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 49.6 cm2 and the density of gold is 19.3g/cm3. A current of 3.20 A is applied to a solution that contains gold in the +3 oxidation state.

Calculate the time required to deposit an even layer of gold 1.00×10⁻³ cm thick on the object.

Answer 1

area = 49.6 cm^2

t = 1.00*10^-3 cm

volume = area * t

= 49.6 cm^2 * 1.00*10^-3 cm

= 0.0496 cm^3

density = 19.3 g/cm^3

so,

mass of gold = density * volume

= 19.3 g/cm^3 * 0.0496 cm^3

= 0.957 g

Electrolysis equation is:

Au3+ + 3e- ------> Au

1 mol of Au requires 3 mol of electron

1 mol of electron = 96485 C

So,1 mol of Au requires 289455 C

let us calculate mol of element deposited:

use:

number of mol, n = mass/molar mass

= 0.957/1.97*10^2

= 4.858*10^-3 mol

total charge = mol of element deposited * charge required for 1 mol

= 4.858*10^-3*2.895*10^5

= 1.406*10^3 C

use:

time = Q/i

= 1.406*10^3/3.2

= 4.394*10^2 seconds

= 4.394*10^2/60 min

= 7.324 min

Answer: 7.32 min