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Chapter 15: Problem 8

Explain what we mean by preoxidation and prereduction. Why is it important to be able to destroy the reagents used for these purposes?

Short Answer

Expert verified
Preoxidation and prereduction prepare substances for reactions by adjusting their oxidation states. Destroying their reagents ensures safety and process purity.

Step by step solution

01

Understanding Preoxidation and Prereduction

Preoxidation and prereduction refer to chemical processes often used to modify the state of a substance before a primary reaction. Preoxidation involves increasing the oxidation state of a substance, typically using an oxidizing agent. This can make certain elements more reactive or prepare them for further reactions. Prereduction, on the other hand, reduces the oxidation state using a reducing agent, which can also make species more suitable for subsequent chemical reactions.
02

Importance of Destroying Reagents

The reagents used in preoxidation and prereduction can sometimes be hazardous or reactive if left in the reaction environment after their intended use. Destroying these reagents helps ensure that they do not interfere with further steps of the process, do not contribute to unwanted side reactions, and do not pose environmental or safety hazards. Proper disposal and destruction of these reagents maintain the integrity and safety of subsequent operations.

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Key Concepts

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

Preoxidation
Preoxidation is a chemical process where the oxidation state of a substance is increased before it goes through a primary reaction. This is done by adding an oxidizing agent, which helps prepare the substance for further chemical changes.
Oxidizing agents, like oxygen or chlorine, interact with the substance and modify its chemical nature.
  • Increases the reactivity of elements.
  • Serves as a preparatory step in many industrial and laboratory processes.
  • Important in processes like wastewater treatment and metal refining.
Understanding preoxidation is crucial because it directly impacts the effectiveness and success of chemical reactions. By preoxidizing a substance, chemists can control the reaction pathway and enhance desired outcomes.
Prereduction
Prereduction, much like preoxidation, prepares a chemical substance for further reactions.
It involves reducing the oxidation state of a substance using a reducing agent, making it more reactive for subsequent reactions.
Reducing agents such as hydrogen or carbon donate electrons to the substance, altering its chemical state. Some of the benefits of prereduction include:
  • Enable specific reaction pathways that are otherwise inactive.
  • Preferred in processes like metal extraction and refining.
  • Essential in organic chemistry for synthesizing various compounds.
Prereduction is vital in many industrial applications, allowing precise control over chemical changes and ensuring that the desired reactions can occur efficiently.
Oxidation State
The oxidation state is a concept that helps chemists understand the electron donation or acceptance of an element in a compound.
This determines the element's ability to engage in chemical reactions.
Each element in a compound has a specific oxidation state, which can be positive, negative, or zero:
  • A positive oxidation state means an element has lost electrons.
  • A negative oxidation state indicates electron gain.
  • A zero state means the element is neutral in terms of electron exchange.
Recognizing the oxidation state is crucial in predicting the behavior of compounds in reactions.
It helps chemists devise suitable methods for preoxidation or prereduction, aligning the reagents to achieve the desired chemical transformation efficiently.
Reagents
Reagents are substances added to a chemical reaction to cause a transformation or achieve a desired result.
In preoxidation and prereduction processes, they are fundamental in altering the chemical state of the substances involved.
Reagents like oxidizing and reducing agents play specific roles:
  • Oxidizing agents increase the oxidation state, preparing substances for further reactions.
  • Reducing agents decrease the oxidation state, making substances more reactive.
  • Other reagents may regulate pH, act as solvents, or speed up reactions as catalysts.
Selecting the right reagent is critical, as it impacts the efficiency and safety of the reaction. This choice can significantly influence the success of the intended chemical transformation.
Chemical Safety
Chemical safety involves the practices and precautions necessary to handle, store, and dispose of chemicals safely.
This is especially crucial when dealing with preoxidation and prereduction reagents because of their reactive nature.
Safety considerations include:
  • Proper labeling and storage of chemicals to prevent accidental misuse or spills.
  • Using personal protective equipment (PPE) like gloves and goggles to protect against exposure.
  • Having proper ventilation and safety protocols to handle hazardous fumes or reactions.
  • Implementing protocols for the destruction and disposal of reactive reagents to prevent environmental harm.
A strong understanding of chemical safety ensures that processes are conducted safely, protecting both the individuals involved and the environment from potential chemical hazards.

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Most popular questions from this chapter

Here is a description of an analytical procedure for superconductors containing unknown quantities of \(\mathrm{Cu}(\mathrm{I}), \mathrm{Cu}(\mathrm{II})\), \(\mathrm{Cu}(\mathrm{III})\), and peroxide \(\left(\mathrm{O}_{2}^{2-}\right):^{32}\) " The possible trivalent copper and/or peroxide-type oxygen are reduced by \(\mathrm{Cu}(\mathrm{I})\) when dissolving the sample (ca. \(50 \mathrm{mg}\) ) in deoxygenated \(\mathrm{HCl}\) solution (1 \(\mathrm{M}\) ) containing a known excess of monovalent copper ions ( \(\mathrm{ca} .25 \mathrm{mg} \mathrm{CuCl}\) ). On the other hand, if the sample itself contained monovalent copper, the amount of \(\mathrm{Cu}(\mathrm{I})\) in the solution would increase upon dissolving the sample. The excess \(\mathrm{Cu}(\mathrm{I})\) was then determined by coulometric back-titration ... in an argon atmosphere." The abbreviation "ca." means "approximately." Coulometry is an electrochemical method in which the electrons liberated in the reaction \(\mathrm{Cu}^{+} \rightarrow \mathrm{Cu}^{2+}+\mathrm{e}^{-}\) are measured from the charge flowing through an electrode. Explain with your own words and equations how this analysis works.

