Chemistry
Chemistry
4th Edition
ISBN: 9780393919370
Author: Thomas R. Gilbert
Publisher: NORTON
Question
Book Icon
Chapter 4, Problem 4.140AP

(a)

Interpretation Introduction

Interpretation: The questions corresponding to the given reaction are to be answered.

Concept introduction: In oxidation there is loss of electrons occur which increases the oxidation state of an atom whereas in reduction process there is gaining of electrons occur which decreases the oxidation state of an atom.

The reactants and the products in a. given ionic equation must be balanced for the given number of ions or atoms of each element.

To determine: The oxidation numbers of nitrogen in the reactants and products of each reaction.

(a)

Expert Solution
Check Mark

Answer to Problem 4.140AP

Solution

The oxidation number of molecular N2 is always be 0_ and the oxidation number of N in NH3 is -3_ in the first reaction of nitrogen fixation.

The oxidation number of N in NH4+ is -3_ and the oxidation number of N in NO2 is +3_ in the nitrification reaction.

Explanation of Solution

Explanation

The first given reaction of nitrogen fixation is,

N2(g)+H+(aq)+M2+(aq)NH3(aq)+H2(g)+M3+(aq)

Where,

  • M is the transition element such as iron.

The oxidation state of molecular N2 is always be 0_ .

The oxidation number nitrogen in the given compound NH3 is calculated as follows.

The oxidation state hydrogen is usually +1 . The oxidation number of N in NH3 is assumed to be x and is calculated by using the formula,

Charge on NH3=[(Number ofNatoms×oxidation number of N)+(Number of H atoms×oxidation number of H)]

Since, the charge on NH3 is 0 .

Substitute the number of atoms and their oxidation number in the above formula to calculate the oxidation number of nitrogen.

Charge on NH3=[(x)+(3×(1))]0=[x+3]x=-3_

Thus, the oxidation number of N in NH3 is -3_ .

Hence, the oxidation number of molecular N2 is always be 0_ and the oxidation number of N in NH3 is -3_ in the first reaction of nitrogen fixation.

The second nitrification reaction is,

NH4+(aq)+M3+(aq)NO2(aq)+M2+(aq)

Where,

  • M is the transition element such as iron.

The oxidation number nitrogen in the given compound NH4+ is calculated as follows.

The oxidation state hydrogen is usually +1 .

The oxidation number of N in NH4+ is assumed to be x and is calculated by using the formula,

Charge on NH4+=[(Number ofNatoms×oxidation number of N)+(Number of H atoms×oxidation number of H)]

Since, the charge on NH4+ is +1 .

Substitute the number of atoms and their oxidation number in the above formula to calculate the oxidation number of nitrogen.

Charge on NH4+=[(x)+(4×(1))]+1=[x+4]x=-3_

Thus, the oxidation number of N in NH4+ is -3_ .

The oxidation number nitrogen in the given compound NO2 is calculated as follows.

The common oxidation state oxygen is usually 2 . The oxidation number of N in NO2 is assumed to be x and is calculated by using the formula,

Charge on NO2=[(Number ofNatoms×oxidation number of N)+(Number of O atoms×oxidation number of O)]

Since, the charge on NO2 is 1 .

Substitute the number of atoms and their oxidation number in the above formula to calculate the oxidation number of nitrogen.

Charge on NO2=[(x)+(2×(2))]1=[x4]x=+3_

Thus, the oxidation number of N in NO2 is +3_ .

Hence, the oxidation number of N in NH4+ is -3_ and the oxidation number of N in NO2 is +3_ in the second reaction of nitrogen fixation.

Therefore, The oxidation number of molecular N2 is always be 0_ and the oxidation number of N in NH3 is -3_ in the first reaction of nitrogen fixation. The oxidation number of N in NH4+ is -3_ and the oxidation number of N in NO2 is +3_ in the second nitrification reaction.

(b)

Interpretation Introduction

To determine: The compounds or ions that are being reduced in each reaction.

