The pair of half-equations below show how potassium iodate(V), $\text{KIO}_3$, in hydrochloric acid solution oxidises iodine to $\text{ICl}_2^-$. $\text{IO}_3^- + 2\text{Cl}^- + 6\text{H}^+ + 4\text{e}^- \rightarrow \text{ICl}_2^- + 3\text{H}_2\text{O}$ $\text{I}_2 + 4\text{Cl}^- \rightarrow 2\text{ICl}_2^- + 2\text{e}^-$ In the balanced equation for the total reaction, what is the ratio of $\text{IO}_3^-$ to $\text{I}_2$?
- A1:1
- B1:2
- C1:4
- D2:1