An aqueous glycerol solution weighing \(100.0 \mathrm{mg}\) was treated with \(50.0 \mathrm{~mL}\) of \(0.0837 \mathrm{M} \mathrm{Ce}^{4+}\) in \(4 \mathrm{M} \mathrm{HClO}_{4}\) at \(60^{\circ} \mathrm{C}\) for \(15 \mathrm{~min}\) to oxidize the glycerol to formic acid: CC(C)(C)O \(\mathrm{HCO}_{2} \mathrm{H}\) FM \(92.095\) Formic acid The excess \(\mathrm{Ce}^{4+}\) required \(12.11 \mathrm{~mL}\) of \(0.0448 \mathrm{M} \mathrm{Fe}^{2+}\) to reach a ferroin end point. What is the weight percent of glycerol in the unknown?

A \(25.00-\mathrm{mL}\) volume of commercial hydrogen peroxide solution was diluted to \(250.0 \mathrm{~mL}\) in a volumetric flask. Then \(25.00 \mathrm{~mL}\) of the diluted solution were mixed with \(200 \mathrm{~mL}\) of water and \(20 \mathrm{~mL}\) of \(3 \mathrm{MH}_{2} \mathrm{SO}_{4}\) and titrated with \(0.02123 \mathrm{M} \mathrm{KMnO}_{4}\). The first pink color was observed with \(27.66 \mathrm{~mL}\) of titrant. A blank prepared from water in place of \(\mathrm{H}_{2} \mathrm{O}_{2}\) required \(0.04 \mathrm{~mL}\) to give visible pink color. Using the \(\mathrm{H}_{2} \mathrm{O}_{2}\) reaction in Table \(15-3\), find the molarity of the commercial \(\mathrm{H}_{2} \mathrm{O}_{2}\).

Write balanced half-reactions in which \(\mathrm{MnO}_{4}^{-}\)acts as an oxidant at (a) \(\mathrm{pH}=0\); (b) \(\mathrm{pH}=10\); (c) \(\mathrm{pH}=15\).

Calcium fluorapatite \(\left(\mathrm{Ca}_{10}\left(\mathrm{PO}_{4}\right)_{6} \mathrm{~F}_{2}, \mathrm{FM} 1008.6\right)\) laser crystals were doped with chromium to improve their efficiency. It was suspected that the chromium could be in the \(+4\) oxidation state. 1\. To measure the total oxidizing power of chromium in the material, a crystal was dissolved in \(2.9 \mathrm{M} \mathrm{HClO}_{4}\) at \(100^{\circ} \mathrm{C}\), cooled to \(20^{\circ} \mathrm{C}\), and titrated with standard \(\mathrm{Fe}^{2+}\), using \(\mathrm{Pt}\) and \(\mathrm{Ag} \mid \mathrm{AgCl}\) electrodes to find the end point. Chromium oxidized above the \(+3\) state should oxidize an equivalent amount of \(\mathrm{Fe}^{2+}\) in this step. That is, \(\mathrm{Cr}^{4+}\) would consume one \(\mathrm{Fe}^{2+}\), and each atom of \(\mathrm{Cr}^{6+}\) in \(\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}\) would consume three \(\mathrm{Fe}^{2+}\) : $$ \begin{aligned} \mathrm{Cr}^{4+}+\mathrm{Fe}^{2+} & \rightarrow \mathrm{Cr}^{3+}+\mathrm{Fe}^{3+} \\ \frac{1}{2} \mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}+3 \mathrm{Fe}^{2+} & \rightarrow \mathrm{Cr}^{3+}+3 \mathrm{Fe}^{3+} \end{aligned} $$ 2\. In a second step, the total chromium content was measured by dissolving a crystal in \(2.9 \mathrm{M} \mathrm{HClO}_{4}\) at \(100^{\circ} \mathrm{C}\) and cooling to \(20^{\circ} \mathrm{C}\). Excess \(\mathrm{S}_{2} \mathrm{O}_{8}^{2-}\) and \(\mathrm{Ag}^{+}\)were then added to oxidize all chromium to \(\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}\). Unreacted \(\mathrm{S}_{2} \mathrm{O}_{8}^{2-}\) was destroyed by boiling, and the remaining solution was titrated with standard \(\mathrm{Fe}^{2+}\). In this step, each \(\mathrm{Cr}\) in the original unknown reacts with three \(\mathrm{Fe}^{2+}\). $$ \begin{aligned} \mathrm{Cr}^{x+} & \stackrel{\mathrm{S}_{2} \mathrm{O}_{8}^{2-}}{\longrightarrow} \mathrm{Cr}_{2} \mathrm{O}_{7}^{2-} \\ \frac{1}{2} \mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}+3 \mathrm{Fe}^{2+} & \rightarrow \mathrm{Cr}^{3+}+3 \mathrm{Fe}^{3+} \end{aligned} $$ In step \(1,0.4375 \mathrm{~g}\) of laser crystal required \(0.498 \mathrm{~mL}\) of \(2.786 \mathrm{mM}\) \(\mathrm{Fe}^{2+}\) (prepared by dissolving \(\mathrm{Fe}\left(\mathrm{NH}_{4}\right)_{2}\left(\mathrm{SO}_{4}\right)_{2} \cdot 6 \mathrm{H}_{2} \mathrm{O}\) in \(2 \mathrm{M} \mathrm{HClO}_{4}\) ). In step \(2,0.1566 \mathrm{~g}\) of crystal required \(0.703 \mathrm{~mL}\) of the same \(\mathrm{Fe}^{2+}\) solution. Find the average oxidation number of \(\mathrm{Cr}\) in the crystal and find the total micrograms of \(\mathrm{Cr}\) per gram of crystal.
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