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Three Faradays of electricity are passed through molten Al2O3, aqueous solution of CuSO4 and molten NaCl taken in three different electrolytic cells. The amount of Al, Cu and Na deposited at the cathodes will be in the ratio of-
- a)1 mole : 2 mole : 3 mole
- b)1 mole : 1.5 mole : 3 mole
- c)3 mole : 2 mole : 1 mole
- d)1 mole : 1.5 mole : 2 mole
Correct answer is option 'B'. Can you explain this answer?
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Verified Answer
Three Faradays of electricity are passed through molten Al2O3, aqueous...
gm equivalent of Al = gm eq. of Cu = gm eq. of Na
3 = 3 = 3
3/3/ = 3/2 = 1
1 : 1.5 : 3
3 = 3 = 3
3/3/ = 3/2 = 1
1 : 1.5 : 3
Most Upvoted Answer
Three Faradays of electricity are passed through molten Al2O3, aqueous...
The ratio of the amount of Al, Cu, and Na deposited at the cathodes can be determined by considering the Faraday's laws of electrolysis and the stoichiometry of the reactions involved.
1. Calculation for molten Al2O3:
- The electrolysis of molten Al2O3 involves the reduction of aluminum ions (Al3+) at the cathode.
- The balanced equation for the reduction of Al3+ is: 2Al3+ + 6e- → 2Al
- According to Faraday's law, the amount of substance deposited at the cathode is directly proportional to the charge passed through the electrolyte.
- Since 3 Faradays of electricity are passed, 6 moles of electrons are involved in the reduction of Al3+.
- Therefore, 1 mole of Al is deposited at the cathode.
2. Calculation for aqueous CuSO4:
- The electrolysis of the aqueous solution of CuSO4 involves the reduction of copper ions (Cu2+) at the cathode.
- The balanced equation for the reduction of Cu2+ is: Cu2+ + 2e- → Cu
- According to Faraday's law, the amount of substance deposited at the cathode is directly proportional to the charge passed through the electrolyte.
- Since 3 Faradays of electricity are passed, 6 moles of electrons are involved in the reduction of Cu2+.
- Therefore, 1 mole of Cu is deposited at the cathode.
3. Calculation for molten NaCl:
- The electrolysis of molten NaCl involves the reduction of sodium ions (Na+) at the cathode.
- The balanced equation for the reduction of Na+ is: Na+ + e- → Na
- According to Faraday's law, the amount of substance deposited at the cathode is directly proportional to the charge passed through the electrolyte.
- Since 3 Faradays of electricity are passed, 3 moles of electrons are involved in the reduction of Na+.
- Therefore, 3 moles of Na are deposited at the cathode.
Therefore, the ratio of Al : Cu : Na deposited at the cathodes is 1 mole : 1 mole : 3 moles, which corresponds to option B.
1. Calculation for molten Al2O3:
- The electrolysis of molten Al2O3 involves the reduction of aluminum ions (Al3+) at the cathode.
- The balanced equation for the reduction of Al3+ is: 2Al3+ + 6e- → 2Al
- According to Faraday's law, the amount of substance deposited at the cathode is directly proportional to the charge passed through the electrolyte.
- Since 3 Faradays of electricity are passed, 6 moles of electrons are involved in the reduction of Al3+.
- Therefore, 1 mole of Al is deposited at the cathode.
2. Calculation for aqueous CuSO4:
- The electrolysis of the aqueous solution of CuSO4 involves the reduction of copper ions (Cu2+) at the cathode.
- The balanced equation for the reduction of Cu2+ is: Cu2+ + 2e- → Cu
- According to Faraday's law, the amount of substance deposited at the cathode is directly proportional to the charge passed through the electrolyte.
- Since 3 Faradays of electricity are passed, 6 moles of electrons are involved in the reduction of Cu2+.
- Therefore, 1 mole of Cu is deposited at the cathode.
3. Calculation for molten NaCl:
- The electrolysis of molten NaCl involves the reduction of sodium ions (Na+) at the cathode.
- The balanced equation for the reduction of Na+ is: Na+ + e- → Na
- According to Faraday's law, the amount of substance deposited at the cathode is directly proportional to the charge passed through the electrolyte.
- Since 3 Faradays of electricity are passed, 3 moles of electrons are involved in the reduction of Na+.
