Answer:
35.7 A
Explanation:
Let's consider the reduction half-reaction of Al³⁺.
Al³⁺ + 3 e⁻ ⇒ Al
We will calculate the charge required to produce 18.0 g of Al using the following conversion factors.
- 1 mole of Al has a mass of 27.0 g
- 1 mole of Al is formed upon the circulation of 3 moles of electrons
- 1 mole of electrons has a charge of 96486 C (Faraday's constant)
[tex]18.0gAl \times \frac{1molAl}{27.0gAl} \times \frac{3mole^{-} }{1molAl} \times \frac{96486C}{1mole^{-} } = 1.93 \times 10^{5} C[/tex]
1.93 × 10⁵ C circulate in 1.50 hours. The intensity is:
[tex]I = \frac{1.93 \times 10^{5} C}{1.50h} \times \frac{1h}{3600s} = 35.7 A[/tex]
