Question

Copper can be electroplated at the cathode of an electrolysis cell by the half-reaction. Cu2+(aq)+2e−→Cu(s)Cu2+(aq)+2e−→Cu(s)

Part A How much time would it take for 338 mgmg of copper to be plated at a current of 7.1 AA ?

Answer 1

Given:

current (I) =7.1 A = 7.1 C/s

Time (t) =?

F = 96485 C/ mol

Mass of Cu deposited = 338 mg = 0.338 g

We have relation, No. of moles = Mass / Molar Mass

No. of moles of Cu = 0.338 g / ( 63.54 g / mol ) = 0.005319 mol

Consider reaction, Cu ²⁺ (aq) + 2 e ^- Cu (s)

According to reaction , for deposition 1 mole Cu 2 mole electrons are required.

Moles of Cu deposited= 1/2 (moles of electrons)

Moles of electrons used in the reaction = 2 ( Moles of Cu deposited ) = 2 ( 0.005319 mol ) = 0.01064 mol

We have relation, Moles of electrons = I t / F

time (t) = F Moles of electrons / I

time (t) required to deposit 338 mg Cu = 96485 C / mol 0.01064 mol / 7.1 C /s

time (t) required to deposit 338 mg Cu = 144.56 s

ANSWER : time (t) required to deposit 338 mg Cu = 1.4 10 ² s