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In the electrolytic extraction of Al at cathode Al metal and at anode F2 gas is liberated. If the molar ratio of Al and F2 is 2 : x, here value of is x
Correct answer is '3'. Can you explain this answer?
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In the electrolytic extraction of Al at cathode Al metal and at anode ...
The molar ratio of Al and F2 is 2 : 3.
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In the electrolytic extraction of Al at cathode Al metal and at anode ...
The electrolytic extraction of aluminum involves the reduction of Al3+ ions at the cathode and the oxidation of F- ions at the anode. This process takes place in a molten mixture of aluminum oxide (Al2O3) and cryolite (Na3AlF6), which acts as a solvent and lowers the melting point of the mixture.
The overall reaction at the cathode can be represented as:
2Al3+ + 6e- -> 2Al
At the anode, the oxidation of F- ions takes place, leading to the liberation of F2 gas:
6F- -> 3F2 + 6e-
From the balanced equations, we can see that for every 2 moles of Al reduced at the cathode, 6 moles of electrons are required. Similarly, for every 6 moles of F- ions oxidized at the anode, 6 moles of electrons are released.
Let's calculate the molar ratio of Al to F2:
For every 2 moles of Al reduced, 6 moles of electrons are required.
For every 3 moles of F2 liberated, 6 moles of electrons are released.
To find the ratio, we can divide the coefficients of Al and F2 by 3:
Al: F2 = 2/3 : 3/3
Al: F2 = 2 : 3
Therefore, the molar ratio of Al to F2 is 2:3. This means that for every 2 moles of aluminum reduced at the cathode, 3 moles of F2 gas are liberated at the anode.
In summary, during the electrolytic extraction of aluminum, Al metal is liberated at the cathode, while F2 gas is liberated at the anode. The molar ratio of Al to F2 is 2:3, meaning that for every 2 moles of Al reduced, 3 moles of F2 gas are produced.
The overall reaction at the cathode can be represented as:
2Al3+ + 6e- -> 2Al
At the anode, the oxidation of F- ions takes place, leading to the liberation of F2 gas:
6F- -> 3F2 + 6e-
From the balanced equations, we can see that for every 2 moles of Al reduced at the cathode, 6 moles of electrons are required. Similarly, for every 6 moles of F- ions oxidized at the anode, 6 moles of electrons are released.
Let's calculate the molar ratio of Al to F2:
For every 2 moles of Al reduced, 6 moles of electrons are required.
For every 3 moles of F2 liberated, 6 moles of electrons are released.
To find the ratio, we can divide the coefficients of Al and F2 by 3:
Al: F2 = 2/3 : 3/3
Al: F2 = 2 : 3
Therefore, the molar ratio of Al to F2 is 2:3. This means that for every 2 moles of aluminum reduced at the cathode, 3 moles of F2 gas are liberated at the anode.
In summary, during the electrolytic extraction of aluminum, Al metal is liberated at the cathode, while F2 gas is liberated at the anode. The molar ratio of Al to F2 is 2:3, meaning that for every 2 moles of Al reduced, 3 moles of F2 gas are produced.
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In the electrolytic extraction of Al at cathode Al metal and at anode F2 gas is liberated. If the molar ratio of Al and F2 is 2 : x, here value of is xCorrect answer is '3'. Can you explain this answer?
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In the electrolytic extraction of Al at cathode Al metal and at anode F2 gas is liberated. If the molar ratio of Al and F2 is 2 : x, here value of is xCorrect answer is '3'. 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 In the electrolytic extraction of Al at cathode Al metal and at anode F2 gas is liberated. If the molar ratio of Al and F2 is 2 : x, here value of is xCorrect answer is '3'. 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 In the electrolytic extraction of Al at cathode Al metal and at anode F2 gas is liberated. If the molar ratio of Al and F2 is 2 : x, here value of is xCorrect answer is '3'. Can you explain this answer?.
In the electrolytic extraction of Al at cathode Al metal and at anode F2 gas is liberated. If the molar ratio of Al and F2 is 2 : x, here value of is xCorrect answer is '3'. 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 In the electrolytic extraction of Al at cathode Al metal and at anode F2 gas is liberated. If the molar ratio of Al and F2 is 2 : x, here value of is xCorrect answer is '3'. 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 In the electrolytic extraction of Al at cathode Al metal and at anode F2 gas is liberated. If the molar ratio of Al and F2 is 2 : x, here value of is xCorrect answer is '3'. Can you explain this answer?.
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