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 15QRT

(a)

Interpretation Introduction

Interpretation:

The overall balanced equation by adding half reactions for the given reaction has 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.  The galvanic cell, voltaic cell electrolytic cell are the types of electrochemical cells.  In each cell, the oxidation half-reaction takes place at the anode while half-reaction reaction takes place at the cathode.

(a)

Expert Solution
Check Mark

Answer to Problem 15QRT

The overall equation is as follows.

    Fe(s)+Br2(l)FeBr2(s)

Explanation of Solution

The given reaction is shown below.

    Fe(s)+Br2(l)FeBr2(s)

The oxidation number of any element in its native form is zero.  So, the oxidation number of Fe(s) and Br2(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.

In the given reaction, the oxidation number of Fe changes from 0 to +2.  So, Fe(s) gets oxidized.

The oxidation number of Br changes from 0 to 1.  So, Br2(l) gets reduced.

The half reaction that represents oxidation is as follows.

    Fe(s)Fe2+(aq)+2e        (1)

The half reaction that represents reduction is as follows.

    12Br2(l)+eBr(aq)        (2)

Multiply equation (2) and add to equation (1) as shown below.

    Fe(s)Fe2+(aq)+2e+Br2(l)+2e2Br(aq)_Fe(s)+Br2(l)Fe2+(aq)+2Br(aq)

The overall equation obtained is as follows.

    Fe(s)+Br2(l)FeBr2(s)

(b)

Interpretation Introduction

Interpretation:

The overall balanced equation by adding half reactions for the given reaction has to be stated.

Concept Introduction:

Refer to part (a).

(b)

Expert Solution
Check Mark

Answer to Problem 15QRT

The overall reaction is shown below.

    2Al(s)+3Cl2(g)2AlCl3(s)

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.

In the given reaction, the oxidation number of Al changes from 0 to +3.  So, Al(s) gets oxidized.

The oxidation number of Cl changes from 0 to 1.  So, Cl2(g) gets reduced.

The half reaction that represents oxidation is as follows.

    Al(s)Al3+(aq)+3e        (3)

The half reaction that represents reduction is as follows.

    12Cl2(g)+eCl(aq)        (4)

Multiply equation (4) with three and add to equation (3) as shown below.

    Al(s)Al3+(aq)+3e+32Cl2(g)+3e3Cl(aq)_Al(s)+32Cl2(g)3Cl(aq)+Al3+(aq)

Multiply the above equation to get final reaction as follows.

    2Al(s)+3Cl2(g)6Cl(aq)+2Al3+(aq)

Therefore, the overall reaction is shown below.

    2Al(s)+3Cl2(g)2AlCl3(s)

(c)

Interpretation Introduction

Interpretation:

The overall balanced equation by adding half reactions for the given reaction has to be stated.

Concept Introduction:

Refer to part (a).

(c)

Expert Solution
Check Mark

Answer to Problem 15QRT

The overall reaction is shown below.

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

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.

In the given reaction, the oxidation number of sulfur changes from +6 to 2.  So, H2SO4 gets reduced.

The oxidation number of I changes from 1 to 0.  So, HI gets oxidized.

The half reaction that represents oxidation is as follows.

    I(aq)12I2(s)+e        (5)

The half reaction that represents reduction is as follows.

    H2SO4(aq)H2S(aq)

The number of oxygen atoms is balanced by adding four molecules of water on product side as shown below.

    H2SO4(aq)H2S(aq)+4H2O(l)

In acidic medium, the number of hydrogen atoms is balanced by adding H+ ions on reactant side as shown below.

    H2SO4(aq)+8H+(aq)H2S(aq)+4H2O(l)

The charge on reactant side is +8 and on product side is 0.  Therefore, in order to balance the charge, eight electrons are added on reactant side as follows.

    H2SO4(aq)+8H+(aq)+8eH2S(aq)+4H2O(l)

The balanced half cell reaction represents reduction is shown below.

    H2SO4(aq)+8H+(aq)+8eH2S(aq)+4H2O(l)        (6)

Multiply equation (5) with eight and add to equation (6) as shown below.

    H2SO4(aq)+8H+(aq)+8eH2S(aq)+4H2O(l)+8I(aq)4I2(s)+8e_H2SO4(aq)+8H+(aq)+8I(aq)H2S(aq)+4H2O(l)+4I2(s)

Therefore, the overall reaction is shown below.

