Q1(f). Role of mercuric chloride in iron titration

Answer in brief

HgCl₂ removes the small excess of SnCl₂ used to reduce Fe(III), preventing that reductant from consuming dichromate.

Key points

• First reduce iron quantitatively: 2Fe³⁺ + Sn²⁺ → 2Fe²⁺ + Sn⁴⁺. A slight excess of SnCl₂ ensures completion.
• Remove the excess: SnCl₂ + 2HgCl₂ → SnCl₄ + Hg₂Cl₂(s). A small white, silky calomel precipitate is acceptable.
• A large excess of Sn²⁺ can reduce Hg₂Cl₂ further to grey metallic Hg, causing interference; the determination should be repeated.
• The remaining Fe²⁺ is titrated: Cr₂O₇²⁻ + 14H⁺ + 6Fe²⁺ → 2Cr³⁺ + 7H₂O + 6Fe³⁺. HgCl₂ is a pretreatment reagent, not the indicator.