Chemistry: The Molecular Science
Chemistry: The Molecular Science
5th Edition
ISBN: 9781285199047
Author: John W. Moore, Conrad L. Stanitski
Publisher: Cengage Learning
bartleby

Videos

Question
Book Icon
Chapter 17, Problem 16QRT

(a)

Interpretation Introduction

Interpretation:

The given redox reaction is to be balanced and the oxidizing agent and reducing agent are to be identified.

Concept Introduction:

The chemical reaction that involves the electron transfer between two species is referred to as redox reaction.  It is any chemical reaction in which change in the oxidation number of a molecule, atom or ion takes place by either gain of electrons or loss of electron.  The specie which gets oxidized in the redox reaction is referred to as reducing agent whereas the specie which gets reduced is termed as oxidizing agent.

(a)

Expert Solution
Check Mark

Answer to Problem 16QRT

The balanced redox reaction is shown below.

  3CO(g)+O3(g)3CO2(g)

In this reaction, CO act as reducing agent and O3 act as oxidizing agent.

Explanation of Solution

The given unbalanced redox reaction is shown below.

    CO(g)+O3(g)CO2(g)

The carbon atoms in the given redox reaction are balanced.  The oxygen atoms are balanced by putting 13 as prefix to ozone.

  CO(g)+13O3(g)CO2(g)

The above balanced reaction is multiplied by 3 so as to get the balanced reaction in integers.

  3CO(g)+O3(g)3CO2(g)

Thus, the balanced chemical reaction is shown below.

  3CO(g)+O3(g)3CO2(g)

The oxidation number of carbon atom in CO and CO2 is +2 and +4 respectively.  Since there is increase in oxidation number, therefore, CO is oxidized to CO2 and CO act as reducing agent.

The oxidation number of oxygen atom in O3 and H2O is 0 and 2 respectively.  Since there is decrease in oxidation number, therefore, O3 is reduced to H2O and O3 act as oxidizing agent.

(b)

Interpretation Introduction

Interpretation:

The given redox reaction is to be balanced and the oxidizing agent and reducing agent are to be identified.

Concept Introduction:

Same as part (a).

(b)

Expert Solution
Check Mark

Answer to Problem 16QRT

The balanced redox reaction is shown below.

  H2(g)+Cl2(g)2HCl(g)

In this reaction, H2 act as reducing agent and Cl2 act as oxidizing agent.

Explanation of Solution

The given unbalanced redox reaction is shown below.

    H2(g)+Cl2(g)HCl(g)

The hydrogen atoms and chlorine atoms are balanced by putting 12 as prefix to both hydrogen and chlorine.

  12H2(g)+12Cl2(g)HCl(g)

The above balanced reaction is multiplied by 2 so as to get the balanced reaction in integers.

  H2(g)+Cl2(g)2HCl(g)

Thus, the balanced chemical reaction is shown below.

  H2(g)+Cl2(g)2HCl(g)

The oxidation number of hydrogen atom in H2 and H2O is 0 and +1 respectively.  Since there is increase in oxidation number, therefore, H2 is oxidized to H2O and H2 act as reducing agent.

The oxidation number of chlorine atom in Cl2 and HCl is 0 and 1 respectively.  Since there is decrease in oxidation number, therefore, Cl2 is reduced to HCl and Cl2 act as oxidizing agent.

(c)

Interpretation Introduction

Interpretation:

The given redox reaction is to be balanced in acidic medium and the oxidizing agent and reducing agent are to be identified.

Concept Introduction:

Same as part (a).

(c)

Expert Solution
Check Mark

Answer to Problem 16QRT

The balanced redox reaction is shown below.

  H2O2(aq)+Ti2+(aq)TiO2(s)+2H+(aq)

In this reaction, Ti2+ act as reducing agent and H2O2 act as oxidizing agent.

Explanation of Solution

The given unbalanced redox reaction is shown below.

