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 7QRT

(a)

Interpretation Introduction

Interpretation:

The oxidation number has to be assigned to each atom of the given reaction.  The substance that is oxidized, reduced, oxidizing agent and reducing agent have to be stated.

Concept Introduction:

Redox reactions are the reactions in which oxidation and reduction occur simultaneously.  Loss of electrons or gain of oxygen is termed as oxidation.  On the other hand, addition of electrons or hydrogen is known as reduction.  Increase in oxidation number signifies oxidation and decrease in oxidation number signifies reduction.

(a)

Expert Solution
Check Mark

Answer to Problem 7QRT

The oxidation number of Fe changes from 0_ to +2_.  So, Fe(s) gets oxidized and acts as a reducing agent.

The oxidation number of Br changes from 0_ to 1_.  So, Br(l) gets reduced and acts as an oxidizing agent

Explanation of Solution

The given reaction is shown below.

    Fe(s)+Br(l)FeBr2(s)

The oxidation number of any element in its native form is zero.  So, the oxidation number of Fe(s) and Br(l) is zero.

In almost all compounds oxidation number of bromine is 1.  Assume the oxidation number of iron to be x.

Apply charge balance formula in FeBr2 as shown below.

  (2×OxidationstateofBr)+OxidationstateofFe=OverallchargeonFeBr22(1)+x=0x=+2.

Therefore, the oxidation number of iron in FeBr2 is +2_.

The element that gets oxidized reduces the other one and is known as reducing agent.  The element that gets reduced oxidizes the other one and is known as oxidizing agent.

In the given reaction, the oxidation number of Fe changes from 0 to +2.  So, Fe(s) gets oxidized and acts as a reducing agent.

The oxidation number of Br changes from 0 to 1.  So, Br(l) gets reduced and acts as an oxidizing agent.

(b)

Interpretation Introduction

Interpretation:

The oxidation number has to be assigned to each atom of the given reaction.  The substance that is oxidized, reduced, oxidizing agent and reducing agent have to be stated.

Concept Introduction:

Refer to part (a).

(b)

Expert Solution
Check Mark

Answer to Problem 7QRT

The oxidation number of Al changes from 0_ to +3_.  So, Al(s) gets oxidized and acts as a reducing agent.

The oxidation number of Cl changes from 0_ to 1_.  So, Cl2(g) gets reduced and acts as an oxidizing agent

Explanation of Solution

The given reaction is shown below.

    2Al(s)+3Cl2(g)2AlCl3(s)

The oxidation number of any element in its native form is zero.  So, the oxidation number of Al(s) and Cl2(g) is zero.

In almost all compounds oxidation number of chlorine is 1.  Assume the oxidation number of aluminium to be x.

Apply charge balance formula in AlCl3 as shown below.

  (3×OxidationstateofCl)+OxidationstateofAl=OverallchargeonAlCl33(1)+x=0x=+3.

Therefore, the oxidation number of aluminium in AlCl3 is +3_.

The element that gets oxidized reduces the other one and is known as reducing agent.  The element that gets reduced oxidizes the other one and is known as oxidizing agent.

In the given reaction, the oxidation number of Al changes from 0 to +3.  So, Al(s) gets oxidized and acts as a reducing agent.

The oxidation number of Cl changes from 0 to 1.  So, Cl2(g) gets reduced and acts as an oxidizing agent.

(c)

Interpretation Introduction

Interpretation:

The oxidation number has to be assigned to each atom of the given reaction.  The substance that is oxidized, reduced, oxidizing agent and reducing agent have to be stated.

Concept Introduction:

Refer to part (a).

(c)

Expert Solution
Check Mark

Answer to Problem 7QRT

The oxidation number of sulfur changes from +6_ to 2_.  So, H2SO4 gets reduced and acts as an oxidizing agent.

The oxidation number of I changes from 1_ to 0_.  So, HI gets oxidized and acts as a reducing agent.

Explanation of Solution

The given reaction is shown below.

    8HI(aq)+H2SO4(aq)H2S(aq)+4I2(s)+4H2O(l)

The oxidation number of any element in its native form is zero.  So, the oxidation number of I2(s) is zero.

In almost all compounds oxidation number of iodine is 1, hydrogen is +1 and oxygen is 2.  Assume the oxidation number of sulfur to be x.

Apply charge balance formula in H2SO4 as shown below.

  ((2×OxidationstateofH)+4×OxidationstateofO+OxidationstateofS)=OverallchargeonH2SO42(+1)+4(2)+x=0x=82=6.

Therefore, the oxidation number of sulfur in H2SO4 is +6_.

Apply charge balance formula in H2S as shown below.

  ((2×OxidationstateofH)+OxidationstateofS)=OverallchargeonH2S2(+1)+x=0x=2.

Therefore, the oxidation number of sulfur in H2S is 2_.

