Logout Open In App
Open In App
Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

Chapter 18: Problem 1

Which of the following statements is(are) true? Explain. a. Oxidation and reduction cannot occur independently of each other. b. Oxidation and reduction accompany all chemical reactions. c. A substance that reacts with oxygen gas will always be oxidized.

Short Answer

Expert verified
Statement a is true. Oxidation and reduction are complementary processes that occur simultaneously in redox reactions, where one substance loses electrons (oxidation) and another gains electrons (reduction). Statement b is false. Not all chemical reactions involve oxidation and reduction. There are other types of reactions, such as acid-base, precipitation, and complexation reactions, that do not involve electron transfer. Statement c is true. When a substance reacts with oxygen gas, it typically loses electrons to the oxygen molecule and gets oxidized, while oxygen itself gets reduced by gaining electrons. This is a characteristic of redox reactions involving oxygen.

Step by step solution

01

Statement a: Oxidation and reduction cannot occur independently of each other.

The statement is true. In a redox (reduction-oxidation) reaction, one substance is oxidized by losing electrons, while the other substance is reduced by gaining electrons. Therefore, they occur simultaneously, and an oxidation process can't take place without its corresponding reduction process happening and vice versa.
02

Statement b: Oxidation and reduction accompany all chemical reactions.

The statement is false. While oxidation and reduction participate in many chemical reactions, they are not involved in every chemical reaction. There are other types of reactions, such as acid-base reactions, precipitation reactions, and complexation reactions, that do not involve the transfer of electrons.
03

Statement c: A substance that reacts with oxygen gas will always be oxidized.

The statement is true. In a redox reaction, the presence of oxygen will typically oxidize other substances by taking their electrons. When a substance reacts with oxygen, it loses its electrons to the oxygen molecule, hence it gets oxidized. Oxygen itself gets reduced by gaining electrons. This is a typical characteristic of a redox reaction involving oxygen.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

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

Oxidation
Oxidation is a fundamental concept in chemistry and plays a critical role in redox reactions. Originally, oxidation referred to the combination of a substance with oxygen. However, the definition has expanded over time. Currently, it means the loss of electrons by a molecule, atom, or ion.

When an element undergoes oxidation, it increases in oxidation state, implying that it loses electrons. For instance, when iron rusts, it reacts with oxygen and loses electrons, becoming iron oxide.

In any oxidation process, something must simultaneously be reduced. This is because electrons cannot exist freely; they must be transferred to another substance.
  • Loss of electrons
  • Increase in oxidation state
  • Occurs simultaneously with reduction
Reduction
Reduction is another key concept in redox reactions, complementing the process of oxidation. The term 'reduction' refers to the gain of electrons by a molecule, atom, or ion.

When a substance is reduced, it decreases in oxidation state, meaning it has gained electrons. A common example is the reduction of oxygen in water formation, whereby oxygen molecules gain electrons from hydrogen.

This process can’t occur alone; if one substance is reduced, another must be oxidized, perpetuating the cycle of electron transfer in redox reactions.
  • Gain of electrons
  • Decrease in oxidation state
  • Occurs alongside oxidation
Electron Transfer
Electron transfer is the mechanism that underlies redox reactions. In these reactions, electrons move from one substance to another.

This movement is what connects oxidation and reduction, since the electron lost by the oxidized substance is the one gained by the reduced substance. The exchange of electrons can involve simple ions or complex molecules.

Understanding electron transfer is crucial because it enables various applications, from generating energy in batteries to facilitating metabolic processes in living organisms.
  • Foundation of redox reactions
  • Involves movement of electrons between substances
  • Essential in energy production and biochemical processes

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Calculate \(\mathscr{E}^{\circ}\) for the following half-reaction: $$\mathrm{AgI}(s)+\mathrm{e}^{-} \longrightarrow \mathrm{Ag}(s)+\mathrm{I}^{-}(a q)$$ (Hint: Reference the \(K_{\mathrm{sp}}\) value for AgI and the standard reduction potential for \(\mathrm{Ag}^{+}\).)

