Chemistry: The Molecular Science
Chemistry: The Molecular Science
5th Edition
ISBN: 9781285199047
Author: John W. Moore, Conrad L. Stanitski
Publisher: Cengage Learning
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Chapter 17, Problem 17QRT

(a)

Interpretation Introduction

Interpretation:

The given redox reaction is to be balanced and the oxidizing agent and reducing agent are to be identified.

Concept Introduction:

The chemical reaction that involves the electron transfer between two species is referred to as redox reaction.  It is any chemical reaction in which change in the oxidation number of a molecule, atom or ion takes place by either gain of electrons or loss of electrons.  The species which gets oxidized in the redox reaction is referred to as reducing agent whereas the specie which gets reduced is termed as oxidizing agent.

(a)

Expert Solution
Check Mark

Answer to Problem 17QRT

The balanced redox reaction is shown below.

  6FeO(s)+O3(g)3Fe2O3(s)

In this reaction, FeO act as reducing agent and O3 act as oxidizing agent.

Explanation of Solution

The given unbalanced redox reaction is shown below.

    FeO(s)+O3(g)Fe2O3(s)

The iron atom and oxygen atom are balanced by putting 2 and 13 as prefix to FeO and ozone respectively.

  2FeO(s)+13O3(g)Fe2O3(s)

The above balanced reaction is multiplied by 3 so as to get the balanced reaction in integers.

  6FeO(s)+O3(g)3Fe2O3(s)

Thus, the balanced chemical reaction is shown below.

  6FeO(s)+O3(g)3Fe2O3(s)

The oxidation number of iron atom in FeO and Fe2O3 is +2 and +3 respectively.  Since there is increase in oxidation number, therefore, FeO is oxidized to Fe2O3 and FeO act as reducing agent.

The oxidation number of oxygen atom in O3 and Fe2O3 is 0 and 2 respectively.  Since there is decrease in oxidation number, therefore, O3 is reduced to Fe2O3 and O3 act as oxidizing agent.

(b)

Interpretation Introduction

Interpretation:

The given redox reaction is to be balanced and the oxidizing agent and reducing agent are to be identified.

Concept Introduction:

Same as part (a).

(b)

Expert Solution
Check Mark

Answer to Problem 17QRT

The balanced redox reaction is shown below.

    P4(s)+10Br2(l)4PBr5(l)

In this reaction, P4 act as reducing agent and Br2 act as oxidizing agent.

Explanation of Solution

The given unbalanced redox reaction is shown below.

    P4(s)+Br2(l)PBr5(l)

The phosphorous atoms and bromine atoms are balanced by putting 14 and 52 as prefix to P4 and Br2 respectively.

  14P4(s)+52Br2(l)PBr5(l)

The above balanced reaction is multiplied by 4 so as to get the balanced reaction in integers.

  P4(s)+10Br2(l)4PBr5(l)

Thus, the balanced chemical reaction is shown below.

  P4(s)+10Br2(l)4PBr5(l)

The oxidation number of phosphorous atom in P4 and PBr5 is 0 and +5 respectively.  Since there is increase in oxidation number, therefore, P4 is oxidized to PBr5 and P4 act as reducing agent.

The oxidation number of bromine atom in Br2 and PBr5 is 0 and 1 respectively.  Since there is decrease in oxidation number, therefore, Br2 is reduced to PBr5 and Br2 act as oxidizing agent.

(c)

Interpretation Introduction

Interpretation:

The given redox reaction is to be balanced in acidic medium and the oxidizing agent and reducing agent are to be identified.

Concept Introduction:

Same as part (a).

(c)

Expert Solution
Check Mark

Answer to Problem 17QRT

The balanced redox reaction is shown below.

  2Co2+(aq)+H2O2(aq)+2H+(aq)2Co3+(aq)+2H2O(l)

In this reaction, Co2+ act as reducing agent and H2O2 act as oxidizing agent.

Explanation of Solution

The given unbalanced redox reaction is shown below.

