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Chapter 20: Problem 2

You may have heard that "antioxidants" are good for your health. Is an "antioxidant" an oxidizing agent or a reducing agent?

Short Answer

Expert verified
An antioxidant is a reducing agent, as it prevents oxidation of other molecules by donating electrons, neutralizing free radicals, and stabilizing them in the process.

Step by step solution

01

Define Antioxidant

An antioxidant is a molecule that inhibits or prevents the oxidation of other molecules by donating an electron to a free radical, thus neutralizing it. Oxidation is a chemical reaction that involves the loss of electrons or an increase in oxidation state, while reduction is a chemical reaction that involves the gain of electrons or a decrease in oxidation state.
02

Understand Oxidizing Agent and Reducing Agent

An oxidizing agent is a substance that gains electrons or reduces its oxidation state during the redox (reduction-oxidation) reaction, whereas a reducing agent is a substance that loses electrons or increases its oxidation state during the redox reaction.
03

Identify the Role of Antioxidants in Redox Reactions

Since antioxidants prevent oxidation by donating electrons, they act as a reducing agent, which loses electrons during the redox reaction. By donating electrons, they decrease the oxidation state of the free radicals, thus stabilizing them.
04

Conclusion

Based on the provided information, an antioxidant acts as a reducing agent鈥攕ince it inhibits or prevents oxidation of other molecules by donating electrons, neutralizing free radicals and stabilizing them in the process.

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Key Concepts

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

Understanding Oxidizing Agents
An oxidizing agent is a chemical substance that plays a vital role in oxidation and reduction reactions. It is responsible for accepting electrons from another substance. This causes the other substance to oxidize. In simple terms, when a molecule loses electrons, it is being oxidized, and the oxidizing agent is the one helping it achieve that. Oxidizing agents themselves undergo a reduction as they gain electrons.
  • In a chemical reaction, the oxidizing agent gains electrons and becomes reduced.
  • They are crucial in processes such as combustion, respiration, and even in the bleaching of fabrics.
  • Common examples include oxygen, chlorine, and hydrogen peroxide.
These agents facilitate the electron transfer, making them essential in many chemical and biological processes.
All About Reducing Agents
Just as oxidizing agents play a critical role in chemical reactions by accepting electrons, reducing agents are equally important because they donate electrons. In chemical reactions, especially oxidation and reduction (redox) reactions, a reducing agent gives away its electrons to another substance, allowing that substance to be reduced.
  • Reducing agents, by donating electrons, increase the oxidation state of themselves while decreasing it in the other substance.
  • These agents are essential in processes like metal extraction and in biological pathways where energy is transferred.
  • Common reducing agents include hydrogen, carbon, and metals like zinc.
So, whenever you see a reducing agent in action, realize it is essentially giving something (electrons) to help stabilize other substances.
Oxidation and Reduction Reactions Unveiled
Oxidation and reduction reactions, often referred to as redox reactions, are fascinating processes where electrons are transferred between substances. This electron transfer is what causes changes in oxidation states鈥攂ut what exactly is happening? Let's break it down:
- **Oxidation** involves the loss of electrons or an increase in the oxidation state of a molecule or atom. Think of it as giving away little pieces of yourself (electrons) and becoming more positive. - **Reduction** is the opposite, where there's a gain of electrons or a decrease in oxidation state. Imagine receiving electrons to become less positive, more stable.
  • These reactions are essential in various fields, from energy production in power plants to metabolic reactions in our body.
  • Understanding redox reactions helps explain how batteries work and why we need antioxidants for health.
  • A classic example is rusting, where iron reacts with oxygen (oxidizing agent) to form rust.
By understanding these processes, we grasp the friendly and often essential exchanges happening in chemistry.

