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When quantity of electricity is passed through cuso4 solution 0.16 g of copper gets deposited. If the same quantity of electricity passed through acidulated water. Then the volume of H2 libarateted at STP will be?
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When quantity of electricity is passed through cuso4 solution 0.16 g o...
Calculation of the amount of electricity passed through CuSO4 solution:
Given, mass of copper deposited = 0.16 g
The molar mass of Cu = 63.55 g/mol
Therefore, the number of moles of Cu deposited = 0.16/63.55 = 0.00252 mol
The equation for the reaction is Cu2+ + 2e- → Cu
Hence, the number of electrons passed = 2 x 0.00252 = 0.00504 C
Calculation of the volume of H2 liberated at STP:
The equation for the reaction is 2H+ + 2e- → H2
From Faraday’s law, the amount of charge required to liberate 1 mole of H2 is 2F or 2 x 96500 C
Therefore, the number of moles of H2 liberated = 0.00504/(2 x 96500) = 2.61 x 10^-5 mol
The molar volume of an ideal gas at STP is 22.4 L/mol
Therefore, the volume of H2 liberated at STP = 2.61 x 10^-5 x 22.4 = 5.84 x 10^-4 L or 0.584 mL
Explanation:
When electricity is passed through CuSO4 solution, copper ions get reduced at the cathode to form copper metal. The amount of copper deposited is directly proportional to the amount of electricity passed. Hence, the mass of copper deposited is used to calculate the amount of electricity passed.
When electricity is passed through acidulated water, hydrogen ions get reduced at the cathode to form hydrogen gas. The amount of hydrogen gas liberated is directly proportional to the amount of electricity passed. From Faraday’s law, the amount of charge required to liberate 1 mole of H2 is known. Hence, the number of moles of H2 liberated can be calculated from the amount of charge passed. The volume of H2 liberated can be calculated using the molar volume of an ideal gas at STP.
Given, mass of copper deposited = 0.16 g
The molar mass of Cu = 63.55 g/mol
Therefore, the number of moles of Cu deposited = 0.16/63.55 = 0.00252 mol
The equation for the reaction is Cu2+ + 2e- → Cu
Hence, the number of electrons passed = 2 x 0.00252 = 0.00504 C
Calculation of the volume of H2 liberated at STP:
The equation for the reaction is 2H+ + 2e- → H2
From Faraday’s law, the amount of charge required to liberate 1 mole of H2 is 2F or 2 x 96500 C
Therefore, the number of moles of H2 liberated = 0.00504/(2 x 96500) = 2.61 x 10^-5 mol
The molar volume of an ideal gas at STP is 22.4 L/mol
Therefore, the volume of H2 liberated at STP = 2.61 x 10^-5 x 22.4 = 5.84 x 10^-4 L or 0.584 mL
Explanation:
When electricity is passed through CuSO4 solution, copper ions get reduced at the cathode to form copper metal. The amount of copper deposited is directly proportional to the amount of electricity passed. Hence, the mass of copper deposited is used to calculate the amount of electricity passed.
When electricity is passed through acidulated water, hydrogen ions get reduced at the cathode to form hydrogen gas. The amount of hydrogen gas liberated is directly proportional to the amount of electricity passed. From Faraday’s law, the amount of charge required to liberate 1 mole of H2 is known. Hence, the number of moles of H2 liberated can be calculated from the amount of charge passed. The volume of H2 liberated can be calculated using the molar volume of an ideal gas at STP.
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When quantity of electricity is passed through cuso4 solution 0.16 g o...
56cm3
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When quantity of electricity is passed through cuso4 solution 0.16 g of copper gets deposited. If the same quantity of electricity passed through acidulated water. Then the volume of H2 libarateted at STP will be?
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When quantity of electricity is passed through cuso4 solution 0.16 g of copper gets deposited. If the same quantity of electricity passed through acidulated water. Then the volume of H2 libarateted at STP will be? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about When quantity of electricity is passed through cuso4 solution 0.16 g of copper gets deposited. If the same quantity of electricity passed through acidulated water. Then the volume of H2 libarateted at STP will be? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for When quantity of electricity is passed through cuso4 solution 0.16 g of copper gets deposited. If the same quantity of electricity passed through acidulated water. Then the volume of H2 libarateted at STP will be?.
When quantity of electricity is passed through cuso4 solution 0.16 g of copper gets deposited. If the same quantity of electricity passed through acidulated water. Then the volume of H2 libarateted at STP will be? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about When quantity of electricity is passed through cuso4 solution 0.16 g of copper gets deposited. If the same quantity of electricity passed through acidulated water. Then the volume of H2 libarateted at STP will be? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for When quantity of electricity is passed through cuso4 solution 0.16 g of copper gets deposited. If the same quantity of electricity passed through acidulated water. Then the volume of H2 libarateted at STP will be?.
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