  H2O2(aq)+Co2+(aq)H2O(l)+Co3+(aq)

The oxidation number of cobalt atom in Co2+ and Co3+ is +2 and +3 respectively.  Since there is increase in oxidation number, therefore, Co2+ is oxidized to Co3+ and Co2+ act as reducing agent.

The oxidation number of oxygen atom in H2O2 and H2O is 1 and 2 respectively.  Since there is decrease in oxidation number, therefore, H2O2 is reduced to H2O and H2O2 act as oxidizing agent.

Therefore, the unbalanced half reactions are shown below.

Oxidation: Co2+Co3+

Reduction: H2O2H2O

The given reaction is to be balanced in acidic medium.  Therefore, oxygen atoms are balanced by adding water molecules to the deficient side.

Oxidation: Co2+Co3+

Reduction: H2O22H2O

After balancing oxygen atoms, now hydrogen atoms are balanced by adding protons (H+) to the deficient side.

Oxidation: Co2+Co3+

Reduction: H2O2+2H+2H2O

The charge is on both sides is balanced by adding electrons to the more positive side of the half- reaction to equal the less positive side of the half- reaction.

Oxidation: Co2+Co3++e

Reduction: H2O2+2H++2e2H2O

Since, to make the equal loss and gain of electron, the oxidation reaction is multiplied by two.  Therefore, the overall cell reaction is obtained by addition the above two half-cell reactions.

    2Co2++H2O2+2H++2e2Co3++2e+2H2O

The simplified chemical equation after removing the chemical species of the similar kind is shown below.

    2Co2+(aq)+H2O2(aq)+2H+(aq)2Co3+(aq)+2H2O(l)

Thus, the balanced chemical reaction is shown below.

  2Co2+(aq)+H2O2(aq)+2H+(aq)2Co3+(aq)+2H2O(l)

(d)

Interpretation Introduction

Interpretation:

The given redox reaction is to be balanced in acidic medium and the oxidizing agent and reducing agent are to be identified.

Concept Introduction:

Same as part (a).

(d)

Expert Solution
Check Mark

Answer to Problem 17QRT

The balanced redox reaction is shown below.

  Cr2O72(aq)+6Cl+14H+(aq)2Cr3+(aq)+3Cl2(g)+7H2O(l)

In this reaction, Cl act as reducing agent and Cr2O72 act as oxidizing agent.

Explanation of Solution

The given unbalanced redox reaction is shown below.

    Cl(aq)+Cr2O72(aq)Cl2(g)+Cr3+(aq)

The oxidation number of chlorine atom in Cl and Cl2 is 1 and 0 respectively.  Since there is increase in oxidation number, therefore, Cl is oxidized to Cl2 and Cl act as reducing agent.

The oxidation number of manganese atom in Cr2O72 and Cr3+ is +6 and +3 respectively.  Since there is decrease in oxidation number, therefore, Cr2O72 is reduced to Cr3+ and Cr2O72 act as oxidizing agent.

Therefore, the unbalanced half reactions are shown below.

Oxidation: Cl Cl2

Reduction: Cr2O72Cr3+

The given reaction is to be balanced in acidic medium.  Therefore, oxygen atoms are balanced by adding water molecules to the deficient side.

Oxidation: 2Cl Cl2

Reduction: Cr2O722Cr3++7H2O

After balancing oxygen atoms, now hydrogen atoms are balanced by adding protons (H+) to the deficient side.

Oxidation: 2Cl Cl2

Reduction: Cr2O72+14H+2Cr3++7H2O

The charge is on both sides is balanced by adding electrons to the more positive side of the half- reaction to equal the less positive side of the half- reaction.

Oxidation: 2Cl Cl2+2e

Reduction: Cr2O72+14H++6e2Cr3++7H2O

Since the electron gain is not equivalent to electron lost.  Thus, multiply the oxidation half reaction by 3 and reduction half reaction by 1.

Oxidation: 6Cl3Cl2+6e

Reduction: Cr2O72+14H++6e2Cr3++7H2O

Now, there is the equal loss and gain of electron in the above two half-cell reactions.  Therefore, the overall cell reaction is get by addition the above two half-cell reactions.

