Question

Gallium is produced by the electrolysis of a solution made by dissolving gallium oxide in concentrated NaOH(aq). Calculate the amount of Ga(s) that can be deposited from a Ga(III) solution using a current of 0.710 A that flows for 60.0 min.

Answer 1

Answer:

0.00883 mol

Explanation:

Let's consider the reduction reaction of gallium.

Ga³⁺(aq) + 3 e⁻ ⇒ Ga(s)

We can establish the following relationships:

• 1 min = 60 s

• 1 A = 1 C/s

• 1 mole of electrons have a charge of 96486 C (Faraday's constant)

• When 3 moles of electrons circulate, 1 mole of Ga is deposited

The amount of Ga deposited using a current of 0.710 A that flows for 60.0 min is:

$60.0min \times \frac{60s}{1min} \times \frac{0.710C}{s} \times \frac{1mole^{-} }{96486C} \times \frac{1molGa}{3mole^{-}} = 0.00883 molGa$