When chlorate(V) ions, $\text{ClO}_3^-$, are reduced by $\text{SO}_2$, the only products formed are chlorine dioxide, $\text{ClO}_2$, and sulfate ions, $\text{SO}_4^{2-}$. Construct an equation for this reaction.
Chlorine dioxide, $\text{ClO}_2$, disproportionates with hydroxide ions, $\text{OH}^-\text{(aq)}$, to give a mixture of $\text{ClO}_2^-$ and $\text{ClO}_3^-$ ions. $2\text{ClO}_2 + 2\text{OH}^- \rightarrow \text{ClO}_2^- + \text{ClO}_3^- + \text{H}_2\text{O}$ Explain, using this reaction as an example, what is meant by disproportionation.
Deduce the ionic half-equations that correspond to the reaction in part (b)(i).
A lithium-iodine electrochemical cell can be used to generate electricity for a heart pacemaker. The cell has a lithium electrode and an inert electrode immersed in body fluids. As 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.
Using the Data Booklet, calculate the $E^\circ_{\text{cell}}$ for this cell.
A current of $2.5 \times 10^{-5}\,\text{A}$ is drawn from this cell. Calculate the time needed for $0.10\,\text{g}$ of lithium electrode to be used up. Assume the current stays constant throughout this period.