/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 15 Indicate whether each of the fol... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 15: Problem 15

Indicate whether each of the following describes the oxidizing agent or the reducing agent in an oxidation-reduction reaction: a. the substance that is oxidized b. the substance that gains electrons

Short Answer

Expert verified
a. Reducing agentb. Oxidizing agent

Step by step solution

01

- Understand Oxidation and Reduction

Oxidation is the process where a substance loses electrons. Reduction is the process where a substance gains electrons. In a redox reaction, one substance is oxidized (loses electrons) and another is reduced (gains electrons).
02

- Identify the Oxidizing Agent

The oxidizing agent is the substance that gains electrons. It causes the oxidation of another substance and is reduced itself in the process.
03

- Identify the Reducing Agent

The reducing agent is the substance that loses electrons. It causes the reduction of another substance and is oxidized itself in the process.
04

- Apply to Given Description

a. The substance that is oxidized: Since it loses electrons, it is the reducing agent. b. The substance that gains electrons: Since it gains electrons, it is the 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.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

Oxidizing Agent
In a redox (oxidation-reduction) reaction, the oxidizing agent plays a crucial role. It is the substance that gains electrons. When it gains these electrons, it undergoes a chemical change and is reduced in the process.
An easy way to remember this is with the phrase 'OIL RIG' - Oxidation Is Loss (of electrons), Reduction Is Gain (of electrons). The oxidizing agent causes another substance to lose electrons (be oxidized), while it itself gains those electrons and is reduced.
  • Example: In the reaction between hydrogen and oxygen to form water, oxygen acts as the oxidizing agent. It gains electrons from hydrogen and is reduced to form water (H2O).
Reducing Agent
The reducing agent in a redox reaction is the substance that loses electrons. It donates these electrons to another substance, causing that substance to be reduced, while it itself is oxidized. Again, referring back to the 'OIL RIG' phrase, the reducing agent is undergoing oxidation because it is losing electrons.
When identifying the reducing agent, look for the substance that donates electrons.
  • Example: Continuing with our water formation reaction, hydrogen acts as the reducing agent. It loses electrons to oxygen and is oxidized to form water (H2O).
Electrons
Electrons are subatomic particles with a negative charge. They play a critical role in oxidation-reduction reactions. Whether a substance gains or loses electrons determines if it is oxidized or reduced.
In a redox reaction:
  • Oxidation involves the loss of electrons.
  • Reduction involves the gain of electrons.
Tracking the flow of electrons in a reaction helps in identifying both the oxidizing and reducing agents.
Understanding how electrons transfer during these reactions can help you predict the behavior of different substances in redox reactions and correctly identify oxidizing and reducing agents.
  • Example: In the formation of water, electrons move from hydrogen atoms (reducing agent) to oxygen atoms (oxidizing agent).

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

Describe the voltaic cell and half-cell components and write the shorthand notation for the following oxidation-reduction reactions: a. \(\mathrm{Cd}(s)+\mathrm{Sn}^{2+}(a q) \longrightarrow \mathrm{Cd}^{2+}(a q)+\operatorname{Sn}(s)\) b. \(\mathrm{Zn}(s)+\mathrm{Cl}_{2}(g) \longrightarrow \mathrm{Zn}^{2+}(a q)+2 \mathrm{Cl}^{-}(a q) \mathrm{C}\) (electrode)

Assign oxidation numbers to all the elements in each of the following reactions, and identify the reactant that is oxidized, the reactant that is reduced, the oxidizing agent, and the reducing agent: (15.1) a. \(2 \mathrm{FeCl}_{2}(a q)+\mathrm{Cl}_{2}(g) \longrightarrow 2 \mathrm{FeCl}_{3}(a q)\) b. \(2 \mathrm{H}_{2} \mathrm{~S}(g)+3 \mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{H}_{2} \mathrm{O}(I)+2 \mathrm{SO}_{2}(g)\) c. \(\mathrm{P}_{2} \mathrm{O}_{5}(s)+5 \mathrm{C}(s) \longrightarrow 2 \mathrm{P}(s)+5 \mathrm{CO}(g)\)

The following half-reaction takes place in a nickel-cadmium battery used in a cordless drill: $$ \mathrm{Cd}(s)+2 \mathrm{OH}^{-}(a q) \longrightarrow \mathrm{Cd}(\mathrm{OH})_{2}(s)+2 e^{-} $$ a. Is the half-reaction an oxidation or a reduction? b. What substance is oxidized or reduced? c. At which electrode would this half-reaction occur?

A concentrated nitric acid solution is used to dissolve copper(II) sulfide. \((12.6,15.2)\) $$ \mathrm{CuS}(s)+\mathrm{HNO}_{3}(a q) \longrightarrow \mathrm{CuSO}_{4}(a q)+\mathrm{NO}(g)+\mathrm{H}_{2} \mathrm{O}(l) $$ a. Write the balanced equation. b. How many milliliters of a \(16.0 \mathrm{M} \mathrm{HNO}_{3}\) solution are needed to dissolve \(24.8 \mathrm{~g}\) of \(\mathrm{CuS}\) ?

What we call "tin cans" are really iron cans coated with a thin layer of tin. The anode is a bar of tin and the cathode is the iron can. An electrical current is used to oxidize the Sn to \(\mathrm{Sn}^{2+}\) in solution, which is reduced to produce a thin coating of \(\mathrm{Sn}\) on the can. a. What half-reaction takes place to tin plate an iron can? b. Why is the iron can the cathode? c. Why is the tin bar the anode?
See all solutions

Recommended explanations on Chemistry Textbooks

Ionic and Molecular Compounds

Read Explanation

Physical Chemistry

Read Explanation

Chemical Reactions

Read Explanation

Kinetics

Read Explanation

The Earths Atmosphere

Read Explanation

Making Measurements

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.