The oxidation state of I may be obtained by keeping the solution acidic (a reducing environment). E.g take Cu2Cl2 add some conc HCl to the solution prepared. ( I am not sure if this answer helps you as it is not clear what you intend to use it for)
Cuprous salts are not very soluble without some complexing agent (e.g. ammonium). They can be prepared from the corresponding Cu(II) salt by reduction with metallic copper. In contact with atmospheric oxygen they quickly oxidize back to Cu(II).
For copper-catalysed click chemistry reactions such as the Huisgen cycloaddition, the copper(I) catalyst is often prepared by using a reductant such as ascorbic acid (vitamin C). Perhaps you could also make use of this approach.
Forming copper(I) complexes (other than the one with water as a ligand) also stabilizes the copper(I) oxidation state. For example, both [Cu(NH3)2]+ and [CuCl2]- are copper(I) complexes which don't disproportionate.
The chlorine-containing complex is formed if copper(I) oxide is dissolved in concentrated hydrochloric acid. You can think of this happening in two stages. First, you get copper(I) chloride formed:
Cu2O(s)+2HCl(aq)→2CuCl(s)+H2O(l)
But in the presence of excess chloride ions from the HCl, this reacts to give a stable, soluble copper(I) complex.
CuCl(s)+Cl−(aq)→[CuCl2]−(aq)
You can get the white precipitate of copper(I) chloride (mentioned above) by adding water to this solution. This reverses the last reaction by stripping off the extra chloride ion. Remember Cu(I) solution usually clear in color.
If you need to use copper (I) solution for cell studies, i suggest you to put the copper (I) (for example CuCl) in an aqueous solution containing ascorbate. The ascorbate is a natural reducing agent for copper in cells.
Cu(I) can be prepared in the present of SO3(2-), meaning that you can prepare H2SO3 by addition of HCl to Na2SO3 and solve CuCl2 to H2SO3 solution. then a white precipitate (CuCl) is formed. remember that CuCl is unstable in exposure to air so keep it under H2SO3 solution.