The best oxidizing agent of the following oxides is (a) \(\mathrm{Li}_{2} \mathrm{O} ;(\mathrm{b}) \mathrm{MgO} ;(\mathrm{c}) \mathrm{Al}_{2} \mathrm{O}_{3} ;(\mathrm{d}) \mathrm{CO}_{2} ;(\mathrm{e}) \mathrm{SnO}_{2} ;(\mathrm{f}) \mathrm{PbO}_{2}.\)

Short Answer

Expert verified
\(\mathrm{PbO}_{2}\) is the best oxidizing agent among the given options.

Step by step solution

01

Analyze all Given Oxides

In our case, the given oxides include \(\mathrm{Li}_{2}\mathrm{O}, \mathrm{MgO}, \mathrm{Al}_{2}\mathrm{O}_{3}, \mathrm{CO}_{2}, \mathrm{SnO}_{2}, \mathrm{PbO}_{2}\). We need to analyze these oxides and note their various potentials.
02

Check Electrochemical Series

The electrochemical series indicates the reactivity series of metals, which shows the reactivity order of various metals in terms of their oxidation or reduction. It is arranged in increasing standard electrode potentials. The species higher up in the series have a greater tendency to be reduced, and hence are better oxidizing agents.
03

Find the Best Oxidizing Agent

By looking at the electrochemical series, it can be seen that \(\mathrm{PbO}_{2}\) has the highest potential and therefore exhibits the greatest tendency to gain electrons. Hence, \(\mathrm{PbO}_{2}\) is the best oxidizing agent among the given oxides.

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Key Concepts

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

Understanding the Electrochemical Series
The electrochemical series, also known as the reactivity series, is a powerful tool in chemistry that organizes elements according to their standard electrode potentials. What does this mean for students of chemistry? Simply put, it's a ranked list showing which elements are more likely to lose or gain electrons during a chemical reaction.

This is particularly important when considering redox reactions, where one species is oxidized (loses electrons) and another is reduced (gains electrons). The electrochemical series lists elements in such a way that those with the highest tendency to lose electrons (and thus get oxidized) are placed at the top, while those that easily gain electrons (get reduced) are at the bottom.

Understanding where an element sits within this series can help predict its behavior as an oxidizing or reducing agent. An element with a high standard electrode potential is a strong oxidizing agent because it has a greater tendency to gain electrons. Therefore, when comparing different elements or their compounds, such as oxides, in terms of their oxidizing power, the electrochemical series serves as an invaluable reference point for determining which species will be the best oxidizer under standard conditions.
Standard Electrode Potentials
Standard electrode potentials (often abbreviated as E°) are crucial in understanding the driving force behind redox reactions. These potentials are measured under standard conditions, which is at a temperature of 298 K, with 1 M concentrations of all aqueous species, a pressure of 1 bar for any gases involved, and a pure solid or liquid as the state of metals.

The more positive the standard electrode potential, the greater the substance's ability to gain electrons and thus be reduced. For students, this means that a list of standard electrode potentials provides the information needed to rank oxidizing agents: substances with higher standard electrode potentials are stronger oxidizers.

When examining an exercise that asks you to determine the best oxidizing agent among a group of compounds, looking at their respective standard electrode potentials gives you a clear and quantifiable way to answer. By checking a standard electrode potential table or chart, which is a staple in every chemistry textbook, you can identify which compound has the highest potential and thereby select it as the strongest oxidizing agent.
Reactivity of Metals
The reactivity of metals is a topic that often confuses students, but it's actually quite logical when you think of it in terms of electron exchange. Metals that are highly reactive have a strong tendency to lose their valence electrons and form positive ions (cations). This reactivity can be influenced by various factors, including the metal's position in the periodic table and the structure of its electron shells.

Why is the metal reactivity important? For one, it helps predict how a metal will behave when it comes into contact with other substances. A highly reactive metal, like lithium or potassium, eagerly participates in chemical reactions, often violently and exothermically. On the opposite side, less reactivity means more stability, as seen with metals like platinum and gold, which barely react with other substances under normal conditions.

In the context of redox chemistry, a highly reactive metal is typically a poor oxidizing agent because it doesn't easily accept electrons. Instead, these metals are excellent reducing agents, as they tend to lose electrons readily. Understanding reactivity is essential not only in identifying which metal is a better reducing agent but also in safety protocols and industrial applications where the reactivity of metals determines their suitability for specific processes and reactions.

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Most popular questions from this chapter

A chemical dictionary gives the following descriptions of the production of some compounds. Write plausible chemical equations based on these descriptions. (a) lead(II) carbonate: adding a solution of sodium bicarbonate to a solution of lead nitrate. (b) lithium carbonate: reaction of lithium oxide with ammonium carbonate solution. (c) hydrogen peroxide: by the action of dilute sulfuric acid on barium peroxide. (d) lead(IV) oxide: action of an alkaline solution of calcium hypochlorite on lead(II) oxide.

