Chemistry
Chemistry
4th Edition
ISBN: 9780393919370
Author: Thomas R. Gilbert
Publisher: NORTON
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Chapter 4, Problem 4.103QP

(a)

Interpretation Introduction

Interpretation: The oxidation numbers of all the elements in the reactants and products of the given equations are to be determined and the elements that are oxidized and reduced in the given reactions 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,

  • The oxidation number present on a neutral compound is always zero.
  • The oxidation number of atoms present in a element in the pure form is zero.
  • 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.
  • The common oxidation number of oxygen in most compounds is 2 and oxidation number of hydrogen is +1 .

To determine: The oxidation numbers of all the elements in the reactants and products of the given equation and the identification of the elements that are oxidized and reduced in the given reaction.

(a)

Expert Solution
Check Mark

Answer to Problem 4.103QP

Solution

The oxidation number of all the reactants and products in the given reaction are,

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

In Fe3O4 , oxidation number of Fe is +83_ and oxygen has oxidation number 2_ .

In Fe2SiO4 , oxidation number of Fe is +2_ , oxygen has oxidation number 2_ and silicon has oxidation number +4_ .

In O2 , oxidation number of oxygen is 0_ .

Iron has reduced and oxygen has oxidized in the reaction.

Explanation of Solution

Explanation

The given reaction is,

3SiO2(s)+2Fe3O4(s)3Fe2SiO4(s)+O2(g)

The reactants in the reaction are SiO2 and Fe3O4(s) .

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 .

The reactant, Fe3O4 contain iron (Fe) and oxygen.

The oxidation number of oxygen is 2 .

The oxidation number of Fe in Fe3O4 is calculated by the formula,

(SumofoxidationnumberofallelementinFe3O4)=Netchargeonwholemolecule

The oxidation number of Fe in Fe3O4 is assumed to be x .

Net charge on Fe3O4 is zero.

Substitute the value of oxidation number in the above expression.

3(Oxidationnumberofiron)+4(Oxidationnumberofoxygen)=03x+[4×(2)]=03x8=0x=+83

The oxidation number of Fe in the given reactant is +83 .

The products formed in the above reaction are Fe2SiO4 and O2 .

The product, Fe2SiO4 contain iron, silicon and oxygen.

The oxidation number of oxygen is 2 and the oxidation number of silicon in SiO2 is +4 as calculated above.

The oxidation number of Fe in Fe2SiO4 is calculated by the formula,

(SumofoxidationnumberofallelementinFe2SiO4)=Netchargeonwholemolecule

The oxidation number of Fe in Fe2SiO4 is assumed to be x .

Net charge on Fe2SiO4 is zero.

Substitute the value of oxidation number in the above expression.

2(Oxidationnumberofiron)+(Oxidationnumberofsilicon)+4(Oxidationnumberofoxygen)=02x+4+(4(2))=02x+48=0x=+2

The oxidation number of Fe in Fe2SiO4 is +2 .

The oxidation number of oxygen is zero in O2 because it is in pure form.

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 Fe3O4 , oxidation number of Fe is +83_ and oxygen has oxidation number 2_ .

In Fe2SiO4 , oxidation number of Fe is +2_ , oxygen has oxidation number 2_ and silicon has oxidation number +4_ .

In O2 , oxidation number of oxygen is 0_ .

Therefore, iron has reduced from +83 to +2 oxidation state and oxygen has oxidized from 2 to 0 oxidation state.

(b)

Interpretation Introduction

To determine: The oxidation numbers of all the elements in the reactants and products of the given equation and the identification of the elements that are oxidized and reduced in the given reaction.

(b)

Expert Solution
Check Mark

Answer to Problem 4.103QP

Solution

The oxidation number of all the reactants and products in the given reaction are,

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

In Fe , the oxidation number of Fe is 0_ .

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

In Fe2SiO4 , The oxidation number of Fe is +2_ , silicon has oxidation number +4_ and oxygen has oxidation number 2 .

Iron has oxidized and oxygen has reduced in the reaction.

Explanation of Solution

Explanation

The given reaction is,

SiO2(s)+2Fe(s)+O2(g)Fe2SiO4(s)

The reactant in the reaction are SiO2 , Fe and O2 .

