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 11QRT

(a)

Interpretation Introduction

Interpretation:

The half cell reaction for reduction of MnO4 to Mn2+ ion in acidic solution 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 11QRT

The half cell reaction for reduction of MnO4 to Mn2+ ion in acidic solution is shown below.

    MnO4(aq)+8H+(aq)+5eMn2+(aq)+4H2O(l)

Explanation of Solution

The reaction for the conversion of MnO4 to Mn2+ ion is shown below.

    MnO4(aq)Mn2+(aq)

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

    MnO4(aq)Mn2+(aq)+4H2O(l)

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

    MnO4(aq)+8H+(aq)Mn2+(aq)+4H2O(l)

The charge on reactant side is +7 and on product side is +2.  Therefore, in order to balance the charge, five electrons are added on reactant side as follows.

    MnO4(aq)+8H+(aq)+5eMn2+(aq)+4H2O(l)

The balanced half cell reaction represents reduction is shown below.

    MnO4(aq)+8H+(aq)+5eMn2+(aq)+4H2O(l)

(b)

Interpretation Introduction

Interpretation:

The half cell reaction for reduction of Cr2O72 to Cr3+ ion in acidic solution has to be stated.

Concept Introduction:

Refer to part (a).

(b)

Expert Solution
Check Mark

Answer to Problem 11QRT

The half cell reaction for reduction of Cr2O72 to Cr3+ ion in acidic solution is shown below.

    Cr2O72(aq)+14H+(aq)+9eCr3+(aq)+7H2O(l)

Explanation of Solution

The reaction for the conversion of Cr2O72 to Cr3+ ion is shown below.

    Cr2O72(aq)Cr3+(aq)

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

    Cr2O72(aq)Cr3+(aq)+7H2O(l)

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

    Cr2O72(aq)+14H+(aq)Cr3+(aq)+7H2O(l)

The charge on reactant side is +12 and on product side is +3.  Therefore, in order to balance the charge, nine electrons are added on reactant side as follows.

    Cr2O72(aq)+14H+(aq)+9eCr3+(aq)+7H2O(l)

The balanced half cell reaction represents reduction is shown below.

    Cr2O72(aq)+14H+(aq)+9eCr3+(aq)+7H2O(l)

(c)

Interpretation Introduction

Interpretation:

The half cell reaction for oxidation of chlorine gas to ClO ions has to be stated.

Concept Introduction:

Refer to part (a).

(c)

Expert Solution
Check Mark

Answer to Problem 11QRT

The half cell reaction for oxidation of chlorine gas to ClO ions is shown below.

    12Cl2(g)+H2O(l)ClO(aq)+2H+(aq)+e

Explanation of Solution

The reaction for the conversion of chlorine gas to ClO ions is shown below.

    12Cl2(g)ClO(aq)

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

    12Cl2(g)+H2O(l)ClO(aq)

The number of hydrogen atoms is balanced by adding H+ ions on product side as shown below.

    12Cl2(g)+H2O(l)ClO(aq)+2H+(aq)

The charge on reactant side is 0 and on product side is +1.  Therefore, in order to balance the charge, one electron is added on product side as follows.

    12Cl2(g)+H2O(l)ClO(aq)+2H+(aq)+e

The balanced half cell reaction represents oxidation is shown below.

    12Cl2(g)+H2O(l)ClO(aq)+2H+(aq)+e

(d)

Interpretation Introduction

Interpretation:

The half cell reaction for reduction of H2O2 to H2O in acidic solution has to be stated.

Concept Introduction:

Refer to part (a).

(d)

Expert Solution
Check Mark

Answer to Problem 11QRT

The half cell reaction for reduction of H2O2 to H2O in acidic solution is shown below.

    H2O2(aq)+2H+(aq)+2e2H2O(l)

Explanation of Solution

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)

(e)

Interpretation Introduction

Interpretation:

The half cell reaction for oxidation of HNO2 to NO3 ion in acidic condition has to be stated.

Concept Introduction:

Refer to part (a).

(e)

Expert Solution
Check Mark

Answer to Problem 11QRT

The half cell reaction for oxidation of HNO2 to NO3 ion in acidic condition is shown below.

    HNO2+H2ONO3+3H++2e

Explanation of Solution

The reaction for the conversion of HNO2 to NO3 ion is shown below.

    HNO2NO3

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

    HNO2+H2ONO3

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

    HNO2+H2ONO3+3H+

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

    HNO2+H2ONO3+3H++2e

The half cell reaction for oxidation of HNO2 to NO3 ion in acidic condition is shown below.

    HNO2+H2ONO3+3H++2e

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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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