Question

A constant current was passed through a solution of AuCl4
-
ions between gold electrodes. After a period of 10.00 minutes the cathode increased in weight by 1.314grams.
How much charge was passed and what was the current ?

Answer 1

Solution.

We write the Faraday law for electrolysis:

$$m = \frac{Q \times M}{F \times z}$$

$$Q = I \times \Delta t$$

$$Q = \frac{m \times F \times z}{M}$$

F = 96484 C/mole

z = +3, because the charge of the gold atom in this ion is +3.

M(Au) = 196.967 g/mole

Q = 1930.98 C

$$I = \frac{Q}{\Delta t}$$

I = 3.22 A

Answer:

Q = 1930.98 C

I = 3.22 A