Chemistry 9701 · AS & A Level · Electrochemistry

Electrochemistry — practice question

At room temperature, the solid compounds sodium chloride and sodium iodide each undergo reaction with concentrated sulfuric acid. For $\text{NaCl}$, the products formed are $\text{NaHSO}_4$ and $\text{HCl}$. For $\text{NaI}$, the products formed are $\text{NaHSO}_4$, $\text{HI}$, $\text{I}_2$, $\text{SO}_2$, $\text{H}_2\text{O}$, $\text{S}$ and $\text{H}_2\text{S}$. What explains this difference in products?

  • AChloride ions will displace iodine from the solution.
  • BHydrogen chloride is more volatile than hydrogen iodide.
  • CIodide ions are better reducing agents than chloride ions.
  • DSulfuric acid is able to act as a dehydrating agent with $\text{NaI}$.

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