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Chapter 4: Problem 132

Hydrochloric acid is not an oxidizing agent in the sense that sulfuric acid and nitric acid are. Explain why the chloride ion is not a strong oxidizing agent like \(\mathrm{SO}_{4}^{2-}\) and \(\mathrm{NO}_{3}^{-}\).

Short Answer

Expert verified
The reason why chloride ion is not a strong oxidizing agent like sulfate and nitrate ions are due to its lower standard electrode potential and its stability with a fully-filled outermost shell after gaining one electron.

Step by step solution

01

Understanding Redox Reactions

In a redox (reduction-oxidation) reaction, the oxidizing agent is the substance that accepts electrons, getting reduced in the process, while the reducing agent is the substance that loses electrons, getting oxidized.
02

Standard Electrode Potentials

Standard electrode potential is a measure of the tendency of a redox couple (a half-reaction) to get reduced. An oxidizing agent with a higher (more positive) standard electrode potential has more tendency to gain electrons and get reduced.
03

Comparing Electrode Potentials

The standard electrode potentials for sulfate ion \(\mathrm{SO}_{4}^{2-}\) and nitrate ion \(\mathrm{NO}_{3}^{-}\) are significantly higher than that of chloride ion \(\mathrm{Cl}^{-}\). Therefore, \(\mathrm{SO}_{4}^{2-}\) and \(\mathrm{NO}_{3}^{-}\) have higher tendencies to accept electrons and act as oxidizing agents.
04

Stability of Chloride Ion

\(\mathrm{Cl}^{-}\) ion is stable and does not have a tendency to get reduced by accepting electrons. This is due to the fully filled outermost electron shell in \(\mathrm{Cl}^{-}\) ion after gaining one electron, following the octet rule.

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Key Concepts

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

Redox Reactions
Redox reactions, short for reduction-oxidation reactions, involve the transfer of electrons between two substances. In these reactions, there are two main players: the oxidizing agent and the reducing agent. The oxidizing agent is the one that accepts electrons, which means it gets reduced. On the other hand, the reducing agent loses electrons and becomes oxidized.
To better grasp this concept, it鈥檚 helpful to visualize the process:
  • The oxidizing agent 'gains' the electrons, thus it undergoes reduction.
  • The reducing agent 'loses' the electrons, thus it puts them into the oxidizing agent and undergoes oxidation.
Understanding who plays what role in a redox reaction is crucial when predicting chemical behavior, especially when discussing the strength of oxidizing agents like sulfate and nitrate versus chloride ions.
Standard Electrode Potential
The standard electrode potential is a numerical way to express the propensity of a substance to lose or gain electrons in a redox reaction. Each half-reaction has a standard electrode potential, often denoted as E掳.
This value is measured in volts and reflects how easily a substance can either donate or receive electrons:
  • A more positive E掳 value indicates a stronger oxidizing agent, as it means the substance is more inclined to gain electrons.
  • A more negative E掳 value suggests a propensity to lose electrons, common among reducing agents.
When comparing different substances, such as the chloride ion and others like sulfate and nitrate, the differences in E掳 values explain why some ions serve as stronger oxidizing agents than others.
Stability of Chloride Ion
The stability of the chloride ion, \( \mathrm{Cl}^{-} \), plays a significant role in its behavior as an oxidizing agent, or rather, its lack thereof. Chloride ion has a completely filled outer electron shell following the octet rule, achieving a very stable electronic configuration similar to noble gases.
This stability means:
  • The chloride ion does not have a tendency to gain additional electrons.
  • With its stable, filled electron shell, it remains inert in situations where more reactive, stronger oxidizing agents might try to accept electrons.
In contrast, sulfate and nitrate ions, with their higher electron affinities and more positive electrode potentials, are more readily reduced, making them stronger oxidizing agents than chloride.

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Most popular questions from this chapter

Which of the following are redox processes? (a) \(\mathrm{CO}_{2} \longrightarrow \mathrm{CO}_{3}^{2-}\) (b) \(\mathrm{VO}_{3} \longrightarrow \mathrm{VO}_{2}\) (c) \(\mathrm{SO}_{3} \longrightarrow \mathrm{SO}_{4}^{2-}\) (d) \(\mathrm{NO}_{2}^{-} \longrightarrow \mathrm{NO}_{3}^{-}\) (e) \(\mathrm{Cr}^{3+} \longrightarrow \mathrm{CrO}_{4}^{2-}\)

Calculate the concentration of the acid (or base) remaining in solution when \(10.7 \mathrm{~mL}\) of \(0.211 \mathrm{M}\) \(\mathrm{HNO}_{3}\) are added to \(16.3 \mathrm{~mL}\) of \(0.258 \mathrm{M} \mathrm{NaOH}\).

Hydrogen halides (HF, HCl, HBr, HI) are highly reactive compounds that have many industrial and laboratory uses. (a) In the laboratory, HF and \(\mathrm{HCl}\) can be generated by reacting \(\mathrm{CaF}_{2}\) and \(\mathrm{NaCl}\) with concentrated sulfuric acid. Write appropriate equations for the reactions. (Hint: These are not redox reactions.) (b) Why is it that HBr and HI cannot be prepared similarly-that is, by reacting NaBr and NaI with concentrated sulfuric acid? (Hint: \(\mathrm{H}_{2} \mathrm{SO}_{4}\) is a stronger oxidizing agent than both \(\mathrm{Br}_{2}\) and \(\mathrm{I}_{2} .\) ) (c) HBr can be prepared by reacting phosphorus tribromide \(\left(\mathrm{PBr}_{3}\right)\) with water. Write an equation for this reaction.

A \(0.8870-\mathrm{g}\) sample of a mixture of \(\mathrm{NaCl}\) and \(\mathrm{KCl}\) is dissolved in water, and the solution is then treated with an excess of \(\mathrm{AgNO}_{3}\) to yield \(1.913 \mathrm{~g}\) of \(\mathrm{AgCl}\). Calculate the percent by mass of each compound in the mixture.

The following "cycle of copper" experiment is performed in some general chemistry laboratories. The series of reactions starts with copper and ends with metallic copper. The steps are as follows: (1) A piece of copper wire of known mass is allowed to react with concentrated nitric acid [the products are copper(II) nitrate, nitrogen dioxide, and water]. (2) The copper(II) nitrate is treated with a sodium hydroxide solution to form copper(II) hydroxide precipitate. (3) On heating, copper(II) hydroxide decomposes to yield copper(II) oxide. (4) The copper(II) oxide is reacted with concentrated sulfuric acid to yield copper(II) sulfate. (5) Copper(II) sulfate is treated with an excess of zinc metal to form metallic copper. (6) The remaining zinc metal is removed by treatment with hydrochloric acid, and metallic copper is filtered, dried, and weighed. (a) Write a balanced equation for each step and classify the reactions. (b) Assuming that a student started with \(65.6 \mathrm{~g}\) of copper, calculate the theoretical yield at each step. (c) Considering the nature of the steps, comment on why it is possible to recover most of the copper used at the start.
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