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If 270.0 g of water is electrolysed during an experiment performed by miss abhilasha with 75% current efficiency then
- a)168 L of O2 (g) will be evolved at anode at 1 atm & 273 K
- b)Total 504 L gases will be produced at 1 atm & 273 K.
- c)336 L of H2(g) will be evolved at anode at 1 atm & 273 K
- d)45 F electricity will be consumed
Correct answer is option 'A,B'. Can you explain this answer?
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Verified Answer
If 270.0 g of water is electrolysed during an experiment performed by ...
Cell reaction
equivalent = 15 × 2 = 30 equivalent
Total volume of gas produced
(c) H2 produced = 30 × 22.4 = 336 L
at cathode current efficilency = 75%
(d) for 30 F electricity consumed
equivalent = 15 × 2 = 30 equivalent
Total volume of gas produced
(c) H2 produced = 30 × 22.4 = 336 L
at cathode current efficilency = 75%
(d) for 30 F electricity consumed
Most Upvoted Answer
If 270.0 g of water is electrolysed during an experiment performed by ...
Electrolysis of Water Experiment
Given:
Mass of water (m) = 270.0 g
Current efficiency (η) = 75%
To find:
a) Volume of O2 evolved
b) Total volume of gases produced
c) Volume of H2 evolved
d) Electrical energy consumed
Solution:
1. Calculate the number of moles of water electrolyzed:
Molar mass of water = 18.015 g/mol
Number of moles of water = Mass of water/Molar mass of water
= 270.0 g/18.015 g/mol
= 14.99 mol
2. Calculate the number of moles of electrons that are required for the electrolysis of water:
2H2O → 2H2 + O2
2 moles of electrons are required to produce 1 mole of O2
4 moles of electrons are required to produce 2 moles of H2
Number of moles of electrons required = 4 × Number of moles of water
= 4 × 14.99 mol
= 59.96 mol
3. Calculate the number of moles of O2 evolved:
Using Faraday's Law:
1 Faraday (F) = 96500 C/mol of electrons
Charge passed (Q) = Current (I) × Time (t)
Number of Faradays (n) = Charge passed (Q) / (1 F)
= I × t / (96500 C)
Number of moles of O2 evolved = Number of Faradays (n) × (1 mole of O2/4 moles of electrons)
= I × t / (96500 C) × (1 mole of O2/4 moles of electrons)
= I × t / (38600 C/mol) × (0.5 mol of O2/mol of electrons)
4. Calculate the volume of O2 evolved:
Volume of O2 (V) = Number of moles of O2 × Molar volume of gas at STP
= Number of moles of O2 × 22.4 L/mol
= I × t / (38600 C/mol) × (0.5 mol of O2/mol of electrons) × 22.4 L/mol
= I × t / 1723.2 C/L
Volume of O2 evolved = I × t / (1723.2 C/L) at 1 atm and 273 K
5. Calculate the volume of H2 evolved:
Using the same method as above, the volume of H2 evolved can be calculated as:
Volume of H2 evolved = I × t / (1723.2 C/L) × 2 L/mol at 1 atm and 273 K
6. Calculate the total volume of gases produced:
Total volume of gases produced = Volume of H2 evolved + Volume of O2 evolved
7. Calculate the electrical energy consumed:
Electrical energy consumed = Voltage (V) × Charge passed (Q)
= Voltage (V) × Current (I) × Time (t)
Therefore, the correct options are A and B.
Given:
Mass of water (m) = 270.0 g
Current efficiency (η) = 75%
To find:
a) Volume of O2 evolved
b) Total volume of gases produced
c) Volume of H2 evolved
d) Electrical energy consumed
Solution:
1. Calculate the number of moles of water electrolyzed:
Molar mass of water = 18.015 g/mol
Number of moles of water = Mass of water/Molar mass of water
= 270.0 g/18.015 g/mol
= 14.99 mol
2. Calculate the number of moles of electrons that are required for the electrolysis of water:
2H2O → 2H2 + O2
2 moles of electrons are required to produce 1 mole of O2
4 moles of electrons are required to produce 2 moles of H2
Number of moles of electrons required = 4 × Number of moles of water
= 4 × 14.99 mol
= 59.96 mol
3. Calculate the number of moles of O2 evolved:
Using Faraday's Law:
1 Faraday (F) = 96500 C/mol of electrons
Charge passed (Q) = Current (I) × Time (t)
Number of Faradays (n) = Charge passed (Q) / (1 F)
= I × t / (96500 C)
Number of moles of O2 evolved = Number of Faradays (n) × (1 mole of O2/4 moles of electrons)
= I × t / (96500 C) × (1 mole of O2/4 moles of electrons)
= I × t / (38600 C/mol) × (0.5 mol of O2/mol of electrons)
4. Calculate the volume of O2 evolved:
Volume of O2 (V) = Number of moles of O2 × Molar volume of gas at STP
= Number of moles of O2 × 22.4 L/mol
= I × t / (38600 C/mol) × (0.5 mol of O2/mol of electrons) × 22.4 L/mol
= I × t / 1723.2 C/L
Volume of O2 evolved = I × t / (1723.2 C/L) at 1 atm and 273 K
5. Calculate the volume of H2 evolved:
Using the same method as above, the volume of H2 evolved can be calculated as:
Volume of H2 evolved = I × t / (1723.2 C/L) × 2 L/mol at 1 atm and 273 K
6. Calculate the total volume of gases produced:
Total volume of gases produced = Volume of H2 evolved + Volume of O2 evolved
7. Calculate the electrical energy consumed:
Electrical energy consumed = Voltage (V) × Charge passed (Q)
= Voltage (V) × Current (I) × Time (t)
Therefore, the correct options are A and B.
