Question

How many gram of Cu will be obtained by passing 4.5 A current for 1.5 hours through 1.5 L 0,6 M CuCl2
aqueous solution by dipping inert electrodes ? What will be the molarity of the solution left behind after the
electrolysis ? {At. Wt. of Cu = 63.5 g/mol​

Answer 1

Given info : A 4.5 A current passing through 1.5 L , 0.6 M CuCl2 aqueous solution for 1.5 hrs.

To find : the amount of cu will be obtained and the molarity of solution left behind after the electrolysis.

Solution : charge = current × time

= 4.5 A × 1.5 × 3600 sec

= 24300 C

But we know, 1 mole of electrons is equivalent to 96500 C

So, 24300 C charge = 24300/96500 mol of electrons = 0.25 mol

Here, CuCl2 ⇒Cu²⁺ + 2Cl¯

Cu²⁺ + 2e ¯ ⇒Cu

So, 2 moles of electrons required to deposit 1 mole of Cu.

so, 0.25 mol of electrons deposits 0.25/2 = 0.125 mol of Cu.

Now mass of 0.125 mol of Cu = 0.125 × 63.5 ≈ 8 g

No of moles of CuCl2 = 1.5 × 0.6 = 0.9 mol

No of moles of CuCl2 left behind after the electrolysis = 0.9 - 0.125 = 0.775 mol

So, molarity = 0.775/1.5 = 0.5167 M

Therefore the amount of Cu will be obtained is 8g and molarity of the solution left behind after the electrolysis is 0.5167 M