    H2O2(aq)+Ti2+(aq)H2O(l)+TiO2(s)

The oxidation number of titanium atom in Ti2+ and TiO2 is +2 and +4 respectively.  Since there is increase in oxidation number, therefore, Ti2+ is oxidized to TiO2 and Ti2+ act as reducing agent.

The oxidation number of oxygen atom in H2O2 and H2O is 1 and 2 respectively.  Since there is decrease in oxidation number, therefore, H2O2 is reduced to H2O and H2O2 act as oxidizing agent.

The given unbalanced redox reaction is shown below.

  H2O2(aq)+Ti2+(aq)H2O(l)+TiO2(s)

The oxidation number of titanium atom in Ti2+ and TiO2 is +2 and +4 respectively.  Since there is increase in oxidation number, therefore, Ti2+ is oxidized to TiO2 and Ti2+ act as reducing agent.

The oxidation number of oxygen atom in H2O2 and H2O is 1 and 2 respectively.  Since there is decrease in oxidation number, therefore, H2O2 is reduced to H2O and H2O2 act as oxidizing agent.

Therefore, the unbalanced half reactions are shown below.

Oxidation: Ti2+TiO2

Reduction: H2O2H2O

The given reaction is to be balanced in acidic medium.  Therefore, oxygen atoms are balanced by adding water molecules to the deficient side.

Oxidation: Ti2++2H2OTiO2

Reduction: H2O22H2O

After balancing oxygen atoms, now hydrogen atoms are balanced by adding protons (H+) to the deficient side.

Oxidation: Ti2++2H2OTiO2+4H+

Reduction: H2O2+2H+2H2O

The charge is on both sides is balanced by adding electrons to the more positive side of the half- reaction to equal the less positive side of the half- reaction.

Oxidation: Ti2++2H2OTiO2+4H++2e

Reduction: H2O2+2H++2e2H2O

Since, there is the equal loss and gain of electron in the above two half-cell reactions.  Therefore, the overall cell reaction is get by addition the above two half-cell reactions.

    Ti2++2H2O+H2O2+2H++2eTiO2+4H++2e+2H2O

The simplified chemical equation after removing the chemical species of the similar kind is shown below.

    H2O2(aq)+Ti2+(aq)TiO2(s)+2H+(aq)

Thus, the balanced chemical reaction is shown below.

  H2O2(aq)+Ti2+(aq)TiO2(s)+2H+(aq)

(d)

Interpretation Introduction

Interpretation:

The given redox reaction is to be balanced in acidic medium and the oxidizing agent and reducing agent are to be identified.

Concept Introduction:

Same as part (a).

(d)

Expert Solution
Check Mark

Answer to Problem 16QRT

The balanced redox reaction is shown below.

  6Cl(aq)+2MnO4(aq)+8H+(aq)3Cl2(g)+2MnO2(s)+4H2O(l)

In this reaction, Cl act as reducing agent and MnO4 act as oxidizing agent.

Explanation of Solution

The given unbalanced redox reaction is shown below.

    Cl(aq)+MnO4(aq)Cl2(g)+MnO2(s)

The oxidation number of chlorine atom in Cl and Cl2 is 1 and 0 respectively.  Since there is increase in oxidation number, therefore, Cl is oxidized to Cl2 and Cl act as reducing agent.

The oxidation number of manganese atom in MnO4 and MnO2 is +7 and +4 respectively.  Since there is decrease in oxidation number, therefore, MnO4 is reduced to MnO2 and MnO4 act as oxidizing agent.

Therefore, the unbalanced half reactions are shown below.

Oxidation: Cl Cl2

Reduction: MnO4MnO2

The given reaction is to be balanced in acidic medium.  Therefore, oxygen atoms are balanced by adding water molecules to the deficient side.

Oxidation: 2Cl Cl2

Reduction: MnO4MnO2+2H2O

After balancing oxygen atoms, now hydrogen atoms are balanced by adding protons (H+) to the deficient side.