- Therefore, 3 moles of Na are deposited at the cathode.
Therefore, the ratio of Al : Cu : Na deposited at the cathodes is 1 mole : 1 mole : 3 moles, which corresponds to option B.
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Three Faradays of electricity are passed through molten Al2O3, aqueous solution of CuSO4and molten NaCl taken in three different electrolytic cells. The amount of Al, Cu and Na deposited at the cathodes will be in the ratio of-a)1 mole : 2 mole : 3 moleb)1 mole : 1.5 mole : 3 molec)3 mole : 2 mole : 1 moled)1 mole : 1.5 mole : 2 moleCorrect answer is option 'B'. Can you explain this answer?
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Three Faradays of electricity are passed through molten Al2O3, aqueous solution of CuSO4and molten NaCl taken in three different electrolytic cells. The amount of Al, Cu and Na deposited at the cathodes will be in the ratio of-a)1 mole : 2 mole : 3 moleb)1 mole : 1.5 mole : 3 molec)3 mole : 2 mole : 1 moled)1 mole : 1.5 mole : 2 moleCorrect answer is option 'B'. Can you explain this answer? for Class 12 2024 is part of Class 12 preparation. The Question and answers have been prepared according to the Class 12 exam syllabus. Information about Three Faradays of electricity are passed through molten Al2O3, aqueous solution of CuSO4and molten NaCl taken in three different electrolytic cells. The amount of Al, Cu and Na deposited at the cathodes will be in the ratio of-a)1 mole : 2 mole : 3 moleb)1 mole : 1.5 mole : 3 molec)3 mole : 2 mole : 1 moled)1 mole : 1.5 mole : 2 moleCorrect answer is option 'B'. Can you explain this answer? covers all topics & solutions for Class 12 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Three Faradays of electricity are passed through molten Al2O3, aqueous solution of CuSO4and molten NaCl taken in three different electrolytic cells. The amount of Al, Cu and Na deposited at the cathodes will be in the ratio of-a)1 mole : 2 mole : 3 moleb)1 mole : 1.5 mole : 3 molec)3 mole : 2 mole : 1 moled)1 mole : 1.5 mole : 2 moleCorrect answer is option 'B'. Can you explain this answer?.
Three Faradays of electricity are passed through molten Al2O3, aqueous solution of CuSO4and molten NaCl taken in three different electrolytic cells. The amount of Al, Cu and Na deposited at the cathodes will be in the ratio of-a)1 mole : 2 mole : 3 moleb)1 mole : 1.5 mole : 3 molec)3 mole : 2 mole : 1 moled)1 mole : 1.5 mole : 2 moleCorrect answer is option 'B'. Can you explain this answer? for Class 12 2024 is part of Class 12 preparation. The Question and answers have been prepared according to the Class 12 exam syllabus. Information about Three Faradays of electricity are passed through molten Al2O3, aqueous solution of CuSO4and molten NaCl taken in three different electrolytic cells. The amount of Al, Cu and Na deposited at the cathodes will be in the ratio of-a)1 mole : 2 mole : 3 moleb)1 mole : 1.5 mole : 3 molec)3 mole : 2 mole : 1 moled)1 mole : 1.5 mole : 2 moleCorrect answer is option 'B'. Can you explain this answer? covers all topics & solutions for Class 12 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Three Faradays of electricity are passed through molten Al2O3, aqueous solution of CuSO4and molten NaCl taken in three different electrolytic cells. The amount of Al, Cu and Na deposited at the cathodes will be in the ratio of-a)1 mole : 2 mole : 3 moleb)1 mole : 1.5 mole : 3 molec)3 mole : 2 mole : 1 moled)1 mole : 1.5 mole : 2 moleCorrect answer is option 'B'. Can you explain this answer?.
Solutions for Three Faradays of electricity are passed through molten Al2O3, aqueous solution of CuSO4and molten NaCl taken in three different electrolytic cells. The amount of Al, Cu and Na deposited at the cathodes will be in the ratio of-a)1 mole : 2 mole : 3 moleb)1 mole : 1.5 mole : 3 molec)3 mole : 2 mole : 1 moled)1 mole : 1.5 mole : 2 moleCorrect answer is option 'B'. Can you explain this answer? in English & in Hindi are available as part of our courses for Class 12.
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