Chemical Principles: The Quest for Insight
Chemical Principles: The Quest for Insight
7th Edition
ISBN: 9781464183959
Author: Peter Atkins, Loretta Jones, Leroy Laverman
Publisher: W. H. Freeman
Question
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Chapter 6, Problem 6K.6E

(a)

Interpretation Introduction

Interpretation:

The following skeletal equation has to be balanced using oxidation and reduction half reactions and also the oxidizing agent, reducing agent has to be identified.

  Cl2O7(g)+H2O2(aq)ClO2-(aq)+O2(g)

Concept Introduction:

Net ionic equation:

Net ionic equation is defined as the specific species that only involves to a particular reaction. This type of equations is generally used in acid-base neutralization reactions and redox reactions.

Oxidizing agent:

The material which gains electron in a chemical reaction is called oxidizing agent. In this reaction, the oxidation number will be decreased.

Reducing agent:

The material, which loses electrons in a chemical reaction, is called reducing agent. In this reaction, the oxidation number will be increased.

(a)

Expert Solution
Check Mark

Answer to Problem 6K.6E

The balanced reaction for the production of chlorite ions from dichlorine heptoxide by reaction with hydrogen peroxide solution is given below,

  4H2O2(aq)+Cl2O7(g)+2OH-(aq)4O2(g)+2ClO2-(aq)+5H2O(l)

Here, the oxidizing agent is Cl2O7 and reducing agent is H2O2.

Explanation of Solution

The unbalanced skeletal equation for the reaction is

  Cl2O7(g)+H2O2(aq)ClO2-(aq)+O2(g)

Oxidation half-reaction:

The oxidation number of O is increased from 1 to 0, therefore this is oxidation reaction.

  H2O2(aq)O2(g)

Balance the H atom by adding H2O molecules to the right side and add the same amount OH- ions to the left side

  H2O2(aq)+2OH-(aq)O2(g)+2H2O(l)

Balance the net charges by adding the electrons.

Here, in left side, the net charge is 2 and the right side the net charge is 0, therefore it required 2 electrons on the right side to reduce the charge from 0 to 2.

  H2O2(aq)+2OH-(aq)O2(g)+2H2O(l)+2e

Therefore, the balanced oxidation half-reaction is

  H2O2(aq)+2OH-(aq)O2(g)+2H2O(l)+2e

Here, H2O2 act as loses of electrons, so it is a reducing agent.

Reduction half-reaction:

The oxidation number of Cl is decreased from +7 to +3, therefore this is reduction reaction.

  Cl2O7(g)ClO2-(aq)

Balance the equation except H and O.

  Cl2O7(g)2ClO2-(aq)

Balance the O atom by adding H2O left side

  Cl2O7(g)2ClO2-(aq)+3H2O(l)

Balance the H atom by adding H2O molecules to the right side and add the same amount OH- ions to the left side

  Cl2O7(g)+6H2O(l)2ClO2-(aq)+3H2O(l)+6OH-(aq)

Balance the net charges by adding the electrons.

Here, in left side, the net charge is 0 and the right side the net charge is 8, therefore it required 8 electrons on the left side to reduce the charge from 0 to 8.

  Cl2O7(g)+6H2O(l)+8e2ClO2-(aq)+3H2O(l)+6OH-(aq)

Therefore, the balanced reduction half-reaction is

  Cl2O7(g)+6H2O(l)+8e2ClO2-(aq)+3H2O(l)+6OH-(aq)

Here, the Cl2O7 act as gaining of electrons, so it is a oxidizing agent.

Now add the two half reactions together. Match the number of electrons in each side.  Because in oxidation half reaction 2 electrons are lost and in reduction half reaction 8 electron is gained. Therefore, multiply the oxidation half reaction by 8 and the reduction half reaction by 2.

  8H2O2(aq)+16OH-(aq)8O2(g)+16H2O(l)+16e2Cl2O7(g)+12H2O(l)+16e4ClO2-(aq)+6H2O(l)+12OH-(aq)

Add the equation and cancel the common ions, electrons and water molecules in each side of the arrow.

  8H2O2(aq)+2Cl2O7(g)+4OH-(aq)8O2(g)+4ClO2-(aq)+10H2O(l)

Divide by 2 each side of the arrow.

  4H2O2(aq)+Cl2O7(g)+2OH-(aq)4O2(g)+2ClO2-(aq)+5H2O(l)

Therefore, the balanced net ionic equation the above reaction is

  4H2O2(aq)+Cl2O7(g)+2OH-(aq)4O2(g)+2ClO2-(aq)+5H2O(l)

(b)

Interpretation Introduction

Interpretation:

The following skeletal equation has to be balanced using oxidation and reduction half reactions and also the oxidizing agent, reducing agent has to be identified.

  MnO4-(aq)+S2-(aq)S(s)+MnO2(s)

Concept introduction:

Refer to part (a).

