What mass of aluminum can be deposited by the passage of a constant current of 5.00 A through a Al(NO₃)₃(aq)solution for 2.00 hours?

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Quizplus · Accepted Answer

The mass of aluminum deposited can be calculated using Faraday's laws of electrolysis. The molar mass of aluminum (Al) is 26.98 g/mol, and it has a valency of 3, meaning 3 moles of electrons are required to deposit 1 mole of aluminum. The total charge (Q) passed through the solution is current (I) times time (t), which is $5.00 \, \text{A} \times 2.00 \times 3600 \, \text{s} = 36000 \, \text{C}$. Since 1 mole of electrons corresponds to 96500 C (Faraday constant), the moles of electrons used are $36000 \, \text{C} / 96500 \, \text{C/mol} = 0.373 \, \text{mol}$. Since it takes 3 moles of electrons to deposit 1 mole of Al, the moles of Al deposited are $0.373 \, \text{mol} / 3 = 0.124 \, \text{mol}$. Therefore, the mass of Al deposited is $0.124 \, \text{mol} \times 26.98 \, \text{g/mol} = 3.36 \, \text{g}$.

What Mass of Aluminum Can Be Deposited by the Passage