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 104QRT

(a)

Interpretation Introduction

Interpretation:

The given redox reaction are to be balanced.

Concept Introduction:

A redox reaction is a kind of reaction in which one reacting species gets oxidize and another reacting species gets reduced simultaneously.  In the redox titration, either an oxidizing agent is titrated with reducing agent or a reducing agent is titrated with an oxidizing agent.  The standard reduction potential of a galvanic cell may be calculated in terms of standard reduction potential of cathode and anode as the relation mentioned below:

  Ecello=EcathodeoEanodeo

The relation between standard cell potential (Ecell°) and standard free energy change (ΔrG) is shown below.

    ΔrG=nFEcell

If the value of Gibbs free energy change (ΔrG) is negative for a chemical reaction, then that chemical reaction is product-favored whereas the positive value indicates that the reaction is reactant-favored.

(a)

Expert Solution
Check Mark

Answer to Problem 104QRT

The first and second balanced chemical reactions are shown below.

  2NO3(aq)+8H+(aq)+6Hg(l)3Hg22+(aq)+2NO(g)+4H2O(l)

  2Hg2+(aq)+2Br(aq)Hg22+(aq)+Br2(l)

Explanation of Solution

The given first unbalanced redox reaction is shown below.

    NO3(aq)+H+(aq)+Hg(l)Hg22+(aq)+NO(g)+H2O(l)

The oxidation number of mercury atom in Hg and Hg22+ is 0 and +1 respectively.  Since there is increase in oxidation number, therefore, Hg is oxidized to Hg22+ and Hg act as reducing agent.

The oxidation number of nitrogen atom in NO3 and NO is +5 and +2 respectively.  Since there is decrease in oxidation number, therefore, NO3 is reduced to NO and NO3 act as oxidizing agent.

Therefore, the unbalanced half reactions are shown below.

Oxidation: HgHg22+

Reduction: NO3NO

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

Oxidation: 2HgHg22+

Reduction: NO3NO+2H2O

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

Oxidation: 2HgHg22+

Reduction: NO3+4H+NO+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: 2HgHg22++2e

Reduction: NO3+4H++3eNO+2H2O

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: 6Hg3Hg22++6e

Reduction: 2NO3+8H++6e2NO+4H2O

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.

    6Hg+2NO3+8H++6e2NO+4H2O+3Hg22++6e

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

  2NO3(aq)+8H+(aq)+6Hg(l)3Hg22+(aq)+2NO(g)+4H2O(l)

Thus, the first balanced chemical reaction is shown below.

  2NO3(aq)+8H+(aq)+6Hg(l) 3Hg22+(aq)+2NO(g)+4H2O(l)

The second unbalanced redox reaction is shown below.

    Hg2+(aq)+Br(aq)Hg22+(aq)+Br2(l)

The oxidation number of mercury atom in Br and Br2 is 1 and 0 respectively.  Since there is increase in oxidation number, therefore, Br is oxidized to Br2 and Br act as reducing agent.

The oxidation number of nitrogen atom in Hg2+ and Hg22+ is +2 and +1 respectively.  Since there is decrease in oxidation number, therefore, Hg2+ is reduced to Hg22+ and Hg2+ act as oxidizing agent.

Therefore, the unbalanced half reactions are shown below.

Oxidation: BrBr2

Reduction: HgHg22+

The given reaction is to be balanced in acidic medium.  Therefore, the atoms other than oxygen and hydrogen are balanced first.

Oxidation: 2BrBr2

Reduction: 2HgHg22+

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: 2BrBr2+2e

Reduction: 2Hg2++2eHg22+

Since, 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.

    2Hg2++2e+2BrHg22++Br2+2e

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

  2Hg2+(aq)+2Br(aq)Hg22+(aq)+Br2(l)

Thus, the second balanced chemical reaction is shown below.

  2Hg2+(aq)+2Br(aq)Hg22+(aq)+Br2(l)

(b)

Interpretation Introduction

Interpretation:

The cell diagram for the given redox reactions are to be drawn.

Concept Introduction:

Same as part (a).

(b)

Expert Solution
Check Mark

Answer to Problem 104QRT

The cell representation for thefirst and second redox reaction is shown below.

  Hg(l)|Hg22+(aq)NO3(aq)|NO(g)|Pt

  Pt|Br(aq)|Br2(l)Hg2+(aq)|Hg22+(aq)|Pt

Explanation of Solution

The first balanced redox reactions are shown below.

  2NO3(aq)+8H+(aq)+6Hg(l)3Hg22+(aq)+2NO(g)+4H2O(l)

The half-cell reactions which take place at cathode and anode in the first redox reaction are shown below.

  Cathode:2NO3+8H++6e2NO+4H2OAnode:6Hg3Hg22++6e

Thus, the cell representation for thefirst redox reaction is shown below.

  Hg(l)|Hg22+(aq)NO3(aq)|NO(g)|Pt

The second balanced redox reactions are shown below.

  2Hg2+(aq)+2Br(aq)Hg22+(aq)+Br2(l)

The half-cell reactions which take place at cathode and anode in the second redox reaction are shown below.

  Cathode:2Hg2++2eHg22+Anode:2BrBr2+2e

Thus, the cell representation for thesecond redox reaction is shown below.