Oxidation: 2Cl Cl2

Reduction: MnO4+4H+MnO2+2H2O

The charge is on both sides is balanced by adding electrons to the more positive side of the half- reaction to equal the less positive side of the half- reaction.

Oxidation: 2Cl Cl2+2e

Reduction: MnO4+4H++3eMnO2+2H2O

Since the electron gain is not equivalent to electron lost.  Thus, multiply the oxidation half reaction by 3 and reduction half reaction by 2.

Oxidation: 6Cl3Cl2+6e

Reduction: 2MnO4+8H++6e2MnO2+4H2O

Now, there is the equal loss and gain of electron in the above two half-cell reactions.  Therefore, the overall cell reaction is get by addition the above two half-cell reactions.

    6Cl(aq)+2MnO4(aq)+8H+(aq)+6e3Cl2(g)+6e+2MnO2(s)+4H2O(l)

The simplified chemical equation after removing the chemical species of the similar kind is shown below.

  6Cl(aq)+2MnO4(aq)+8H+(aq)3Cl2(g)+2MnO2(s)+4H2O(l)

Thus, the balanced chemical reaction is shown below.

  6Cl(aq)+2MnO4(aq)+8H+(aq)3Cl2(g)+2MnO2(s)+4H2O(l)

(e)

Interpretation Introduction

Interpretation:

The given redox reaction is to be balanced and the oxidizing agent and reducing agent are to be identified.

Concept Introduction:

Same as part (a).

(e)

Expert Solution
Check Mark

Answer to Problem 16QRT

The balanced redox reaction is shown below.

  4FeS2(s)+11O2(g)2Fe2O3(s)+8SO2(g)

In this reaction, FeS2 act as reducing agent and O2 act as oxidizing agent.

Explanation of Solution

The given unbalanced redox reaction is shown below.

    FeS2(s)+O2(g)Fe2O3(s)+SO2(g)

The oxidation number of iron atom in FeS2 and Fe2O3 is +2 and +3 respectively.  Since there is increase in oxidation number, therefore, FeS2 is oxidized to Fe2O3 and FeS2 act as reducing agent.

The oxidation number of oxygen atom in O2 and H2O is 0 and 2 respectively.  Since there is decrease in oxidation number, therefore, O2 is reduced to H2O and O2 act as oxidizing agent.

Therefore, the unbalanced half reactions are shown below.

Oxidation: FeS2Fe2O3+SO2

Reduction: O2H2O

The iron atom and sulphur atoms are balanced by putting 2 and 4 as prefix to FeS2 and SO2 respectively.

Oxidation: 2FeS2Fe2O3+4SO2

Reduction: O2H2O

The given reaction is to be balanced in acidic medium.  Therefore, oxygen atoms are balanced by adding water molecules to the deficient side.

Oxidation: 2FeS2+11H2OFe2O3+4SO2

Reduction: O22H2O

After balancing oxygen atoms, now hydrogen atoms are balanced by adding protons (H+) to the deficient side.

Oxidation: 2FeS2+11H2OFe2O3+4SO2+22H+

Reduction: O2+4H+2H2O

The charge is on both sides is balanced by adding electrons to the more positive side of the half- reaction to equal the less positive side of the half- reaction.

Oxidation: 2FeS2+11H2OFe2O3+4SO2+22H++22e

Reduction: O2+4H++4e2H2O

Since the electron gain is not equivalent to electron lost.  Thus, multiply the oxidation half reaction by 2 and reduction half reaction by 11.

Oxidation: 4FeS2+22H2O2Fe2O3+8SO2+44H++44e

Reduction: 11O2+44H++44e22H2O

Now, there is the equal loss and gain of electron in the above two half-cell reactions.  Therefore, the overall cell reaction is get by addition the above two half-cell reactions.

    4FeS2+22H2O+11O2+44H++44e2Fe2O3+8SO2+44H++44e+22H2O

The simplified chemical equation after removing the chemical species of the similar kind is shown below.

