Question

Calculate the current in amperes required to produce 18.0g of aluminium in 1.50hrs (al=27.og )

Answer 1

35.7 A

Step-by-step 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)

`18.0gAl * (1molAl)/(27.0gAl) * (3mole^(-) )/(1molAl) * (96486C)/(1mole^(-) ) = 1.93 * 10^(5) C`

1.93 × 10⁵ C circulate in 1.50 hours. The intensity is:

`I = (1.93 * 10^(5) C)/(1.50h) * (1h)/(3600s) = 35.7 A`