Question

If 12.3 g of Cu is deposited at the cathode of an electrolytic cell after 5.50 h, what was the current used?​

Answer 1

1.88 A

Step-by-step explanation:

Let's consider the reduction of copper in an electrolytic cell.

Cu²⁺ + 2 e⁻ ⇒ Cu

We can calculate the charge used to deposit 12.3 g of Cu using the following relations.

• The molar mass of Cu is 63.55 g/mol.

• 1 mole of Cu is deposited when 2 moles of electrons circulate.

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

The charge used is:

`12.3 g * (1 molCu)/(63.55gCu) * (2molElectron)/(1molCu) * (96486C)/(1molElectron) = 3.73 * 10^(4) C`

We can convert 5.50 h to seconds using the conversion factor 1 h = 3600 s.

5.50 h × 3600 s/1 h = 1.98 × 10⁴ s

The current used is:

I = q/t = 3.73 × 10⁴ C/1.98 × 10⁴ s = 1.88 A