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

Short Answer

Expert verified
In summary, SO₂ can act as a reducing agent because it has the potential to be further oxidized, as its sulfur atom has an oxidation state of +4, which can increase to +6, forming SO₃. However, SO₃ cannot act as a reducing agent because its sulfur atom is already at its maximum oxidation state of +6, making it unable to donate electrons to other species.

Step by step solution

01

Understanding Reducing Agents

A reducing agent is a species that donates electrons to another species in a redox reaction, thereby getting oxidized itself. The reducing agent typically has a low oxidation state, with the potential for further oxidation.
02

Identifying Oxidation States

In SO₂, the sulfur atom has an oxidation state of +4 as each of the two oxygen atoms has an oxidation state of -2. In contrast, the sulfur atom in SO₃ has an oxidation state of +6 as each of the three oxygen atoms has an oxidation state of -2.
03

Evaluating the Redox Potential

Examining the oxidation states, it is clear that SO₂ has the potential to be further oxidized since sulfur is only at a +4 oxidation state. The maximum oxidation state sulfur can reach is +6, which is the same oxidation state sulfur has in SO₃. This means that if SO₂ undergoes further oxidation, it can form SO₃, as seen in the balanced equation below: \[2\,\text{SO}_2 + \text{O}_2 \rightarrow 2\,\text{SO}_3\]
04

Comparing Redox Stability

For SO₂, since it can be further oxidized, it has the ability to donate electrons to other species, hence, acting as a reducing agent. In contrast, SO₃ cannot act as a reducing agent because the sulfur atom has already reached its maximum oxidation state (+6), and it does not have the ability to donate electrons to other species.
05

Conclusion

Based on the oxidation state comparison and the redox potential of SO₂ and SO₃, we can conclude that SO₂ can act as a reducing agent, as it can donate electrons and be further oxidized to SO₃. However, SO₃ cannot act as a reducing agent because its sulfur atom is already at its maximum oxidation state, and it does not have the ability to donate electrons to other species.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Understanding Oxidation States
Getting to grips with oxidation states is crucial for anyone tackling redox chemistry. Oxidation states indicate the degree of oxidation of an atom within a compound. Think of it like a team jersey number that helps pinpoint each atom's role in a redox process.

An atom's oxidation state can often clue you into its potential to act in a reaction. Generally speaking, an atom with a lower oxidation state has more electrons that it 'could' lose, making it a prime candidate to be a reducing agent—a chemical Samaritan that gives electrons away to another atom in need. In our exercise, sulfur dioxide (SO₂) has a sulfur oxidation state of +4, which means it's halfway up the ladder—it can climb up to a max of +6 or step down. On the other side, sulfur trioxide (SO₃) has a sulfur oxidation state of +6, basically shouting 'I'm all out of electrons to give!'
Redox Reactions Demystified
Ever seen a party where some guests take the snacks, and others bring them? That’s kind of how redox reactions work. These reactions involve the transfer of electrons between substances. Reducing agents are the generous ones offering some of their electrons, while oxidizing agents are those taking the electrons.

Let's make it super simple: reducing agents reduce the charge of other atoms by giving out electrons, while they themselves get oxidized, or lose electrons. Caught in a redox reaction, sulfur dioxide (SO₂) can hand over electrons to become sulfur trioxide (SO₃) as shown in the chemical equation from the exercise. SO₂ is like the person at the party who brings an extra pizza for everyone—it's a hit because it's so giving!
Sulfur Dioxide as a Reducing Agent
Considering sulfur dioxide as a reducing agent isn't just about chemistry; it's like understanding why a half-charged battery still has juice to power a device. SO₂ is electron-rich — its +4 oxidation state of sulfur means it's prepared to give up electrons and climb to a +6 state, transforming into SO₃.

Why can't SO₃ do the same? Think of SO₃ as the fully charged battery; it has nowhere to go — no more electrons to give. SO₂ is ready to be the superhero, swinging into the reaction to rescue other atoms by donating electrons. Meanwhile, SO₃ is the retired hero, having already given its all. Remember, a great reducing agent likes to live life on the edge, ready to lose some electrons and jump to a higher oxidation state, just like SO₂ is set to do.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

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 \(\mathrm{HF}\) is much higher than those of the other hydrogen halides. (d) The halogens decrease in oxidizing power in the order \(\mathrm{F}_{2}>\mathrm{Cl}_{2}>\mathrm{Br}_{2}>\mathrm{I}_{2}\)

Give the chemical formula for (a) hydrocyanic acid, (b) nickel tetracarbonyl, (c) barium bicarbonate, (d) calcium acetylide (e) potassium carbonate.

(a) What is the oxidation state of \(\mathrm{P}\) in \(\mathrm{PO}_{4}^{3-}\) and of \(\mathrm{N}\) in \(\mathrm{NO}_{3}^{-} ?(\mathbf{b})\) Why doesn't \(\mathrm{N}\) form a stable \(\mathrm{NO}_{4}^{3-}\) ion analogous to \(\mathrm{P} ?\)

Ultrapure germanium, like silicon, is used in semiconductors. Germanium of "ordinary" purity is prepared by the hightemperature reduction of \(\mathrm{GeO}_{2}\) with carbon. The Ge is converted to \(\mathrm{GeCl}_{4}\) by treatment with \(\mathrm{Cl}_{2}\) and then purified by distillation; \(\mathrm{GeCl}_{4}\) is then hydrolyzed in water to \(\mathrm{GeO}_{2}\) and reduced to the elemental form with \(\mathrm{H}_{2}\). The element is then zone refined. Write a balanced chemical equation for each of the chemical transformations in the course of forming ultrapure Ge from \(\mathrm{GeO}_{2}\)

Write the chemical formula for each of the following \(\mathrm{com}^{-}\) pounds, and indicate the oxidation state of the group \(6 \mathrm{~A}\) element in each: (a) selenous acid, (b) potassium hydrogen sulfite, (c) hydrogen telluride, (d) carbon disulfide, (e) calcium sulfate, (f) cadmium sulfide, (g) zinc telluride.

See all solutions

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free