  4FeS2(s)+11O2(g)2Fe2O3(s)+8SO2(g)

Thus, the balanced chemical reaction is shown below.

  4FeS2(s)+11O2(g)2Fe2O3(s)+8SO2(g)

(f)

Interpretation Introduction

Interpretation:

The given redox reaction is to be balanced and the oxidizing agent and reducing agent are to be identified.

Concept Introduction:

Same as part (a).

(f)

Expert Solution
Check Mark

Answer to Problem 16QRT

The balanced redox reaction is shown below.

  O3(g)+NO(g)O2(g)+NO2(g)

In this reaction, NO act as reducing agent and O3 act as oxidizing agent.

Explanation of Solution

The given redox reaction is shown below.

  O3(g)+NO(g)O2(g)+NO2(g)

The given redox reaction is already balanced and there are equal numbers of each atom both the sides.

The oxidation number of nitrogen atom in NO and NO2 is +2 and +4 respectively.  Since there is increase in oxidation number, therefore, NO is oxidized to NO2 and NO act as reducing agent.

The oxidation number of oxygen atom in O3 and NO2 is 0 and 2 respectively.  Since there is decrease in oxidation number, therefore, O3 is reduced to NO2 and O3 act as oxidizing agent.

(g)

Interpretation Introduction

Interpretation:

The given redox reaction is to be balanced in basic medium and the oxidizing agent and reducing agent are to be identified.

Concept Introduction:

Same as part (a).

(g)

Expert Solution
Check Mark

Answer to Problem 16QRT

The balanced redox reaction is shown below.

  Zn(s)+HgO(s)+H2O(l)Zn(OH)2(s)+Hg(l)

In this reaction, Zn act as reducing agent and HgO act as oxidizing agent.

Explanation of Solution

The given unbalanced redox reaction is shown below.

  Zn(s)+HgO(s)Zn(OH)2(s)+Hg(l)

The oxidation number of zinc atom in Zn and Zn(OH)2 is 0 and +2 respectively.  Since there is increase in oxidation number, therefore, Zn is oxidized to Zn(OH)2 and Zn act as reducing agent.

The oxidation number of mercury atom in HgO and Hg is +2 and 0 respectively.  Since there is decrease in oxidation number, therefore, HgO is reduced to Hg and HgO act as oxidizing agent.

Therefore, the unbalanced half reactions are shown below.

Oxidation: ZnZn(OH)2

Reduction: HgOHg

The given reaction is to be balanced in basic medium.  Therefore, oxygen atoms are balanced by adding water molecules to the deficient side.

Oxidation: Zn+2H2OZn(OH)2

Reduction: HgOHg+H2O

After balancing oxygen atoms, hydrogen atoms are balanced by adding water molecules (H2O) to the deficient side and equal number of hydroxide ions (OH) to the other side.

Oxidation: Zn+2H2O+2OHZn(OH)2+2H2O

Reduction: HgO+2H2OHg+H2O+2OH

The charge is on both sides is balanced by adding electrons to the more positive side of the half- reaction to equal the less positive side of the half- reaction.

Oxidation: Zn+2H2O+2OHZn(OH)2+2H2O+2e

Reduction: HgO+2H2O+2eHg+H2O+2OH

Now, there is the equal loss and gain of electron in the above two half-cell reactions.  Therefore, the overall cell reaction is get by addition the above two half-cell reactions.

Zn+2H2O+2OH+HgO+2H2O+2eZn(OH)2+2H2O+2e+Hg+H2O+2OH

The simplified chemical equation after removing the chemical species of the similar kind is shown below.

  Zn(s)+HgO(s)+H2O(l)Zn(OH)2(s)+Hg(l)

Thus, the balanced chemical reaction is shown below.

