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In terms of Faraday, how much electricity is required to produce : (i) From molten CaCl2, 20.0 g of Ca. (ii) From molten Al2O3, 40.0 g of Al.?
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Electricity required for producing Ca and Al
Introduction: Faraday's laws of electrolysis state that the amount of substance produced at an electrode during electrolysis is proportional to the quantity of electricity passed through the electrolyte.
Calculation:
(i) For the production of 1 mole of Ca, 2 moles of electrons are required. The atomic mass of Ca is 40 g/mol. Therefore, the mass of Ca produced is given by:
Mass of Ca = (20.0 g / 40 g/mol) = 0.5 mol
The amount of electricity required to produce 0.5 mol of Ca is given by:
Amount of electricity = 2 × 0.5 × 96500 C/mol = 96500 C
(ii) For the production of 1 mole of Al, 3 moles of electrons are required. The atomic mass of Al is 27 g/mol. Therefore, the mass of Al produced is given by:
Mass of Al = (40.0 g / 27 g/mol) = 1.48 mol
The amount of electricity required to produce 1.48 mol of Al is given by:
Amount of electricity = 3 × 1.48 × 96500 C/mol = 4.26 × 10^5 C
Conclusion: Thus, the amount of electricity required to produce 20.0 g of Ca and 40.0 g of Al is 96500 C and 4.26 × 10^5 C, respectively, according to Faraday's laws of electrolysis.
Introduction: Faraday's laws of electrolysis state that the amount of substance produced at an electrode during electrolysis is proportional to the quantity of electricity passed through the electrolyte.
Calculation:
(i) For the production of 1 mole of Ca, 2 moles of electrons are required. The atomic mass of Ca is 40 g/mol. Therefore, the mass of Ca produced is given by:
Mass of Ca = (20.0 g / 40 g/mol) = 0.5 mol
The amount of electricity required to produce 0.5 mol of Ca is given by:
Amount of electricity = 2 × 0.5 × 96500 C/mol = 96500 C
(ii) For the production of 1 mole of Al, 3 moles of electrons are required. The atomic mass of Al is 27 g/mol. Therefore, the mass of Al produced is given by:
Mass of Al = (40.0 g / 27 g/mol) = 1.48 mol
The amount of electricity required to produce 1.48 mol of Al is given by:
Amount of electricity = 3 × 1.48 × 96500 C/mol = 4.26 × 10^5 C
Conclusion: Thus, the amount of electricity required to produce 20.0 g of Ca and 40.0 g of Al is 96500 C and 4.26 × 10^5 C, respectively, according to Faraday's laws of electrolysis.
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In terms of Faraday, how much electricity is required to produce : (i) From molten CaCl2, 20.0 g of Ca. (ii) From molten Al2O3, 40.0 g of Al.?
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In terms of Faraday, how much electricity is required to produce : (i) From molten CaCl2, 20.0 g of Ca. (ii) From molten Al2O3, 40.0 g of Al.? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about In terms of Faraday, how much electricity is required to produce : (i) From molten CaCl2, 20.0 g of Ca. (ii) From molten Al2O3, 40.0 g of Al.? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for In terms of Faraday, how much electricity is required to produce : (i) From molten CaCl2, 20.0 g of Ca. (ii) From molten Al2O3, 40.0 g of Al.?.
In terms of Faraday, how much electricity is required to produce : (i) From molten CaCl2, 20.0 g of Ca. (ii) From molten Al2O3, 40.0 g of Al.? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about In terms of Faraday, how much electricity is required to produce : (i) From molten CaCl2, 20.0 g of Ca. (ii) From molten Al2O3, 40.0 g of Al.? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for In terms of Faraday, how much electricity is required to produce : (i) From molten CaCl2, 20.0 g of Ca. (ii) From molten Al2O3, 40.0 g of Al.?.
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