Total Reagents Capable of Oxidising Aqueous Iodide to Iodine

To determine the total number of reagents capable of oxidising aqueous iodide ions ($\text{I}^-$) to iodine ($\text{I}_2$), we analyze the redox behavior of each given reagent:

1. Acidified $\text{K}_2\text{Cr}_2\text{O}_7$: Dichromate ion ($\text{Cr}_2\text{O}_7^{2-}$) in an acidic medium is a strong oxidizing agent and oxidizes $\text{I}^-$ to $\text{I}_2$: $\text{Cr}_2\text{O}_7^{2-} + 14\text{H}^+ + 6\text{I}^- \longrightarrow 2\text{Cr}^{3+} + 3\text{I}_2 + 7\text{H}_2\text{O}$ $\rightarrow$ Capable

2. Alkaline $\text{KMnO}_4$: In an alkaline or neutral medium, permanganate ion ($\text{MnO}_4^-$) oxidizes $\text{I}^-$ further to iodate ions ($\text{IO}_3^-$) rather than iodine ($\text{I}_2$): $2\text{MnO}_4^- + \text{H}_2\text{O} + \text{I}^- \longrightarrow 2\text{MnO}_2 + 2\text{OH}^- + \text{IO}_3^-$ $\rightarrow$ Not capable (produces $\text{IO}_3^-$ instead of $\text{I}_2$)

3. $\text{CuSO}_4$: $\text{Cu}^{2+}$ ions oxidise iodide ions to iodine, precipitating cuprous iodide ($\text{Cu}_2\text{I}_2$): $2\text{Cu}^{2+} + 4\text{I}^- \longrightarrow \text{Cu}_2\text{I}_2 + \text{I}_2$ $\rightarrow$ Capable

4. $\text{H}_2\text{O}_2$: Hydrogen peroxide oxidizes iodide ions to iodine in acidic/neutral media: $\text{H}_2\text{O}_2 + 2\text{I}^- + 2\text{H}^+ \longrightarrow \text{I}_2 + 2\text{H}_2\text{O}$ $\rightarrow$ Capable

5. $\text{Cl}_2$: Chlorine is a stronger oxidizing agent than iodine ($E^\circ_{\text{Cl}_2/\text{Cl}^-} = +1.36\text{ V} > E^\circ_{\text{I}_2/\text{I}^-} = +0.54\text{ V}$) and displaces iodide to yield iodine: $\text{Cl}_2 + 2\text{I}^- \longrightarrow 2\text{Cl}^- + \text{I}_2$ $\rightarrow$ Capable

6. $\text{O}_3$: Ozone oxidizes iodide ions in the presence of water to produce iodine: $\text{O}_3 + 2\text{I}^- + \text{H}_2\text{O} \longrightarrow \text{I}_2 + \text{O}_2 + 2\text{OH}^-$ $\rightarrow$ Capable

7. $\text{FeCl}_3$: Ferric ions ($\text{Fe}^{3+}$) have a standard reduction potential ($E^\circ_{\text{Fe}^{3+}/\text{Fe}^{2+}} = +0.77\text{ V}$) greater than that of $\text{I}_2/\text{I}^-$ ($+0.54\text{ V}$), so $\text{Fe}^{3+}$ oxidizes $\text{I}^-$ to $\text{I}_2$: $2\text{Fe}^{3+} + 2\text{I}^- \longrightarrow 2\text{Fe}^{2+} + \text{I}_2$ $\rightarrow$ Capable

8. $\text{HNO}_3$: Nitric acid is a strong oxidizing agent that readily oxidizes $\text{I}^-$ to $\text{I}_2$: $2\text{HNO}_3 + 2\text{I}^- + 2\text{H}^+ \longrightarrow \text{I}_2 + 2\text{NO}_2 + 2\text{H}_2\text{O}$ $\rightarrow$ Capable

9. $\text{Na}_2\text{S}_2\text{O}_3$: Sodium thiosulfate acts as a reducing agent (reducing $\text{I}_2$ to $\text{I}^-$) and cannot oxidize iodide ions. $\rightarrow$ Not capable

---

Conclusion:

The 7 reagents capable of oxidising aqueous iodide to iodine are:

1. Acidified $\text{K}_2\text{Cr}_2\text{O}_7$
2. $\text{CuSO}_4$
3. $\text{H}_2\text{O}_2$
4. $\text{Cl}_2$
5. $\text{O}_3$
6. $\text{FeCl}_3$
7. $\text{HNO}_3$

Thus, the total number of reagents is 7.