Stoichiometry and Redox Analysis of Chlorate and Iodide Reaction

To determine the correct statement(s) regarding the given chemical reaction, we first balance the reaction using the ion-electron (half-reaction) method.

1. Balancing the Redox Reaction

Oxidation Half-Reaction: Iodide ions ($\text{I}^-$) are oxidized to molecular iodine ($\text{I}_2$). $\text{I}^- \rightarrow \text{I}_2$ Balancing atoms and electrons: $2\text{I}^- \rightarrow \text{I}_2 + 2e^- \quad \text{--- (1)}$ Here, the oxidation state of iodine increases from $-1$ in $\text{I}^-$ to $0$ in $\text{I}_2$, which confirms that iodide ions undergo oxidation.

Reduction Half-Reaction: Chlorate ions ($\text{ClO}_3^-$) are reduced to chloride ions ($\text{Cl}^-$) in an acidic medium. $\text{ClO}_3^- \rightarrow \text{Cl}^-$ Balancing oxygen atoms by adding $\text{H}_2\text{O}$: $\text{ClO}_3^- \rightarrow \text{Cl}^- + 3\text{H}_2\text{O}$ Balancing hydrogen atoms by adding $\text{H}^+$: $\text{ClO}_3^- + 6\text{H}^+ \rightarrow \text{Cl}^- + 3\text{H}_2\text{O}$ Balancing charge by adding electrons: $\text{ClO}_3^- + 6\text{H}^+ + 6e^- \rightarrow \text{Cl}^- + 3\text{H}_2\text{O} \quad \text{--- (2)}$

Overall Ionic Reaction: Multiplying equation (1) by $3$ and adding it to equation (2) to equalize the number of electrons: $3 \times (2\text{I}^- \rightarrow \text{I}_2 + 2e^-)$ $6\text{I}^- + \text{ClO}_3^- + 6\text{H}^+ \rightarrow 3\text{I}_2 + \text{Cl}^- + 3\text{H}_2\text{O}$

2. Expressing in Molecular/Full Equation Form

Since $\text{H}_2\text{SO}_4$ acts as the source of $\text{H}^+$ ions according to the reaction scheme and dissociates partially as: $\text{H}_2\text{SO}_4 \rightarrow \text{H}^+ + \text{HSO}_4^-$

We replace $6\text{H}^+$ with $6\text{H}_2\text{SO}_4$, which yields $6\text{HSO}_4^-$ on the product side: $6\text{I}^- + \text{ClO}_3^- + 6\text{H}_2\text{SO}_4 \rightarrow 3\text{I}_2 + \text{Cl}^- + 6\text{HSO}_4^- + 3\text{H}_2\text{O}$

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3. Verification of Options

• Option A: In the balanced equation, the stoichiometric coefficient of $\text{HSO}_4^-$ is $6$. (Correct)
• Option B: Iodide ($\text{I}^-$) loses electrons and its oxidation state increases from $-1$ to $0$, so iodide ions undergo oxidation. (Correct)
• Option C: The oxidation state of sulfur in both $\text{H}_2\text{SO}_4$ and $\text{HSO}_4^-$ is $+6$. Therefore, sulfur does not undergo reduction. (Incorrect)
• Option D: Water ($\text{H}_2\text{O}$) is formed as a product to balance the oxygen and hydrogen atoms. (Correct)

Thus, the correct options are A, B, and D.