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

0.00883 mol

Step-by-step 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 * (60s)/(1min) * (0.710C)/(s) * (1mole^(-) )/(96486C) * (1molGa)/(3mole^(-)) = 0.00883 molGa`