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

(d)

Interpretation Introduction

Interpretation:

The overall balanced equation by adding half reactions for the given reaction has to be stated.

Concept Introduction:

Refer to part (a).

(d)

Expert Solution
Check Mark

Answer to Problem 15QRT

The overall reaction is as follows.

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

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.

In the given reaction, the oxidation number of Fe2+ changes from +2 to +3.  So, Fe2+ gets oxidized.

The oxidation number of O changes from 1 to 2.  So, H2O2 gets reduced.

The half reaction that represents oxidation is as follows.

    Fe2+(aq)Fe3+(aq)+e        (7)

The reaction for the conversion of H2O2 to H2O is shown below.

    H2O2(aq)H2O(l)

The number of oxygen atoms is balanced by adding one molecule of water on product side as shown below.

    H2O2(aq)H2O(l)+H2O(l)

In acidic medium, the number of hydrogen atoms is balanced by adding H+ ions on reactant side as shown below.

    H2O2(aq)+2H+(aq)H2O(l)+H2O(l)

The charge on reactant side is +2 and on product side is 0.  Therefore, in order to balance the charge, two electrons are added on reactant side as follows.

    H2O2(aq)+2H+(aq)+2eH2O(l)+H2O(l)

The balanced half cell reaction represents reduction is shown below.

    H2O2(aq)+2H+(aq)+2e2H2O(l)        (8)

Multiply equation (7) with two and add to equation (8) as shown below.

    H2O2(aq)+2H+(aq)+2e2H2O(l)+2Fe2+(aq)2Fe3+(aq)+2e_H2O2(aq)+2H+(aq)+2Fe2+(aq)2Fe3+(aq)+2H2O(l)

Therefore, the overall reaction is as follows.

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

(e)

Interpretation Introduction

Interpretation:

The overall balanced equation by adding half reactions for the given reaction has to be stated.

Concept Introduction:

Refer to part (a).

(e)

Expert Solution
Check Mark

Answer to Problem 15QRT

The, the overall reaction is as follows.

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

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.

In the given reaction, the oxidation number of Fe2+ changes from +2 to +3.  So, FeS gets oxidized.

In the given reaction, the oxidation number of S changes from 2 to +6.  So, FeS gets oxidized.

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

The reaction for the oxidation of FeS is shown below.

    FeS(s)Fe3+(aq)+SO42(aq)

The number of oxygen atoms is balanced by adding four molecules of water on reactant side as shown below.

    FeS(s)+4H2O(l)Fe3+(aq)+SO42(aq)

In acidic medium, the number of hydrogen atoms is balanced by adding H+ ions on product side as shown below.

    FeS(s)+4H2O(l)Fe3+(aq)+SO42(aq)+8H+

The charge on reactant side is zero and the charge on product side is +9.  Therefore, in order to balance the charge, nine electrons are added on product side as follows.

    FeS(s)+4H2O(l)Fe3+(aq)+SO42(aq)+8H++9e

The half cell reaction for oxidation of FeS is shown below.

    FeS(s)+4H2O(l)Fe3+(aq)+SO42(aq)+8H++9e        (9)

The reaction for the reduction of NO3 ion is shown below.

    NO3(aq)NO(g)

The number of oxygen atoms is balanced by adding two molecules of water on product side as shown below.

    NO3(aq)NO(g)+2H2O(l)

In acidic medium, the number of hydrogen atoms is balanced by adding H+ ions on product side as shown below.

    NO3(aq)+4H+(aq)NO(g)+2H2O(l)

The charge on reactant side is +3 and on product side is zero.  Therefore, in order to balance the charge, three electrons are added on reactant side as follows.

    NO3(aq)+4H+(aq)+3eNO(g)+2H2O(l)

The half cell reaction for reduction of NO3 ion is shown below.

    NO3(aq)+4H+(aq)+3eNO(g)+2H2O(l)        (10)

Multiply equation (10) with three and add to equation (10) as shown below.

  3NO3(aq)+12H+(aq)+9e3NO(g)+6H2O(l)+FeS(s)+4H2O(l)Fe3+(aq)+SO42(aq)+8H++9e_3NO3(aq)+12H+(aq)+FeS(s)+4H2O(l)3NO(g)+6H2O(l)+Fe3+(aq)+SO42(aq)+8H+

Therefore, the overall reaction is as follows.

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

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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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