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Direction : This section contains 3 paragraphs, each describing theory, experiments, data, etc. Six questions related to the paragraph have been given. Each question has only one correct answer among the four given options (a), (b), (c) and (d).
Passage I
The electrochemical cell shown below is a concentration cell :
M/Mz+ (saturated solution of a sparingly soluble salt(MX2) || M2+(0.001 mol dm-3)|M
The EMF of the cell depends on the difference in concentration of M2+ ions at the two electrodes .The EMF of the cell at 298 K is 0.059 V.
Q.
Ksp mol3 dm-9 (the solubility product) of MX2 at 298 K is
(2.303 RT/F = 0.059 V at 298 K)
- a)1 x 10-15
- b)4 x 10-15
- c)1 x 10-12
- d)4 x 10-12
Correct answer is option 'B'. Can you explain this answer?
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Answer:
Explanation:
Given, Emf of the cell (E) = 0.059 V at 298 K
We know that for the given cell reaction,
M + 2X- <-> MX2 (1)
M2 + 2X- <-> M(MX2)2 (2)
The equation for the Emf of the cell is given by the Nernst equation as follows.
E = (RT/nF) ln Q
where R is the gas constant (8.314 J/K/mol), T is the temperature in Kelvin, F is the Faraday constant (96500 C/mol), n is the number of electrons transferred, and Q is the reaction quotient.
The reaction quotient for the given cell reaction can be written as
Q = [M2]/[M]
At equilibrium, the concentration of M2+ ions at the anode is 0.001 mol dm-3, and the concentration of M2+ ions at the cathode is equal to the solubility of MX2. The concentration of M+ ions at both electrodes can be assumed to be constant and equal to the concentration of M+ ions in the saturated solution of MX2.
So, Q = [M2]/[M] = (s/[M])^2
where s is the solubility of MX2.
Substituting the values in the Nernst equation, we get
0.059 = (2.303RT/2F) log (s/[M])^2
0.059 = (0.059/2) log (s/[M])^2 (at 298 K, 2.303RT/F = 0.059 V)
log (s/[M])^2 = 1
log s - 2 log [M] = 0.5
log s = 2 log [M] + 0.5
s = 10^(2 log [M] + 0.5)
s = 10^0.5 [M]^2
s = 3.98 x 10^-15 mol3 dm-9
Therefore, Ksp of MX2 at 298 K is given by
Ksp = [M][X]^2 = [M](s/[M]^2) = s/M = 3.98 x 10^-15/1 x 10^-3 = 3.98 x 10^-12 mol3 dm-9
Hence, the correct answer is option (b) 4 x 10^-15.
Explanation:
Given, Emf of the cell (E) = 0.059 V at 298 K
We know that for the given cell reaction,
M + 2X- <-> MX2 (1)
M2 + 2X- <-> M(MX2)2 (2)
The equation for the Emf of the cell is given by the Nernst equation as follows.
E = (RT/nF) ln Q
where R is the gas constant (8.314 J/K/mol), T is the temperature in Kelvin, F is the Faraday constant (96500 C/mol), n is the number of electrons transferred, and Q is the reaction quotient.
The reaction quotient for the given cell reaction can be written as
Q = [M2]/[M]
At equilibrium, the concentration of M2+ ions at the anode is 0.001 mol dm-3, and the concentration of M2+ ions at the cathode is equal to the solubility of MX2. The concentration of M+ ions at both electrodes can be assumed to be constant and equal to the concentration of M+ ions in the saturated solution of MX2.
So, Q = [M2]/[M] = (s/[M])^2
where s is the solubility of MX2.
Substituting the values in the Nernst equation, we get
0.059 = (2.303RT/2F) log (s/[M])^2
0.059 = (0.059/2) log (s/[M])^2 (at 298 K, 2.303RT/F = 0.059 V)
log (s/[M])^2 = 1
log s - 2 log [M] = 0.5
log s = 2 log [M] + 0.5
s = 10^(2 log [M] + 0.5)
s = 10^0.5 [M]^2
s = 3.98 x 10^-15 mol3 dm-9
Therefore, Ksp of MX2 at 298 K is given by
Ksp = [M][X]^2 = [M](s/[M]^2) = s/M = 3.98 x 10^-15/1 x 10^-3 = 3.98 x 10^-12 mol3 dm-9
Hence, the correct answer is option (b) 4 x 10^-15.
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Comprehension TypeDirection : This section contains 3 paragraphs, each describing theory, experiments, data, etc. Six questions related to the paragraph have been given. Each question has only one correct answer among the four given options (a), (b), (c) and (d).Passage IThe electrochemical cell shown below is a concentration cell :M/Mz+(saturated solution of a sparingly soluble salt(MX2) || M2+(0.001 mol dm-3)|MThe EMF of the cell depends on the difference in concentration of M2+ ions at the two electrodes .The EMF of the cell at 298 K is 0.059 V.Q. Ksp mol3 dm-9 (the solubility product) of MX2 at 298 K is (2.303 RT/F = 0.059 V at 298 K) a)1 x 10-15b)4x 10-15c)1 x 10-12d)4x 10-12Correct answer is option 'B'. Can you explain this answer?