  Pt|Br(aq)|Br2(l)Hg2+(aq)|Hg22+(aq)|Pt

(c)

Interpretation Introduction

Interpretation:

The standard cell potential for the given cell reaction has to be determined.

Concept Introduction:

Same as part (a).

(c)

Expert Solution
Check Mark

Answer to Problem 104QRT

The standard cell potential of first and second redox reaction is 0.164V_ and -1.0085V_ respectively.

Explanation of Solution

The first balanced chemical reaction is shown below.

  2NO3(aq)+8H+(aq)+6Hg(l)3Hg22+(aq)+2NO(g)+4H2O(l)

The half-cell reactions of Hg electrode and NO3 along with their standard reduction potentials are shown below.

Oxidation half reaction (Anode): 6Hg3Hg22++6e    Eo=0.7960V

Reduction half reaction (Cathode): 2NO3+8H++6e2NO+4H2O    Eo=0.96V

The value of Eanodeo is same as that of EHg/Hg22+o which is 0.7960V.

The value of Ecathodeo is same as that of ENO3/NOo which is 0.96V.

The standard cell potential is calculated as shown below.

    Ecello=EcathodeoEanodeo        (1)

Substitute the value of Ecathodeo and Eanodeo in equation (1).

    Ecello=ENO3/NOoEHg/Hg22+o=0.96V0.7960V=0.164V

The second balanced redox reaction is shown below.

  2Hg2+(aq)+2Br(aq)Hg22+(aq)+Br2(l)

The half-cell reactions of Hg2+ electrode and Br along with their standard reduction potentials are shown below.

Oxidation half reaction (Anode): 2BrBr2+2e    Eo=1.066V

Reduction half reaction (Cathode): 2Hg2++2eHg22+    Eo=0.0575V

The value of Eanodeo is same as that of EBr2/Bro which is 1.066V.

The value of Ecathodeo is same as that of EHg2+/Hg22+o which is 0.0575V.

Substitute the value of Ecathodeo and Eanodeo in equation (1).

    Ecello=EHg2+/Hg22+oEBr2/Bro=0.0575V1.066V=1.0085V

Thus, the standard cell potential of first and second redox reaction is 0.164V_ and -1.0085V_ respectively.

(d)

Interpretation Introduction

Interpretation:

The standard change in Gibbs free energy for the given redox reactions has to be determined.

Concept Introduction:

Same as part (a).

(d)

Expert Solution
Check Mark

Answer to Problem 104QRT

The ΔrG for the first and second redox reaction is 94.94124kJ/mol_ and 194.610245kJ/mol_ respectively.

Explanation of Solution

The half-cell reactions which take place at cathode and anode in the first redox reaction are shown below.

  Cathode:2NO3+8H++6e2NO+4H2OAnode:6Hg3Hg22++6e

The number of electron change in the half cell reactions of first redox reaction is 6

As per the given data the standard cell potential, Ecell° for the first redox reaction is 0.164V.

The relation between standard cell potential and standard free energy change is shown below.

    ΔrG=nFEcell        (1)

Where,

  • ΔrG is the standard free energy change.
  • Ecell is the standard cell potential.
  • F is the Faradays constant.
  • n is the number of electron change in the half cell reaction.

The value of n is 6.

The value of Ecell is 0.164V.

The value of F is 96485C/mole.

Substitute the value of n, Ecell and F for the first redox reaction in equation (1).

  ΔrG=6e×96485C1mole×(0.164V)=94941.24J1mol×1kJ1000J=94.94124kJ/mol

Thus, the ΔrG for the first redox reaction is -94.94124kJ/mol_.

The half-cell reactions which take place at cathode and anode in the second redox reaction are shown below.

  Cathode:2Hg2++2eHg22+Anode:2BrBr2+2e

The number of electron change in the half cell reactions of second redox reaction is 2

As per the given data the standard cell potential, Ecell° for the second redox reaction is 1.0085V.

The value of n is 2.

The value of Ecell is 1.0085V.

The value of F is 96485C/mole.

Substitute the value of n, Ecell and F for the second redox reaction in equation (1).

  ΔrG=2e×96485C1mole×(1.0085V)=194610.245J1mol×1kJ1000J=194.610245kJ/mol

Thus, the ΔrG for the second redox reaction is 194.610245kJ/mol_.

(e)

Interpretation Introduction

Interpretation:

The given redox reaction are product favoured has to be determined.

Concept Introduction:

Same as part (a).

(e)

Expert Solution
Check Mark

Answer to Problem 104QRT

The first redox reactions is the product favored reaction.

Explanation of Solution

The calculated values of the standard cell potential (Ecell°) in the part (c), for the first and second redox reaction is 0.164V and 1.0085V respectively.

The calculated values of the standard Gibbs free energy change (ΔrG) in the part (d), for the first and second redox reaction is 94.94124kJ/mol and 194.610245kJ/mol respectively.

The value of Gibbs free energy change (ΔrG) for the first redox reaction is negative.  It indicates that the overall reaction is spontaneous and product favored.

The value of Gibbs free energy change (ΔrG) for the second redox reaction is positive.  It indicates that the overall reaction is non-spontaneous and reactant favored.

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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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