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Passage for Q. Nos. (4-5)
The electrochemical cell shown below is a concentration cell. M | M2+ (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 298K is 0.059 V.
Q.
The solubility product (Ksp ; mol3dm-9) of MX2 at 298 K based on the information available the given concentration cell is (take 2.303x R x 298 / F= 0.059 V)
- 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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Passage for Q. Nos. (4-5)The electrochemical cell shown below is a con...
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Passage for Q. Nos. (4-5)The electrochemical cell shown below is a con...
Answer:
To find the solubility product (Ksp) of MX2 at 298 K, we need to understand the concept of concentration cells and their relation to the emf of the cell.
Concept of Concentration Cells:
A concentration cell is an electrochemical cell that consists of two half-cells with the same electrode but different concentrations of the same ion. The emf of the cell is dependent on the difference in concentration of the ion at the two electrodes.
In the given concentration cell, M and M2 represent the same electrode (solid M) but with different concentrations of M2 ions. The left half-cell consists of a saturated solution of the sparingly soluble salt MX2, while the right half-cell consists of a solution with a concentration of 0.001 mol dm-3 of M2 ions.
Determination of Emf:
The emf of the cell is given as 0.059 V at 298 K. This value is related to the Nernst equation, which is given as:
Ecell = E°cell - (0.0592/n)logQ
Here, Ecell is the cell potential, E°cell is the standard cell potential, n is the number of moles of electrons transferred, and Q is the reaction quotient.
In the given concentration cell, the reaction quotient (Q) can be expressed as the ratio of the concentration of M2 ions at the left electrode (saturated solution) to the concentration of M2 ions at the right electrode (0.001 mol dm-3):
Q = [M2]left / [M2]right
Since the emf of the cell depends on the difference in concentration of M2 ions at the two electrodes, we can assume that the concentration of M2 ions at the left electrode is greater than the concentration at the right electrode. Therefore, Q is greater than 1.
Relation between Emf and Ksp:
The emf of the cell is related to Ksp through the Nernst equation. At equilibrium, the reaction quotient Q is equal to the equilibrium constant (Ksp):
Ecell = E°cell - (0.0592/n)logKsp
Comparing this equation with the given equation (0.059 V = 0.0592 logKsp), we can see that E°cell is equal to 0.059 V and n is equal to 2 (since MX2 is a sparingly soluble salt).
Calculation of Ksp:
Using the given equation, we can calculate Ksp as follows:
0.059 V = 0.0592/n logKsp
0.059 V = 0.0592/2 logKsp
0.059 V = 0.0296 logKsp
Taking the antilogarithm on both sides:
10^(0.059 V) = 10^(0.0296 logKsp)
1.149 = (10^(0.0296))^logKsp
1.149 = (1.0326)^logKsp
Taking the logarithm on both sides:
log1.149 = log[(1.0326)^logKsp]
0.0593 = logKsp * log(1.0326)
Solving for logKsp:
logKsp = 0.0593 / log
To find the solubility product (Ksp) of MX2 at 298 K, we need to understand the concept of concentration cells and their relation to the emf of the cell.
Concept of Concentration Cells:
A concentration cell is an electrochemical cell that consists of two half-cells with the same electrode but different concentrations of the same ion. The emf of the cell is dependent on the difference in concentration of the ion at the two electrodes.
In the given concentration cell, M and M2 represent the same electrode (solid M) but with different concentrations of M2 ions. The left half-cell consists of a saturated solution of the sparingly soluble salt MX2, while the right half-cell consists of a solution with a concentration of 0.001 mol dm-3 of M2 ions.
Determination of Emf:
The emf of the cell is given as 0.059 V at 298 K. This value is related to the Nernst equation, which is given as:
Ecell = E°cell - (0.0592/n)logQ
Here, Ecell is the cell potential, E°cell is the standard cell potential, n is the number of moles of electrons transferred, and Q is the reaction quotient.
In the given concentration cell, the reaction quotient (Q) can be expressed as the ratio of the concentration of M2 ions at the left electrode (saturated solution) to the concentration of M2 ions at the right electrode (0.001 mol dm-3):
Q = [M2]left / [M2]right
Since the emf of the cell depends on the difference in concentration of M2 ions at the two electrodes, we can assume that the concentration of M2 ions at the left electrode is greater than the concentration at the right electrode. Therefore, Q is greater than 1.
Relation between Emf and Ksp:
The emf of the cell is related to Ksp through the Nernst equation. At equilibrium, the reaction quotient Q is equal to the equilibrium constant (Ksp):
Ecell = E°cell - (0.0592/n)logKsp
Comparing this equation with the given equation (0.059 V = 0.0592 logKsp), we can see that E°cell is equal to 0.059 V and n is equal to 2 (since MX2 is a sparingly soluble salt).
