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If 0.5 amp current is passed through acidified silver nitrate solution for 100 minutes. The mass of silver deposited on cathode, is (eq.wt.of silver nitrate = 108) [1995]
- a)2.3523 g
- b)3.3575 g
- c)5.3578 g
- d)6.3575 g
Correct answer is option 'B'. Can you explain this answer?
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Verified Answer
If 0.5 amp current is passed through acidified silver nitrate solution...
Given current (i) = 0.5 amp;
Time (t) = 100 minutes × 60 = 6000 sec
Equivalent weight of silver nitrate (E) = 108.
According to Faraday's first law of electrolysis
Time (t) = 100 minutes × 60 = 6000 sec
Equivalent weight of silver nitrate (E) = 108.
According to Faraday's first law of electrolysis
Most Upvoted Answer
If 0.5 amp current is passed through acidified silver nitrate solution...
Given:
- Current (I) = 0.5 A
- Time (t) = 100 minutes
- Equivalent weight of silver nitrate (AgNO3) = 108 g/mol
Calculating the amount of charge passed:
- Charge (Q) = Current (I) × Time (t)
- Substituting the given values, Q = 0.5 A × 100 minutes
Converting time to seconds:
- 1 minute = 60 seconds
- Therefore, 100 minutes = 100 × 60 seconds = 6000 seconds
Substituting the values:
- Q = 0.5 A × 6000 seconds
Calculating the amount of silver deposited:
- The amount of charge passed (Q) is directly proportional to the amount of silver deposited.
- The ratio of charge to the amount of silver deposited is given by the Faraday's law of electrolysis:
- Q = n × F
- Where n is the number of moles of electrons transferred and F is Faraday's constant.
- The Faraday's constant is given by 1 Faraday = 96485 C/mol.
Calculating the number of moles of electrons transferred:
- n = Q / F
- Substituting the values, n = (0.5 A × 6000 seconds) / 96485 C/mol
Calculating the mass of silver deposited:
- The number of moles (n) is directly proportional to the mass of silver deposited.
- The ratio of moles to mass is given by the molar mass of silver (Ag).
- The molar mass of silver is 107.87 g/mol.
Calculating the mass of silver:
- Mass of silver = n × molar mass of silver
- Substituting the values, Mass of silver = (0.5 A × 6000 seconds / 96485 C/mol) × 107.87 g/mol
Simplifying the equation:
- Mass of silver = (0.5 A × 6000 seconds × 107.87 g) / 96485 C
Calculating the mass of silver (approximated to four decimal places):
- Mass of silver ≈ 3.3575 g
The mass of silver deposited on the cathode is approximately 3.3575 g. Therefore, the correct answer is option (B).
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If 0.5 amp current is passed through acidified silver nitrate solution for 100 minutes. The mass of silver deposited on cathode, is (eq.wt.of silver nitrate = 108) [1995]a)2.3523 gb)3.3575 gc)5.3578 gd)6.3575 gCorrect answer is option 'B'. Can you explain this answer?
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If 0.5 amp current is passed through acidified silver nitrate solution for 100 minutes. The mass of silver deposited on cathode, is (eq.wt.of silver nitrate = 108) [1995]a)2.3523 gb)3.3575 gc)5.3578 gd)6.3575 gCorrect 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 If 0.5 amp current is passed through acidified silver nitrate solution for 100 minutes. The mass of silver deposited on cathode, is (eq.wt.of silver nitrate = 108) [1995]a)2.3523 gb)3.3575 gc)5.3578 gd)6.3575 gCorrect 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 If 0.5 amp current is passed through acidified silver nitrate solution for 100 minutes. The mass of silver deposited on cathode, is (eq.wt.of silver nitrate = 108) [1995]a)2.3523 gb)3.3575 gc)5.3578 gd)6.3575 gCorrect answer is option 'B'. Can you explain this answer?.
If 0.5 amp current is passed through acidified silver nitrate solution for 100 minutes. The mass of silver deposited on cathode, is (eq.wt.of silver nitrate = 108) [1995]a)2.3523 gb)3.3575 gc)5.3578 gd)6.3575 gCorrect 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 If 0.5 amp current is passed through acidified silver nitrate solution for 100 minutes. The mass of silver deposited on cathode, is (eq.wt.of silver nitrate = 108) [1995]a)2.3523 gb)3.3575 gc)5.3578 gd)6.3575 gCorrect 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 If 0.5 amp current is passed through acidified silver nitrate solution for 100 minutes. The mass of silver deposited on cathode, is (eq.wt.of silver nitrate = 108) [1995]a)2.3523 gb)3.3575 gc)5.3578 gd)6.3575 gCorrect answer is option 'B'. Can you explain this answer?.
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