An industrial chemist treats solid \(\mathrm{NaCl}\) with concentrated \(\mathrm{H}_{2} \mathrm{SO}_{4}\) and obtains gaseous \(\mathrm{HCl}\) and \(\mathrm{NaHSO}_{4}\). When she substitutes solid NaI for \(\mathrm{NaCl}\), she obtains gaseous \(\mathrm{H}_{2} \mathrm{~S},\) solid \(\mathrm{I}_{2},\) and \(\mathrm{S}_{8},\) but no \(\mathrm{HI}\). (a) What type of reaction did the \(\mathrm{H}_{2} \mathrm{SO}_{4}\) undergo with NaI? (b) Why does NaI, but not \(\mathrm{NaCl}\), cause this type of reaction? (c) To produce HI \((g)\) by reacting NaI with an acid, how does the acid have to differ from sulfuric acid?

A redox reaction is short for reduction-oxidation reaction. This type of chemical reaction involves the transfer of electrons between two substances. One substance gets oxidized (loses electrons) while the other gets reduced (gains electrons).

The key to identifying a redox reaction is to look for changes in oxidation states. In the case of NaI and concentrated \(\text{H}_{2} \text{SO}_{4}\), iodide ions (I-) act as a strong reducing agent. They lose electrons and are oxidized to make solid iodine (I2) and even sulfur (S8). Meanwhile, the sulfuric acid (\tex{H}_{2} \tex{SO}_{4}) accepts these electrons and is reduced, producing H2S gas.

So, in the initial reaction with NaCl, there is no reduction happening to the \(\text{H}_{2} \text{SO}_{4}\). Chloride ions (Cl-) are not strong enough as reducing agents.

An acid-base reaction is a type of chemical reaction where an acid reacts with a base to produce salt and water. These reactions do not involve electron transfer like redox reactions. Instead, they involve the transfer of protons (H+) between the acid and the base.

In the reaction involving NaCl and \(\text{H}_{2} \text{SO}_{4}\), only an acid-base reaction takes place. The strong acid (\tex{H}_{2} \tex{SO}_{4}) donates protons (H+) to Cl-, forming gaseous HCl and solid NaHSO4. This is a straightforward acid-base reaction with no changes in oxidation states for either element.

However, with NaI, the iodide ions go beyond just receiving protons. They start participating as reducing agents, making the interaction with \(\text{H}_{2} \text{SO}_{4}\) more complex and converting it into a redox reaction.

Iodide ions (I-) are known for being good reducing agents, meaning they can easily donate electrons to other substances. In the presence of a strong oxidizing acid like \(\text{H}_{2} \text{SO}_{4}\), iodide ions undergo a redox reaction.

In the reaction: \[8 \tex{NaI}_{(s)} + \tex{H}_{2} \tex{SO}_{4(aq)} \rightarrow \tex{H}_{2} \tex{S}_{(g)} + 4 \tex{I}_{2(s)} + \tex{S}_{8(s)} + \tex{NaHSO}_{4(s)}\], iodide ions (\tex{I}^{-}) are oxidized to form solid iodine (\tex{I}_2) and sulfur (\tex{S}_8). The \(\text{H}_{2} \text{SO}_{4}\) gets reduced to \tex{H}_{2}\tex{S} gas.

Not all acids can facilitate this reaction. For industrial purposes, where HI gas is desired, a less oxidizing acid like phosphoric acid (\tex{H}_3\tex{PO}_4) is used. This prevents the iodide from being oxidized further and allows HI to form.