During the electrolysis of the Cucl2 solution, 6.4 g of the substance was formed at the cathode.

During the electrolysis of the Cucl2 solution, 6.4 g of the substance was formed at the cathode. How many liters and what substance was formed at the anode, if its yield is 85% of the theoretical?

Solution:
CuCl2 (solution) = (Decomposition) Cu (+2) ++ 2Cl⁻
K (-) Cu²⁺ + 2e⁻ = Cu
A (+) 2Cl⁻-2e⁻ = Cl₂
CuCl2 = Cu + Cl2
Let us find its volume of theoretical output from the proportion, since gas – chlorine – was released at the anode:
1) V theoretical (Cl2) = m (Cu) * Vm / M (Cu) = 6.4 g * 22.4 l / mol / 64 g / mol = 2.24 l.
Let’s find the volume of the practical yield of chlorine:
2) V practical (Cl2) = V (Cl2) * 85% / 100% = 2.24 l * 85% / 100% = 1.904 l.
Answer: 1.904 liters were formed at the anode. chlorine.