Calculation of Ksp:
Using the given equation, we can calculate Ksp as follows:
0.059 V = 0.0592/n logKsp
0.059 V = 0.0592/2 logKsp
0.059 V = 0.0296 logKsp
Taking the antilogarithm on both sides:
10^(0.059 V) = 10^(0.0296 logKsp)
1.149 = (10^(0.0296))^logKsp
1.149 = (1.0326)^logKsp
Taking the logarithm on both sides:
log1.149 = log[(1.0326)^logKsp]
0.0593 = logKsp * log(1.0326)
Solving for logKsp:
logKsp = 0.0593 / log
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Passage for Q. Nos. (4-5)The electrochemical cell shown below is a concentration cell. M |M2+ (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 298K is 0.059 V.Q.The solubility product (Ksp ; mol3dm-9) of MX2 at 298 K based on the information available the given concentration cell is (take 2.303x R x 298 / F= 0.059 V)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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Passage for Q. Nos. (4-5)The electrochemical cell shown below is a concentration cell. M |M2+ (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 298K is 0.059 V.Q.The solubility product (Ksp ; mol3dm-9) of MX2 at 298 K based on the information available the given concentration cell is (take 2.303x R x 298 / F= 0.059 V)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 JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about Passage for Q. Nos. (4-5)The electrochemical cell shown below is a concentration cell. M |M2+ (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 298K is 0.059 V.Q.The solubility product (Ksp ; mol3dm-9) of MX2 at 298 K based on the information available the given concentration cell is (take 2.303x R x 298 / F= 0.059 V)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 JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Passage for Q. Nos. (4-5)The electrochemical cell shown below is a concentration cell. M |M2+ (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 298K is 0.059 V.Q.The solubility product (Ksp ; mol3dm-9) of MX2 at 298 K based on the information available the given concentration cell is (take 2.303x R x 298 / F= 0.059 V)a)1 x 10-15b)4x 10-15c)1 x 10-12d)4x 10-12Correct answer is option 'B'. Can you explain this answer?.
Passage for Q. Nos. (4-5)The electrochemical cell shown below is a concentration cell. M |M2+ (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 298K is 0.059 V.Q.The solubility product (Ksp ; mol3dm-9) of MX2 at 298 K based on the information available the given concentration cell is (take 2.303x R x 298 / F= 0.059 V)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 JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about Passage for Q. Nos. (4-5)The electrochemical cell shown below is a concentration cell. M |M2+ (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 298K is 0.059 V.Q.The solubility product (Ksp ; mol3dm-9) of MX2 at 298 K based on the information available the given concentration cell is (take 2.303x R x 298 / F= 0.059 V)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 JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Passage for Q. Nos. (4-5)The electrochemical cell shown below is a concentration cell. M |M2+ (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 298K is 0.059 V.Q.The solubility product (Ksp ; mol3dm-9) of MX2 at 298 K based on the information available the given concentration cell is (take 2.303x R x 298 / F= 0.059 V)a)1 x 10-15b)4x 10-15c)1 x 10-12d)4x 10-12Correct answer is option 'B'. Can you explain this answer?.
Solutions for Passage for Q. Nos. (4-5)The electrochemical cell shown below is a concentration cell. M |M2+ (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 298K is 0.059 V.Q.The solubility product (Ksp ; mol3dm-9) of MX2 at 298 K based on the information available the given concentration cell is (take 2.303x R x 298 / F= 0.059 V)a)1 x 10-15b)4x 10-15c)1 x 10-12d)4x 10-12Correct answer is option 'B'. Can you explain this answer? in English & in Hindi are available as part of our courses for JEE.
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Here you can find the meaning of Passage for Q. Nos. (4-5)The electrochemical cell shown below is a concentration cell. M |M2+ (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 298K is 0.059 V.Q.The solubility product (Ksp ; mol3dm-9) of MX2 at 298 K based on the information available the given concentration cell is (take 2.303x R x 298 / F= 0.059 V)a)1 x 10-15b)4x 10-15c)1 x 10-12d)4x 10-12Correct answer is option 'B'. Can you explain this answer? defined & explained in the simplest way possible. Besides giving the explanation of
Passage for Q. Nos. (4-5)The electrochemical cell shown below is a concentration cell. M |M2+ (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 298K is 0.059 V.Q.The solubility product (Ksp ; mol3dm-9) of MX2 at 298 K based on the information available the given concentration cell is (take 2.303x R x 298 / F= 0.059 V)a)1 x 10-15b)4x 10-15c)1 x 10-12d)4x 10-12Correct answer is option 'B'. Can you explain this answer?, a detailed solution for Passage for Q. Nos. (4-5)The electrochemical cell shown below is a concentration cell. M |M2+ (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 298K is 0.059 V.Q.The solubility product (Ksp ; mol3dm-9) of MX2 at 298 K based on the information available the given concentration cell is (take 2.303x R x 298 / F= 0.059 V)a)1 x 10-15b)4x 10-15c)1 x 10-12d)4x 10-12Correct answer is option 'B'. Can you explain this answer? has been provided alongside types of Passage for Q. Nos. (4-5)The electrochemical cell shown below is a concentration cell. M |M2+ (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 298K is 0.059 V.Q.The solubility product (Ksp ; mol3dm-9) of MX2 at 298 K based on the information available the given concentration cell is (take 2.303x R x 298 / F= 0.059 V)a)1 x 10-15b)4x 10-15c)1 x 10-12d)4x 10-12Correct answer is option 'B'. Can you explain this answer? theory, EduRev gives you an
ample number of questions to practice Passage for Q. Nos. (4-5)The electrochemical cell shown below is a concentration cell. M |M2+ (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 298K is 0.059 V.Q.The solubility product (Ksp ; mol3dm-9) of MX2 at 298 K based on the information available the given concentration cell is (take 2.303x R x 298 / F= 0.059 V)a)1 x 10-15b)4x 10-15c)1 x 10-12d)4x 10-12Correct answer is option 'B'. Can you explain this answer? tests, examples and also practice JEE tests.
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