Chemistry: The Science in Context (Fifth Edition)
Chemistry: The Science in Context (Fifth Edition)
5th Edition
ISBN: 9780393614046
Author: Thomas R. Gilbert, Rein V. Kirss, Natalie Foster, Stacey Lowery Bretz, Geoffrey Davies
Publisher: W. W. Norton & Company
bartleby

Videos

Question
Book Icon
Chapter 18, Problem 18.19QP

(a)

Interpretation Introduction

Interpretation: The net ionic equations for the given reactions are to be written with the help of the appropriate half reactions from Appendix 6.

Concept introduction: Voltaic cell is a type of electrochemical cell which converts the chemical energy to electrical energy. This chemical energy generates by processing the redox reaction. Redox reaction is the combination of two half reactions that are oxidation half and reduction half. Oxidation is a process in which loss of electrons take place, while in reduction gain of electron take place.

In voltaic cell during the process of redox reaction electrons are generated on anode and then transfer to the cathode by the external circuit. In this way we can say that the process of oxidation occurs at the anode and the process of reduction occurs at cathode and hence, electron flows.

The species with higher reduction potential always undergoes reduction. The species with lower reduction potential always undergoes oxidation.

In a voltaic cell, for the spontaneous cell reaction the overall cell potential must be positive.

To determine: The net ionic equation for the given reaction between Aluminum metal and Fe3+ ions in solution that produces dissolved Al3+ and Fe2+ .

(a)

Expert Solution
Check Mark

Answer to Problem 18.19QP

Solution

The net ionic equation for the given reaction is given as,

Al(s)+3Fe3+(aq)Al3+(aq)+3Fe2+(aq)

Explanation of Solution

Explanation

As given above that the net ionic reaction is the combination of two half reactions that are oxidation half and reduction half.

With the help of reduction values given in Appendix 6, the reactions are written as,

For aluminum,

Al3+(aq)+3eAl(s)Eο=1.66

For Fe3+ ,

Fe3+(aq)+eFe2+(aq)Eο=+0.77

The first reaction undergoes oxidation due to lower value of reduction potential. It is represented as,

Al(s)Al3+(aq)+3eEο=+1.66 (1)

The sign of value of reduction potential is revered if the reaction is represented in reverse manner. Therefore, for the first reaction the sigh has been reversed.

In second reaction undergoes reduction due to higher value of reduction potential. It is represented as it is.

Fe3+(aq)+eFe2+(aq)Eο=+0.77 (2)

To balance the reactions, equation (2) is multiplied by three. It is represented as,

3Fe3+(aq)+3e3Fe2+(aq)Eο=+0.77 (3)

Now, add the equation (1) and (3). It is represented as,

Al(s)Al3+(aq)+3eEο=+1.663Fe3+(aq)+3e3Fe2+(aq)Eο=+0.77Al(s)+3Fe3+(aq)Al3+(aq)+3Fe2+(aq) (4)

The three electrons present in both the equations are cancelled with each other. Therefore, the equation (4) is the net ionic balanced equation.

(b)

Interpretation Introduction

To determine: The net ionic equation for the given reaction between I2 and NO2 ions in an alkaline solutions that produces I and NO3 ions.

(b)

Expert Solution
Check Mark

Answer to Problem 18.19QP

Solution

The net ionic equation for the given reaction is given as,

NO2(aq)+I2(aq)+2OHNO3(aq)+2I(aq)+H2O(l)

Explanation of Solution

Explanation

As given above that the net ionic reaction is the combination of two half reactions that are oxidation half and reduction half.

The molecule of I2 is converted to the I ion and NO2 is converted to the NO3 ions. Therefore, I2 undergoes reduction and NO2 undergoes oxidation.

With the help of reduction values given in Appendix 6, the reactions are written as,

For I2 ,

I2(aq)+2e2I(aq) (1)

For NO2 ,

NO2(aq)+2OHNO3(aq)+H2O(l)+2e (2)

In both the reactions there are two electrons are involved. Therefore, on the addition of the reactions both the electrons are cancelled with each other.

The net ionic reaction is written as,

NO2(aq)+I2(aq)+2OHNO3(aq)+2I(aq)+H2O(l)

(c)

Interpretation Introduction

To determine: The net ionic equation for the given reaction between MnO4 and Cr3+ ions in an acidic solution that produces Mn2+ and Cr2O72 ions.

