Chemistry
Chemistry
4th Edition
ISBN: 9780393919370
Author: Thomas R. Gilbert
Publisher: NORTON
bartleby

Concept explainers

bartleby

Videos

Question
Book Icon
Chapter 4, Problem 4.104QP

(a)

Interpretation Introduction

Interpretation: The oxidation numbers of each of the elements in the given reactions are to be stated and the elements that are oxidized or reduced in the reaction are to be identified.

Concept introduction: The oxidation state or number in a compound is the charge present on the individual atom if the all bonds are assumed to be ionic. The charge present on the ionic compound is equal to the some of the charges of all the atoms present in that ionic compound.

The oxidation number is assigned by the following set of rules,

  1. 1. The oxidation number present on a neutral compound is always zero.
  2. 2. The oxidation number of atoms present in a element in the pure form is zero.
  3. 3. In the compounds, if halogens are present the oxidation number of halogens is 1 but when oxygen is present with them they have different oxidation number.
  4. 4. The common oxidation number of oxygen in most compounds is 2 and oxidation number of hydrogen is +1 .

To determine: The oxidation number of each of the elements in the given reaction and identification of the elements that are oxidized or reduced in the reaction.

(a)

Expert Solution
Check Mark

Answer to Problem 4.104QP

Solution

The oxidation number of the elements that are reactants and products in the reaction are,

In SiO2 , silicon has oxidation number +4_ and oxygen has oxidation number 2_ .

In H2O , the oxidation number of oxygen is 2_ and oxidation number of hydrogen is +1_ .

In H4SiO4 , oxidation number of hydrogen is +1_ , silicon has oxidation number +4_ and oxygen has oxidation number 2_ .

No elements is oxidized or reduced in the reaction.

Explanation of Solution

Explanation

The given reaction is,

SiO2(s)+H2O(l)H4SiO4(aq)

The reactant in the reaction are SiO2 and H2O .

The reactant, SiO2 contain silicon and oxygen elements.

The oxidation number of silicone in SiO2 is calculated by the formula,

(SumofoxidationnumberofallelementinSiO2)=Netchargeonwholemolecule

The oxidation number of silicon in SiO2 is assumed to be x .

The oxidation number of oxygen is 2 .

Net charge on SiO2 is zero.

Substitute the value of oxidation number in the above expression.

(Oxidationnumberofsilicon)+2(Oxidationnumberofoxygen)=0x+[2×(2)]=0x4=0x=+4

The oxidation number of silicon in the given reactant SiO2 is +4 .

In H2O , the oxidation number of oxygen is 2 and oxidation number of hydrogen is +1 .

The product formed in the above reaction is H4SiO4 .

The product H4SiO4 contains silicon, hydrogen and oxygen.

The oxidation number of hydrogen is +1 and oxidation number of oxygen is 2 .

The oxidation number of silicon in H4SiO4 is calculated by the formula,

(SumofoxidationnumberofallelementinH4SiO4)=Netchargeonwholemolecule

The oxidation number of silicon in H4SiO4 is assumed to be x .

Net charge on H4SiO4 is zero.

Substitute the value of oxidation number in the above expression.

4(Oxidationnumberofhydrogen)+(Oxidationnumberofsilicon)+4(Oxidationnumberofoxygen)=04(+1)+x+[4×(2)]=0x=84x=+4

The oxidation number of silicon in H4SiO4 is +4 .

Therefore, the oxidation number of the elements that are reactants and products in the reaction are as follows

In SiO2 , silicon has oxidation number +4_ and oxygen has oxidation number 2_ .

In H2O , the oxidation number of oxygen is 2_ and oxidation number of hydrogen is +1_ .

In H4SiO4 , oxidation number of hydrogen is +1_ , silicon has oxidation number +4_ and oxygen has oxidation number 2_ .

No elements is oxidized or reduced in the reaction because oxidation state of all the elements remains the same.

