14   6 Redox reactions can be understood in terms of electron transfer or in terms of changes of oxidation number. (a) The equation for a redox reaction is shown. 8KI + 5H2SO4 → 4I2 + 4K2SO4 + H2S + 4H2O (i) State, in terms of electron transfer, what you understand by the terms oxidation and reduction. (1) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... (ii) Explain, in terms of electron transfer, which of the species is the oxidising agent and which is the reducing agent. (2) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... Turn over 15   (b) Bromine reacts with sodium hydroxide solution in a disproportionation reaction to form BrO3 –. Complete the equation for this disproportionation. Justify the balancing of the equation in terms of the changes in oxidation number. (4) ...................Br2 + ...................OH– → ...................BrO3 – + ................... + ................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... (Total for Question 6 = 7 marks)