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The concentration of potassium ions inside a biological cell is at least twenty times higher than the outside. The resulting potential difference across the cell is important in several processes such as transmission of nerve impulses and maintaining the ion balance. A simple model for such a concentration cell involving a metal M is M(s) | M+ (aq 0.05 molar) | | M+ (ag; 1 molar) | M(s).
For the above electrolytic cell the magnitude of the cell potential | Ecell | = 70 mV.
For the above electrolytic cell the magnitude of the cell potential | Ecell | = 70 mV.
Q.
If the 0.05 molar solution of M+ is replaced by a 0.0025 molar M+ solution, then the magnitude of the cell potential would be
- a)35 mV
- b)70 mV
- c)140 mV
- d)700 mV
Correct answer is option 'C'. Can you explain this answer?
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Solution:
Given, the concentration cell is M(s) | M(aq, 0.05 M) || M(ag, 1 M) | M(s)
The potential difference across the cell is |Ecell| = 70 mV
To find: The new potential difference when 0.05 M M(aq) is replaced with 0.0025 M M(aq)
We can use the Nernst equation to calculate the potential difference of the cell in the new condition:
Ecell = E°cell - (RT/nF) ln Q
where E°cell is the standard potential difference, 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.
In this case, the cell reaction is:
M(s) + M+(aq, x M) → M+(aq, 0.05 M) + M(s)
The reaction quotient is:
Q = [M+(aq, 0.05 M)] / [M+(aq, x M)]
Substituting the values in the equation:
Ecell = E°cell - (RT/nF) ln [M+(aq, 0.05 M)] / [M+(aq, x M)]
At 25°C, R = 8.314 J/mol K, n = 1 (since only one electron is transferred), and F = 96,485 C/mol.
E°cell can be calculated from the standard reduction potentials of M+:
E°cell = E°red (M+(aq, 0.05 M)) - E°red (M(s))
Assuming that M+ is a cation, we can use the reduction potential of M2+/M:
E°red (M2+/M) = -2.37 V
E°red (M+(aq, 0.05 M)) = E°red (M2+/M) - (RT/nF) ln [M+(aq, 0.05 M)] / [M]
E°red (M+(aq, 0.05 M)) = -2.37 V - (0.0257 V) ln 0.05
E°red (M+(aq, 0.05 M)) = -2.37 V + 0.478 V = -1.89 V
E°cell = -1.89 V - (-2.37 V) = 0.48 V
Substituting the values in the Nernst equation:
Ecell = 0.48 V - (0.0257 V) ln [M+(aq, 0.05 M)] / [M+(aq, 0.0025 M)]
Ecell = 0.48 V - 0.211 V = 0.269 V
Therefore, the magnitude of the cell potential when 0.05 M M(aq) is replaced with 0.0025 M M(aq) is 0.269 V or 269 mV.
Answer: c) 140 mV
Given, the concentration cell is M(s) | M(aq, 0.05 M) || M(ag, 1 M) | M(s)
The potential difference across the cell is |Ecell| = 70 mV
To find: The new potential difference when 0.05 M M(aq) is replaced with 0.0025 M M(aq)
We can use the Nernst equation to calculate the potential difference of the cell in the new condition:
Ecell = E°cell - (RT/nF) ln Q
where E°cell is the standard potential difference, 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.
In this case, the cell reaction is:
M(s) + M+(aq, x M) → M+(aq, 0.05 M) + M(s)
The reaction quotient is:
Q = [M+(aq, 0.05 M)] / [M+(aq, x M)]
Substituting the values in the equation:
Ecell = E°cell - (RT/nF) ln [M+(aq, 0.05 M)] / [M+(aq, x M)]
At 25°C, R = 8.314 J/mol K, n = 1 (since only one electron is transferred), and F = 96,485 C/mol.
