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Chapter 24: Problem 23

Potassium superoxide, \(\mathrm{KO}_{2}(s)\), is a strong oxidizing agent. Explain why.

Short Answer

Expert verified
Potassium superoxide is a strong oxidizing agent due to the reactive superoxide ion ( ext{O}_{2}^{-}) it contains, which can readily donate electrons.

Step by step solution

01

Composition of Potassium Superoxide

Potassium superoxide ( ext{KO}_{2}) is composed of potassium ( ext{K}) and the superoxide ion ( ext{O}_{2}^{-}). The presence of the superoxide ion is critical to its oxidizing properties.
02

Electron Configuration and Oxidation States

The superoxide ion ( ext{O}_{2}^{-}) has an oxidation state of -1/2 for each oxygen. This is unusual compared to other oxygen-containing compounds, where oxygen is typically -2. The high energy of the unpaired electron in this ion makes it highly reactive.
03

Mechanism of Oxidation

Potassium superoxide can readily donate its extra electron to other substances, especially those that are electron-deficient, thus undergoing a reduction process and causing the other substance to be oxidized.
04

Applications in Chemistry

Because of its ability to produce oxygen and act as an oxidizing agent, ext{KO}_{2} is valuable in applications such as breathing apparatus in submarines and space stations. It can regenerate oxygen, which is highly useful in closed systems.

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Key Concepts

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

Oxidizing Agents
Oxidizing agents are substances that have the ability to acquire electrons from other molecules during chemical reactions. In simpler terms, they cause other substances to lose electrons. Oxidizing agents themselves gain electrons in the process, undergoing reduction. Potassium superoxide (\(\text{KO}_{2}\)) is considered a potent oxidizing agent due to the presence of the superoxide ion (\(\text{O}_{2}^{-}\)). This ion plays a crucial role in its oxidizing properties.
  • The superoxide ion is eager to accept electrons, making \(\text{KO}_{2}\) effective at oxidizing other materials.
  • This ability makes it valuable for various applications, such as regenerating oxygen in closed environments.
Superoxide Ion
The superoxide ion (\(\text{O}_{2}^{-}\)) is an unusual and interesting component. It consists of two oxygen atoms bonded together and has a single negative charge. What makes the superoxide ion stand out is its unique electron configuration and oxidation state.Unlike most other oxygen compounds where oxygen has an oxidation state of -2, the oxygen atoms within the superoxide ion each have an oxidation state of -1/2.
  • The presence of an unpaired electron gives the superoxide ion high reactivity and instability.
  • This unpaired electron is also responsible for its ability to act as a strong oxidizing agent.
Understanding why superoxide ions are so reactive helps to explain their effectiveness as oxidizers.
Electron Configuration
Electron configuration refers to the arrangement of electrons around an atom鈥檚 nucleus. It鈥檚 a way of understanding how these electrons are distributed in shells and subshells. The superoxide ion鈥檚 high reactivity is linked to its electron configuration.In its natural state, oxygen tends to form stable molecules, such as \(\text{O}_{2}\). Once it becomes superoxide, the presence of an unpaired electron disrupts the typical stability.The odd number of electrons leads to a condition known as paramagnetism, where the ion is attracted to magnetic fields, which can highlight its reactive nature.
  • The superoxide ion eagerly seeks to pair its unpaired electron, thus facilitating oxidation reactions.
  • This pursuit of electron pairing can drive the reactions \(\text{KO}_{2}\) participates in, contributing to its role as a formidable oxidizing agent.

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Most popular questions from this chapter

A rock sample is to be assayed for its tin content by an oxidation-reduction titration with \(\mathrm{I}_{3}^{-}(a q)\). A \(10.0\) -gram sample of the rock is crushed, dissolved in sulfuric acid, and passed over a reducing agent so that all the tin is in the form \(\mathrm{Sn}^{2+}(a q)\). The \(\mathrm{Sn}^{2+}(a q)\) is completely oxidized by \(34.6 \mathrm{~mL}\) of a \(0.556\) -M solution of \(\operatorname{NaI}_{3}(a q)\). The unbalanced equation for the reaction is $$ \mathrm{I}_{3}^{-}(a q)+\mathrm{Sn}^{2+}(a q) \rightarrow \mathrm{Sn}^{4+}(a q)+\mathrm{I}^{-}(a q) $$ Calculate the amount of tin in the sample and its mass percentage in the rock.

The quantity of antimony in a sample can be determined by an oxidation- reduction titration with an oxidizing agent. A 9.62-gram sample of stibnite, an ore of antimony, is dissolved in hot, concentrated \(\mathrm{HCl}(a q)\) and passed over a reducing agent so that all the antimony is in the form \(\mathrm{Sb}^{3+}(a q)\). The \(\mathrm{Sb}^{3+}(a q)\) is completely oxidized by \(43.7 \mathrm{~mL}\) of a \(0.125\) -M aqueous solution of \(\mathrm{KBrO}_{3}(a q)\). The unbalanced equation for the reaction is $$ \operatorname{BrO}_{3}^{-}(a q)+\mathrm{Sb}^{3+}(a q) \rightarrow \mathrm{Br}^{-}(a q)+\mathrm{Sb}^{5+}(a q) $$ Calculate the amount of antimony in the sample and its percentage in the ore.

Sodium chlorite, \(\mathrm{NaClO}_{2}(s)\), is a powerful but stable oxidizing agent used in the paper industry, especially for the final whitening of paper. Sodium chlorite is capable of bleaching materials containing cellulose without oxidizing the cellulose. Sodium chlorite is made by the reaction described by $$ \begin{array}{r} \mathrm{NaOH}(a q)+\mathrm{Ca}(\mathrm{OH})_{2}(s)+\mathrm{C}(s)+\mathrm{ClO}_{2}(g) \rightarrow \\ \mathrm{NaClO}_{2}(a q)+\mathrm{CaCO}_{3}(s) \end{array} $$ Balance this equation and calculate the number of kilograms of \(\mathrm{ClO}_{2}(g)\) required to make \(1.00\) metric ton of \(\mathrm{NaClO}_{2}\)

24-36. Complete and balance the following equations: (a) \(\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}(a q)+\mathrm{I}^{-}(a q) \rightarrow \mathrm{Cr}^{3+}(a q)+\mathrm{I}_{2}(s) \quad\) (acidic) (b) \(\mathrm{CuS}(s)+\mathrm{NO}_{3}^{-}(a q) \rightarrow\) \(\mathrm{Cu}^{2+}(a q)+\mathrm{S}(s)+\mathrm{NO}(g) \quad\) (acidic)

Sodium nitrite, an important chemical in the dye industry, is manufactured by the reaction between sodium nitrate and lead according to $$ \mathrm{NaNO}_{3}(a q)+\mathrm{Pb}(s) \rightarrow \mathrm{NaNO}_{2}(a q)+\mathrm{PbO}(s) $$ Identify the oxidizing and reducing agents in this equation.
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