16 6 This question is about the reactions of the halogens and their salts. (a) The potassium halides react with concentrated sulfuric acid to form hydrogen halides. (i) The equation for this reaction for potassium chloride can be written KCl + H2SO4 → HCl + KHSO4 The hydrogen chloride does not react further. State why this reaction is not a redox reaction. (1) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... Turn over 17 (ii) On descending Group 7, the hydrogen halides become better reducing agents. Explain how the reactions of potassium chloride, potassium bromide and potassium iodide with concentrated sulfuric acid provide evidence for this statement. No explanation of the trend is required. (3) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... 18 (b) The reaction that occurs between chlorine and sodium hydroxide depends on the temperature. (i) At room temperature the reaction that occurs is Cl2 + NaOH → NaClO + NaCl Explain, with reference to oxidation numbers, why this is a disproportionation reaction. (2) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... (ii) With hot sodium hydroxide solution, a different disproportionation reaction occurs. Sodium chlorate(V) is one of the products. Complete the equation for this reaction. State symbols are not required. (2) ………………….Cl2 + ………………….NaOH → Turn over 19 (c) Chlorine is used as a bleach in the textiles industry. Any excess chlorine can be removed by reduction to chloride ions. The half-equation for the reaction of chlorine is Cl2 + 2e– → 2Cl– In one reaction, 768 cm3 of chlorine gas was reduced. (i) Calculate the number of moles of electrons gained by chlorine molecules during this reaction. [Under these conditions one mole of gas occupies 24 dm3] (2) (ii) The reducing agent was a solution containing thiosulfate ions, S2O3 2–. The chlorine reacted with 40 cm3 of a 0.20 mol dm–3 solution of these ions. Deduce the number of moles of electrons lost by each atom of sulfur in the thiosulfate ion, and hence the final oxidation state of the sulfur in the product. (3) (Total for Question 6 = 13 marks)