Class 12 Exam > Class 12 Questions > Consider the cell reaction:Cd(s) | Cd2+(1.0 M...
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Consider the cell reaction:
Cd(s) | Cd2+ (1.0 M) || Cu2+ (1.0 m) | Cu (s)
If we wish to make a cell with more positive voltage using the same substances, we should:
Cd(s) | Cd2+ (1.0 M) || Cu2+ (1.0 m) | Cu (s)
If we wish to make a cell with more positive voltage using the same substances, we should:
- a)Increase [Cd2+] as well as [Cu2+] to 2.0 M
- b)Increase only [Cu2+] to 2.0 M
- c)Reduce only [Cd2+] to 0.1 M
- d)Decreases [Cd2+] to 0.1M and increases [Cu2+] to 1.0M
Correct answer is option 'D'. Can you explain this answer?
Most Upvoted Answer
Consider the cell reaction:Cd(s) | Cd2+(1.0 M) || Cu2+(1.0 m) | Cu (s)...
The correct answer is Option D.
Redox reaction:
Cd(s)→Cd2++2e
Cu2++2e→Cu(s)
Ecell = E°cell − (0.059/2) log ([Cd2+]/ [Cu2+])
Decreases [Cd2+] to 0.1M and increases [Cu2+] to 1.0M
Cd(s)→Cd2++2e
Cu2++2e→Cu(s)
Ecell = E°cell − (0.059/2) log ([Cd2+]/ [Cu2+])
Decreases [Cd2+] to 0.1M and increases [Cu2+] to 1.0M
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Community Answer
Consider the cell reaction:Cd(s) | Cd2+(1.0 M) || Cu2+(1.0 m) | Cu (s)...
To understand why option D is the correct answer, let's analyze the given cell reaction and how changing the concentrations of the substances affects the cell voltage.
Given cell reaction:
Cd(s) | Cd2+(1.0 M) || Cu2+(1.0 M) | Cu(s)
1. Understand the Cell Reaction:
The given cell reaction consists of two half-reactions:
- An oxidation half-reaction at the anode: Cd(s) → Cd2+(aq) + 2e^-
- A reduction half-reaction at the cathode: Cu2+(aq) + 2e^- → Cu(s)
2. Nernst Equation:
The cell voltage (Ecell) can be calculated using the Nernst equation, which relates the cell potential to the activities (concentrations) of the reactants and products. The Nernst equation for the given cell reaction is:
Ecell = E°cell - (RT/nF) * ln(Q)
Where:
- Ecell is the cell potential
- E°cell is the standard cell potential
- R is the gas constant (8.314 J/mol·K)
- T is the temperature in Kelvin
- n is the number of electrons transferred in the balanced cell reaction
- F is Faraday's constant (96,485 C/mol)
- Q is the reaction quotient (based on the concentrations of the reactants and products)
3. Effect of Concentration on Cell Voltage:
Based on the Nernst equation, we can observe that changing the concentrations of the substances involved in the cell reaction can affect the cell voltage. Specifically, increasing the concentration of a reactant or decreasing the concentration of a product will shift the cell potential to a more positive value.
4. Applying the Concept to the Given Cell Reaction:
In the given cell reaction, we have Cd2+(1.0 M) and Cu2+(1.0 M) as the reactants. To increase the cell voltage and make it more positive, we need to shift the reaction to the right (towards the product side) by increasing the concentration of the reactant Cd2+ and decreasing the concentration of the product Cu2+.
Hence, the correct answer is option D:
- Decrease [Cd2+] to 0.1 M: This reduces the concentration of the product Cd2+ and shifts the reaction towards the product side.
- Increase [Cu2+] to 1.0 M: This increases the concentration of the reactant Cu2+ and shifts the reaction towards the product side.
By adjusting the concentrations as described in option D, we can increase the cell voltage and make it more positive.
Given cell reaction:
Cd(s) | Cd2+(1.0 M) || Cu2+(1.0 M) | Cu(s)
1. Understand the Cell Reaction:
The given cell reaction consists of two half-reactions:
- An oxidation half-reaction at the anode: Cd(s) → Cd2+(aq) + 2e^-
- A reduction half-reaction at the cathode: Cu2+(aq) + 2e^- → Cu(s)
2. Nernst Equation:
The cell voltage (Ecell) can be calculated using the Nernst equation, which relates the cell potential to the activities (concentrations) of the reactants and products. The Nernst equation for the given cell reaction is:
Ecell = E°cell - (RT/nF) * ln(Q)
Where:
- Ecell is the cell potential
- E°cell is the standard cell potential
- R is the gas constant (8.314 J/mol·K)
- T is the temperature in Kelvin
- n is the number of electrons transferred in the balanced cell reaction
- F is Faraday's constant (96,485 C/mol)
- Q is the reaction quotient (based on the concentrations of the reactants and products)
3. Effect of Concentration on Cell Voltage:
Based on the Nernst equation, we can observe that changing the concentrations of the substances involved in the cell reaction can affect the cell voltage. Specifically, increasing the concentration of a reactant or decreasing the concentration of a product will shift the cell potential to a more positive value.
