Example Question - 3

Question: Explain the different products formed when solid sodium chloride and solid sodium iodide react with concentrated sulfuric acid.Sodium chloride: only an acid-base reaction occurs, giving steamy fumes of hydrogen chloride. NaCl + H₂SO₄ → NaHSO₄ + HCl. Chloride ions are not strong enough reducing agents to reduce the sulfuric acid.Sodium iodide: HI forms first, then iodide ions reduce the sulfuric acid. Sulfur is reduced to SO₂ (+4), then to S (0) and finally to H₂S (-2). Observations include purple fumes or a black solid of iodine, a yellow solid of sulfur and a bad egg smell of hydrogen sulfide.For example: 8H⁺ + 8I^- + H₂SO₄ → 4I₂ + H₂S + 4H₂OThe reducing ability of the halide ions increases down the group because the ions get larger and the outer electrons are further from the nucleus and more shielded, so they are lost more easily.