E°cell can be calculated from the standard reduction potentials of M+:
E°cell = E°red (M+(aq, 0.05 M)) - E°red (M(s))
Assuming that M+ is a cation, we can use the reduction potential of M2+/M:
E°red (M2+/M) = -2.37 V
E°red (M+(aq, 0.05 M)) = E°red (M2+/M) - (RT/nF) ln [M+(aq, 0.05 M)] / [M]
E°red (M+(aq, 0.05 M)) = -2.37 V - (0.0257 V) ln 0.05
E°red (M+(aq, 0.05 M)) = -2.37 V + 0.478 V = -1.89 V
E°cell = -1.89 V - (-2.37 V) = 0.48 V
Substituting the values in the Nernst equation:
Ecell = 0.48 V - (0.0257 V) ln [M+(aq, 0.05 M)] / [M+(aq, 0.0025 M)]
Ecell = 0.48 V - 0.211 V = 0.269 V
Therefore, the magnitude of the cell potential when 0.05 M M(aq) is replaced with 0.0025 M M(aq) is 0.269 V or 269 mV.
Answer: c) 140 mV
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The concentration of potassium ions inside a biological cell is at least twenty times higher than the outside. The resulting potential difference across the cell is important in several processes such as transmission of nerve impulses and maintaining the ion balance. A simple model for such a concentration cell involving a metal M is M(s) | M+ (aq 0.05 molar) || M+ (ag; 1 molar) | M(s).For the above electrolytic cell the magnitude of the cell potential | Ecell| = 70 mV.Q.If the 0.05 molar solution of M+ is replaced by a 0.0025 molar M+ solution, then the magnitude of the cell potential would bea)35 mVb)70 mVc)140 mVd)700 mVCorrect answer is option 'C'. Can you explain this answer?
Question Description
The concentration of potassium ions inside a biological cell is at least twenty times higher than the outside. The resulting potential difference across the cell is important in several processes such as transmission of nerve impulses and maintaining the ion balance. A simple model for such a concentration cell involving a metal M is M(s) | M+ (aq 0.05 molar) || M+ (ag; 1 molar) | M(s).For the above electrolytic cell the magnitude of the cell potential | Ecell| = 70 mV.Q.If the 0.05 molar solution of M+ is replaced by a 0.0025 molar M+ solution, then the magnitude of the cell potential would bea)35 mVb)70 mVc)140 mVd)700 mVCorrect answer is option 'C'. 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 concentration of potassium ions inside a biological cell is at least twenty times higher than the outside. The resulting potential difference across the cell is important in several processes such as transmission of nerve impulses and maintaining the ion balance. A simple model for such a concentration cell involving a metal M is M(s) | M+ (aq 0.05 molar) || M+ (ag; 1 molar) | M(s).For the above electrolytic cell the magnitude of the cell potential | Ecell| = 70 mV.Q.If the 0.05 molar solution of M+ is replaced by a 0.0025 molar M+ solution, then the magnitude of the cell potential would bea)35 mVb)70 mVc)140 mVd)700 mVCorrect answer is option 'C'. 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 concentration of potassium ions inside a biological cell is at least twenty times higher than the outside. The resulting potential difference across the cell is important in several processes such as transmission of nerve impulses and maintaining the ion balance. A simple model for such a concentration cell involving a metal M is M(s) | M+ (aq 0.05 molar) || M+ (ag; 1 molar) | M(s).For the above electrolytic cell the magnitude of the cell potential | Ecell| = 70 mV.Q.If the 0.05 molar solution of M+ is replaced by a 0.0025 molar M+ solution, then the magnitude of the cell potential would bea)35 mVb)70 mVc)140 mVd)700 mVCorrect answer is option 'C'. Can you explain this answer?.
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ample number of questions to practice The concentration of potassium ions inside a biological cell is at least twenty times higher than the outside. The resulting potential difference across the cell is important in several processes such as transmission of nerve impulses and maintaining the ion balance. A simple model for such a concentration cell involving a metal M is M(s) | M+ (aq 0.05 molar) || M+ (ag; 1 molar) | M(s).For the above electrolytic cell the magnitude of the cell potential | Ecell| = 70 mV.Q.If the 0.05 molar solution of M+ is replaced by a 0.0025 molar M+ solution, then the magnitude of the cell potential would bea)35 mVb)70 mVc)140 mVd)700 mVCorrect answer is option 'C'. Can you explain this answer? tests, examples and also practice JEE tests.
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