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
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Chapter 4, Problem 4.124AP

(a)

Interpretation Introduction

Interpretation: The oxidation numbers to the elements are to be assigned and the redox reaction is to be balanced. The net ionic equation describing the reaction

For formation of chlorine is to be stated. The equation for the conversion of Cl2 to HCl and HOCl is to be balanced.

Concept introduction: The reaction in which reduction and oxidation process occurs simultaneously is termed as redox reaction. Redox reaction involves all the chemical reactions in which atoms will change their oxidation states.

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.

Oxidation number is defined as the charge present on ions. When an atom forms a chemical bond with another atom, the number of electrons that an atom gain or loses is the oxidation number. It is also known as oxidation state.

To determine: The oxidation number of the elements in each compound of the given reaction and balanced redox reaction in acid solution.

(a)

Expert Solution
Check Mark

Answer to Problem 4.124AP

Solution

The oxidation numbers of the elements in each compound have been rightfully stated.

The balanced redox reaction in acid solution is as follows:

4NaCl(aq)+2H2SO4(aq)+MnO2(aq)2Na2SO4(aq)+MnCl2(aq)+H2O(l)+Cl2(g)

Explanation of Solution

Explanation

The given element is MnO2 in the redox reaction.

The oxidation state of oxygen is usually 2 as it requires two more electrons in order to achieve stable electronic configuration.

The oxidation number of Mn is assumed to be x and is calculated by using the formula,

Charge on MnO2=[(Number of Mnatoms×Oxidation number of Mn)+(Number of O atoms×Oxidation number of O)]

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

Charge on MnO2=[(1×x)+(2×(2))]0=x4x=+4

Hence, the oxidation number of Mn in MnO2 is +4_ .

The oxidation numbers of the elements in Na2SO4 is as follows:

The given element is Na2SO4 in the redox reaction.

The oxidation state of oxygen is usually 2 as it requires two more electrons in order to achieve stable electronic configuration and Na+ has oxidation state of +1 Na2SO4 .

The oxidation number of S in Na2SO4 is assumed to be x and is calculated by using the formula,

Charge on Na2SO4=[(Number of Naatoms×oxidation number of Na)+(Number ofSatoms×oxidation number of S)+(Number of O atoms×oxidation number of O)]

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

Charge on Na2SO4=[(2×(+1))+(x×1)+(4×(2))]0=[(2)+(x)+(8)]x=28x=+6

Hence, the oxidation number of S in Na2SO4 is +6_ .

The oxidation numbers of the elements in MnCl2 is as follows

The given element is MnCl2 in the redox reaction. The oxidation state of chlorine is usually 1 as it requires one more electrons in order to achieve stable electronic configuration.

The oxidation number of Mn is assumed to be x and is calculated by using the formula,

Charge on MnCl2=[(Number of Mnatoms×oxidation number of Mn)+(Number of Cl atoms×oxidation number of Cl)]

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

Charge on MnCl2=[(1×x)+(2×(1))]0=x2x=+2

Hence, the oxidation number of Mn in MnCl2 is +2_ .

The given unbalanced redox reaction is,

NaCl(aq)+H2SO4(aq)+MnO2(aq)Na2SO4(aq)+MnCl2(aq)+H2O(l)+Cl2(g)

To balance the above redox reaction the half reactions must be shown.

The reduction half reaction in which Mn is reduced is,

MnO2(aq)+2eMnCl2(g)

The oxidation half reaction is in which Cl is oxidized is,

NaCl(aq)+1eCl2(g)

Now balance the hydrogen atoms on both the sides by adding 4H+ ions and 2H2O molecule to the reduction half reaction,

MnO2(aq)+2e+4H+(aq)MnCl2(g)+2H2O(l)

The oxidation half reaction is multiplied by 2 to balance the charge .

2NaCl(aq)2Cl2(g)+2e

To balance Cl atoms add coefficient of 2 at the reactant side of oxidation half reaction,

4NaCl(aq)2Cl2(g)+2e

Now, add both the two half reactions which is shown as follows;

4NaCl(aq)2Cl2(g)+2eMnO2(aq)+2e+4H+(aq)MnCl2(g)+2H2O(l)MnO2(aq)+4NaCl(aq)+4H+(aq)MnCl2(g)+2H2O(l)+2Cl2(g)

As given the reaction is taking place in acid solution, the redox reaction will be balanced by replacing 4H+ ions with 2H2SO4(aq) because 4H+ ions is equals to 2H2SO4(aq) .