The element that gets oxidized reduces the other one and is known as reducing agent.  The element that gets reduced oxidizes the other one and is known as oxidizing agent.

In the given reaction, the oxidation number of sulfur changes from +6 to 2.  So, H2SO4 gets reduced and acts as an oxidizing agent.

The oxidation number of I changes from 1 to 0.  So, HI gets oxidized and acts as a reducing agent.

(d)

Interpretation Introduction

Interpretation:

The oxidation number has to be assigned to each atom of the given reaction.  The substance that is oxidized, reduced, oxidizing agent and reducing agent have to be stated.

Concept Introduction:

Refer to part (a).

(d)

Expert Solution
Check Mark

Answer to Problem 7QRT

The oxidation number of Fe2+ changes from +2_ to +3_.  So, Fe2+ gets oxidized and acts as a reducing agent.

The oxidation number of O changes from 1_ to 2_.  So, H2O2 gets reduced and acts as an oxidizing agent.

Explanation of Solution

The given reaction is shown below.

    H2O2(aq)+2Fe2+(aq)+2H+(aq)2Fe3+(aq)+2H2O(l)

The oxidation number of any element carrying charge is equal to that charge.  So, the oxidation number of H+, Fe2+ and Fe3+ is +1_,+2_ and +3_ respectively.

In almost all compounds oxidation number of hydrogen is +1.  Assume the oxidation number of oxygen to be x.

Apply charge balance formula in H2O2 as shown below.

  ((2×OxidationstateofH)+2×OxidationstateofO)=OverallchargeonH2O22(+1)+2x=0x=22=1.

Therefore, the oxidation number of oxygen in H2O2 is 1_.

Apply charge balance formula in H2O as shown below.

  ((2×OxidationstateofH)+OxidationstateofO)=OverallchargeonH2O2(+1)+x=0x=2.

Therefore, the oxidation number of oxygen in H2O is 2_.

The element that gets oxidized reduces the other one and is known as reducing agent.  The element that gets reduced oxidizes the other one and is known as oxidizing agent.

In the given reaction, the oxidation number of Fe2+ changes from +2 to +3.  So, Fe2+ gets oxidized and acts as a reducing agent.

The oxidation number of O changes from 1 to 2.  So, H2O2 gets reduced and acts as an oxidizing agent.

(e)

Interpretation Introduction

Interpretation:

The oxidation number has to be assigned to each atom of the given reaction.  The substance that is oxidized, reduced, oxidizing agent and reducing agent have to be stated.

Concept Introduction:

Refer to part (a).

(e)

Expert Solution
Check Mark

Answer to Problem 7QRT

The oxidation number of Fe2+ changes from +2_ to +3_.  So, FeS gets oxidized and acts as a reducing agent.

In the given reaction, the oxidation number of S changes from 2_ to +6_.  So, FeS gets oxidized and acts as a reducing agent.

The oxidation number of N changes from +3_ to +2_.  So, NO3 gets reduced and acts as an oxidizing agent.

Explanation of Solution

The given reaction is shown below.

    FeS(s)+3NO3(aq)+4H+(aq)3NO(g)+SO42(aq)+Fe3+(aq)+2H2O(l)

The oxidation number of any element carrying charge is equal to that charge.  So, the oxidation number of H+ and Fe3+ is +1_ and +3_ respectively.

In almost all compounds oxidation number of hydrogen is +1 and oxygen is 2.  Assume the oxidation number of nitrogen to be x.

Apply charge balance formula in NO3 as shown below.

  (OxidationstateofN+3×OxidationstateofO)=OverallchargeonNO3x+2(2)=1x=1+4=3.

Therefore, the oxidation number of nitrogen in NO3 is +3_.

Apply charge balance formula in NO as shown below.

  ((OxidationstateofN)+OxidationstateofO)=OverallchargeonNOx+(2)=0x=2.

Therefore, the oxidation number of nitrogen in NO is +2_.

Apply charge balance formula in SO42 as shown below.

  (OxidationstateofS+4×OxidationstateofO)=OverallchargeonSO42x+4(2)=2x=2+8=6.

Therefore, the oxidation number of nitrogen in SO42 is +6_.

Apply charge balance formula in FeS as shown below.

  ((OxidationstateofFe)+OxidationstateofS)=OverallchargeonFeSx+(2)=0x=2.

Therefore, the oxidation number of iron in FeS is +2_ and that of sulfur is 2_.

The element that gets oxidized reduces the other one and is known as reducing agent.  The element that gets reduced oxidizes the other one and is known as oxidizing agent.

In the given reaction, the oxidation number of Fe2+ changes from +2 to +3.  So, FeS gets oxidized and acts as a reducing agent.

In the given reaction, the oxidation number of S changes from 2 to +6.  So, FeS gets oxidized and acts as a reducing agent.

The oxidation number of N changes from +3 to +2.  So, NO3 gets reduced and acts as an 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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