis -0.786V.Q. For the determination of solubility product of Fe(OH)3 (Ksp) the appropriate cell representation and its EMF respectively are a)Fe | Fe(OH)3(s) || OH-. (aq) Fe3+(aq) | Fe -0.750 Vb)Fe | Fe3+(aq), OH-(aq) || Fe(OH)3(s) | | Fe -0.750 Vc)Fe|Fe(OH3)(s) || OH-(aq), Fe3+(aq) | Fe +0.750Vd)Fe|Fe3+(aq), OH-(aq)| Fe(OH3)(s)|Fe -0.822VCorrect answer is option 'B'. 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. For the determination of solubility product of Fe(OH)3 (Ksp) the appropriate cell representation and its EMF respectively are a)Fe | Fe(OH)3(s) || OH-. (aq) Fe3+(aq) | Fe -0.750 Vb)Fe | Fe3+(aq), OH-(aq) || Fe(OH)3(s) | | Fe -0.750 Vc)Fe|Fe(OH3)(s) || OH-(aq), Fe3+(aq) | Fe +0.750Vd)Fe|Fe3+(aq), OH-(aq)| Fe(OH3)(s)|Fe -0.822VCorrect answer is option 'B'. 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. For the determination of solubility product of Fe(OH)3 (Ksp) the appropriate cell representation and its EMF respectively are a)Fe | Fe(OH)3(s) || OH-. (aq) Fe3+(aq) | Fe -0.750 Vb)Fe | Fe3+(aq), OH-(aq) || Fe(OH)3(s) | | Fe -0.750 Vc)Fe|Fe(OH3)(s) || OH-(aq), Fe3+(aq) | Fe +0.750Vd)Fe|Fe3+(aq), OH-(aq)| Fe(OH3)(s)|Fe -0.822VCorrect answer is option 'B'. Can you explain this answer? theory, EduRev gives you an
ample number of questions to practice Passage IIThe standard half-cell reduction potentialof Fe3+(aq) | Fe is -0.036Vand that ofOH- | Fe(OH3)(s) | Feis -0.786V.Q. For the determination of solubility product of Fe(OH)3 (Ksp) the appropriate cell representation and its EMF respectively are a)Fe | Fe(OH)3(s) || OH-. (aq) Fe3+(aq) | Fe -0.750 Vb)Fe | Fe3+(aq), OH-(aq) || Fe(OH)3(s) | | Fe -0.750 Vc)Fe|Fe(OH3)(s) || OH-(aq), Fe3+(aq) | Fe +0.750Vd)Fe|Fe3+(aq), OH-(aq)| Fe(OH3)(s)|Fe -0.822VCorrect answer is option 'B'. Can you explain this answer? tests, examples and also practice Class 12 tests.
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