4. Applying the Concept to the Given Cell Reaction:
In the given cell reaction, we have Cd2+(1.0 M) and Cu2+(1.0 M) as the reactants. To increase the cell voltage and make it more positive, we need to shift the reaction to the right (towards the product side) by increasing the concentration of the reactant Cd2+ and decreasing the concentration of the product Cu2+.
Hence, the correct answer is option D:
- Decrease [Cd2+] to 0.1 M: This reduces the concentration of the product Cd2+ and shifts the reaction towards the product side.
- Increase [Cu2+] to 1.0 M: This increases the concentration of the reactant Cu2+ and shifts the reaction towards the product side.
By adjusting the concentrations as described in option D, we can increase the cell voltage and make it more positive.
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Consider the cell reaction:Cd(s) | Cd2+(1.0 M) || Cu2+(1.0 m) | Cu (s)If we wish to make a cell with more positive voltage using the same substances, we should:a)Increase [Cd2+] as well as [Cu2+] to 2.0 Mb)Increase only [Cu2+] to 2.0 Mc)Reduce only [Cd2+] to 0.1 Md)Decreases [Cd2+] to 0.1M and increases [Cu2+] to 1.0MCorrect answer is option 'D'. Can you explain this answer? for Class 12 2025 is part of Class 12 preparation. The Question and answers have been prepared according to the Class 12 exam syllabus. Information about Consider the cell reaction:Cd(s) | Cd2+(1.0 M) || Cu2+(1.0 m) | Cu (s)If we wish to make a cell with more positive voltage using the same substances, we should:a)Increase [Cd2+] as well as [Cu2+] to 2.0 Mb)Increase only [Cu2+] to 2.0 Mc)Reduce only [Cd2+] to 0.1 Md)Decreases [Cd2+] to 0.1M and increases [Cu2+] to 1.0MCorrect answer is option 'D'. Can you explain this answer? covers all topics & solutions for Class 12 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Consider the cell reaction:Cd(s) | Cd2+(1.0 M) || Cu2+(1.0 m) | Cu (s)If we wish to make a cell with more positive voltage using the same substances, we should:a)Increase [Cd2+] as well as [Cu2+] to 2.0 Mb)Increase only [Cu2+] to 2.0 Mc)Reduce only [Cd2+] to 0.1 Md)Decreases [Cd2+] to 0.1M and increases [Cu2+] to 1.0MCorrect answer is option 'D'. Can you explain this answer?.
Consider the cell reaction:Cd(s) | Cd2+(1.0 M) || Cu2+(1.0 m) | Cu (s)If we wish to make a cell with more positive voltage using the same substances, we should:a)Increase [Cd2+] as well as [Cu2+] to 2.0 Mb)Increase only [Cu2+] to 2.0 Mc)Reduce only [Cd2+] to 0.1 Md)Decreases [Cd2+] to 0.1M and increases [Cu2+] to 1.0MCorrect answer is option 'D'. Can you explain this answer? for Class 12 2025 is part of Class 12 preparation. The Question and answers have been prepared according to the Class 12 exam syllabus. Information about Consider the cell reaction:Cd(s) | Cd2+(1.0 M) || Cu2+(1.0 m) | Cu (s)If we wish to make a cell with more positive voltage using the same substances, we should:a)Increase [Cd2+] as well as [Cu2+] to 2.0 Mb)Increase only [Cu2+] to 2.0 Mc)Reduce only [Cd2+] to 0.1 Md)Decreases [Cd2+] to 0.1M and increases [Cu2+] to 1.0MCorrect answer is option 'D'. Can you explain this answer? covers all topics & solutions for Class 12 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Consider the cell reaction:Cd(s) | Cd2+(1.0 M) || Cu2+(1.0 m) | Cu (s)If we wish to make a cell with more positive voltage using the same substances, we should:a)Increase [Cd2+] as well as [Cu2+] to 2.0 Mb)Increase only [Cu2+] to 2.0 Mc)Reduce only [Cd2+] to 0.1 Md)Decreases [Cd2+] to 0.1M and increases [Cu2+] to 1.0MCorrect answer is option 'D'. Can you explain this answer?.
Solutions for Consider the cell reaction:Cd(s) | Cd2+(1.0 M) || Cu2+(1.0 m) | Cu (s)If we wish to make a cell with more positive voltage using the same substances, we should:a)Increase [Cd2+] as well as [Cu2+] to 2.0 Mb)Increase only [Cu2+] to 2.0 Mc)Reduce only [Cd2+] to 0.1 Md)Decreases [Cd2+] to 0.1M and increases [Cu2+] to 1.0MCorrect answer is option 'D'. Can you explain this answer? in English & in Hindi are available as part of our courses for Class 12.
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ample number of questions to practice Consider the cell reaction:Cd(s) | Cd2+(1.0 M) || Cu2+(1.0 m) | Cu (s)If we wish to make a cell with more positive voltage using the same substances, we should:a)Increase [Cd2+] as well as [Cu2+] to 2.0 Mb)Increase only [Cu2+] to 2.0 Mc)Reduce only [Cd2+] to 0.1 Md)Decreases [Cd2+] to 0.1M and increases [Cu2+] to 1.0MCorrect answer is option 'D'. Can you explain this answer? tests, examples and also practice Class 12 tests.
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