MnO2(aq)+4NaCl(aq)+2H2SO4(aq)MnCl2(g)+2H2O(l)+2Cl2(g)

Now balance the chlorine atom by removing one chlorine atom from the product side of the reaction then the reaction becomes as,

MnO2(aq)+4NaCl(aq)+2H2SO4(aq)MnCl2(g)+2H2O(l)+Cl2(g)

In the above redox reaction, there are four sodium atoms and four sulfate atoms are placed on the left hand side  so we need to add four sodium atoms and four sulfate atoms on the right hand side of the equation in the form of Na2SO4 which will be shown as follows,

MnO2(aq)+4NaCl(aq)+2H2SO4(aq)2Na2SO4(aq)+MnCl2(g)+2H2O(l)+Cl2(g)

Thus, the complete and balanced redox reaction is,

MnO2(aq)+4NaCl(aq)+2H2SO4(aq)2Na2SO4(aq)+MnCl2(g)+2H2O(l)+Cl2(g)

(b)

Interpretation Introduction

To determine: A net ionic equation describing the formation of chlorine.

(b)

Expert Solution
Check Mark

Answer to Problem 4.124AP

Solution

A net ionic equation describing the formation of chlorine is,

Mn4+(aq)+4Cl(aq)Mn2+(g)+2Cl2(g)

Explanation of Solution

Explanation

The given unbalanced redox reaction is,

NaCl(aq)+H2SO4(aq)+MnO2(aq)Na2SO4(aq)+MnCl2(aq)+H2O(l)+Cl2(g)

In the above reaction the formation of chlorine atom occurs in the reduction half reaction which is,

MnO2(aq)+2eMnCl2(g)

The oxidation state of Mn in MnO2 is +4_ whereas the oxidation state of Mn in MnCl2 is +2_ . The reaction will remove the ions that will be the same that is oxygen atoms will be vanished from the reactant side and will add appropriate amount of chlorine atoms to equalize the charge on manganese ions to both the sides of the reaction so the reaction will be shown as

Mn4+(aq)+4Cl(aq)Mn2+(g)+2Cl2(g)

Therefore, a net ionic equation describing the formation of chlorine is,

Mn4+(aq)+4Cl(aq)Mn2+(g)+2Cl2(g)

(c)

Interpretation Introduction

To determine: The balanced equation for the conversion of Cl2 to HCl and HOCl

(c)

Expert Solution
Check Mark

Answer to Problem 4.124AP

Solution

The balanced equation for the conversion of Cl2 to HCl and HOCl is,

3Cl2(g)+3H2O(l)5HCl(aq)+HClO3(aq)

Explanation of Solution

Explanation

The reaction of water with Cl2(g) will give HCl(aq) and HOCl(aq)  the equation as ,

Cl2(g)+3H2O(l)HCl(aq)+HClO3(aq)

To balance the equation the amount of chlorine add coefficient of 2 in front of Cl2 and add coefficient 5 in front of HCl on the product side, therefore the balanced reaction will be,

2Cl2(g)+3H2O(l)5HCl(aq)+HClO3(aq)

Therefore The balanced equation for the conversion of Cl2 to HCl and HOCl is

2Cl2(g)+3H2O(l)5HCl(aq)+HClO3(aq)

Conclusion

We conclude that

  1. a) The oxidation numbers of the elements in each compound of the given reaction are ,

    Na+=+1_,Cl=-1,_H=+1_,O=-2_,MninMnO2=+4_,SinNa2SO4=+6_andMninMnCl2=+2_ The balanced redox reaction in acid solution is as follows:

    4NaCl(aq)+2H2SO4(aq)+MnO2(aq)2Na2SO4(aq)+MnCl2(aq)+H2O(l)+Cl2(g)

  2. b) A net ionic equation describing the formation of chlorine is,

    Mn4+(aq)+4Cl(aq)Mn2+(g)+2Cl2(g)

  3. c) The balanced equation for the conversion of Cl2 to HCl and HOCl is, 3Cl2(g)+3H2O(l)5HCl(aq)+HClO3(aq)

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Chapter 4 Solutions

Chemistry: The Science in Context (Fifth Edition)

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. 4.121APCh. 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. 4.144APCh. 4 - Prob. 4.145AP
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