(b)

Expert Solution
Check Mark

Answer to Problem 4.140AP

Solution

In the first reaction of nitrogen fixation the molecule N2(g) reduced to NH3(aq) and H+(aq) ion reduced to H2(g) molecule.

In the nitrification reaction M3+ ion reduced to M2+ ion.

Explanation of Solution

Explanation

The first given reaction of nitrogen fixation is,

N2(g)+H+(aq)+M2+(aq)NH3(aq)+H2(g)+M3+(aq)

Where,

  • M is the transition element such as iron.

In the above reaction the oxidation number of molecular N2 is 0 and the oxidation number of N in NH3 is 3 as shown above.

Thus, there is a decrease in the oxidation state. Hence, N2(g) reduced to NH3(aq) .

The oxidation state of H+ ion is +1 and the oxidation state of molecular H2 is always 0 .

Thus, there is a decrease in the oxidation state. Hence, H+(aq) ion reduced to H2(g) molecule.

The second given nitrification reaction is,

NH4+(aq)+M3+(aq)NO2(aq)+M2+(aq)

Where,

  • M is the transition element such as iron.

Here, the oxidation number of N in NH4+ is 3 and the oxidation number of N in NO2 is +3 in the second reaction of nitrogen fixation.

Thus, there is increase in the oxidation state. Hence, oxidation process is occurring.

The oxidation state of M3+ is +3 and oxidation state of M2+ is +2 . Thus, there is decrease in the oxidation state of M . Hence, M3+ ion reduced to M2+ ion.

Therefore, in the first reaction of nitrogen fixation the molecule N2(g) reduced to NH3(aq) and H+(aq) ion reduced to H2(g) molecule. In the nitrification reaction M3+ ion reduced to M2+ ion.

(c)

Interpretation Introduction

To determine: The balanced equations of nitrogen fixation.

(c)

Expert Solution
Check Mark

Answer to Problem 4.140AP

Solution

The balanced equation of nitrogen fixation is,

N2(g)+8H+(aq)+M2+(aq)+7e2NH3(aq)+H2(g)+M3+(aq)

The balanced equation of nitrification reaction is,

NH4+(aq)+2H2O(aq)+M3+(aq)+3eNO2(aq)+M2+(aq)+4H2(g)

Explanation of Solution

Explanation

The first given unbalanced reaction of nitrogen fixation is,

N2(g)+H+(aq)+M2+(aq)NH3(aq)+H2(g)+M3+(aq)

Where,

  • M is the transition element such as iron.

The number of nitrogen atom is balanced by placing coefficient of 2 before NH3(aq) to the product side of the equation. The equation becomes as,

N2(g)+H+(aq)+M2+(aq)2NH3(aq)+H2(g)+M3+(aq)

Now, to balance the hydrogen atom to both sides place coefficient of 8 in front of H+ ions to the reactant side of the equation as,

N2(g)+8H+(aq)+M2+(aq)2NH3(aq)+H2(g)+M3+(aq)

There is unequal net charge on the equation. There is total +10 charge on the reactant side from eight H+ ions and two from M2+ ion and on the product side there is total of only +3 charge from M3+ ion. Therefore, to balance the charge add 7 electrons to the reactant side of the equation as,

N2(g)+8H+(aq)+M2+(aq)+7e2NH3(aq)+H2(g)+M3+(aq)

Hence, the balanced equation of nitrogen fixation is,

N2(g)+8H+(aq)+M2+(aq)+7e2NH3(aq)+H2(g)+M3+(aq)

The second given nitrification reaction is,

NH4+(aq)+M3+(aq)NO2(aq)+M2+(aq)

Where,

  • M is the transition element such as iron.