A galvanic cell is based on the following half-reactions: $$\begin{array}{ll}{\mathrm{Fe}^{2+}+2 \mathrm{e}^{-} \longrightarrow \mathrm{Fe}(s)} & {\mathscr{E}^{\circ}=-0.440 \mathrm{V}} \\ {2 \mathrm{H}^{+}+2 \mathrm{e}^{-} \longrightarrow \mathrm{H}_{2}(g)} & {\mathscr{E}^{\circ}=0.000 \mathrm{V}}\end{array}$$ where the iron compartment contains an iron electrode and \(\left[\mathrm{Fe}^{2+}\right]=1.00 \times 10^{-3} M\) and the hydrogen compartment contains a platinum electrode, \(P_{\mathrm{H}_{2}}=1.00 \mathrm{atm},\) and a weak acid, HA, at an initial concentration of 1.00 \(\mathrm{M}\) . If the observed cell potential is 0.333 \(\mathrm{V}\) at \(25^{\circ} \mathrm{C},\) calculate the \(K_{\mathrm{a}}\) value for the weak acid HA.

Define oxidation and reduction in terms of both change in oxidation number and electron loss or gain.

You have a concentration cell with Cu electrodes and \(\left[\mathrm{Cu}^{2+}\right]=1.00 M(\text { right side })\) and \(1.0 \times 10^{-4} M(\text { left side })\) a. Calculate the potential for this cell at \(25^{\circ} \mathrm{C}\) b. The \(\mathrm{Cu}^{2+}\) ion reacts with \(\mathrm{NH}_{3}\) to form \(\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}^{2+}\) by the following equation: $$\begin{aligned} \mathrm{Cu}^{2+}(a q)+4 \mathrm{NH}_{3}(a q) \rightleftharpoons \mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}^{2+}(a q) & \\\ & K=1.0 \times 10^{13} \end{aligned}$$ Calculate the new cell potential after enough \(\mathrm{NH}_{3}\) is added to the left cell compartment such that at equilibrium \(\left[\mathrm{NH}_{3}\right]=2.0 \mathrm{M} .\)

A chemist wishes to determine the concentration of \(\mathrm{CrO}_{4}^{2-}\) electrochemically. A cell is constructed consisting of a saturated calomel electrode (SCE; see Exercise 111\()\) and a silver wire coated with \(\mathrm{Ag}_{2} \mathrm{CrO}_{4}\) . The \(\mathscr{E}^{\circ}\) value for the following half-reaction is 0.446 \(\mathrm{V}\) relative to the standard hydrogen electrode: $$\mathrm{Ag}_{2} \mathrm{CrO}_{4}+2 \mathrm{e}^{-} \longrightarrow 2 \mathrm{Ag}+\mathrm{CrO}_{4}^{2-}$$ a. Calculate \(\mathscr{E}_{\text { cell } \text { and }} \Delta G\) at \(25^{\circ} \mathrm{C}\) for the cell reaction when \(\left[\mathrm{CrO}_{4}^{2-}\right]=1.00 \mathrm{mol} / \mathrm{L}\) . b. Write the Nernst equation for the cell. Assume that the SCE concentrations are constant. c. If the coated silver wire is placed in a solution (at \(25^{\circ} \mathrm{C} )\) in which \(\left[\mathrm{CrO}_{4}^{2-}\right]=1.00 \times 10^{-5} M,\) what is the expected cell potential? d. The measured cell potential at \(25^{\circ} \mathrm{C}\) is 0.504 \(\mathrm{V}\) when the coated wire is dipped into a solution of unknown \(\left[\mathrm{CrO}_{4}^{2-}\right] .\) What is \(\left[\mathrm{CrO}_{4}^{2-}\right]\) for this solution? e. Using data from this problem and from Table \(18.1,\) calculate the solubility product \(\left(K_{\mathrm{sp}}\right)\) for \(\mathrm{Ag}_{2} \mathrm{CrO}_{4}\).
See all solutions

Recommended explanations on Chemistry Textbooks

Chemistry Branches

Read Explanation

Chemical Reactions

Read Explanation

Making Measurements

Read Explanation

Ionic and Molecular Compounds

Read Explanation

Inorganic Chemistry

Read Explanation

Chemical Analysis

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free