(b)

Expert Solution
Check Mark

Answer to Problem 6K.6E

The balanced reaction of permanganate ions with sulfide ions is given below,

  3S2-(aq)+2MnO4-(aq)+4H2O(l)3S(s)+2MnO2(s)++8OH-(aq)

Here, the oxidizing agent is MnO4- and reducing agent is S2-.

Explanation of Solution

The unbalanced skeletal equation for the reaction is

  MnO4-(aq)+S2-(aq)S(s)+MnO2(s)

Oxidation half-reaction:

The oxidation number of S is increased from 2 to 0, therefore this is oxidation reaction.

  S2-(aq)S(s)

Balance the net charges by adding the electrons.

Here, in left side, the net charge is 2 and the right side the net charge is 0, therefore it required 2 electrons on the right side to reduce the charge from 0 to 2.

  S2-(aq)S(s)+2e

Therefore, the balanced oxidation half-reaction is

  S2-(aq)S(s)+2e

Here, S2- act as loses of electrons, so it is a reducing agent.

Reduction half-reaction:

The oxidation number of Mn is decreased from +7 to +4, therefore this is reduction reaction.

  MnO4-(aq)MnO2(s)

Balance the equation except H and O.

  MnO4-(aq)MnO2(s)

Balance the O atom by adding H2O right side

  MnO4-(aq)MnO2(s)+2H2O(l)

Balance the H atom by adding H2O molecules to the left side and add the same amount OH- ions to the right side

  MnO4-(aq)+4H2O(l)MnO2(s)+2H2O(l)+4OH-(aq)

Balance the net charges by adding the electrons.

Here, in left side, the net charge is 1 and the right side the net charge is 4, therefore it required 3 electrons on the left side to reduce the charge from 1 to 4.

  MnO4-(aq)+4H2O(l)+3eMnO2(s)+2H2O(l)+4OH-(aq)

Here, the MnO4- act as gaining of electrons, so it is an oxidizing agent.

Now add the two half reactions together. Match the number of electrons in each side.  Because in oxidation half reaction 2 electrons are lost and in reduction half reaction 2 electrons is gained. Therefore, multiply the oxidation half reaction by 2 and the reduction half reaction by 2.

  3S2-(aq)3S(s)+6e2MnO4-(aq)+8H2O(l)+6e2MnO2(s)+4H2O(l)+8OH-(aq)

Add the equation and cancel the common ions, electrons and water molecules in each side of the arrow.

  3S2-(aq)+2MnO4-(aq)+4H2O(l)3S(s)+2MnO2(s)++8OH-(aq)

Therefore, the balanced net ionic equation the above reaction is

  3S2-(aq)+2MnO4-(aq)+4H2O(l)3S(s)+2MnO2(s)++8OH-(aq)

(c)

Interpretation Introduction

Interpretation:

The following skeletal equation has to be balanced using oxidation and reduction half reactions and also the oxidizing agent, reducing agent has to be identified.

  N2H4(g)+ClO3-(aq)NO(g)+Cl-(aq)

Concept introduction:

Refer to part (a).

(c)

Expert Solution
Check Mark

Answer to Problem 6K.6E

The balanced reaction of hydrazine with chlorate ions is given below,

  3N2H4(g)+2ClO3-(aq)6NO(g)+2Cl-(aq)

Here, the oxidizing agent is ClO3- and reducing agent is N2H4.

Explanation of Solution

The unbalanced skeletal equation for the reaction is

  N2H4(g)+ClO3-(aq)NO(g)+Cl-(aq)

Oxidation half-reaction:

The oxidation number of N is increased from 2 to +2, therefore this is oxidation reaction.

  N2H4(g)NO(g)

Balance the equation except H and O.

  N2H4(g)2NO(g)

Balance the O atom by adding H2O left side

  N2H4(g)+2H2O(l)2NO(g)

Balance the H atom by adding H2O molecules to the right side and add the same amount OH- ions to the left side

  N2H4(g)+2H2O(l)+4OH-(aq)2NO(g)+4H2O(l)

Balance the net charges by adding the electrons.

Here, in left side, the net charge is 4 and the right side the net charge is 0, therefore it required 4 electron on the right side to reduce the charge from 0 to 4.

  N2H4(g)+2H2O(l)+4OH-(aq)2NO(g)+4H2O(l)+4e

Therefore, the balanced oxidation half-reaction is

  N2H4(g)+2H2O(l)+4OH-(aq)2NO(g)+4H2O(l)+4e

Here, N2H4 lost 4 electrons, so it is a reducing agent.

Reduction half-reaction:

The oxidation number of Cl is decreased from +5 to 1, therefore this is reduction reaction.

  ClO3-(aq)Cl-(aq)

Balance the O atom by adding H2O right side

  ClO3-(aq)Cl-(aq)+3H2O(l)

Balance the H atom by adding H2O molecules to the left side and add the same amount OH- ions to the right side

  ClO3-(aq)+6H2O(l)Cl-(aq)+3H2O(l)+6OH-(aq)

Balance the net charges by adding the electrons.

