1. We need to count how many reagents from the list can oxidise aqueous iodide, $I^-$, to iodine, $I_2$.

2. The oxidation half-reaction is: $2I^- \rightarrow I_2 + 2e^-$ Its standard reduction potential in reverse form is based on: $I_2 + 2e^- \rightarrow 2I^- \qquad E^\circ = +0.54\text{ V}$ So, any reagent with sufficiently positive oxidising ability under the given conditions can oxidise $I^-$ to $I_2$.

3. Check each reagent one by one.

(i) Acidified $K_2Cr_2O_7$

In acidic medium, dichromate is a strong oxidising agent: $Cr_2O_7^{2-} + 14H^+ + 6e^- \rightarrow 2Cr^{3+} + 7H_2O$ It oxidises iodide to iodine: $Cr_2O_7^{2-} + 14H^+ + 6I^- \rightarrow 2Cr^{3+} + 7H_2O + 3I_2$ So this does oxidise $I^-$.

✅ Count = 1

(ii) Alkaline $KMnO_4$

In alkaline medium, permanganate is reduced to $MnO_2$ or manganate and still acts as an oxidising agent. It oxidises iodide to iodine (or further to iodate in strongly alkaline conditions), so oxidation of $I^-$ certainly occurs. For example: $2MnO_4^- + 6I^- + 4H_2O \rightarrow 2MnO_2 + 3I_2 + 8OH^-$ So this does oxidise $I^-$.

✅ Count = 2

(iii) $CuSO_4$

$Cu^{2+}$ can oxidise iodide: $2Cu^{2+} + 4I^- \rightarrow 2CuI(s) + I_2$ This reaction occurs because $CuI$ precipitates, driving the reaction forward. So $CuSO_4$ does oxidise $I^-$.

✅ Count = 3

(iv) $H_2O_2$

Hydrogen peroxide can act as an oxidising agent: $H_2O_2 + 2I^- + 2H^+ \rightarrow I_2 + 2H_2O$ So it does oxidise iodide in acidic medium.

✅ Count = 4

(v) $Cl_2$

Chlorine is a stronger oxidising agent than iodine: $Cl_2 + 2I^- \rightarrow 2Cl^- + I_2$ So it does oxidise iodide.

✅ Count = 5

(vi) $O_3$

Ozone is a very strong oxidising agent and oxidises iodide to iodine: $O_3 + 2I^- + H_2O \rightarrow I_2 + O_2 + 2OH^-$ So it does oxidise iodide.

✅ Count = 6

(vii) $FeCl_3$

$Fe^{3+}$ can oxidise iodide: $2Fe^{3+} + 2I^- \rightarrow 2Fe^{2+} + I_2$ So $FeCl_3$ does oxidise iodide.

✅ Count = 7

(viii) $HNO_3$

Nitric acid is an oxidising acid and can oxidise iodide/HI to iodine (often further depending on concentration, but oxidation to iodine certainly occurs). A representative reaction is: $2NO_3^- + 8H^+ + 6I^- \rightarrow 2NO + 4H_2O + 3I_2$ So $HNO_3$ does oxidise iodide.

✅ Count = 8

(ix) $Na_2S_2O_3$

Thiosulfate is a reducing agent, not an oxidising agent toward iodide. In fact, it reduces iodine to iodide: $2S_2O_3^{2-} + I_2 \rightarrow S_4O_6^{2-} + 2I^-$ So it does not oxidise iodide.

❌ Count remains 8

4. Therefore, the total number of reagents that can oxidise aqueous iodide to iodine is: $\boxed{8}$

5. Comparison with stored correct answer:

• Stored answer = $7$
• Derived answer = $8$

The likely reason for the discrepancy is that some solutions exclude $HNO_3$, but chemically $HNO_3$ is an oxidising agent and does oxidise iodide/HI to iodine. Hence $HNO_3$ should be counted.