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Comprehension TypeDirection : This section contains 3 paragraphs, each describing theory, experiments, data, etc. Six questions related to the paragraph have been given. Each question has only one correct answer among the four given options (a), (b), (c) and (d).Passage IThe electrochemical cell shown below is a concentration cell :M/Mz+(saturated solution of a sparingly soluble salt(MX2) || M2+(0.001 mol dm-3)|MThe EMF of the cell depends on the difference in concentration of M2+ ions at the two electrodes .The EMF of the cell at 298 K is 0.059 V.Q. Ksp mol3 dm-9 (the solubility product) of MX2 at 298 K is (2.303 RT/F = 0.059 V at 298 K) a)1 x 10-15b)4x 10-15c)1 x 10-12d)4x 10-12Correct 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 Comprehension TypeDirection : This section contains 3 paragraphs, each describing theory, experiments, data, etc. Six questions related to the paragraph have been given. Each question has only one correct answer among the four given options (a), (b), (c) and (d).Passage IThe electrochemical cell shown below is a concentration cell :M/Mz+(saturated solution of a sparingly soluble salt(MX2) || M2+(0.001 mol dm-3)|MThe EMF of the cell depends on the difference in concentration of M2+ ions at the two electrodes .The EMF of the cell at 298 K is 0.059 V.Q. Ksp mol3 dm-9 (the solubility product) of MX2 at 298 K is (2.303 RT/F = 0.059 V at 298 K) a)1 x 10-15b)4x 10-15c)1 x 10-12d)4x 10-12Correct 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 Comprehension TypeDirection : This section contains 3 paragraphs, each describing theory, experiments, data, etc. Six questions related to the paragraph have been given. Each question has only one correct answer among the four given options (a), (b), (c) and (d).Passage IThe electrochemical cell shown below is a concentration cell :M/Mz+(saturated solution of a sparingly soluble salt(MX2) || M2+(0.001 mol dm-3)|MThe EMF of the cell depends on the difference in concentration of M2+ ions at the two electrodes .The EMF of the cell at 298 K is 0.059 V.Q. Ksp mol3 dm-9 (the solubility product) of MX2 at 298 K is (2.303 RT/F = 0.059 V at 298 K) a)1 x 10-15b)4x 10-15c)1 x 10-12d)4x 10-12Correct answer is option 'B'. Can you explain this answer?.
Comprehension TypeDirection : This section contains 3 paragraphs, each describing theory, experiments, data, etc. Six questions related to the paragraph have been given. Each question has only one correct answer among the four given options (a), (b), (c) and (d).Passage IThe electrochemical cell shown below is a concentration cell :M/Mz+(saturated solution of a sparingly soluble salt(MX2) || M2+(0.001 mol dm-3)|MThe EMF of the cell depends on the difference in concentration of M2+ ions at the two electrodes .The EMF of the cell at 298 K is 0.059 V.Q. Ksp mol3 dm-9 (the solubility product) of MX2 at 298 K is (2.303 RT/F = 0.059 V at 298 K) a)1 x 10-15b)4x 10-15c)1 x 10-12d)4x 10-12Correct 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 Comprehension TypeDirection : This section contains 3 paragraphs, each describing theory, experiments, data, etc. Six questions related to the paragraph have been given. Each question has only one correct answer among the four given options (a), (b), (c) and (d).Passage IThe electrochemical cell shown below is a concentration cell :M/Mz+(saturated solution of a sparingly soluble salt(MX2) || M2+(0.001 mol dm-3)|MThe EMF of the cell depends on the difference in concentration of M2+ ions at the two electrodes .The EMF of the cell at 298 K is 0.059 V.Q. Ksp mol3 dm-9 (the solubility product) of MX2 at 298 K is (2.303 RT/F = 0.059 V at 298 K) a)1 x 10-15b)4x 10-15c)1 x 10-12d)4x 10-12Correct 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 Comprehension TypeDirection : This section contains 3 paragraphs, each describing theory, experiments, data, etc. Six questions related to the paragraph have been given. Each question has only one correct answer among the four given options (a), (b), (c) and (d).Passage IThe electrochemical cell shown below is a concentration cell :M/Mz+(saturated solution of a sparingly soluble salt(MX2) || M2+(0.001 mol dm-3)|MThe EMF of the cell depends on the difference in concentration of M2+ ions at the two electrodes .The EMF of the cell at 298 K is 0.059 V.Q. Ksp mol3 dm-9 (the solubility product) of MX2 at 298 K is (2.303 RT/F = 0.059 V at 298 K) a)1 x 10-15b)4x 10-15c)1 x 10-12d)4x 10-12Correct answer is option 'B'. Can you explain this answer?.
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