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If 270.0 g of water is electrolysed during an experiment performed by miss abhilasha with 75% current efficiency thena)168 L of O2(g) will be evolved at anode at 1 atm & 273 Kb)Total 504 L gases will be produced at 1 atm & 273 K.c)336 L of H2(g) will be evolved at anode at 1 atm & 273 Kd)45 F electricity will be consumedCorrect answer is option 'A,B'. Can you explain this answer?
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If 270.0 g of water is electrolysed during an experiment performed by miss abhilasha with 75% current efficiency thena)168 L of O2(g) will be evolved at anode at 1 atm & 273 Kb)Total 504 L gases will be produced at 1 atm & 273 K.c)336 L of H2(g) will be evolved at anode at 1 atm & 273 Kd)45 F electricity will be consumedCorrect answer is option 'A,B'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about If 270.0 g of water is electrolysed during an experiment performed by miss abhilasha with 75% current efficiency thena)168 L of O2(g) will be evolved at anode at 1 atm & 273 Kb)Total 504 L gases will be produced at 1 atm & 273 K.c)336 L of H2(g) will be evolved at anode at 1 atm & 273 Kd)45 F electricity will be consumedCorrect answer is option 'A,B'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for If 270.0 g of water is electrolysed during an experiment performed by miss abhilasha with 75% current efficiency thena)168 L of O2(g) will be evolved at anode at 1 atm & 273 Kb)Total 504 L gases will be produced at 1 atm & 273 K.c)336 L of H2(g) will be evolved at anode at 1 atm & 273 Kd)45 F electricity will be consumedCorrect answer is option 'A,B'. Can you explain this answer?.
If 270.0 g of water is electrolysed during an experiment performed by miss abhilasha with 75% current efficiency thena)168 L of O2(g) will be evolved at anode at 1 atm & 273 Kb)Total 504 L gases will be produced at 1 atm & 273 K.c)336 L of H2(g) will be evolved at anode at 1 atm & 273 Kd)45 F electricity will be consumedCorrect answer is option 'A,B'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about If 270.0 g of water is electrolysed during an experiment performed by miss abhilasha with 75% current efficiency thena)168 L of O2(g) will be evolved at anode at 1 atm & 273 Kb)Total 504 L gases will be produced at 1 atm & 273 K.c)336 L of H2(g) will be evolved at anode at 1 atm & 273 Kd)45 F electricity will be consumedCorrect answer is option 'A,B'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for If 270.0 g of water is electrolysed during an experiment performed by miss abhilasha with 75% current efficiency thena)168 L of O2(g) will be evolved at anode at 1 atm & 273 Kb)Total 504 L gases will be produced at 1 atm & 273 K.c)336 L of H2(g) will be evolved at anode at 1 atm & 273 Kd)45 F electricity will be consumedCorrect answer is option 'A,B'. Can you explain this answer?.
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If 270.0 g of water is electrolysed during an experiment performed by miss abhilasha with 75% current efficiency thena)168 L of O2(g) will be evolved at anode at 1 atm & 273 Kb)Total 504 L gases will be produced at 1 atm & 273 K.c)336 L of H2(g) will be evolved at anode at 1 atm & 273 Kd)45 F electricity will be consumedCorrect answer is option 'A,B'. Can you explain this answer?, a detailed solution for If 270.0 g of water is electrolysed during an experiment performed by miss abhilasha with 75% current efficiency thena)168 L of O2(g) will be evolved at anode at 1 atm & 273 Kb)Total 504 L gases will be produced at 1 atm & 273 K.c)336 L of H2(g) will be evolved at anode at 1 atm & 273 Kd)45 F electricity will be consumedCorrect answer is option 'A,B'. Can you explain this answer? has been provided alongside types of If 270.0 g of water is electrolysed during an experiment performed by miss abhilasha with 75% current efficiency thena)168 L of O2(g) will be evolved at anode at 1 atm & 273 Kb)Total 504 L gases will be produced at 1 atm & 273 K.c)336 L of H2(g) will be evolved at anode at 1 atm & 273 Kd)45 F electricity will be consumedCorrect answer is option 'A,B'. Can you explain this answer? theory, EduRev gives you an
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