A current of 0.5 A is passed through molten AlCl3for 40 min the mass of aluminium deposited at the cathode is _____g (Al = 27)Correct answer is between '0.110,0.115'. Can you explain this answer?

Question

Answer

To calculate the mass of aluminum deposited at the cathode during electrolysis, we can use Faraday's laws of electrolysis. Here’s a detailed breakdown:
1. Understanding the Reaction
- In the electrolysis of molten AlCl3, aluminum ions (Al³⁺) gain electrons to form aluminum metal:
$$ \text{Al}^{3+} + 3e^- \rightarrow \text{Al} $$
2. Determine Charge Passed
- To find out how much charge is passed, use the formula:
$$ Q = I \times t $$
where:
- $ I = 0.5 \, A $ (current)
- $ t = 40 \, \text{min} = 40 \times 60 = 2400 \, \text{s} $
- Calculating the charge:
$$ Q = 0.5 \, A \times 2400 \, s = 1200 \, C $$
3. Calculate Moles of Electrons
- Using Faraday's constant ($ F \approx 96500 \, C/mol $), calculate the moles of electrons:
$$ n_{e^-} = \frac{Q}{F} = \frac{1200}{96500} \approx 0.01242 \, mol $$
4. Relate Moles of Aluminum to Moles of Electrons
- From the reaction, 3 moles of electrons are required to deposit 1 mole of aluminum:
$$ n_{Al} = \frac{n_{e^-}}{3} = \frac{0.01242}{3} \approx 0.00414 \, mol $$
5. Calculate Mass of Aluminum
- Finally, using the molar mass of aluminum (27 g/mol):
$$ \text{mass} = n_{Al} \times \text{molar mass} = 0.00414 \times 27 \approx 0.11178 \, g $$
Thus, the mass of aluminum deposited at the cathode is approximately 0.112 g, which falls within the specified range of 0.110 to 0.115 g.

Similar IIT JAM Doubts

An electric currentof 0.965 ampere is passed for 2000 seconds through a solution containing [Cu(CH3CN)4]+ and metallic copper is deposited at the cathode. The amount of Cu deposited is

Aluminium (Al) can occur in both tetrahedral and octahedral co- ordinations in silicates. The amount of octahedral Al in a pyroxene crystal of composition MG1.4 Fe0.4 Al0.4 Si1.8 O6 is ____ (give answer in one decimal plac e). Correct answer is '0.2'. Can you explain this answer?

(Ag ) e==Ag, Cu 2 2e =Cu, Eo= 0.50 V; Eo = 0.34 V A 100 ml solution is 1080 mg with respect to Agt and 635 mg with respect to Cu2. If 0.1mg Ag left in the solution is considered to be the complete deposition of Ag , the cathode potential, so that no copper is deposited during the process is?

The statement that is correct for both electrochemical (galvani c) cells and electrolytic cell is

Top Courses for IIT JAM

Suggested Free Tests

Related Content

Schools

Competitive Examinations

Quick Access Links

Technical Exams