The cathode half-equation is $2\text{H}_2\text{O(l)} + 2\text{e}^- \rightarrow \text{H}_2\text{(g)} + 2\text{OH}^-\text{(aq)}$. Beginning with neutral $\text{NaCl(aq)}$, write equations showing the anode formation of $\text{O}_2\text{(g)}$.
Starting from neutral $\text{NaCl(aq)}$, write equations showing the anode production of $\text{Cl}_2\text{(g)}$.
For electrolysis to take place, the applied voltage must be at least equal to the $E^\circ_{\text{cell}}$ value obtained from standard electrode potentials. Use the Data Booklet to work out $E^\circ_{\text{cell}}$ for the anode production of $\text{O}_2\text{(g)}$.
Use the Data Booklet to work out $E^\circ_{\text{cell}}$ for the anode production of $\text{Cl}_2\text{(g)}$.
Using one of the phrases more positive, less positive or no change, use the equations you wrote in (a) to infer the effect of increasing $[\text{Cl}^-\text{(aq)}]$ on: • the $E_{\text{anode}}$ for the production of $\text{O}_2\text{(g)}$, • the $E_{\text{anode}}$ for the production of $\text{Cl}_2\text{(g)}$.
Hence explain why the $\text{Cl}_2\text{(g)} : \text{O}_2\text{(g)}$ ratio becomes larger as $\text{NaCl(aq)}$ increases.
Sodium chlorate(V) is made on a large scale by electrolysing $\text{NaCl(aq)}$ in a cell that lets the anode and cathode electrolytes mix. The cathode reaction is unchanged from that given in (a). The anode equation is $\text{Cl}^-\text{(aq)} + 6\text{OH}^-\text{(aq)} - 6\text{e}^- \rightarrow \text{ClO}_3^-\text{(aq)} + 3\text{H}_2\text{O(l)}$. Construct an ionic equation for the overall reaction.
Calculate the mass of $\text{NaClO}_3$ that is obtained when a current of $250\ \text{A}$ is passed through the cell for $60$ minutes.