(c)

Expert Solution
Check Mark

Answer to Problem 18.19QP

Solution

The net ionic equation for the given reaction is given as,

6MnO4(aq)+10Cr3+(aq)+11H2O(l)6Mn2+(s)+5Cr2O72(aq)+22H+

Explanation of Solution

Explanation

The molecular ion MnO4 is converted to Mn2+ ions and Cr3+ ions are converted to Cr2O72 ions in acidic condition.

The above reaction is a redox reaction. With the help of reduction values given in Appendix 6, the reactions are written as,

Chemistry: The Science in Context (Fifth Edition), Chapter 18, Problem 18.19QP MnO4(aq)Mn2+(s)Cr3+(aq)Cr2O72(aq)

The above two reactions are the simplest representation of the complete redox reaction.

Now, to get the balanced and complete reaction, both the half reactions are balanced separately and then added together.

In this method the following steps are followed,

  • The atoms other than oxygen and hydrogen are balanced.
  • After that to balance oxygen and hydrogen water molecules are added.
  • After that in acidic condition hydrogen ions are added to balance the hydrogen.
  • After that charges are balanced by the addition of electrons.
  • After the balancing of both the half reactions, both are added together.

Now, apply these rules on first half reaction that is reduction half,

MnO4(aq)Mn2+(s)

Both the side same number of manganese atom are present therefore, no coefficient is added either side. The equation is represented as,

MnO4(aq)Mn2+(s)

Now, oxygen atoms are balanced by putting the two water molecule to the right side. Equation is represented as,

MnO4(aq)Mn2+(s)+4H2O(l)

Now, add eight hydrogen ions to the left side to balance the hydrogen atom. The equation is represented as,

MnO4(aq)+8H+Mn2+(s)+4H2O(l)

Now, left side has plus seven charge and right side has plus two charge. Therefore, to balance the charge five electrons are added to the left side. The equation is represented as,

MnO4(aq)+8H+(aq)+5eMn2+(s)+4H2O(l) (1)

Now, apply these rules on second half reaction that is oxidation half,

Cr3+(aq)Cr2O72(aq)

Add the coefficient 2 on the left side. The equation is represented as,

2Cr3+(aq)Cr2O72(aq)

Now, oxygen atoms are balanced by the addition of seven water molecules to the left side. Equation is represented as,

2Cr3+(aq)+7H2O(l)Cr2O72(aq)

Now, add fourteen hydrogen ions to the right side to balance the hydrogen atom. The equation is represented as,

2Cr3+(aq)+7H2O(l)Cr2O72(aq)+14H+

Now, right side has plus twelve charge and left side has six positive charge. Therefore, to balance the charge six electrons are added to the right side. The equation is represented as,

2Cr3+(aq)+7H2O(l)Cr2O72(aq)+14H++6e (2)

Now, to balance the electrons in equation (1) and (2), the equation (1) is multiplied by the coefficient six and equation (2) is multiplied by the coefficient five. In combined it is represented as,

6MnO4(aq)+48H+(aq)+30e6Mn2+(s)+24H2O(l)10Cr3+(aq)+35H2O(l)5Cr2O72(aq)+70H++30e

Now, add both the above equations. All the electrons are cancelled with each other. Overall reaction is represented as,

6MnO4(aq)+10Cr3+(aq)+11H2O(l)6Mn2+(s)+5Cr2O72(aq)+22H+

This is the net ionic equation.

Conclusion

The net ionic equations for the given reactions are given as,

  1. a) Al(s)+3Fe3+(aq)Al3+(aq)+3Fe2+(aq)
  2. b) NO2(aq)+I2(aq)+2OHNO3(aq)+2I(aq)+H2O(l)
  3. c) 6MnO4(aq)+10Cr3+(aq)+11H2O(l)6Mn2+(s)+5Cr2O72(aq)+22H+

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 18 Solutions

Chemistry: The Science in Context (Fifth Edition)

