Question

a thin circular-disk earning,5.00 cm in diameter is plated with a coating of gold 0.20 mm thick from an Au+3 bath.

a) how many days does it take to deposit the gold on one side of this earring if the current is 0.010 Ampere (d of Au = 19.3 g/cm3)

b) how many days does it take to deposit the gold on both sides of a pair of these earrings

c) if the price of Au is $320 per troy ounce 31.10 grams), what is the total cost of the gold plating?

Answer 1

The electroplating technique is used for the coating of gold on the circular disk.

The circular disk is of 5cm & we want to have a gold coating of 0.2 mm thick on the disk.

The overall calculation & phenomenon is explained by Faradays Law.

Faradays Law is as below;

W = (I x t x A) / (n x F) ------------------ (1)

Where, W = weight of plated metal in grams.

I = current in coulombs per second or Ampere

t = time in seconds.

A = atomic weight of the metal in grams per mole.

n = valence of the dissolved metal in solution in equivalents per mole.

F = Faraday's constant in coulombs per equivalent. F = 96485.309 coulombs/equivalent

But the data we have is the thickness of the coating we want to have i.e. 0.20 mm

T = (W ) / (rho x S) ------------- (2)

Where, T = thickness in cm.

rho = density in grams per cubic centimeter.

S = surface area of the plated part in square centimeters.

  

A) In this question we want to find out the time require to coat on one side

Surface area of the circular disk of 5 cm will be (pi x r^2) 78.5375 cm2

If we combine equation 1 & 2, then we get

T = (I x t x A) / (n x F x rho x S)

Using this formula & data given in the question we can calculate the "t" in seconds

0.020 = (0.010 x t x 196.96) / (3 x 96485.309 x 19.3 x 78.5375)

t = 4455216.25 sec = 74253.60 min = 1237.56 hours = 51.565 days

B) The time taken to deposit the gold on both sides of the earring will twice of the time taken for depositing on the single side i.e. 51.565 x 2 = 103.13 days

C) The amount of the gold required for the coating can be calculated using the equation 1.

W = (0.010 x (103.13 x 24 x 60 x 60) x 196.96) / (3 x 96485.309)

W = 60.63 grams = 1.9495 troy

So, the total cost required for the gold plating will be $ 623.85 i.e. $ 624

Please fill feel free to give the feedback & any improvements for the answering technique.