  Zn(s)+HgO(s)+H2O(l)Zn(OH)2(s)+Hg(l)

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Chapter 17 Solutions

Chemistry: The Molecular Science

Ch. 17.4 - Given this reaction, its standard potential, and...Ch. 17.5 - Prob. 17.5PSPCh. 17.5 - Prob. 17.8CECh. 17.5 - Prob. 17.9CECh. 17.5 - Prob. 17.10CECh. 17.6 - Prob. 17.6PSPCh. 17.6 - Prob. 17.11ECh. 17.6 - Prob. 17.7PSPCh. 17.7 - Calculate the cell potential for the Zn(s) +...Ch. 17.7 - Prob. 17.9PSPCh. 17.8 - Prob. 17.12ECh. 17.8 - Prob. 17.13ECh. 17.8 - Prob. 17.14ECh. 17.10 - Predict the results of passing a direct electrical...Ch. 17.10 - In 1886. Henri Moissan was the first to prepare...Ch. 17.11 - In the commercial production of sodium metal by...Ch. 17.11 - Prob. 17.16CECh. 17.11 - Prob. 17.17ECh. 17.11 - Prob. 17.18CECh. 17.11 - Prob. 17.19ECh. 17.12 - Prob. 17.20CECh. 17.12 - Prob. 17.21CECh. 17 - Prob. 2SPCh. 17 - Prob. 1QRTCh. 17 - Prob. 2QRTCh. 17 - Prob. 3QRTCh. 17 - Prob. 4QRTCh. 17 - Identify each statement as true or false. Rewrite...Ch. 17 - Prob. 6QRTCh. 17 - Prob. 7QRTCh. 17 - Prob. 8QRTCh. 17 - Answer Question 8 again, but this time find a...Ch. 17 - Prob. 10QRTCh. 17 - Prob. 11QRTCh. 17 - For the reaction in Question 6, write balanced...Ch. 17 - Prob. 13QRTCh. 17 - Prob. 14QRTCh. 17 - Prob. 15QRTCh. 17 - Prob. 16QRTCh. 17 - Prob. 17QRTCh. 17 - For the reaction Cu2+(aq) + Zn(s) → Cu(s) + Zn2+...Ch. 17 - Prob. 19QRTCh. 17 - Prob. 20QRTCh. 17 - Prob. 21QRTCh. 17 - Prob. 22QRTCh. 17 - Draw a diagram of each cell. Label the anode, the...Ch. 17 - Prob. 24QRTCh. 17 - Prob. 25QRTCh. 17 - Prob. 26QRTCh. 17 - Prob. 27QRTCh. 17 - Prob. 28QRTCh. 17 - Prob. 29QRTCh. 17 - Prob. 30QRTCh. 17 - Prob. 31QRTCh. 17 - Consider these half-reactions: (a) Which is the...Ch. 17 - Consider these half-reactions: (a) Which is the...Ch. 17 - In principle, a battery could be made from...Ch. 17 - Prob. 35QRTCh. 17 - Hydrazine, N2H4, can be used as the reducing agent...Ch. 17 - Prob. 37QRTCh. 17 - Prob. 38QRTCh. 17 - Prob. 39QRTCh. 17 - Prob. 40QRTCh. 17 - Prob. 41QRTCh. 17 - Prob. 42QRTCh. 17 - Prob. 43QRTCh. 17 - Prob. 44QRTCh. 17 - Prob. 45QRTCh. 17 - Prob. 46QRTCh. 17 - Consider the voltaic cell 2 Ag+(aq) + Cd(s) 2...Ch. 17 - Consider a voltaic cell with the reaction H2(g) +...Ch. 17 - Calculate the cell potential of a concentration...Ch. 17 - Prob. 50QRTCh. 17 - Prob. 51QRTCh. 17 - Prob. 52QRTCh. 17 - Prob. 53QRTCh. 17 - NiCad batteries are rechargeable and are commonly...Ch. 17 - Prob. 55QRTCh. 17 - Prob. 56QRTCh. 17 - Prob. 57QRTCh. 17 - Hydrazine, N2H4, has been proposed as the fuel in...Ch. 17 - Consider the electrolysis of water in the presence...Ch. 17 - Prob. 60QRTCh. 