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Aluminium displaces hydrogen from acids,but copper does not. A galvanic cell prepared by combining Cu/Cu2+ and Al/Al3+ has an EMF of 2.0 V at 298 K. If the potential of copper electrode is + 0.34 V,that of aluminium electrode is
- a)-2.3 V
- b)+ 2.34 V
- c)- 1.66 V
- d)1.66 V
Correct answer is option 'C'. Can you explain this answer?
Most Upvoted Answer
Aluminium displaces hydrogen from acids,but copper does not. A galvani...
Given:
- Aluminium displaces hydrogen from acids, but copper does not.
- Galvanic cell: Cu/Cu2 and Al/Al3
- EMF of 2.0 V at 298 K
- Potential of copper electrode is 0.34 V
To find:
- Potential of aluminium electrode
Solution:
1. Write the half-reactions for the oxidation and reduction that occur in the galvanic cell:
- Oxidation half-reaction: Cu → Cu2+ + 2e-
- Reduction half-reaction: 2Al3+ + 6e- → 2Al
2. Write the overall reaction for the galvanic cell:
- Cu + 2Al3+ → Cu2+ + 2Al
3. Determine the standard reduction potentials for the half-reactions:
- Cu2+/Cu: E° = +0.34 V
- Al3+/Al: E° = -1.66 V
4. Use the Nernst equation to calculate the potential of the aluminium electrode:
- Ecell = E°cell - (RT/nF)lnQ
- Ecell = 2.0 V (given)
- E°cell = E°reduction - E°oxidation
- E°cell = (-1.66 V) - (+0.34 V) = -2.00 V
- R = 8.314 J/K/mol (gas constant)
- T = 298 K (temperature)
- n = 6 (number of electrons transferred)
- F = 96,485 C/mol (Faraday constant)
- Q = [Al]2/[Cu2+][Al3+] (reaction quotient)
Solving for Q:
- Q = [Al]2/[Cu2+][Al3+]
- Q = (1/[Al3+]2) / ([Cu2+]/[Al3+])
- Q = [Al3+] / [Cu2+]
Substituting the values:
- Ecell = -2.00 V - (RT/6F)ln([Al3+]/[Cu2+])
- 2.00 V = (8.314 J/K/mol x 298 K / 6 x 96,485 C/mol)ln([Al3+]/[Cu2+])
- ln([Al3+]/[Cu2+]) = -0.632
- [Al3+]/[Cu2+] = e^-0.632
- [Al3+]/[Cu2+] = 0.531
5. Use the Nernst equation again to calculate the potential of the aluminium electrode:
- E = E° - (RT/nF)ln([Al3+]/[Al])
- E° = -1.66 V (given)
- R, T, n, and F are the same as before
- [Al3+] = 0.531 (calculated above)
- [Al] = 1 (assumed, since it is a solid)
Solving for E:
- E = -1.66 V - (8.314 J/K/mol x 298 K / 3 x 96,485 C/mol)ln(0.531)
-
- Aluminium displaces hydrogen from acids, but copper does not.
