Chlorate(V) ions are powerful oxidising agents.

1. (a)The reduction of chlorate(V) ions, ClO3⁻, with SO2 forms chlorine dioxide, ClO2, and sulfate ions, SO42-, as the only products. Construct an equation for this reaction.[1]
2. (b)(i)Chlorine dioxide, ClO2, disproportionates with hydroxide ions, OH⁻(aq), to form a mixture of ClO2⁻ and ClO3⁻ ions. 2ClO2 + 2OH⁻ → ClO2⁻ + ClO3⁻ + H₂O Explain, using this reaction as an example, what is meant by disproportionation.[1]
3. (b)(ii)Deduce the ionic half-equations for the reaction in (b)(i).[2]
4. (c)(i)A lithium-iodine electrochemical cell can be used to generate electricity for a heart pacemaker. The cell consists of a lithium electrode and an inert electrode immersed in body fluids. When current flows lithium is oxidised and iodine is reduced. Use the Data Booklet to write half-equations for the reactions taking place at the two electrodes. Hence write the overall equation for when a current flows. • • overall equation[2]
5. (c)(ii)Use the Data Booklet to calculate the Ecell for this cell. Ecell =[1]
6. (c)(iii)A current of 2.5 × 10⁻⁵A is drawn from this cell. Calculate the time taken for 0.10 g of lithium electrode to be used up. Assume the current remains constant throughout this period. time =[3]