17 - Prob. 61QRTCh. 17 - Prob. 62QRTCh. 17 - Identify the products of the electrolysis of a 1-M...Ch. 17 - Prob. 64QRTCh. 17 - Prob. 65QRTCh. 17 - Prob. 66QRTCh. 17 - Prob. 67QRTCh. 17 - Prob. 68QRTCh. 17 - Prob. 69QRTCh. 17 - Prob. 70QRTCh. 17 - Prob. 71QRTCh. 17 - Prob. 72QRTCh. 17 - Prob. 73QRTCh. 17 - Prob. 74QRTCh. 17 - Calculate how long it would take to electroplate a...Ch. 17 - Prob. 76QRTCh. 17 - Prob. 77QRTCh. 17 - Prob. 78QRTCh. 17 - Prob. 79QRTCh. 17 - Prob. 80QRTCh. 17 - Prob. 81QRTCh. 17 - Prob. 82QRTCh. 17 - Prob. 83QRTCh. 17 - Prob. 84QRTCh. 17 - Prob. 85QRTCh. 17 - Prob. 86QRTCh. 17 - Prob. 87QRTCh. 17 - Prob. 88QRTCh. 17 - You wish to electroplate a copper surface having...Ch. 17 - Prob. 90QRTCh. 17 - Prob. 91QRTCh. 17 - Prob. 92QRTCh. 17 - Prob. 93QRTCh. 17 - An electrolytic cell is set up with Cd(s) in...Ch. 17 - Prob. 95QRTCh. 17 - Prob. 96QRTCh. 17 - Prob. 97QRTCh. 17 - Prob. 98QRTCh. 17 - Prob. 99QRTCh. 17 - Prob. 100QRTCh. 17 - Prob. 101QRTCh. 17 - Prob. 102QRTCh. 17 - Prob. 103QRTCh. 17 - Prob. 104QRTCh. 17 - Prob. 105QRTCh. 17 - Prob. 106QRTCh. 17 - Prob. 107QRTCh. 17 - Prob. 108QRTCh. 17 - Prob. 109QRTCh. 17 - Prob. 110QRTCh. 17 - Prob. 111QRTCh. 17 - Prob. 17.ACPCh. 17 - Prob. 17.BCP
Knowledge Booster
Background pattern image
Chemistry
Learn more about
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, chemistry and related others by exploring similar questions and additional content below.
Similar questions
SEE MORE QUESTIONS
Recommended textbooks for you
Text book image
Chemistry: The Molecular Science
Chemistry
ISBN:9781285199047
Author:John W. Moore, Conrad L. Stanitski
Publisher:Cengage Learning
Text book image
Chemistry
Chemistry
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Cengage Learning
Text book image
Introduction to General, Organic and Biochemistry
Chemistry
ISBN:9781285869759
Author:Frederick A. Bettelheim, William H. Brown, Mary K. Campbell, Shawn O. Farrell, Omar Torres
Publisher:Cengage Learning
Text book image
Chemistry: Principles and Practice
Chemistry
ISBN:9780534420123
Author:Daniel L. Reger, Scott R. Goode, David W. Ball, Edward Mercer
Publisher:Cengage Learning
Text book image
Chemistry & Chemical Reactivity
Chemistry
ISBN:9781133949640
Author:John C. Kotz, Paul M. Treichel, John Townsend, David Treichel
Publisher:Cengage Learning
Text book image
Chemistry: An Atoms First Approach
Chemistry
ISBN:9781305079243
Author:Steven S. Zumdahl, Susan A. Zumdahl
Publisher:Cengage Learning
Balancing Redox Reactions in Acidic and Basic Conditions; Author: Professor Dave Explains;https://www.youtube.com/watch?v=N6ivvu6xlog;License: Standard YouTube License, CC-BY