    Cr2O72+14H++6e+6Cl2Cr3++7H2O+3Cl2+6e

The simplified chemical equation after removing the chemical species of the similar kind is shown below.

  Cr2O72(aq)+6Cl+14H+(aq)2Cr3+(aq)+3Cl2(g)+7H2O(l)

Thus, the balanced chemical reaction is shown below.

  Cr2O72(aq)+6Cl+14H+(aq)2Cr3+(aq)+3Cl2(g)+7H2O(l)

(e)

Interpretation Introduction

Interpretation:

The given redox reaction is to be balanced and the oxidizing agent and reducing agent are to be identified.

Concept Introduction:

Same as part (a).

(e)

Expert Solution
Check Mark

Answer to Problem 17QRT

The balanced redox reaction is shown below.

  3Zn(s)+2MnO4(aq)+4H2O(l)3Zn(OH)2(s)+2MnO2(s)+2OH(aq)

In this reaction, Zn act as reducing agent and MnO4 act as oxidizing agent.

Explanation of Solution

The given unbalanced redox reaction is shown below.

  MnO4(aq)+Zn(s)MnO2(s)+Zn(OH)2(s)

The oxidation number of zinc atom in Zn and Zn(OH)2 is 0 and +2 respectively.  Since there is increase in oxidation number, therefore, Zn is oxidized to Zn(OH)2 and Zn act as reducing agent.

The oxidation number of manganese atom in MnO4 and MnO2 is +7 and +4 respectively.  Since there is decrease in oxidation number, therefore, MnO4 is reduced to MnO2 and MnO4 act as oxidizing agent.

Therefore, the unbalanced half reactions are shown below.

Oxidation: ZnZn(OH)2

Reduction: MnO4MnO2

The given reaction is to be balanced in basic medium.  Therefore, oxygen atoms are balanced by adding water molecules to the deficient side.

Oxidation: Zn+2H2OZn(OH)2

Reduction: MnO4MnO2+2H2O

After balancing oxygen atoms, hydrogen atoms are balanced by adding water molecules (H2O) to the deficient side and equal number of hydroxide ions (OH) to the other side.

Oxidation: Zn+2H2O+2OHZn(OH)2+2H2O

Reduction: MnO4+2H2OMnO2+4OH

The charge is on both sides is balanced by adding electrons to the more positive side of the half- reaction to equal the less positive side of the half- reaction.

Oxidation: Zn+2H2O+2OHZn(OH)2+2H2O+2e

Reduction: MnO4+3e+2H2OMnO2+4OH

Since the electron gain is not equivalent to electron lost.  Thus, multiply the oxidation half reaction by 3 and reduction half reaction by 2.

Oxidation: 3Zn+6H2O+6OH3Zn(OH)2+6H2O+6e

Reduction: 2MnO4+6e+4H2O2MnO2+8OH

Now, there is the equal loss and gain of electron in the above two half-cell reactions.  Therefore, the overall cell reaction is get by addition the above two half-cell reactions.

    3Zn+6H2O+6OH+2MnO4+6e+4H2O 3Zn(OH)2+6H2O+6e+2MnO2+8OH

The simplified chemical equation after removing the chemical species of the similar kind is shown below.

  3Zn(s)+2MnO4(aq)+4H2O(l)3Zn(OH)2(s)+2MnO2(s)+2OH(aq)

Thus, the balanced chemical reaction is shown below.

  3Zn(s)+2MnO4(aq)+4H2O(l)3Zn(OH)2(s)+2MnO2(s)+2OH(aq)

(f)

Interpretation Introduction

Interpretation:

The given redox reaction is to be balanced and the oxidizing agent and reducing agent are to be identified.

Concept Introduction:

Same as part (a).

(f)

Expert Solution
Check Mark

Answer to Problem 17QRT

The balanced redox reaction is shown below.

  O3(g)+NO(g)O2(g)+NO2(g)

In this reaction, NO act as reducing agent and O3 act as oxidizing agent.

Explanation of Solution

The given redox reaction is shown below.

  O3(g)+NO(g)O2(g)+NO2(g)

The given redox reaction is already balanced and there are equal numbers of each atom both the sides.