Write the simplest chemical equation to represent the reaction of (a) \(\mathrm{K}_{2} \mathrm{CO}_{3}\left(\text { aq) and } \mathrm{Ba}(\mathrm{OH})_{2}(\mathrm{aq})\right.\); (b) \(\operatorname{Mg}\left(\mathrm{HCO}_{3}\right)_{2}(\) aq) on heating; (c) tin(II) oxide when heated with carbon; (d) \(\mathrm{CaF}_{2}(\mathrm{s})\) and \(\mathrm{H}_{2} \mathrm{SO}_{4}(\mathrm{concd}\) aq); (e) \(\mathrm{NaHCO}_{3}(\mathrm{s})\) and \(\mathrm{HCl}(\mathrm{aq}) ;\) (f) \(\mathrm{PbO}_{2}(\mathrm{s})\) and HBr(aq); and (g) the reduction of \(\mathrm{SiF}_{4}\) to pure \(\mathrm{Si},\) by using Na as the reducing agent.

In the purification of bauxite ore, a preliminary step in the production of aluminum, \(\left[\mathrm{Al}(\mathrm{OH})_{4}\right]^{-}(\mathrm{aq})\) can be converted to \(\mathrm{Al}(\mathrm{OH})_{3}(\mathrm{s})\) by passing \(\mathrm{CO}_{2}(\mathrm{g})\) through the solution. Write an equation for the reaction that occurs. Could HCl(aq) be used instead of \(\mathrm{CO}_{2}(\mathrm{g}) ?\) Explain.

The melting point of \(\mathrm{NaCl}(\mathrm{s})\) is \(801^{\circ} \mathrm{C},\) much higher than that of \(\mathrm{NaOH}\left(322^{\circ} \mathrm{C}\right) .\) More energy is consumed to melt and maintain molten NaCl than NaOH. Yet the preferred commercial process for the production of sodium is electrolysis of \(\mathrm{NaCl}(\mathrm{l})\) rather than \(\mathrm{NaOH}(1)\) Give a reason or reasons for this discrepancy.

Lithium superoxide, \(\mathrm{LiO}_{2}(\mathrm{s}),\) has never been isolated. Use ideas from Chapter \(12,\) together with data from this chapter and Appendix \(D\), to estimate \(\Delta H_{f}\) for \(\mathrm{LiO}_{2}(\mathrm{s})\) and assess whether \(\mathrm{LiO}_{2}(\mathrm{s})\) is thermodynamically stable with respect to \(\mathrm{Li}_{2} \mathrm{O}(\mathrm{s})\) and \(\mathrm{O}_{2}(\mathrm{g}).\) (a) Use the Kapustinskii equation, along with appropriate data below, to estimate the lattice energy, \(U,\) for \(\left.\mathrm{LiO}_{2}(\mathrm{s}) . \text { (See exercise } 126 \text { in Chapter } 12 .\right)\) The ionic radii for \(L\) i \(^{+}\) and \(O_{2}^{-}\) are \(73 \mathrm{pm}\) and \(144 \mathrm{pm},\) respectively. (b) Use your result from part (a) in the BornFajans-Haber cycle to estimate \(\Delta H_{\mathrm{f}}^{2}\) for \(\mathrm{LiO}_{2}(\mathrm{s})\) [Hint: For the process \(\mathrm{O}_{2}(\mathrm{g})+\mathrm{e}^{-} \rightarrow \mathrm{O}_{2}^{-}(\mathrm{g}), \Delta H^{\circ}=.\) \(-43 \mathrm{kJ} \mathrm{mol}^{-1} .\) See Table 21.2 and Appendix \(\mathrm{D}\) for the other data that are required.] (c) Use your result from part (b) to calculate the enthalpy change for the decomposition of \(\mathrm{LiO}_{2}(\mathrm{s})\) to \(\mathrm{Li}_{2} \mathrm{O}(\mathrm{s})\) and \(\mathrm{O}_{2}(\mathrm{g}) .\) For \(\mathrm{Li}_{2} \mathrm{O}(\mathrm{s}), \Delta H_{\mathrm{f}}^{\circ}=-598.73\) \(\mathrm{kJmol}^{-1}.\) (d) Use your result from part (c) to decide whether \(\mathrm{LiO}_{2}(\mathrm{s})\) is thermodynamically stable with respect to \(\mathrm{Li}_{2} \mathrm{O}(\mathrm{s})\) and \(\mathrm{O}_{2}(\mathrm{g}) .\) Assume that entropy effects can be neglected.

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