Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 4: Problem 16

In self-reduction, the reducing species is A) \(\mathrm{S}\) B) \(\mathrm{O}^{2-}\) C) \(\mathrm{S}^{2-}\) D) \(\mathrm{SO}_{2}\)

Short Answer

Expert verified
\(\mathrm{SO}_{2}\) is the reducing species in self-reduction.

Step by step solution

01

Understanding Self-Reduction

Self-reduction is a process in which a substance acts both as a reducing agent and as an oxidizing agent. This means the same substance gets oxidized as well as reduced during the reaction.
02

Identifying the Species that Undergoes Self-Reduction

To undergo self-reduction, a species must exist in multiple oxidation states such that it can both accept and donate electrons to itself. Among the given options, only \(\mathrm{SO}_{2}\) has the capability to both get reduced to sulfur (\(\mathrm{S}\)) and oxidized to sulfate (\(\mathrm{SO}_{4}^{2-}\)), acting as a reducing and oxidizing agent.

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.

Redox Reactions
Redox reactions, short for reduction-oxidation reactions, are fundamental chemical processes where the oxidation state of atoms are changed through the transfer of electrons. In these reactions, one species loses electrons and is oxidized, while another gains electrons and is reduced. It's important to understand that reduction and oxidation always occur together—it’s a reciprocal process.

Examples in Everyday Life

Simple examples of redox reactions include the rusting of iron or the browning of fruit. In both cases, the substances involved are undergoing electron transfer leading to a change in their oxidation states.
Oxidation States
Understanding oxidation states is crucial for mastering redox reactions. An oxidation state is a numerical indicator of the degree of oxidation of an atom in a compound. It’s a hypothetical charge that an atom would have if all bonds were purely ionic. This concept is applied as a bookkeeping tool to keep track of electrons in chemical reactions.

Determining Oxidation States

Oxidation states can generally be determined by following a set of rules such as: oxygen usually has an oxidation state of -2 (exception being peroxides); hydrogen is +1 when bonded to non-metals and -1 when bonded to metals; and the sum of the oxidation states for all atoms in a neutral molecule must equal zero.
Chemical Reducing Agents
Chemical reducing agents are substances that bring about reduction by donating electrons to other substances and, in the process, they get oxidized. These agents are vital in redox reactions as they are the source of electrons that another species requires to be reduced.

Characteristics and Examples

Reducing agents have certain characteristics; they are often metals or hydrides, possess a low electronegativity, and have a tendency to form positive ions. Examples of common reducing agents include hydrogen gas, metals like zinc or iron, and compounds like sodium bisulfite. In electrochemistry, reducing agents supply the electrons for the reduction half of a redox reaction in an electrochemical cell.

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

A student uses a simple pendulum of exactly \(1 \mathrm{~m}\) length to determine \(\mathrm{g}\), the acceleration due to gravity. He uses a stop watch with the least count of \(1 \mathrm{sec}\) for this and records 40 seconds for 20 oscillations. For this observation, which of the following statement(s) is (are) true ? A) Error \(\Delta \mathrm{T}\) in measuring \(\mathrm{T}\), the time period, is \(0.05\) seconds B) Error \(\Delta \mathrm{T}\) in measuring \(\mathrm{T}\), the time period, is 1 second C) Percentage error in the determination of \(g\) is \(5 \%\) D) Percentage error in the determination of \(g\) is \(2.5 \%\)

Incandescent bulbs are designed by keeping in mind that the resistance of their filament increases with the increase in temperature. If at room temperature, \(100 \mathrm{~W}, 60 \mathrm{~W}\) and \(40 \mathrm{~W}\) bulbs have filament resistances \(\mathrm{R}_{100}, \mathrm{R}_{60}\) and \(\mathrm{R}_{40}\). respectively, the relation between these resistances is A) \(\frac{1}{\mathrm{R}_{100}}=\frac{1}{\mathrm{R}_{40}}+\frac{1}{\mathrm{R}_{60}}\) B) \(R_{100}=R_{40}+R_{60}\) C) \(\mathrm{R}_{100}>\mathrm{R}_{60}>\mathrm{R}_{40}\) D) \(\frac{1}{\mathrm{R}_{100}}>\frac{1}{\mathrm{R}_{60}}>\frac{1}{\mathrm{R}_{40}}\)

The bond energy (in \(\mathbf{k c a l} \mathbf{m o l}^{-1}\) ) of a C-C single bond is approximately A) 1 B) 10 C) 100 D) 1000

The value of \(\lim _{\mathrm{x} \rightarrow 0} \frac{1}{\mathrm{x}^{3}} \int_{0}^{\mathrm{x}} \frac{\mathrm{t} \ell n(1+\mathrm{t})}{\mathrm{t}^{4}+4} \mathrm{dt}\) is A) 0 B) \(\frac{1}{12}\) C) \(\frac{1}{24}\) D) \(\frac{1}{64}\)
See all solutions

Recommended explanations on English Textbooks

Text Comparison

Read Explanation

Phonetics

Read Explanation

Graphology

Read Explanation

Sociolinguistics

Read Explanation

Linguistic Terms

Read Explanation

5 Paragraph Essay

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