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On passing electric current through moten AlCl3, 11.2 litre of Cl2 is liberated at N.T.P. at anode. The quantity of aluminium deposited at cathode in g will be :- (At. wt. of Al = 27)
Correct answer is '9'. Can you explain this answer?
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On passing electric current through moten AlCl3, 11.2 litre of Cl2is l...
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Understanding the Electrolysis of AlCl3
When electric current passes through molten aluminum chloride (AlCl3), it undergoes electrolysis, leading to the liberation of chlorine gas (Cl2) at the anode and deposition of aluminum (Al) at the cathode.
Step 1: Calculate Moles of Cl2 Produced
- At N.T.P (Normal Temperature and Pressure), 1 mole of a gas occupies 22.4 liters.
- Volume of Cl2 liberated = 11.2 liters.
Calculating Moles:
- Moles of Cl2 = Volume / Molar Volume = 11.2 L / 22.4 L/mol = 0.5 mol.
Step 2: Chlorine Production Reaction
- The reaction at the anode can be represented as:
\[ 2Cl^- \rightarrow Cl_2 + 2e^- \]
- From the equation, 1 mole of Cl2 corresponds to 2 moles of electrons (2F).
Step 3: Relating Electrons to Aluminum Deposition
- The reduction of Al occurs at the cathode:
\[ Al^{3+} + 3e^- \rightarrow Al \]
- This shows that to produce 1 mole of Al, 3 moles of electrons are required.
Step 4: Calculate Moles of Al Deposited
- If 0.5 moles of Cl2 are produced, then the moles of electrons involved = 0.5 mol Cl2 × 2 = 1 mol of electrons.
- Moles of Al produced = Moles of electrons / 3 = 1 mol / 3 = 0.33 mol.
Step 5: Calculate Mass of Aluminum Deposited
- Molar mass of Al = 27 g/mol.
Calculating Mass:
- Mass of Al = Moles × Molar Mass = 0.33 mol × 27 g/mol = 8.91 g, which can be approximated to 9 g.
Conclusion
- Therefore, the quantity of aluminum deposited at the cathode is approximately 9 grams.
When electric current passes through molten aluminum chloride (AlCl3), it undergoes electrolysis, leading to the liberation of chlorine gas (Cl2) at the anode and deposition of aluminum (Al) at the cathode.
Step 1: Calculate Moles of Cl2 Produced
- At N.T.P (Normal Temperature and Pressure), 1 mole of a gas occupies 22.4 liters.
- Volume of Cl2 liberated = 11.2 liters.
Calculating Moles:
- Moles of Cl2 = Volume / Molar Volume = 11.2 L / 22.4 L/mol = 0.5 mol.
Step 2: Chlorine Production Reaction
- The reaction at the anode can be represented as:
\[ 2Cl^- \rightarrow Cl_2 + 2e^- \]
- From the equation, 1 mole of Cl2 corresponds to 2 moles of electrons (2F).
Step 3: Relating Electrons to Aluminum Deposition
- The reduction of Al occurs at the cathode:
\[ Al^{3+} + 3e^- \rightarrow Al \]
- This shows that to produce 1 mole of Al, 3 moles of electrons are required.
Step 4: Calculate Moles of Al Deposited
- If 0.5 moles of Cl2 are produced, then the moles of electrons involved = 0.5 mol Cl2 × 2 = 1 mol of electrons.
- Moles of Al produced = Moles of electrons / 3 = 1 mol / 3 = 0.33 mol.
Step 5: Calculate Mass of Aluminum Deposited
- Molar mass of Al = 27 g/mol.
Calculating Mass:
- Mass of Al = Moles × Molar Mass = 0.33 mol × 27 g/mol = 8.91 g, which can be approximated to 9 g.
Conclusion
- Therefore, the quantity of aluminum deposited at the cathode is approximately 9 grams.
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On passing electric current through moten AlCl3, 11.2 litre of Cl2is liberated at N.T.P. at anode. The quantity of aluminium deposited at cathode in g will be :- (At. wt. of Al = 27)Correct answer is '9'. Can you explain this answer?
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On passing electric current through moten AlCl3, 11.2 litre of Cl2is liberated at N.T.P. at anode. The quantity of aluminium deposited at cathode in g will be :- (At. wt. of Al = 27)Correct answer is '9'. 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 On passing electric current through moten AlCl3, 11.2 litre of Cl2is liberated at N.T.P. at anode. The quantity of aluminium deposited at cathode in g will be :- (At. wt. of Al = 27)Correct answer is '9'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for On passing electric current through moten AlCl3, 11.2 litre of Cl2is liberated at N.T.P. at anode. The quantity of aluminium deposited at cathode in g will be :- (At. wt. of Al = 27)Correct answer is '9'. Can you explain this answer?.
On passing electric current through moten AlCl3, 11.2 litre of Cl2is liberated at N.T.P. at anode. The quantity of aluminium deposited at cathode in g will be :- (At. wt. of Al = 27)Correct answer is '9'. 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 On passing electric current through moten AlCl3, 11.2 litre of Cl2is liberated at N.T.P. at anode. The quantity of aluminium deposited at cathode in g will be :- (At. wt. of Al = 27)Correct answer is '9'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for On passing electric current through moten AlCl3, 11.2 litre of Cl2is liberated at N.T.P. at anode. The quantity of aluminium deposited at cathode in g will be :- (At. wt. of Al = 27)Correct answer is '9'. Can you explain this answer?.
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