(b)

Interpretation Introduction

To determine: The oxidation number of each of the elements in the given reaction and identification of the elements that are oxidized or reduced in the reaction.

(b)

Expert Solution
Check Mark

Answer to Problem 4.104QP

Solution

The oxidation number of the elements that are reactants and products in the reaction are,

In MnCO3 , manganese has oxidation number +2_ , carbon has oxidation number +4_ and oxygen has oxidation number 2_ .

In O2 , the oxidation number of oxygen is 0_ .

In MnO2 manganese has oxidation number +4_ and oxygen has oxidation number 2_ .

In CO2 the oxidation number of oxygen is 2_ and carbon has oxidation number +4_ .

Manganese is oxidized and oxygen is reduced in the reaction.

Explanation of Solution

Explanation

The given reaction is,

2MnCO3(s)+O2(g)2MnO2(s)+2CO2(g)

The reactants in the above reaction are MnCO3 and O2 .

The reactant MnCO3 contain manganese (Mn) , carbon and oxygen elements.

The oxidation number of carbon is +4 and oxidation number of oxygen is 2 .

The oxidation number of manganese (Mn) in MnCO3 is calculated by the formula,

(SumofoxidationnumberofallelementinMnCO3)=Netchargeonwholemolecule

The oxidation number of manganese in MnCO3 is assumed to be x .

Net charge on MnCO3 is zero.

Substitute the value of oxidation number in the above expression.

4(Oxidationnumberofmanganese)+(Oxidationnumberofcarbon)+3(Oxidationnumberofoxygen)=0x+4+[3×(2)]=0x+46=0x=+2

The oxidation number of manganese in MnCO3 is +2 .

The oxidation number of oxygen is 0 in O2 because it is present in the pure form.

The products in the above reaction are MnO2 and CO2 .

The product, MnO2 contain manganese (Mn) and oxygen elements.

The oxidation number of oxygen is 2 .

The oxidation number of manganese in MnO2 is calculated by the formula,

(SumofoxidationnumberofallelementinMnO2)=Netchargeonwholemolecule

The oxidation number of manganese in MnO2 is assumed to be x .

Net charge on MnO2 is zero.

Substitute the value of oxidation number in the above expression.

(Oxidationnumberofmanganese)+2(Oxidationnumberofoxygen)=0x+[2×(2)]=0x4=0x=+4

The oxidation number of manganese in MnO2 is +4 .

The product, CO2 contain carbon and oxygen.

The oxidation number of oxygen is 2 .

The oxidation number of carbon in CO2 is calculated by the formula,

(SumofoxidationnumberofallelementinCO2)=Netchargeonwholemolecule

The oxidation number of carbon in CO2 is assumed to be x .

Net charge on CO2 is zero.

Substitute the value of oxidation number in the above expression.

(Oxidationnumberofcarbon)+2(Oxidationnumberofoxygen)=0x+[2×(2)]=0x4=0x=+4

The oxidation number of carbon in CO2 is +4 .

Therefore, the oxidation number of the elements that are reactants and products in the reaction are as follows

In MnCO3 , manganese has oxidation number +2_ , carbon has oxidation number +4_ and oxygen has oxidation number 2_ .

In O2 , the oxidation number of oxygen is 0_ .

In MnO2 manganese has oxidation number +4_ and oxygen has oxidation number 2_ .

In CO2 the oxidation number of oxygen is 2_ and carbon has oxidation number +4_ .

Therefore, manganese is oxidized from +2 to +4 oxidation state. and oxygen is reduced from 0 to 2 in the reaction.

(c)

Interpretation Introduction

To determine: The oxidation number of each of the elements in the given reaction and identification of the elements that are oxidized or reduced in the reaction.

(c)

Expert Solution
Check Mark

Answer to Problem 4.104QP

Solution

The oxidation number of the elements that are reactants and products in the reaction are,

In NO2 , the oxidation number of nitrogen is +4_ and oxygen has oxidation number 2_ .

