Question

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?

Short answer

NaI undergoes a redox reaction with \(\text{H}_{2} \text{SO}_{4}\), but NaCl does not because iodide ions are stronger reducing agents than chloride ions. To produce HI, a weaker, less oxidizing acid like \(\text{H}_{3} \text{PO}_{4}\) should be used.

Step-by-step solution

Analyzing the Reaction with NaCl

When NaCl is treated with concentrated \(\text{H}_{2} \text{SO}_{4}\), the products are gaseous HCl and solid NaHSO4. The reaction can be written as: \[ \text{NaCl}_{(s)} + \text{H}_{2} \text{SO}_{4(aq)} \rightarrow \text{HCl}_{(g)} + \text{NaHSO}_{4(s)} \]

Analyzing the Reaction with NaI

When NaI is treated with concentrated \(\text{H}_{2} \text{SO}_{4}\), the products are gaseous H2S, solid I2, and S8 with no HI. The reaction involves a reduction of \(\text{H}_{2} \text{SO}_{4}\) by iodide ions: \[ 8 \text{NaI}_{(s)} + \text{H}_{2} \text{SO}_{4(aq)} \rightarrow \text{H}_{2} \text{S}_{(g)} + 4 \text{I}_{2(s)} + \text{S}_{8(s)} + \text{NaHSO}_{4(s)} \]

Identify Type of Reaction for NaI and H2SO4

The reaction between NaI and concentrated \(\text{H}_{2} \text{SO}_{4}\) is a redox reaction. Iodide ions are strong reducing agents and reduce the \(\text{H}_{2} \text{SO}_{4}\) to produce H2S, I2, and S8.

Explain the Difference in Reactions

NaCl does not cause this type of reaction because chloride ions are much weaker reducing agents compared to iodide ions. Hence, in the case of NaCl, \(\text{H}_{2} \text{SO}_{4}\) does not get reduced and only acts as an acid, producing HCl.

Selecting an Appropriate Acid for HI production

To produce HI gas by reacting NaI with an acid, the acid must be weaker or less oxidizing than \(\text{H}_{2} \text{SO}_{4}\). A suitable choice would be phosphoric acid \(\text{H}_{3} \text{PO}_{4}\), which is not a strong oxidizing agent and would not reduce the iodide ions.

Key concepts

redox reaction

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.

acid-base reaction

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 reduction

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.