A current of 0.5 A is passed through molten AlCl3for 40 min the mass o...
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.