In H2O , the oxidation number of oxygen is 2_ and oxidation number of hydrogen is +1_ .

In NO3 , the oxidation number of nitrogen is +5_ and oxygen has oxidation number 2_ .

In NO , the oxidation number of nitrogen is +2_ and oxygen has oxidation number 2_ .

In H+ , oxidation number of hydrogen is +1_ .

Nitrogen is oxidized as well as reduced in the reaction.

Explanation of Solution

Explanation

The given reaction is,

3NO2(g)+H2O(l)2NO3(aq)+NO(g)+2H+(aq)

The reactants in the above reaction are NO2 and H2O .

The reactant, NO2 contain nitrogen (N) and oxygen elements.

The oxidation number of oxygen is 2 .

The oxidation number of nitrogen in NO2 is calculated by the formula,

(SumofoxidationnumberofallelementinNO2)=Netchargeonwholemolecule

The oxidation number of nitrogen in NO2 is assumed to be x .

Net charge on NO2 is zero.

Substitute the value of oxidation number in the above expression.

(Oxidationnumberofnitrogen)+2(Oxidationnumberofoxygen)=0x+[2×(2)]=0x4=0x=+4

The oxidation number of nitrogen in NO2 is +4 .

In H2O , the oxidation number of oxygen is 2 and oxidation number of hydrogen is +1 .

The products in the above reaction are NO3 , NO and H+ .

In the product, H+ oxidation number of hydrogen is +1 .

The product, NO3 contain nitrogen (N) and oxygen elements.

The oxidation number of oxygen is 2 .

The oxidation number of nitrogen in NO3 is calculated by the formula,

(SumofoxidationnumberofallelementinNO3)=Netchargeonwholemolecule

The oxidation number of nitrogen in NO3 is assumed to be x .

Net charge on NO3 is 1 .

Substitute the value of oxidation number in the above expression.

(Oxidationnumberofnitrogen)+3(Oxidationnumberofoxygen)=1x+[3×(2)]=1x6=1x=+5

The oxidation number of nitrogen in NO3 is +5 .

The product, NO contain nitrogen (N) and oxygen elements.

The oxidation number of oxygen is 2 .

The oxidation number of nitrogen in NO is calculated by the formula,

(SumofoxidationnumberofallelementinNO)=Netchargeonwholemolecule

The oxidation number of nitrogen in NO is assumed to be x .

Net charge on NO is zero.

Substitute the value of oxidation number in the above expression.

(Oxidationnumberofnitrogen)+(Oxidationnumberofoxygen)=0x+(2)=0x2=0x=+2

The oxidation number of nitrogen in NO is +2 .

Therefore, the oxidation number of the elements that are reactants and products in the reaction are as follows

In NO2 , the oxidation number of nitrogen is +4_ and oxygen has oxidation number 2_ .

In H2O , the oxidation number of oxygen is 2_ and oxidation number of hydrogen is +1_ .

In NO3 , the oxidation number of nitrogen is +5_ and oxygen has oxidation number 2_ .

In NO , the oxidation number of nitrogen is +2_ and oxygen has oxidation number 2_ .

In H+ , oxidation number of hydrogen is +1_ .

Therefore, nitrogen is oxidized from +4 to +5 oxidation state and nitrogen is also reduced from +4 to +2 oxidation state in the reaction.

Conclusion

  1. a) The oxidation number of each element in the given reaction has been rightfully stated. No elements is oxidized or reduced in the reaction.
  2. b) The oxidation number of each element in the given reaction has been rightfully stated. Manganese is oxidized and oxygen is reduced in the reaction.
  3. c) The oxidation number of each element in the given reaction has been rightfully stated. Nitrogen is oxidized as well as reduced in the reaction

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
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
Acid-Base Titration | Acids, Bases & Alkalis | Chemistry | FuseSchool; Author: FuseSchool - Global Education;https://www.youtube.com/watch?v=yFqx6_Y6c2M;License: Standard YouTube License, CC-BY