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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. [IIT-2012]
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 × R × 298 /F = 0.059 V)
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. [IIT-2012]
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 × R × 298 /F = 0.059 V)
- a)1 × 10–15
- b)4 × 10–15
- c)1 × 10–12
- d)4 × 10–12
Correct answer is option 'B'. Can you explain this answer?
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The electrochemical cell shown below is a concentration cell.M | M2+ (...
The electrochemical cell shown is a concentration cell, meaning that the difference in concentration of the two solutions is driving the cell reaction.
In this cell, the cell reaction occurs between two different metal ions, M and M2, in solution. The two half-cells are connected by a salt bridge or a porous membrane to allow ion flow.
The left half-cell contains a metal M electrode in contact with a saturated solution of a sparingly soluble salt, Mx2. This means that the concentration of M2+ ions in the left half-cell is higher than in the right half-cell.
The right half-cell contains a metal M2 electrode in contact with a solution of M2 with a concentration of 0.001 mol dm^-3. This concentration is lower than in the left half-cell, creating a concentration gradient.
The overall cell reaction can be represented as:
M + Mx2 ⇌ M2 + M(Mx2)
The reaction proceeds in both directions, with M2+ ions moving from the left to the right half-cell, and M ions moving from the right to the left half-cell. This movement of ions equalizes the concentration of M2+ ions in both half-cells.
The cell potential of a concentration cell depends on the difference in concentration between the two half-cells. As the cell reaction proceeds and the concentrations become more equal, the cell potential decreases. Eventually, when the concentrations are equal, the cell potential becomes zero and the reaction reaches equilibrium.
Overall, concentration cells are useful for measuring and comparing the concentration of ions in different solutions. They can also be used to determine solubility product constants or to study the behavior of sparingly soluble salts.
In this cell, the cell reaction occurs between two different metal ions, M and M2, in solution. The two half-cells are connected by a salt bridge or a porous membrane to allow ion flow.
The left half-cell contains a metal M electrode in contact with a saturated solution of a sparingly soluble salt, Mx2. This means that the concentration of M2+ ions in the left half-cell is higher than in the right half-cell.
The right half-cell contains a metal M2 electrode in contact with a solution of M2 with a concentration of 0.001 mol dm^-3. This concentration is lower than in the left half-cell, creating a concentration gradient.
The overall cell reaction can be represented as:
M + Mx2 ⇌ M2 + M(Mx2)
The reaction proceeds in both directions, with M2+ ions moving from the left to the right half-cell, and M ions moving from the right to the left half-cell. This movement of ions equalizes the concentration of M2+ ions in both half-cells.
The cell potential of a concentration cell depends on the difference in concentration between the two half-cells. As the cell reaction proceeds and the concentrations become more equal, the cell potential decreases. Eventually, when the concentrations are equal, the cell potential becomes zero and the reaction reaches equilibrium.
Overall, concentration cells are useful for measuring and comparing the concentration of ions in different solutions. They can also be used to determine solubility product constants or to study the behavior of sparingly soluble salts.
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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 ofthe cell depends on the difference in concentrations of M2+ ions at the two electrodes. The emf of thecell at 298 K is 0.059 V. [IIT-2012]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 × R × 298 /F = 0.059 V)a)1 × 10–15 b)4 × 10–15c)1 × 10–12d)4 × 10–12Correct answer is option 'B'. Can you explain this answer?
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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 ofthe cell depends on the difference in concentrations of M2+ ions at the two electrodes. The emf of thecell at 298 K is 0.059 V. [IIT-2012]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 × R × 298 /F = 0.059 V)a)1 × 10–15 b)4 × 10–15c)1 × 10–12d)4 × 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 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 ofthe cell depends on the difference in concentrations of M2+ ions at the two electrodes. The emf of thecell at 298 K is 0.059 V. [IIT-2012]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 × R × 298 /F = 0.059 V)a)1 × 10–15 b)4 × 10–15c)1 × 10–12d)4 × 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 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 ofthe cell depends on the difference in concentrations of M2+ ions at the two electrodes. The emf of thecell at 298 K is 0.059 V. [IIT-2012]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 × R × 298 /F = 0.059 V)a)1 × 10–15 b)4 × 10–15c)1 × 10–12d)4 × 10–12Correct answer is option 'B'. Can you explain this answer?.
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 ofthe cell depends on the difference in concentrations of M2+ ions at the two electrodes. The emf of thecell at 298 K is 0.059 V. [IIT-2012]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 × R × 298 /F = 0.059 V)a)1 × 10–15 b)4 × 10–15c)1 × 10–12d)4 × 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 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 ofthe cell depends on the difference in concentrations of M2+ ions at the two electrodes. The emf of thecell at 298 K is 0.059 V. [IIT-2012]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 × R × 298 /F = 0.059 V)a)1 × 10–15 b)4 × 10–15c)1 × 10–12d)4 × 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 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 ofthe cell depends on the difference in concentrations of M2+ ions at the two electrodes. The emf of thecell at 298 K is 0.059 V. [IIT-2012]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 × R × 298 /F = 0.059 V)a)1 × 10–15 b)4 × 10–15c)1 × 10–12d)4 × 10–12Correct answer is option 'B'. Can you explain this answer?.
Solutions for 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 ofthe cell depends on the difference in concentrations of M2+ ions at the two electrodes. The emf of thecell at 298 K is 0.059 V. [IIT-2012]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 × R × 298 /F = 0.059 V)a)1 × 10–15 b)4 × 10–15c)1 × 10–12d)4 × 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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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 ofthe cell depends on the difference in concentrations of M2+ ions at the two electrodes. The emf of thecell at 298 K is 0.059 V. [IIT-2012]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 × R × 298 /F = 0.059 V)a)1 × 10–15 b)4 × 10–15c)1 × 10–12d)4 × 10–12Correct answer is option 'B'. Can you explain this answer?, a detailed solution for 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 ofthe cell depends on the difference in concentrations of M2+ ions at the two electrodes. The emf of thecell at 298 K is 0.059 V. [IIT-2012]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 × R × 298 /F = 0.059 V)a)1 × 10–15 b)4 × 10–15c)1 × 10–12d)4 × 10–12Correct answer is option 'B'. Can you explain this answer? has been provided alongside types of 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 ofthe cell depends on the difference in concentrations of M2+ ions at the two electrodes. The emf of thecell at 298 K is 0.059 V. [IIT-2012]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 × R × 298 /F = 0.059 V)a)1 × 10–15 b)4 × 10–15c)1 × 10–12d)4 × 10–12Correct answer is option 'B'. Can you explain this answer? theory, EduRev gives you an
ample number of questions to practice 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 ofthe cell depends on the difference in concentrations of M2+ ions at the two electrodes. The emf of thecell at 298 K is 0.059 V. [IIT-2012]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 × R × 298 /F = 0.059 V)a)1 × 10–15 b)4 × 10–15c)1 × 10–12d)4 × 10–12Correct answer is option 'B'. Can you explain this answer? tests, examples and also practice JEE tests.
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