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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.
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.
The value of ΔG (kJ mol–1) for the given cell is (take 1 F = 96500 C mol–1)
- a)–5.7
- b)5.7
- c)11.4
- d)-11.4
Correct answer is option 'D'. Can you explain this answer?
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Most Upvoted Answer
The electrochemical cell shown below is a concentration cell.M | M2+ (...
-3 solution of Mx2) | M
Explanation:
A concentration cell is a type of electrochemical cell where the same material is present in both half-cells, but at different concentrations. In this case, the material is M2, and it is present in both half-cells, but at different concentrations.
The anode half-cell is M | M2 (saturated solution of a sparingly soluble salt, Mx2). This means that M is the anode, and it is in contact with a saturated solution of Mx2, which means that the concentration of M2 ions is very high. At the anode, M undergoes oxidation and loses electrons to become M2+. This reaction is represented as:
M → M2+ + 2e-
The cathode half-cell is M2 (0.001 mol dm-3 solution of Mx2) | M. This means that M2 is the cathode, and it is in contact with a solution of Mx2, which has a concentration of 0.001 mol dm-3. At the cathode, M2+ undergoes reduction and gains electrons to become M. This reaction is represented as:
M2+ + 2e- → M
Overall, the cell reaction is:
M → M2+ + 2e- (anode half-cell)
M2+ + 2e- → M (cathode half-cell)
-----------------------
M → M (cell reaction)
The cell potential 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, n is the number of electrons transferred in the cell reaction, F is the Faraday constant, and Q is the reaction quotient.
Since this is a concentration cell, the standard cell potential is zero, so E°cell = 0. The reaction quotient can be calculated using the concentrations of M2+ in the two half-cells:
Q = [M2+]cathode / [M2+]anode
= (0.001 mol dm-3) / (very high concentration)
Since the concentration of M2+ in the anode half-cell is very high, we can assume that it is constant and equal to the solubility product of Mx2. Therefore, Q = Ksp.
The solubility product of Mx2 is not given, so we cannot calculate Q or the cell potential. However, we can say that the cell potential will be negative, since the reaction is non-spontaneous (E°cell = 0) and the anode half-cell has a higher concentration of M2+. The cell will therefore require an external voltage source to operate.
Explanation:
A concentration cell is a type of electrochemical cell where the same material is present in both half-cells, but at different concentrations. In this case, the material is M2, and it is present in both half-cells, but at different concentrations.
The anode half-cell is M | M2 (saturated solution of a sparingly soluble salt, Mx2). This means that M is the anode, and it is in contact with a saturated solution of Mx2, which means that the concentration of M2 ions is very high. At the anode, M undergoes oxidation and loses electrons to become M2+. This reaction is represented as:
M → M2+ + 2e-
The cathode half-cell is M2 (0.001 mol dm-3 solution of Mx2) | M. This means that M2 is the cathode, and it is in contact with a solution of Mx2, which has a concentration of 0.001 mol dm-3. At the cathode, M2+ undergoes reduction and gains electrons to become M. This reaction is represented as:
M2+ + 2e- → M
Overall, the cell reaction is:
M → M2+ + 2e- (anode half-cell)
M2+ + 2e- → M (cathode half-cell)
-----------------------
M → M (cell reaction)
The cell potential 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, n is the number of electrons transferred in the cell reaction, F is the Faraday constant, and Q is the reaction quotient.
Since this is a concentration cell, the standard cell potential is zero, so E°cell = 0. The reaction quotient can be calculated using the concentrations of M2+ in the two half-cells:
Q = [M2+]cathode / [M2+]anode
= (0.001 mol dm-3) / (very high concentration)
Since the concentration of M2+ in the anode half-cell is very high, we can assume that it is constant and equal to the solubility product of Mx2. Therefore, Q = Ksp.
The solubility product of Mx2 is not given, so we cannot calculate Q or the cell potential. However, we can say that the cell potential will be negative, since the reaction is non-spontaneous (E°cell = 0) and the anode half-cell has a higher concentration of M2+. The cell will therefore require an external voltage source to operate.
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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.The value of ΔG (kJ mol–1) for the given cell is (take 1 F = 96500 C mol–1)a)–5.7b)5.7c)11.4d)-11.4Correct answer is option 'D'. 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.The value of ΔG (kJ mol–1) for the given cell is (take 1 F = 96500 C mol–1)a)–5.7b)5.7c)11.4d)-11.4Correct answer is option 'D'. 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.The value of ΔG (kJ mol–1) for the given cell is (take 1 F = 96500 C mol–1)a)–5.7b)5.7c)11.4d)-11.4Correct answer is option 'D'. 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.The value of ΔG (kJ mol–1) for the given cell is (take 1 F = 96500 C mol–1)a)–5.7b)5.7c)11.4d)-11.4Correct answer is option 'D'. 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.The value of ΔG (kJ mol–1) for the given cell is (take 1 F = 96500 C mol–1)a)–5.7b)5.7c)11.4d)-11.4Correct answer is option 'D'. 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.The value of ΔG (kJ mol–1) for the given cell is (take 1 F = 96500 C mol–1)a)–5.7b)5.7c)11.4d)-11.4Correct answer is option 'D'. 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.The value of ΔG (kJ mol–1) for the given cell is (take 1 F = 96500 C mol–1)a)–5.7b)5.7c)11.4d)-11.4Correct answer is option 'D'. 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.The value of ΔG (kJ mol–1) for the given cell is (take 1 F = 96500 C mol–1)a)–5.7b)5.7c)11.4d)-11.4Correct answer is option 'D'. Can you explain this answer? in English & in Hindi are available as part of our courses for JEE.
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