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 .

The oxidation number of Fe is 0 because it is present in its elemental form.

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

The product formed in the above reaction is Fe2SiO4 .

The product, Fe2SiO4 contain iron, silicon and oxygen.

The oxidation number of oxygen is 2 and the oxidation number of silicon in SiO2 is +4 as calculated above.

The oxidation number of Fe in Fe2SiO4 is calculated by the formula,

(SumofoxidationnumberofallelementinFe2SiO4)=Netchargeonwholemolecule

The oxidation number of Fe in Fe2SiO4 is assumed to be x .

Net charge on Fe2SiO4 is zero.

Substitute the value of oxidation number in the above expression.

2(Oxidationnumberofiron)+(Oxidationnumberofsilicon)+4(Oxidationnumberofoxygen)=02x+4+(4(2))=02x+48=0x=+2

Therefore, the oxidation number of Fe in Fe2SiO4 is +2 .

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 Fe , the oxidation number of Fe is 0_ .

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

In Fe2SiO4 , The oxidation number of Fe is +2_ , silicon has oxidation number +4_ and oxygen has oxidation number 2 .

Therefore, iron has oxidized from 0 to +2 oxidation state and oxygen has reduced to 2 from 0 oxidation state.

(c)

Interpretation Introduction

To determine: The oxidation numbers of all the elements in the reactants and products of the given equation and the identification of the elements that are oxidized and reduced in the given reaction.

(c)

Expert Solution
Check Mark

Answer to Problem 4.103QP

Solution

The oxidation number of all the reactants and products in the given reaction are,

In FeO , the oxidation number of Fe is +2_ and oxidation number of oxygen is 2_ .

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

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

In Fe(OH)3 , the oxidation number of Fe is +3_ , oxidation number of hydrogen is +1_ and oxidation number of oxygen is 2_ .

Iron has oxidized and oxygen has reduced in the reaction.

Explanation of Solution

Explanation

The given reaction is,

4FeO(s)+O2(g)+6H2O4Fe(OH)3(s)

The reactants are FeO , O2 and H2O .

The reactant, FeO contain iron and oxygen.

The oxidation number of oxygen is 2 .

Net charge on FeO is zero.

The oxidation number of Fe in FeO is calculated by the formula,

(SumofoxidationnumberofallelementinFeO)=Netchargeonwholemolecule

The oxidation number of Fe in FeO is assumed to be x .

Substitute the value of oxidation number in the above expression.

(Oxidationnumberofiron)+(Oxidationnumberofoxygen)=0x+_2=0x=+2

The oxidation number of Fe in the given reactant is +2 .

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

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

The product form in the above reaction is Fe(OH)3 .

The product Fe(OH)3 contains iron, hydrogen and oxygen.

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

The oxidation number of Fe in Fe(OH)3 is calculated by the formula,

(SumofoxidationnumberofallelementinFe(OH)3)=Netchargeonwholemolecule

The oxidation number of Fe in Fe(OH)3 is assumed to be x .

Substitute the value of oxidation number in the above expression.

(Oxidationnumberofiron)+3(Oxidationnumberofoxygen)+3(Oxidationnumberofhydrogen)=0x+3(2)+3(+1)=0x=+3

The oxidation number of Fe in the given reactant is +3 .

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

In FeO , the oxidation number of Fe is +2_ and oxidation number of oxygen is 2_ .

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

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

In Fe(OH)3 , the oxidation number of Fe is +3_ , oxidation number of hydrogen is +1_ and oxidation number of oxygen is 2_ .

Therefore, Iron has oxidized from +2 to +3 oxidation state and oxygen has reduced from 0 to 2 in the reaction.

Conclusion

  1. a) The oxidation number of all the reactants and products in the given reaction has been rightfully stated. Iron has reduced and oxygen has oxidized in the reaction.
  2. b) The oxidation number of all the reactants and products in the given reaction has been rightfully stated. Iron has oxidized and oxygen has reduced in the reaction.
  3. c) The oxidation number of all the reactants and products in the given reaction has been rightfully stated. Iron has oxidized and oxygen has reduced in the reaction

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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
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