The oxidation number of nitrogen atom in NO and NO2 is +2 and +4 respectively.  Since there is increase in oxidation number, therefore, NO is oxidized to NO2 and NO act as reducing agent.

The oxidation number of oxygen atom in O3 and NO2 is 0 and 2 respectively.  Since there is decrease in oxidation number, therefore, O3 is reduced to NO2 and O3 act as oxidizing agent.

(g)

Interpretation Introduction

Interpretation:

The given redox reaction is to be balanced in basic medium and the oxidizing agent and reducing agent are to be identified.

Concept Introduction:

Same as part (a).

(g)

Expert Solution
Check Mark

Answer to Problem 17QRT

The balanced redox reaction is shown below.

  C3H8(g)+5O2(g)3CO2(g)+4H2O(l)

In this reaction, C3H8 act as reducing agent and O2 act as oxidizing agent.

Explanation of Solution

The given unbalanced redox reaction is shown below.

    C3H8(g)+O2(g)CO2(g)+H2O(l)

The phosphorous atoms and bromine atoms are balanced by putting 5, 3 and 4 as prefix to O2, CO2 and H2O respectively.

  C3H8(g)+5O2(g)3CO2(g)+4H2O(l)

Thus, the balanced chemical reaction is shown below.

  C3H8(g)+5O2(g)3CO2(g)+4H2O(l)

The oxidation number of carbon atom in C3H8 and CO2 is +2,+3 and +4 respectively.  Since there is increase in oxidation number, therefore, C3H8 is oxidized to CO2 and C3H8 act as reducing agent.

The oxidation number of oxygen atom in O2 and H2O is 0 and 2 respectively.  Since there is decrease in oxidation number, therefore, O2 is reduced to H2O and O2 act as oxidizing agent.

