Question

On passing H₂S through a solution containing Cu²⁺ and Cd²⁺ ion and excess of KCN solution, only Cd gets precipitated. Explain.

Answer 1

1. Complex Formation with KCN:

• Cu²⁺ reacts readily with CN⁻ ions from KCN to form a highly stable complex ion:
  \[\ce{Cu^2+ + 4CN^- → [Cu(CN)4]^2-}\]
• This complex has a very high stability constant, which significantly reduces the free Cu²⁺ ion concentration in solution.
• On the other hand, Cd²⁺ also forms complexes with CN⁻, but the stability of [Cd(CN)₄]²⁻ is much lower than that of the copper complex. Hence, a relatively higher concentration of free Cd²⁺ ions remains in solution.

2. Precipitation by H₂S:

• On the other hand, Cd²⁺ does not form a stable complex with CN⁻ ions and thus remains as free Cd²⁺ in the solution.
• When H₂S is passed through the solution, H₂S dissociates into HS⁻ and S²⁻ ions:
  \[\ce{H2S ⇌ 2H^+ + S^2-}\]
• The S²⁻ ions can react with free Cd²⁺ ions to form cadmium sulfide (CdS), a yellow precipitate:
  \[\ce{Cd^2+ + S^2- → CdS↓}\]
• Therefore, Cd²⁺ precipitates as CdS, while Cu²⁺ remains in solution as [Cu(CN)₂]²⁻ and does not form a precipitate.