The numbers of oxygen atoms are balanced by adding water molecule to the reactant side as,

NH4+(aq)+H2O(aq)+M3+(aq)NO2(aq)+M2+(aq)

The numbers of oxygen atoms are balanced by placing coefficient of 2 in front of water molecule to the reactant side as,

NH4+(aq)+2H2O(aq)+M3+(aq)NO2(aq)+M2+(aq)

To balance hydrogen ions add four molecules of H2 to the product side of the equation as,

NH4+(aq)+2H2O(aq)+M3+(aq)NO2(aq)+M2+(aq)+4H2(g)

The numbers of all the atoms are balanced now. There is unequal distribution of charge both the side as there is total of +4   charge on the reactant side and there is only +1 charge on the product side. Thus, to balance the charge on both the sides add 3 electrons to the reactant side of the equation as,

NH4+(aq)+2H2O(aq)+M3+(aq)+3eNO2(aq)+M2+(aq)+4H2(g) .

Therefore, the balanced equation of nitrification reaction is,

NH4+(aq)+2H2O(aq)+M3+(aq)+3eNO2(aq)+M2+(aq)+4H2(g)

Conclusion

  1. a) The oxidation number of molecular N2 is always be 0_ and the oxidation number of N in NH3 is -3_ in the first reaction of nitrogen fixation. The oxidation number of N in NH4+ is -3_ and the oxidation number of N in NO2 is +3_ in the second reaction of nitrogen fixation.
  2. b) In the first reaction of nitrogen fixation the molecule N2(g) reduced to NH3(aq) and H+(aq) ion reduced to H2(g) molecule. In the second reaction of nitrogen fixation M3+ ion reduced to M2+ ion.
  3. c) The balanced equation of nitrogen fixation is,

    N2(g)+8H+(aq)+M2+(aq)+7e2NH3(aq)+H2(g)+M3+(aq) .

    The balanced equation of nitrification reaction is,

    NH4+(aq)+2H2O(aq)+M3+(aq)+3eNO2(aq)+M2+(aq)+4H2(g)

