What is the chemical reaction for the formation of diaquabis(ethylenediamine) copper(II) iodide from ethylenediamine, potassium iodide and water?

The chemical equation for the formation of diaquabis(ethylenediamine)copper(II) iodide is

(CH₃COO)₂Cu·H₂O(aq) + 2H₂NCH₂CH₂NH₂(aq) + 2KI(aq) + H₂O(l) → Cu(H₂NCH₂CH₂NH₂)₂(H₂O)₂I₂(s) + 2CH₃COOK(aq)

The product is a purple ionic solid, Cu(en)₂(H₂O)₂²⁺, 2I⁻.

Octahedral diaquabis(ethylenediamine) complexes can exist in two isomeric forms.

In the trans isomer, the bonds to the two water molecules are at 180° to each other.

In the cis isomer, the bonds to the two water molecules are at 90° to each other.

I am not an inorganic chemist, so I am not sure, but I believe the compound prepared in this procedure is the trans isomer.