Here, in left side, the net charge is 1 and the right side the net charge is -7, therefore it required 6 electrons on the left side to reduce the charge from 1 to -7.

  ClO3-(aq)+6H2O(l)+6eCl-(aq)+3H2O(l)+6OH-(aq)

Here, the ClO3- gains 6 electrons, so it is an oxidizing agent.

Now add the two half reactions together. Match the number of electrons in each side.  Because in oxidation half reaction 4 electrons are lost and in reduction half reaction 6 electrons are gained. Therefore, multiply the oxidation half reaction by 6 and the reduction half reaction by 4.

  6N2H4(g)+12H2O(l)+24OH-(aq)12NO(g)+24H2O(l)+24e4ClO3-(aq)+24H2O(l)+24e4Cl-(aq)+12H2O(l)+24OH-(aq)

Add the equation and cancel the common ions, electrons and water molecules in each side of the arrow.

  6N2H4(g)+4ClO3-(aq)12NO(g)+4Cl-(aq)

Divide by 2 each side of the arrow.

  3N2H4(g)+2ClO3-(aq)6NO(g)+2Cl-(aq)

Therefore, the balanced net ionic equation the above reaction is

  3N2H4(g)+2ClO3-(aq)6NO(g)+2Cl-(aq)

(d)

Interpretation Introduction

Interpretation:

The following skeletal equation has to be balanced using oxidation and reduction half reactions and also the oxidizing agent, reducing agent has to be identified.

  Pb(OH)42-(aq)+ClO-(aq)PbO2(s)+Cl(aq)

Concept introduction:

Refer to part (a).

(d)

Expert Solution
Check Mark

Answer to Problem 6K.6E

The balanced reaction of plumblate ions and hypochlorite ions is given below,

  Pb(OH)42-(aq)+ClO-(aq)PbO2(s)+Cl(aq)+H2O(l)+2OH-(aq)

Here, the oxidizing agent is ClO- and reducing agent is Pb(OH)42-.

Explanation of Solution

The unbalanced skeletal equation for the reaction is

  Pb(OH)42-(aq)+ClO-(aq)PbO2(s)+Cl(aq)

Oxidation half-reaction:

The oxidation number of Pb is increased from +2 to +4, therefore this is oxidation reaction.

  Pb(OH)42-(aq)PbO2(s)

Balance the equation except H and O.

  Pb(OH)42-(aq)PbO2(s)

Balance the O atom by adding H2O right side

  Pb(OH)42-(aq)PbO2(s)+2H2O(l)

Balance the net charges by adding the electrons.

Here, in left side, the net charge is 2 and the right side the net charge is 0, therefore it required 2 electrons on the left side to reduce the charge from 0 to 2.

  Pb(OH)42-(aq)PbO2(s)+2H2O(l)+2e

Therefore, the balanced oxidation half-reaction is

  Pb(OH)42-(aq)PbO2(s)+2H2O(l)+2e

Here, Pb(OH)42- act as loses of electrons, so it is a reducing agent.

Reduction half-reaction:

The oxidation number of Cl is decreased from +1 to 1, therefore this is reduction reaction.

  ClO-(aq)Cl(aq)

Balance the equation except H and O.

  ClO-(aq)Cl(aq)

Balance the O atom by adding H2O right side

  ClO-(aq)Cl(aq)+H2O(l)

Balance the H atom by adding H2O molecules to the left side and add the same amount OH- ions to the right side

  ClO-(aq)+2H2O(l)Cl(aq)+H2O(l)+2OH-(aq)

Balance the net charges by adding the electrons.

Here, in left side, the net charge is 1 and the right side the net charge is 3, therefore it required 2 electrons on the left side to reduce the charge from 1 to 3.

  ClO-(aq)+2H2O(l)+2eCl(aq)+H2O(l)+2OH-(aq)

Here, the ClO- act as gaining of electrons, so it is an oxidizing agent.

Now add the two half reactions together. Match the number of electrons in each side.  Because in oxidation half reaction 2 electrons are lost and in reduction half reaction 2 electrons is gained. Therefore, multiply the oxidation half reaction by 2 and the reduction half reaction by 2.

  2Pb(OH)42-(aq)2PbO2(s)+4H2O(l)+4e2ClO-(aq)+4H2O(l)+4e2Cl(aq)+2H2O(l)+4OH-(aq)

Add the equation and cancel the common ions, electrons and water molecules in each side of the arrow.

  2Pb(OH)42-(aq)+2ClO-(aq)2PbO2(s)+2Cl(aq)+2H2O(l)+4OH-(aq)

Divide by 2 each side of the arrow.

  Pb(OH)42-(aq)+ClO-(aq)PbO2(s)+Cl(aq)+H2O(l)+2OH-(aq)

Therefore, the balanced net ionic equation the above reaction is

  Pb(OH)42-(aq)+ClO-(aq)PbO2(s)+Cl(aq)+H2O(l)+2OH-(aq)

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Chapter 6 Solutions

Chemical Principles: The Quest for Insight

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