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 4 Solutions

Chemistry

Ch. 4.7 - Prob. 11PECh. 4.7 - Prob. 12PECh. 4.7 - Prob. 13PECh. 4.7 - Prob. 14PECh. 4.7 - Prob. 15PECh. 4.9 - Prob. 16PECh. 4.9 - Prob. 17PECh. 4 - Prob. 4.1VPCh. 4 - Prob. 4.2VPCh. 4 - Prob. 4.3VPCh. 4 - Prob. 4.4VPCh. 4 - Prob. 4.5VPCh. 4 - Prob. 4.6VPCh. 4 - Prob. 4.7VPCh. 4 - Prob. 4.8VPCh. 4 - Prob. 4.9QPCh. 4 - Prob. 4.10QPCh. 4 - Prob. 4.11QPCh. 4 - Prob. 4.12QPCh. 4 - Prob. 4.13QPCh. 4 - Prob. 4.14QPCh. 4 - Prob. 4.15QPCh. 4 - Prob. 4.16QPCh. 4 - Prob. 4.17QPCh. 4 - Prob. 4.18QPCh. 4 - Prob. 4.19QPCh. 4 - Prob. 4.20QPCh. 4 - Prob. 4.21QPCh. 4 - Prob. 4.22QPCh. 4 - Prob. 4.23QPCh. 4 - Prob. 4.24QPCh. 4 - Prob. 4.25QPCh. 4 - Prob. 4.26QPCh. 4 - Prob. 4.27QPCh. 4 - Prob. 4.28QPCh. 4 - Prob. 4.29QPCh. 4 - Prob. 4.30QPCh. 4 - Prob. 4.31QPCh. 4 - Prob. 4.32QPCh. 4 - Prob. 4.33QPCh. 4 - Prob. 4.34QPCh. 4 - Prob. 4.35QPCh. 4 - Prob. 4.36QPCh. 4 - Prob. 4.37QPCh. 4 - Prob. 4.38QPCh. 4 - Prob. 4.39QPCh. 4 - Prob. 4.40QPCh. 4 - Prob. 4.41QPCh. 4 - Prob. 4.42QPCh. 4 - Prob. 4.43QPCh. 4 - Prob. 4.44QPCh. 4 - Prob. 4.45QPCh. 4 - Prob. 4.46QPCh. 4 - Prob. 4.47QPCh. 4 - Prob. 4.48QPCh. 4 - Prob. 4.49QPCh. 4 - Prob. 4.50QPCh. 4 - Prob. 4.51QPCh. 4 - Prob. 4.52QPCh. 4 - Prob. 4.53QPCh. 4 - Prob. 4.54QPCh. 4 - Prob. 4.55QPCh. 4 - Prob. 4.56QPCh. 4 - Prob. 4.57QPCh. 4 - Prob. 4.58QPCh. 4 - Prob. 4.59QPCh. 4 - Prob. 4.60QPCh. 4 - Prob. 4.61QPCh. 4 - Prob. 4.62QPCh. 4 - Prob. 4.63QPCh. 4 - Prob. 4.64QPCh. 4 - Prob. 4.65QPCh. 4 - Prob. 4.66QPCh. 4 - Prob. 4.67QPCh. 4 - Prob. 4.68QPCh. 4 - Prob. 4.69QPCh. 4 - Prob. 4.70QPCh. 4 - Prob. 4.71QPCh. 4 - Prob. 4.72QPCh. 4 - Prob. 4.73QPCh. 4 - Prob. 4.74QPCh. 4 - Prob. 4.75QPCh. 4 - Prob. 4.76QPCh. 4 - Prob. 4.77QPCh. 4 - Prob. 4.78QPCh. 4 - Prob. 4.79QPCh. 4 - Prob. 4.80QPCh. 4 - Prob. 4.81QPCh. 4 - Prob. 4.82QPCh. 4 - Prob. 4.83QPCh. 4 - Prob. 4.84QPCh. 4 - Prob. 4.85QPCh. 4 - Prob. 4.86QPCh. 4 - Prob. 4.87QPCh. 4 - Prob. 4.88QPCh. 4 - Prob. 4.89QPCh. 4 - Prob. 4.90QPCh. 4 - Prob. 4.91QPCh. 4 - Prob. 4.92QPCh. 4 - Prob. 4.93QPCh. 4 - Prob. 4.94QPCh. 4 - Prob. 4.95QPCh. 4 - Prob. 4.96QPCh. 4 - Prob. 4.97QPCh. 4 - Prob. 4.98QPCh. 4 - Prob. 4.99QPCh. 4 - Prob. 4.100QPCh. 4 - Prob. 4.101QPCh. 4 - Prob. 4.102QPCh. 4 - Prob. 4.103QPCh. 4 - Prob. 4.104QPCh. 4 - Prob. 4.105QPCh. 4 - Prob. 4.106QPCh. 4 - Prob. 4.107QPCh. 4 - Prob. 4.108QPCh. 4 - Prob. 4.109QPCh. 4 - Prob. 4.110QPCh. 4 - Prob. 4.111QPCh. 4 - Prob. 4.112QPCh. 4 - Prob. 4.113QPCh. 4 - Prob. 4.114QPCh. 4 - Prob. 4.115QPCh. 4 - Prob. 4.116QPCh. 4 - Prob. 4.117QPCh. 4 - Prob. 4.118QPCh. 4 - Prob. 4.119QPCh. 4 - Prob. 4.120QPCh. 4 - Prob. 4.122APCh. 4 - Prob. 4.123APCh. 4 - Prob. 4.124APCh. 4 - Prob. 4.125APCh. 4 - Prob. 4.126APCh. 4 - Prob. 4.127APCh. 4 - Prob. 4.128APCh. 4 - Prob. 4.129APCh. 4 - Prob. 4.130APCh. 4 - Prob. 4.131APCh. 4 - Prob. 4.132APCh. 4 - Prob. 4.133APCh. 4 - Prob. 4.134APCh. 4 - Prob. 4.135APCh. 4 - Prob. 4.136APCh. 4 - Prob. 4.137APCh. 4 - Prob. 4.138APCh. 4 - Prob. 4.139APCh. 4 - Prob. 4.140APCh. 4 - Prob. 4.141APCh. 4 - Prob. 4.142APCh. 4 - Prob. 4.143APCh. 4 - Prob. 4.144AP
Knowledge Booster
Background pattern image
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