3. The reaction of halide salts with concentrated sulphuric acid. Explanation of differing reducing power of halides A reducing agent donates electrons. The reducing power of the halides increases down group 7 They have a greater tendency to donate electrons. This is because as the ions get bigger it is easier for the outer electrons to be given away as the pull from the nucleus on them becomes smaller. The Halides show increasing power as reducing agents as one goes down the group. This can be clearly demonstrated in the various reactions of the solid halides with concentrated sulphuric acid. Know the equations and observations of these reactions very well. F- and Clions are not strong enough reducing agents to reduce the S in H2SO4 . No redox reactions occur. Only acid-base reactions occur. Fluoride and Chloride NaF(s) + H2SO4 (l) NaHSO4 (s) + HF(g) Observations: White steamy fumes of HF are evolved. NaCl(s) + H2SO4 (l) NaHSO4 (s) + HCl(g) Observations: White steamy fumes of HCl are evolved. These are acid –base reactions and not redox reactions. H2SO4 plays the role of an acid (proton donor). Br- ions are stronger reducing agents than Cl- and F- and after the initial acidbase reaction reduce the Sulphur in H2SO4 from +6 to + 4 in SO2 Bromide Acid- base step: NaBr(s) + H2SO4 (l) NaHSO4 (s) + HBr(g) Redox step: 2HBr + H2SO4 Br2 (g) + SO2 (g) + 2H2O(l) Observations: White steamy fumes of HBr are evolved. Red fumes of Bromine are also evolved and a colourless, acidic gas SO2 Ox ½ equation 2Br – Br2 + 2eRe ½ equation H2SO4 + 2 H+ + 2 e- → SO2 + 2 H2O Iodide I- ions are the strongest halide reducing agents. They can reduce the Sulphur from +6 in H2SO4 to + 4 in SO2 , to 0 in S and -2 in H2S. NaI(s) + H2SO4 (l) NaHSO4 (s) + HI(g) 2HI + H2SO4 I2 (s) + SO2 (g) + 2H2O(l) 6HI + H2SO4 → 3 I2 + S (s) + 4 H2O (l) 8HI + H2SO4 4I2 (s) + H2S(g) + 4H2O(l) Observations: White steamy fumes of HI are evolved. Black solid and purple fumes of Iodine are also evolved A colourless, acidic gas SO2 A yellow solid of Sulphur H2S (Hydrogen Sulphide), a gas with a bad egg smell, Ox ½ equation 2I – I2 + 2eRe ½ equation H2SO4 + 2 H+ + 2 e- → SO2 + 2 H2O Re ½ equation H2SO4 + 6 H+ + 6 e- → S + 4 H2O Re ½ equation H2SO4 + 8 H+ + 8 e- → H2S + 4 H2O Often in exam questions these redox reactions are worked out after first making the half-equations Reduction product = sulphur dioxide Note the H2SO4 plays the role of acid in the first step producing HBr and then acts as an oxidising agent in the second redox step. Note the H2SO4 plays the role of acid in the first step producing HI and then acts as an oxidising agent in the three redox steps Reduction products = sulphur dioxide, sulphur and hydrogen sulphide
184.108.40.206 Trends in properties
The trend in reducing ability of the halide ions, including the reactions of solid sodium halides with concentrated sulfuric acid.