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Chapter 17: Problem 71

An oxidizing agent causes the (oxidation/reduction) of another species, and the oxidizing agent itself is (oxidized/reduced).

Short Answer

Expert verified
An oxidizing agent causes the \(oxidation\) of another species, and the oxidizing agent itself is \(reduced\).

Step by step solution

01

Understanding the role of an oxidizing agent

In a redox reaction, an oxidizing agent is a substance that has the ability to gain electrons from other species. This forces the other species to lose electrons and become oxidized. The oxidizing agent will become reduced as it gains electrons.
02

Determining the effect on another species

Since the oxidizing agent causes another species to lose electrons, it forces the other species to undergo oxidation.
03

Determining the change in the oxidizing agent itself

As the oxidizing agent accepts electrons from other species, its own oxidation state decreases, meaning that it is reduced. So, based on the analysis and solution steps above: An oxidizing agent causes the \(oxidation\) of another species, and the oxidizing agent itself is \(reduced\).

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Most popular questions from this chapter

In which direction do electrons flow in a galvanic cell, anode to cathode or vice versa?

Why must the sum of all the oxidation states of the atoms in a neutral molecule be zero?

Assign oxidation states to all of the atoms in each of the following: a. \(\mathrm{Na}_{2} \mathrm{CrO}_{4}\) b. \(\mathrm{Na}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}\) c. \(\mathrm{Cr} \mathrm{Cl}_{3}\) d. \(\mathrm{Cr}_{2} \mathrm{O}_{3}\)

What reactions go on during the recharging of an automobile battery?

Carbon compounds containing double bonds (such compounds are called alkenes) react readily with many other reagents. In each of the following reactions, identify which atoms are oxidized and which are reduced, and specify the oxidizing and reducing agents. a. \(\mathrm{CH}_{2}=\mathrm{CH}_{2}(g)+\mathrm{Cl}_{2}(g) \rightarrow \mathrm{ClCH}_{2}-\mathrm{CH}_{2} \mathrm{Cl}(l)\) b. \(\mathrm{CH}_{2}=\mathrm{CH}_{2}(g)+\mathrm{Br}_{2}(g) \rightarrow \mathrm{BrCH}_{2}-\mathrm{CH}_{2} \operatorname{Br}(l)\) c. \(\mathrm{CH}_{2}=\mathrm{CH}_{2}(g)+\mathrm{HBr}(g) \rightarrow \mathrm{CH}_{3}-\mathrm{CH}_{2} \mathrm{Br}(l)\) d. \(\mathrm{CH}_{2}=\mathrm{CH}_{2}(g)+\mathrm{H}_{2}(g) \rightarrow \mathrm{CH}_{3}-\mathrm{CH}_{3}(g)\)
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