The reduction of chlorate(V) ions, $\text{ClO}_3^-$, with $\text{SO}_2$ produces chlorine dioxide, $\text{ClO}_2$, and sulfate ions, $\text{SO}_4^{2-}$, as the only products. Construct an equation for this reaction.
Chlorine dioxide, $\text{ClO}_2$, disproportionates with hydroxide ions, $\text{OH}^-\text{(aq)}$, to produce a mixture of $\text{ClO}_2^-$ and $\text{ClO}_3^-$ ions.\n\n$2\text{ClO}_2 + 2\text{OH}^- \rightarrow \text{ClO}_2^- + \text{ClO}_3^- + \text{H}_2\text{O}$\n\nUsing this reaction as an example, explain what disproportionation means.
Deduce the ionic half-equations that apply to the reaction in part (b)(i).
A lithium-iodine electrochemical cell may be used to supply electricity for a heart pacemaker. The cell contains a lithium electrode and an inert electrode in body fluids. As current flows, lithium is oxidised while iodine is reduced.\n\nUse the Data Booklet to write half-equations for the reactions at the two electrodes. Hence write the overall equation for when current flows.
Use the Data Booklet to work out the $E^\circ_{cell}$ for this cell.
A current of $2.5 \times 10^{-5}\,\text{A}$ is taken from this cell. Calculate the time required for $0.10\,\text{g}$ of the lithium electrode to be consumed. Assume that the current stays constant throughout this period.