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Paragraph for Questions 31 and 32
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 concentrations of M2+ ions at the two electrodes. The emf of the cell at 298 K is 0.059 V.
Q. The solubility product (Ksp; mol3 dm–9) of MX2 at 298 K based on the information available for the given
concentration cell is (take 2.303 x R x 298/F = 0.059 V)
concentration cell is (take 2.303 x 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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Paragraph for Questions 31 and 32The electrochemical cell shown below ...
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Paragraph for Questions 31 and 32The electrochemical cell shown below ...
The electrochemical cell shown above is a concentration cell, which means it operates based on the concentration difference between two solutions. In this case, the cell consists of two half-cells. The left half-cell contains a metal M in contact with a saturated solution of a sparingly soluble salt, MX2. The right half-cell contains the same metal M in contact with a solution of M2 with a concentration of 0.001 mol dm^3.
In this concentration cell, the concentration gradient of the metal cation M2+ is the driving force for the cell's operation. The metal M in the left half-cell is in contact with a saturated solution of MX2, which means that the concentration of M2+ ions in the left half-cell is much higher than in the right half-cell. This concentration difference creates a potential difference between the two half-cells, leading to the flow of electrons from the left half-cell to the right half-cell.
In question 31, we are asked to identify the positive electrode in this concentration cell. The positive electrode is where the oxidation half-reaction occurs, which means that the metal M in the left half-cell undergoes oxidation and loses electrons. Therefore, the positive electrode in this concentration cell is the metal M in the left half-cell.
In question 32, we are asked to explain the change in potential difference when the concentration of M2+ in the right half-cell is increased. Increasing the concentration of M2+ in the right half-cell would decrease the concentration gradient between the two half-cells. As a result, the driving force for the flow of electrons would decrease, leading to a decrease in the potential difference between the two half-cells.
In this concentration cell, the concentration gradient of the metal cation M2+ is the driving force for the cell's operation. The metal M in the left half-cell is in contact with a saturated solution of MX2, which means that the concentration of M2+ ions in the left half-cell is much higher than in the right half-cell. This concentration difference creates a potential difference between the two half-cells, leading to the flow of electrons from the left half-cell to the right half-cell.
In question 31, we are asked to identify the positive electrode in this concentration cell. The positive electrode is where the oxidation half-reaction occurs, which means that the metal M in the left half-cell undergoes oxidation and loses electrons. Therefore, the positive electrode in this concentration cell is the metal M in the left half-cell.
In question 32, we are asked to explain the change in potential difference when the concentration of M2+ in the right half-cell is increased. Increasing the concentration of M2+ in the right half-cell would decrease the concentration gradient between the two half-cells. As a result, the driving force for the flow of electrons would decrease, leading to a decrease in the potential difference between the two half-cells.
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Paragraph for Questions 31 and 32The electrochemical cell shown below is a concentration cell.M|M2+ (saturated solution of a sparingly soluble salt, MX2) || M2+ (0.001 mol dm–3) | MThe emf of the cell depends on the difference in concentrations of M2+ ions at the two electrodes. The emf of the cell at 298 K is 0.059 V.Q.The solubility product (Ksp; mol3 dm–9) of MX2 at 298 K based on the information available for the givenconcentration cell is (take 2.303 xR x298/F = 0.059 V)a)1 x10–15b)4 x10–15c)1 x10–12d)4 x10–12Correct answer is option 'B'. Can you explain this answer?
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Paragraph for Questions 31 and 32The electrochemical cell shown below is a concentration cell.M|M2+ (saturated solution of a sparingly soluble salt, MX2) || M2+ (0.001 mol dm–3) | MThe emf of the cell depends on the difference in concentrations of M2+ ions at the two electrodes. The emf of the cell at 298 K is 0.059 V.Q.The solubility product (Ksp; mol3 dm–9) of MX2 at 298 K based on the information available for the givenconcentration cell is (take 2.303 xR x298/F = 0.059 V)a)1 x10–15b)4 x10–15c)1 x10–12d)4 x10–12Correct 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 Paragraph for Questions 31 and 32The electrochemical cell shown below is a concentration cell.M|M2+ (saturated solution of a sparingly soluble salt, MX2) || M2+ (0.001 mol dm–3) | MThe emf of the cell depends on the difference in concentrations of M2+ ions at the two electrodes. The emf of the cell at 298 K is 0.059 V.Q.The solubility product (Ksp; mol3 dm–9) of MX2 at 298 K based on the information available for the givenconcentration cell is (take 2.303 xR x298/F = 0.059 V)a)1 x10–15b)4 x10–15c)1 x10–12d)4 x10–12Correct 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 Paragraph for Questions 31 and 32The electrochemical cell shown below is a concentration cell.M|M2+ (saturated solution of a sparingly soluble salt, MX2) || M2+ (0.001 mol dm–3) | MThe emf of the cell depends on the difference in concentrations of M2+ ions at the two electrodes. The emf of the cell at 298 K is 0.059 V.Q.The solubility product (Ksp; mol3 dm–9) of MX2 at 298 K based on the information available for the givenconcentration cell is (take 2.303 xR x298/F = 0.059 V)a)1 x10–15b)4 x10–15c)1 x10–12d)4 x10–12Correct answer is option 'B'. Can you explain this answer?.
