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Chapter 22: Problem 88

Explain why \(\mathrm{SO}_{2}\) can be used as a reducing agent but \(\mathrm{SO}_{3}\) cannot.

Short Answer

Expert verified
\(\mathrm{SO}_{2}\) can be used as a reducing agent because sulfur has an oxidation state of +4, making it less stable and able to donate electrons to other substances. In contrast, \(\mathrm{SO}_{3}\) has sulfur in its maximum oxidation state of +6, making it more stable and unable to donate electrons, and thus cannot act as a reducing agent.

Step by step solution

01

Understanding oxidation states of sulfur in \(\mathrm{SO}_{2}\) and \(\mathrm{SO}_{3}\)

In \(\mathrm{SO}_{2}\), the sulfur has an oxidation state of +4, whereas in \(\mathrm{SO}_{3}\), the oxidation state of sulfur is +6. The higher the oxidation state, the more oxidized the element is.
02

Understanding the concept of reducing agents

A reducing agent is a substance that can donate electrons to another substance, consequently reducing the other substance's oxidation state while the reducing agent itself gets oxidized. In other words, the reducing agent must have a strong ability to lose electrons and have a less stable oxidation state.
03

Evaluating the stability of \(\mathrm{SO}_{2}\) and \(\mathrm{SO}_{3}\)

Given the oxidation states of sulfur in \(\mathrm{SO}_{2}\) and \(\mathrm{SO}_{3}\), we can determine which molecule is less stable and has a greater tendency to lose electrons. Since the oxidation state of sulfur in \(\mathrm{SO}_{3}\) is +6, which is the highest positive oxidation state for the element, it is more stable compared to the one in \(\mathrm{SO}_{2}\). Therefore, sulfur in \(\mathrm{SO}_{2}\) has a greater tendency to lose electrons.
04

Reaction mechanism demonstrating \(\mathrm{SO}_{2}\)'s reducing ability

Here is an example of a reaction where \(\mathrm{SO}_{2}\) acts as a reducing agent: \[2 \mathrm{SO}_{2} + O_{2} \longrightarrow 2 \mathrm{SO}_{3}\] In this reaction, \(\mathrm{SO}_{2}\) donates electrons to molecular oxygen, causing the sulfur oxidation state to increase from +4 to +6 (i.e., from \(\mathrm{SO}_{2}\) to \(\mathrm{SO}_{3}\)). This results in the reduction of oxygen from 0 to -2. To summarize:
05

Reason why \(\mathrm{SO}_{2}\) can be used as reducing agent and not \(\mathrm{SO}_{3}\)

The sulfur in \(\mathrm{SO}_{2}\) has a lower oxidation state (+4) compared to that in \(\mathrm{SO}_{3}\), which is at its maximum, (+6). This makes \(\mathrm{SO}_{2}\) less stable, allowing it to act as a reducing agent by donating electrons to other substances. On the other hand, \(\mathrm{SO}_{3}\) is more stable in its oxidation state, and thus it does not have the ability to donate electrons to act as a reducing agent.

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Most popular questions from this chapter

Explain each of the following observations: (a) At room temperature \(\mathrm{I}_{2}\) is a solid, \(\mathrm{Br}_{2}\) is a liquid, and \(\mathrm{Cl}_{2}\) and \(\mathrm{F}_{2}\) are both gases. (b) \(\mathrm{F}_{2}\) cannot be prepared by electrolytic oxidation of aqueous \(\mathrm{F}^{-}\) solutions. (c) The boiling point of HF is much higher than those of the other hydrogen halides. (d) The halogens decrease in oxidizing power in the order \(\mathrm{F}_{2}>\mathrm{Cl}_{2}>\mathrm{Br}_{2}>\mathrm{I}_{2}\)

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