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W g of Ag is deposited at the cathode of one electrolytic cell due to passage of 1A of current for 1h.Time required for passage of current to deposit W g of Mg by the same value of current is
- a)9 h
- b)4.5 h
- c)1 h
- d)3.0 h
Correct answer is option 'A'. Can you explain this answer?
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To determine the time required for the passage of current to deposit a certain amount of Mg, we can use Faraday's laws of electrolysis. According to these laws, the amount of substance deposited or liberated at an electrode during electrolysis is directly proportional to the quantity of electricity passed through the electrolyte.
First, let's calculate the amount of electricity required to deposit W grams of Ag.
1. Calculate the charge (Q) passed through the electrolyte:
Q = I * t
where I is the current (1A) and t is the time (1h = 3600s)
Q = 1A * 3600s = 3600C
2. Calculate the number of moles of Ag deposited:
n(Ag) = Q / F
where F is the Faraday constant (96500C/mol)
n(Ag) = 3600C / 96500C/mol = 0.0373 mol
3. Calculate the molar mass of Ag:
M(Ag) = W / n(Ag)
where W is the mass of Ag (given) and M(Ag) is the molar mass of Ag (107.87 g/mol)
Now, let's calculate the time required to deposit the same amount of Mg using the same current.
1. Calculate the number of moles of Mg:
n(Mg) = W / M(Mg)
where M(Mg) is the molar mass of Mg (24.31 g/mol)
2. Calculate the charge required to deposit n(Mg) moles of Mg:
Q(Mg) = n(Mg) * F
3. Calculate the time required:
t(Mg) = Q(Mg) / I
Substituting the values:
t(Mg) = (n(Mg) * F) / I
t(Mg) = (W / M(Mg)) * F / I
t(Mg) = (W / 24.31) * (96500 / 1)
t(Mg) = 3970W
Therefore, the time required to deposit W grams of Mg is approximately 3970W seconds.
Since the time required for the deposition of Ag is 3600 seconds, which is less than 3970W seconds, the correct answer is option A: 9h.
First, let's calculate the amount of electricity required to deposit W grams of Ag.
1. Calculate the charge (Q) passed through the electrolyte:
Q = I * t
where I is the current (1A) and t is the time (1h = 3600s)
Q = 1A * 3600s = 3600C
2. Calculate the number of moles of Ag deposited:
n(Ag) = Q / F
where F is the Faraday constant (96500C/mol)
n(Ag) = 3600C / 96500C/mol = 0.0373 mol
3. Calculate the molar mass of Ag:
M(Ag) = W / n(Ag)
where W is the mass of Ag (given) and M(Ag) is the molar mass of Ag (107.87 g/mol)
Now, let's calculate the time required to deposit the same amount of Mg using the same current.
1. Calculate the number of moles of Mg:
n(Mg) = W / M(Mg)
where M(Mg) is the molar mass of Mg (24.31 g/mol)
2. Calculate the charge required to deposit n(Mg) moles of Mg:
Q(Mg) = n(Mg) * F
3. Calculate the time required:
t(Mg) = Q(Mg) / I
Substituting the values:
t(Mg) = (n(Mg) * F) / I
t(Mg) = (W / M(Mg)) * F / I
t(Mg) = (W / 24.31) * (96500 / 1)
t(Mg) = 3970W
Therefore, the time required to deposit W grams of Mg is approximately 3970W seconds.
Since the time required for the deposition of Ag is 3600 seconds, which is less than 3970W seconds, the correct answer is option A: 9h.
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Wg of Ag is deposited at the cathode of one electrolytic cell due to passage of 1A of current for 1h.Time required for passage of current to deposit W g of Mg by the same value of current isa)9 hb)4.5 hc)1 hd)3.0 hCorrect answer is option 'A'. 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 Wg of Ag is deposited at the cathode of one electrolytic cell due to passage of 1A of current for 1h.Time required for passage of current to deposit W g of Mg by the same value of current isa)9 hb)4.5 hc)1 hd)3.0 hCorrect answer is option 'A'. 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 Wg of Ag is deposited at the cathode of one electrolytic cell due to passage of 1A of current for 1h.Time required for passage of current to deposit W g of Mg by the same value of current isa)9 hb)4.5 hc)1 hd)3.0 hCorrect answer is option 'A'. Can you explain this answer?.
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