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Most popular questions from this chapter

Gold exists in two common positive oxidation states, +1 and +3 . The standard reduction potentials for these oxidation states are $$ \begin{array}{l} \mathrm{Au}^{+}(a q)+\mathrm{e}^{-} \quad \longrightarrow \mathrm{Au}(s) \quad E_{\mathrm{red}}^{\circ}=+1.69 \mathrm{~V} \\ \mathrm{Au}^{3+}(a q)+3 \mathrm{e}^{-} \longrightarrow \mathrm{Au}(s) E_{\mathrm{red}}^{\circ}=+1.50 \mathrm{~V} \end{array} $$ (a) Can you use these data to explain why gold does not tarnish in the air? (b) Suggest several substances that should be strong enough oxidizing agents to oxidize gold metal. (c) Miners obtain gold by soaking gold-containing ores in an aqueous solution of sodium cyanide. A very soluble complex ion of gold forms in the aqueous solution because of the redox reaction $$ \begin{aligned} 4 \mathrm{Au}(s)+8 \mathrm{NaCN}(a q) &+2 \mathrm{H}_{2} \mathrm{O}(l)+\mathrm{O}_{2}(g) \\ \longrightarrow & 4 \mathrm{Na}\left[\mathrm{Au}(\mathrm{CN})_{2}\right](a q)+4 \mathrm{NaOH}(a q) \end{aligned} $$ What is being oxidized, and what is being reduced in this reaction? (d) Gold miners then react the basic aqueous product solution from part (c) with \(\mathrm{Zn}\) dust to get gold metal. Write a balanced redox reaction for this process. What is being oxidized, and what is being reduced?

Hydrogen gas has the potential for use as a clean fuel in reaction with oxygen. The relevant reaction is $$ 2 \mathrm{H}_{2}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{H}_{2} \mathrm{O}(l) $$ Consider two possible ways of utilizing this reaction as an electrical energy source: (i) Hydrogen and oxygen gases are combusted and used to drive a generator, much as coal is currently used in the electric power industry; (ii) hydrogen and oxygen gases are used to generate electricity directly by using fuel cells that operate at \(85^{\circ} \mathrm{C} .\) (a) Use data in Appendix \(\mathrm{C}\) to calculate \(\Delta H^{\circ}\) and \(\Delta S^{\circ}\) for the reaction. We will assume that these values do not change appreciably with temperature. (b) Based on the values from part (a), what trend would you expect for the magnitude of \(\Delta G\) for the reaction as the temperature increases? (c) What is the significance of the change in the magnitude of \(\Delta G\) with temperature with respect to the utility of hydrogen as a fuel? (d) Based on the analysis here, would it be more efficient to use the combustion method or the fuel-cell method to generate electrical energy from hydrogen?

A disproportionation reaction is an oxidation-reduction reaction in which the same substance is oxidized and reduced. Complete and balance the following disproportionation reactions: (a) \(\mathrm{Fe}^{2+}(a q) \longrightarrow \mathrm{Fe}(s)+\mathrm{Fe}^{3+}(a q)\) (b) \(\mathrm{Br}_{2}(l) \longrightarrow \mathrm{Br}^{-}(a q)+\mathrm{BrO}_{3}^{-}(a q)\) (acidic solution) (c) \(\mathrm{Cr}^{3+}(a q) \longrightarrow \mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}(a q)+\mathrm{Cr}(s)\) (acidic solution) (d) \(\mathrm{NO}(g) \longrightarrow \mathrm{N}_{2}(g)+\mathrm{NO}_{3}^{-}(a q)\) (acidic solution)

In a galvanic cell the cathode is an \(\mathrm{Ag}^{+}(1.00 \mathrm{M}) / \mathrm{Ag}(s)\) half-cell. The anode is a standard hydrogen electrode immersed in a buffer solution containing \(0.10 \mathrm{M}\) benzoic acid \(\left(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{COOH}\right)\) and \(0.050 \mathrm{M}\) sodium benzoate \(\left(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{COO}^{-} \mathrm{Na}^{+}\right)\). The measured cell voltage is \(1.030 \mathrm{~V}\). What is the \(\mathrm{p} K_{\mathrm{a}}\) of benzoic acid?

At \(298 \mathrm{~K}\) a cell reaction has a standard cell potential of \(+0.63 \mathrm{~V}\). The equilibrium constant for the reaction is \(3.8 \times 10^{10}\). What is the value of \(n\) for the reaction?
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