Statement I: The $\mathrm{Cu}^{2+}$ ions can be separated from $\mathrm{Cd}^{2+}$ ions in a mixture

Question

Statement I: The $\mathrm{Cu}^{2+}$ ions can be separated from $\mathrm{Cd}^{2+}$ ions in a mixture of them by adding $\mathrm{KCN}_{(\mathrm{aq})}$ and then by passing $\mathrm{H}_{2} \mathrm{~S}_{(\mathrm{g})}$ through it\ Statement II: The $\mathrm{Cu}^{2+}$ ions are reduced to $\mathrm{Cu}^{+}$ by KCN and $\mathrm{Cu}^{+}$then complexes with CN ; but $\mathrm{Cd}^{2+}$ cannot complex with $\mathrm{CN}^{-}$\

Professor PC Thomas Classes · Answer

The $\mathrm{Cu}^{+}$ion from $\mathrm{K}_{3}\left[\mathrm{Cu}(\mathrm{CN})_{4}ight]$ and $\mathrm{Cd}^{2+}$ from $\mathrm{K}_{2}\left[\mathrm{Cd}(\mathrm{CN})_{4}ight]$ But the latter is decomposed to $\mathrm{CdS}_{(\mathrm{s})}$ while $\mathrm{Cu}^{+}$complex does not.\