Question

How long (hrs) must a current of 2.5 amps flow to produce 10 g of Cu from a 0.1500 M solution of CuCl₂?

PLEASE SHOW WORK AND EXPLAIN

Answer 1

CuCl2 --- > Cu+2 and 2Cl-

Cu+2 + 2e- --- > Cu(s)

I = 2.5 A = 2.5 C/s

Charge of an electron = 1.6*10^-19 C

If we want 10 g of Cu

This is

MW Cu = 63.5 g/gmol

n = mass/MW = 10/63.5 = 0.1575 mol of Cu

1 mol of Cu needs 2 mol of e- to form Cu(s)

Therefore 0.1575 mol of Cu need 2*0.1575 = 0.316 mol of electrons are needed

Calculate the amount of electrons

1 mol of e - = 6.022*10^23

0.316 mol of e- = 1.90*10^23 electrons

Let us calculate the charge then

1 e - = 1.6*10^-19 C

1.90*10^23 e = (1.90*10^23)(1.6*10^-19) C = 30400 C

we can provide 2.5 C/s (from I = 2.5 A)

Therefore....

time needed for that charge is:

30400/2.5 = seconds

time = 12,160 seconds

time = 202.7 min

time = 3.37 hrs