Paragraph for Questions 31 and 32The electrochemical cell shown below is a concentration cell.M|M2+ (saturated solution of a sparingly soluble salt, MX2) || M2+ (0.001 mol dm–3) | MThe emf of the cell depends on the difference in concentrations of M2+ ions at the two electrodes. The emf of the cell at 298 K is 0.059 V.Q.The solubility product (Ksp; mol3 dm–9) of MX2 at 298 K based on the information available for the givenconcentration cell is (take 2.303 xR x298/F = 0.059 V)a)1 x10–15b)4 x10–15c)1 x10–12d)4 x10–12Correct 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 Paragraph for Questions 31 and 32The electrochemical cell shown below is a concentration cell.M|M2+ (saturated solution of a sparingly soluble salt, MX2) || M2+ (0.001 mol dm–3) | MThe emf of the cell depends on the difference in concentrations of M2+ ions at the two electrodes. The emf of the cell at 298 K is 0.059 V.Q.The solubility product (Ksp; mol3 dm–9) of MX2 at 298 K based on the information available for the givenconcentration cell is (take 2.303 xR x298/F = 0.059 V)a)1 x10–15b)4 x10–15c)1 x10–12d)4 x10–12Correct 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 Paragraph for Questions 31 and 32The electrochemical cell shown below is a concentration cell.M|M2+ (saturated solution of a sparingly soluble salt, MX2) || M2+ (0.001 mol dm–3) | MThe emf of the cell depends on the difference in concentrations of M2+ ions at the two electrodes. The emf of the cell at 298 K is 0.059 V.Q.The solubility product (Ksp; mol3 dm–9) of MX2 at 298 K based on the information available for the givenconcentration cell is (take 2.303 xR x298/F = 0.059 V)a)1 x10–15b)4 x10–15c)1 x10–12d)4 x10–12Correct answer is option 'B'. Can you explain this answer?.
Solutions for Paragraph for Questions 31 and 32The electrochemical cell shown below is a concentration cell.M|M2+ (saturated solution of a sparingly soluble salt, MX2) || M2+ (0.001 mol dm–3) | MThe emf of the cell depends on the difference in concentrations of M2+ ions at the two electrodes. The emf of the cell at 298 K is 0.059 V.Q.The solubility product (Ksp; mol3 dm–9) of MX2 at 298 K based on the information available for the givenconcentration cell is (take 2.303 xR x298/F = 0.059 V)a)1 x10–15b)4 x10–15c)1 x10–12d)4 x10–12Correct 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 Paragraph for Questions 31 and 32The electrochemical cell shown below is a concentration cell.M|M2+ (saturated solution of a sparingly soluble salt, MX2) || M2+ (0.001 mol dm–3) | MThe emf of the cell depends on the difference in concentrations of M2+ ions at the two electrodes. The emf of the cell at 298 K is 0.059 V.Q.The solubility product (Ksp; mol3 dm–9) of MX2 at 298 K based on the information available for the givenconcentration cell is (take 2.303 xR x298/F = 0.059 V)a)1 x10–15b)4 x10–15c)1 x10–12d)4 x10–12Correct answer is option 'B'. Can you explain this answer? defined & explained in the simplest way possible. Besides giving the explanation of
Paragraph for Questions 31 and 32The electrochemical cell shown below is a concentration cell.M|M2+ (saturated solution of a sparingly soluble salt, MX2) || M2+ (0.001 mol dm–3) | MThe emf of the cell depends on the difference in concentrations of M2+ ions at the two electrodes. The emf of the cell at 298 K is 0.059 V.Q.The solubility product (Ksp; mol3 dm–9) of MX2 at 298 K based on the information available for the givenconcentration cell is (take 2.303 xR x298/F = 0.059 V)a)1 x10–15b)4 x10–15c)1 x10–12d)4 x10–12Correct answer is option 'B'. Can you explain this answer?, a detailed solution for Paragraph for Questions 31 and 32The electrochemical cell shown below is a concentration cell.M|M2+ (saturated solution of a sparingly soluble salt, MX2) || M2+ (0.001 mol dm–3) | MThe emf of the cell depends on the difference in concentrations of M2+ ions at the two electrodes. The emf of the cell at 298 K is 0.059 V.Q.The solubility product (Ksp; mol3 dm–9) of MX2 at 298 K based on the information available for the givenconcentration cell is (take 2.303 xR x298/F = 0.059 V)a)1 x10–15b)4 x10–15c)1 x10–12d)4 x10–12Correct answer is option 'B'. Can you explain this answer? has been provided alongside types of Paragraph for Questions 31 and 32The electrochemical cell shown below is a concentration cell.M|M2+ (saturated solution of a sparingly soluble salt, MX2) || M2+ (0.001 mol dm–3) | MThe emf of the cell depends on the difference in concentrations of M2+ ions at the two electrodes. The emf of the cell at 298 K is 0.059 V.Q.The solubility product (Ksp; mol3 dm–9) of MX2 at 298 K based on the information available for the givenconcentration cell is (take 2.303 xR x298/F = 0.059 V)a)1 x10–15b)4 x10–15c)1 x10–12d)4 x10–12Correct answer is option 'B'. Can you explain this answer? theory, EduRev gives you an
ample number of questions to practice Paragraph for Questions 31 and 32The electrochemical cell shown below is a concentration cell.M|M2+ (saturated solution of a sparingly soluble salt, MX2) || M2+ (0.001 mol dm–3) | MThe emf of the cell depends on the difference in concentrations of M2+ ions at the two electrodes. The emf of the cell at 298 K is 0.059 V.Q.The solubility product (Ksp; mol3 dm–9) of MX2 at 298 K based on the information available for the givenconcentration cell is (take 2.303 xR x298/F = 0.059 V)a)1 x10–15b)4 x10–15c)1 x10–12d)4 x10–12Correct answer is option 'B'. Can you explain this answer? tests, examples and also practice JEE tests.
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