Ch. 18 - Prob. 18.3VPCh. 18 - Prob. 18.4VPCh. 18 - Prob. 18.5VPCh. 18 - Prob. 18.6VPCh. 18 - Prob. 18.7VPCh. 18 - Prob. 18.8VPCh. 18 - Prob. 18.9VPCh. 18 - Prob. 18.10VPCh. 18 - Prob. 18.11QPCh. 18 - Prob. 18.12QPCh. 18 - Prob. 18.13QPCh. 18 - Prob. 18.14QPCh. 18 - Prob. 18.15QPCh. 18 - Prob. 18.16QPCh. 18 - Prob. 18.17QPCh. 18 - Prob. 18.18QPCh. 18 - Prob. 18.19QPCh. 18 - Prob. 18.20QPCh. 18 - Prob. 18.21QPCh. 18 - Prob. 18.22QPCh. 18 - Prob. 18.23QPCh. 18 - Prob. 18.24QPCh. 18 - Prob. 18.25QPCh. 18 - Prob. 18.26QPCh. 18 - Prob. 18.27QPCh. 18 - Prob. 18.28QPCh. 18 - Prob. 18.29QPCh. 18 - Prob. 18.30QPCh. 18 - Prob. 18.31QPCh. 18 - Prob. 18.32QPCh. 18 - Prob. 18.33QPCh. 18 - Prob. 18.34QPCh. 18 - Prob. 18.35QPCh. 18 - Prob. 18.36QPCh. 18 - Prob. 18.37QPCh. 18 - Prob. 18.38QPCh. 18 - Prob. 18.39QPCh. 18 - Prob. 18.40QPCh. 18 - Prob. 18.41QPCh. 18 - Prob. 18.42QPCh. 18 - Prob. 18.43QPCh. 18 - Prob. 18.44QPCh. 18 - Prob. 18.45QPCh. 18 - Prob. 18.46QPCh. 18 - Prob. 18.47QPCh. 18 - Prob. 18.48QPCh. 18 - Prob. 18.49QPCh. 18 - Prob. 18.50QPCh. 18 - Prob. 18.51QPCh. 18 - Prob. 18.52QPCh. 18 - Prob. 18.53QPCh. 18 - Prob. 18.54QPCh. 18 - Prob. 18.55QPCh. 18 - Prob. 18.56QPCh. 18 - Prob. 18.57QPCh. 18 - Prob. 18.58QPCh. 18 - Prob. 18.59QPCh. 18 - Prob. 18.60QPCh. 18 - Prob. 18.61QPCh. 18 - Prob. 18.62QPCh. 18 - Prob. 18.63QPCh. 18 - Prob. 18.64QPCh. 18 - Prob. 18.65QPCh. 18 - Prob. 18.66QPCh. 18 - Prob. 18.67QPCh. 18 - Prob. 18.68QPCh. 18 - Prob. 18.69QPCh. 18 - Prob. 18.70QPCh. 18 - Prob. 18.71QPCh. 18 - Prob. 18.72QPCh. 18 - Prob. 18.73QPCh. 18 - Prob. 18.74QPCh. 18 - Prob. 18.75QPCh. 18 - Prob. 18.76QPCh. 18 - Prob. 18.77QPCh. 18 - Prob. 18.78QPCh. 18 - Prob. 18.79QPCh. 18 - Prob. 18.80QPCh. 18 - Prob. 18.81QPCh. 18 - Prob. 18.82QPCh. 18 - Prob. 18.83QPCh. 18 - Prob. 18.84QPCh. 18 - Prob. 18.85QPCh. 18 - Prob. 18.86QPCh. 18 - Prob. 18.87QPCh. 18 - Prob. 18.88QPCh. 18 - Prob. 18.89QPCh. 18 - Prob. 18.90QPCh. 18 - Prob. 18.91APCh. 18 - Prob. 18.92APCh. 18 - Prob. 18.93APCh. 18 - Prob. 18.94APCh. 18 - Prob. 18.95APCh. 18 - Prob. 18.96APCh. 18 - Prob. 18.97APCh. 18 - Prob. 18.98APCh. 18 - Prob. 18.99APCh. 18 - Prob. 18.100AP
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.
Recommended textbooks for you
Text book image
Chemistry
Chemistry
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Cengage Learning
Text book image
Chemistry
Chemistry
ISBN:9781259911156
Author:Raymond Chang Dr., Jason Overby Professor
Publisher:McGraw-Hill Education
Text book image
Principles of Instrumental Analysis
Chemistry
ISBN:9781305577213
Author:Douglas A. Skoog, F. James Holler, Stanley R. Crouch
Publisher:Cengage Learning
Text book image
Organic Chemistry
Chemistry
ISBN:9780078021558
Author:Janice Gorzynski Smith Dr.
Publisher:McGraw-Hill Education
Text book image
Chemistry: Principles and Reactions
Chemistry
ISBN:9781305079373
Author:William L. Masterton, Cecile N. Hurley
Publisher:Cengage Learning
Text book image
Elementary Principles of Chemical Processes, Bind...
Chemistry
ISBN:9781118431221
Author:Richard M. Felder, Ronald W. Rousseau, Lisa G. Bullard
Publisher:WILEY
Introduction to Electrochemistry; Author: Tyler DeWitt;https://www.youtube.com/watch?v=teTkvUtW4SA;License: Standard YouTube License, CC-BY