Question

A thin layer of gold can be applied to another material by an electrolytic process. The surface area of an object to be gold plated is 49.9cm2 and the density of gold is 19.3g/cm3. A current of 3.35A is applied to a solution that contains gold in the +3 oxidation state.

Calculate the time (in sec) required to deposit an even layer of gold 1.20

Answer 1

A thin layer of gold can be applied to another material by an electrolytic process. The surface area of an object to be gold plated is 49.9cm2 and the density of gold is 19.3g/cm3. A current of 3.35A is applied to a solution that contains gold in the +3 oxidation state. Calculate the time (in sec) required to deposit an even layer of gold 1.20

A = 49.9 cm2

D = 19.3 g/cm3

I = 3.35 A = 3.35 C/s

Au3+ state...

height required = 1.2 cm

First, calculate total mass required

Volume = h*A = 1.2*49.9 cm3 = 59.88 cm3

Now, calculate mass of gold:

D = mass/Volume

mass = D*V = 59.88 cm3 *19.3 g/cm3= 1,155.684 g required

mol of gold = mass/MW = 1155.684/196.96655 = 5.867 mol of gold

now, 1 mol of Au requires 3 mol of e- due to +3 state

so

5.867 mol o fau = 3x5.867 = 17.601 mol of e- required

Total charge:

1 mol of e- = 96500 C (according to farady)

17.601 mol of e- = 17.601x96500 = 1,698,496.5 C

Calcualte time:

I = C/t

t = C/I

t = (1,698,496.5)/3.35

t = 507,013.88 seconds = 507,013.88/60 mins= 8,450.231 min = 140.837 h