Question

Determine the time required to obtain a Cu deposit of thickness 1 μm when electrodeposition is done at 4 A and 6 A, respectively. The surface area of the substrate is 314 cm2.

Answer 1

We have the following data:

Thickness:

Surface area:

A = 314 cm²

Current:

I₁ = 4 A and I₂ = 6 A.

The volume of Cu deposit can be calculated as:

V = A x h = (314 cm² ) x (1 x 10^-4 cm) = 3.14 x 10^-2 cm³

From the data available in literature, the density of copper can be taken as 8.960 g/cm³.

Thus, the amount of copper deposited will be:

m = density x volume = 8.96 x 3.14 x 10^-2 = 0.2813 g = (0.2813/63.5) = 4.43 x 10^-3 mol

The deposition of copper can be represented by the equation:

Thus, for one mole of Cu deposition, 2 moles of electrons are required.

Therefore, for 4.43 x 10^-3 mol of Cu deposit, number of moles of electrons required is

n _e = 4.43 x 10^-3 x 2 mol = 8.86 x 10^-3 mol

And Faraday's constant is F= 96500 C mole ^-1

Therefore, the amount of charge required for Cu deposit is,

Q = n _e x F = 8.86 x 10^-3 mol x 96500 C mole ^-1 = 854.99 C

Now, for current I₁ = 4 A, time required for deposition is

Similarly, for I₂ = 6 A, time required for deposition is