- Galvanic cell: Cu/Cu2 and Al/Al3
- EMF of 2.0 V at 298 K
- Potential of copper electrode is 0.34 V
To find:
- Potential of aluminium electrode
Solution:
1. Write the half-reactions for the oxidation and reduction that occur in the galvanic cell:
- Oxidation half-reaction: Cu → Cu2+ + 2e-
- Reduction half-reaction: 2Al3+ + 6e- → 2Al
2. Write the overall reaction for the galvanic cell:
- Cu + 2Al3+ → Cu2+ + 2Al
3. Determine the standard reduction potentials for the half-reactions:
- Cu2+/Cu: E° = +0.34 V
- Al3+/Al: E° = -1.66 V
4. Use the Nernst equation to calculate the potential of the aluminium electrode:
- Ecell = E°cell - (RT/nF)lnQ
- Ecell = 2.0 V (given)
- E°cell = E°reduction - E°oxidation
- E°cell = (-1.66 V) - (+0.34 V) = -2.00 V
- R = 8.314 J/K/mol (gas constant)
- T = 298 K (temperature)
- n = 6 (number of electrons transferred)
- F = 96,485 C/mol (Faraday constant)
- Q = [Al]2/[Cu2+][Al3+] (reaction quotient)
Solving for Q:
- Q = [Al]2/[Cu2+][Al3+]
- Q = (1/[Al3+]2) / ([Cu2+]/[Al3+])
- Q = [Al3+] / [Cu2+]
Substituting the values:
- Ecell = -2.00 V - (RT/6F)ln([Al3+]/[Cu2+])
- 2.00 V = (8.314 J/K/mol x 298 K / 6 x 96,485 C/mol)ln([Al3+]/[Cu2+])
- ln([Al3+]/[Cu2+]) = -0.632
- [Al3+]/[Cu2+] = e^-0.632
- [Al3+]/[Cu2+] = 0.531
5. Use the Nernst equation again to calculate the potential of the aluminium electrode:
- E = E° - (RT/nF)ln([Al3+]/[Al])
- E° = -1.66 V (given)
- R, T, n, and F are the same as before
- [Al3+] = 0.531 (calculated above)
- [Al] = 1 (assumed, since it is a solid)
Solving for E:
- E = -1.66 V - (8.314 J/K/mol x 298 K / 3 x 96,485 C/mol)ln(0.531)
-
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Aluminium displaces hydrogen from acids,but copper does not. A galvanic cell prepared by combining Cu/Cu2+ and Al/Al3+ has an EMF of 2.0 V at 298 K. If the potential of copper electrode is + 0.34 V,that of aluminium electrode isa)-2.3 Vb)+ 2.34 Vc)- 1.66 Vd)1.66 VCorrect answer is option 'C'. Can you explain this answer?
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Aluminium displaces hydrogen from acids,but copper does not. A galvanic cell prepared by combining Cu/Cu2+ and Al/Al3+ has an EMF of 2.0 V at 298 K. If the potential of copper electrode is + 0.34 V,that of aluminium electrode isa)-2.3 Vb)+ 2.34 Vc)- 1.66 Vd)1.66 VCorrect answer is option 'C'. Can you explain this answer? for JEE 2025 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about Aluminium displaces hydrogen from acids,but copper does not. A galvanic cell prepared by combining Cu/Cu2+ and Al/Al3+ has an EMF of 2.0 V at 298 K. If the potential of copper electrode is + 0.34 V,that of aluminium electrode isa)-2.3 Vb)+ 2.34 Vc)- 1.66 Vd)1.66 VCorrect answer is option 'C'. Can you explain this answer? covers all topics & solutions for JEE 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Aluminium displaces hydrogen from acids,but copper does not. A galvanic cell prepared by combining Cu/Cu2+ and Al/Al3+ has an EMF of 2.0 V at 298 K. If the potential of copper electrode is + 0.34 V,that of aluminium electrode isa)-2.3 Vb)+ 2.34 Vc)- 1.66 Vd)1.66 VCorrect answer is option 'C'. Can you explain this answer?.
Aluminium displaces hydrogen from acids,but copper does not. A galvanic cell prepared by combining Cu/Cu2+ and Al/Al3+ has an EMF of 2.0 V at 298 K. If the potential of copper electrode is + 0.34 V,that of aluminium electrode isa)-2.3 Vb)+ 2.34 Vc)- 1.66 Vd)1.66 VCorrect answer is option 'C'. Can you explain this answer? for JEE 2025 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about Aluminium displaces hydrogen from acids,but copper does not. A galvanic cell prepared by combining Cu/Cu2+ and Al/Al3+ has an EMF of 2.0 V at 298 K. If the potential of copper electrode is + 0.34 V,that of aluminium electrode isa)-2.3 Vb)+ 2.34 Vc)- 1.66 Vd)1.66 VCorrect answer is option 'C'. Can you explain this answer? covers all topics & solutions for JEE 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Aluminium displaces hydrogen from acids,but copper does not. A galvanic cell prepared by combining Cu/Cu2+ and Al/Al3+ has an EMF of 2.0 V at 298 K. If the potential of copper electrode is + 0.34 V,that of aluminium electrode isa)-2.3 Vb)+ 2.34 Vc)- 1.66 Vd)1.66 VCorrect answer is option 'C'. Can you explain this answer?.
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