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Chapter 17 Solutions

Chemistry: The Molecular Science

Ch. 17.4 - Given this reaction, its standard potential, and...Ch. 17.5 - Prob. 17.5PSPCh. 17.5 - Prob. 17.8CECh. 17.5 - Prob. 17.9CECh. 17.5 - Prob. 17.10CECh. 17.6 - Prob. 17.6PSPCh. 17.6 - Prob. 17.11ECh. 17.6 - Prob. 17.7PSPCh. 17.7 - Calculate the cell potential for the Zn(s) +...Ch. 17.7 - Prob. 17.9PSPCh. 17.8 - Prob. 17.12ECh. 17.8 - Prob. 17.13ECh. 17.8 - Prob. 17.14ECh. 17.10 - Predict the results of passing a direct electrical...Ch. 17.10 - In 1886. Henri Moissan was the first to prepare...Ch. 17.11 - In the commercial production of sodium metal by...Ch. 17.11 - Prob. 17.16CECh. 17.11 - Prob. 17.17ECh. 17.11 - Prob. 17.18CECh. 17.11 - Prob. 17.19ECh. 17.12 - Prob. 17.20CECh. 17.12 - Prob. 17.21CECh. 17 - Prob. 2SPCh. 17 - Prob. 1QRTCh. 17 - Prob. 2QRTCh. 17 - Prob. 3QRTCh. 17 - Prob. 4QRTCh. 17 - Identify each statement as true or false. Rewrite...Ch. 17 - Prob. 6QRTCh. 17 - Prob. 7QRTCh. 17 - Prob. 8QRTCh. 17 - Answer Question 8 again, but this time find a...Ch. 17 - Prob. 10QRTCh. 17 - Prob. 11QRTCh. 17 - For the reaction in Question 6, write balanced...Ch. 17 - Prob. 13QRTCh. 17 - Prob. 14QRTCh. 17 - Prob. 15QRTCh. 17 - Prob. 16QRTCh. 17 - Prob. 17QRTCh. 17 - For the reaction Cu2+(aq) + Zn(s) → Cu(s) + Zn2+...Ch. 17 - Prob. 19QRTCh. 17 - Prob. 20QRTCh. 17 - Prob. 21QRTCh. 17 - Prob. 22QRTCh. 17 - Draw a diagram of each cell. Label the anode, the...Ch. 17 - Prob. 24QRTCh. 17 - Prob. 25QRTCh. 17 - Prob. 26QRTCh. 17 - Prob. 27QRTCh. 17 - Prob. 28QRTCh. 17 - Prob. 29QRTCh. 17 - Prob. 30QRTCh. 17 - Prob. 31QRTCh. 17 - Consider these half-reactions: (a) Which is the...Ch. 17 - Consider these half-reactions: (a) Which is the...Ch. 17 - In principle, a battery could be made from...Ch. 17 - Prob. 35QRTCh. 17 - Hydrazine, N2H4, can be used as the reducing agent...Ch. 17 - Prob. 37QRTCh. 17 - Prob. 38QRTCh. 17 - Prob. 39QRTCh. 17 - Prob. 40QRTCh. 17 - Prob. 41QRTCh. 17 - Prob. 42QRTCh. 17 - Prob. 43QRTCh. 17 - Prob. 44QRTCh. 17 - Prob. 45QRTCh. 17 - Prob. 46QRTCh. 17 - Consider the voltaic cell 2 Ag+(aq) + Cd(s) 2...Ch. 17 - Consider a voltaic cell with the reaction H2(g) +...Ch. 17 - Calculate the cell potential of a concentration...Ch. 17 - Prob. 50QRTCh. 17 - Prob. 51QRTCh. 17 - Prob. 52QRTCh. 17 - Prob. 53QRTCh. 17 - NiCad batteries are rechargeable and are commonly...Ch. 17 - Prob. 55QRTCh. 17 - Prob. 56QRTCh. 17 - Prob. 57QRTCh. 17 - Hydrazine, N2H4, has been proposed as the fuel in...Ch. 17 - Consider the electrolysis of water in the presence...Ch. 17 - Prob. 60QRTCh. 17 - Prob. 61QRTCh. 17 - Prob. 62QRTCh. 17 - Identify the products of the electrolysis of a 1-M...Ch. 17 - Prob. 64QRTCh. 17 - Prob. 65QRTCh. 17 - Prob. 66QRTCh. 17 - Prob. 67QRTCh. 17 - Prob. 68QRTCh. 17 - Prob. 69QRTCh. 17 - Prob. 70QRTCh. 17 - Prob. 71QRTCh. 17 - Prob. 72QRTCh. 17 - Prob. 73QRTCh. 17 - Prob. 74QRTCh. 17 - Calculate how long it would take to electroplate a...Ch. 17 - Prob. 76QRTCh. 17 - Prob. 77QRTCh. 17 - Prob. 78QRTCh. 17 - Prob. 79QRTCh. 17 - Prob. 80QRTCh. 17 - Prob. 81QRTCh. 17 - Prob. 82QRTCh. 17 - Prob. 83QRTCh. 17 - Prob. 84QRTCh. 17 - Prob. 85QRTCh. 17 - Prob. 86QRTCh. 17 - Prob. 87QRTCh. 17 - Prob. 88QRTCh. 17 - You wish to electroplate a copper surface having...Ch. 17 - Prob. 90QRTCh. 17 - Prob. 91QRTCh. 17 - Prob. 92QRTCh. 17 - Prob. 93QRTCh. 17 - An electrolytic cell is set up with Cd(s) in...Ch. 17 - Prob. 95QRTCh. 17 - Prob. 96QRTCh. 17 - Prob. 97QRTCh. 17 - Prob. 98QRTCh. 17 - Prob. 99QRTCh. 17 - Prob. 100QRTCh. 17 - Prob. 101QRTCh. 17 - Prob. 102QRTCh. 17 - Prob. 103QRTCh. 17 - Prob. 104QRTCh. 17 - Prob. 105QRTCh. 17 - Prob. 106QRTCh. 17 - Prob. 107QRTCh. 17 - Prob. 108QRTCh. 17 - Prob. 109QRTCh. 17 - Prob. 110QRTCh. 17 - Prob. 111QRTCh. 17 - Prob. 17.ACPCh. 17 - Prob. 17.BCP
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