For the equilibrium reaction in aqueous medium :

$$ 2 \mathrm{Cu}^{+} \rightleftharpoons \mathrm{Cu}^0+\mathrm{Cu}^{2+} $$

(i) If concentration of copper (I) is reduced, then according to Le-chatelier's principle the equilibrium reaction will shift backward.

(ii) The copper (II) ion reacts with chloride $\left(\mathrm{Cl}^{-}\right)$ion forming copper (III) chloride, which reacts with copper to produce the precipitate of CuCl .

$$ \begin{aligned} & \mathrm{Cu}^{2+}+2 \mathrm{Cl}^{-} \rightarrow \mathrm{CuCl}_2 \\\\ & \mathrm{CuCl}_2+\mathrm{Cu} \rightarrow 2 \mathrm{CuCl} \end{aligned} $$

Since, the end product involves consumption of copper (I) ion to form CuCl , hence, reactive shifts backward.

(iii) The copper (II) ion also reacts with cyanide $\left(\mathrm{CN}^{-}\right)$ion forming copper (II) cyanide which on further reactions form $\left[\mathrm{Cu}(\mathrm{CN})_4\right]^{3-}$ with copper in (I) oxidation state shifting reaction backward.

$$ \begin{aligned} \mathrm{Cu}^{2+}+2 \mathrm{CN}^{-} & \rightarrow \mathrm{Cu}(\mathrm{CO})_2 \\\\ 2 \mathrm{Cu}(\mathrm{CN})_2 & \rightarrow 2 \mathrm{CuCN}+(\mathrm{CN})_2 \\\\ \mathrm{CuCN}+3 \mathrm{CN}^{-} & \rightarrow\left[\mathrm{Cu}(\mathrm{CN})_4\right]^{3-} \end{aligned} $$

(iv) Reaction of copper (II) ion with thiocyanide $\left(\mathrm{SCN}^{-}\right)$gives a complex $\left(\mathrm{Cu}(\mathrm{SCN})_4\right]^{3-}$ with copper in +1 oxidation state. This shifts reaction backward.

$$ \mathrm{Cu}^{2+}+4 \mathrm{SCN}^{-} \rightarrow\left[\mathrm{Cu}(\mathrm{SCN})_4\right]^{3-} $$

Hence, cyanide, thiocyanide and chloride forms complexes with copper in I oxidation state. Hence, reaction shifts backward to generate more copper (I) ions.