Chemistry: Principles and Reactions
Chemistry: Principles and Reactions
8th Edition
ISBN: 9781305079373
Author: William L. Masterton, Cecile N. Hurley
Publisher: Cengage Learning
Question
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Chapter 17, Problem 7QAP
Interpretation Introduction

(a)

Interpretation:

The balanced equations for the following reaction in basic solution needs to be determined.

SO2(g) + I2(aq)SO3(g) + I-(aq)

Concept introduction:

Here are the rules to balance redox reactions in basic medium:

  1. Determine the oxidation numbers of the elements and write the oxidation and reduction half equation.
  2. Balance the atoms of elements other than O and H. Use H2 O to balance O, H+ to balance H, electrons to balance the charges. Equalize the number of electrons on both the half reactions and then add the half reactions.
  3. Add OH- to neutralize the excess protons since this is a basic medium. Add OH- to both side of the equation to balance it.

Expert Solution
Check Mark

Answer to Problem 7QAP

SO2(g)+I2(aq)+ 2OH-SO3(g) + 2I-(aq)+H2O

Explanation of Solution

SO2(g) + I2(aq)SO3(g) + I-(aq)

Determining the oxidation numbers:

The oxidation number of I in I2 and I- is 0 and -1 respectively. As the oxidation number of I is decreasing this is the reduction half reaction.

Reduction half reaction:  I2(aq) I-(aq)

Oxidation half reaction: SO2(g) SO3(g)

To balance the excess O atoms on the product side we add one H2 O on the reactant side, and then we balance the excess H atom on the reactant side by adding two H+ on the product side. Finally, we balance the charge by adding two electrons to the product side of the half reaction.

Thus, the balanced oxidation half reaction:

SO2(g)+H2OSO3(g) + 2H+ + 2e 

Now, balance the charge by adding two electrons to the reactant side of the half reaction.

Thus, the balanced reduction half reaction:

I2(aq)+2e 2I-(aq)

Net reaction:

By adding both the half reaction, the net reaction is obtained as follows:

SO2(g)+I2(aq)+H2OSO3(g) + 2I-(aq)+ 2H+ 

Basic medium:

Since the reaction takes place in basic medium the excess proton on the reactant side of the net reaction is neutralized with OH-. 2 OH- ions are added on both side of the equation to even out the charges. H+ and OH- combines to form H2 O, which is then cancelled out on both side of the net reaction.

Thus, the balanced net reaction in the basic medium will be:

SO2(g)+I2(aq)+H2O+ 2OH-SO3(g) + 2I-(aq)+ 2H+ + 2OH-

Or,

SO2(g)+I2(aq)+ 2OH-SO3(g) + 2I-(aq)+H2O

Interpretation Introduction

(b)

Interpretation:

The balanced equations for the following reaction in basic solution needs to be determined.

 Zn(s) + NO3-(aq)NH3(aq) + Zn2+(aq)

Concept introduction:

Here are the rules to balance redox reactions in basic medium:

  1. Determine the oxidation numbers of the elements and write the oxidation and reduction half equation.
  2. Balance the atoms of elements other than O and H. Use H2 O to balance O, H+ to balance H, electrons to balance the charges. Equalize the number of electrons on both the half reactions and then add the half reactions.
  3. Add OH- to neutralize the excess protons since this is a basic medium. Add OH- to both side of the equation to balance it.

Expert Solution
Check Mark

Answer to Problem 7QAP

 4Zn(s)+NO3-(aq)+6H2O4Zn2+(aq)+NH3(aq)+ 9OH-

Explanation of Solution

 Zn(s) + NO3-(aq)NH3(aq) + Zn2+(aq)

Determining the oxidation numbers:

The oxidation number of Zn in Zn and Zn2+ is 0 and +2 respectively. As the oxidation number of Zn is increasing this is the oxidation half of the reaction.

Oxidation half reaction:  Zn(s)  Zn2+(aq)

Reduction half reaction:  NO3-(aq)NH3(aq)

Balance the charge by adding two electrons to the product side of the half reaction.

Thus, the balanced oxidation half reaction will be:

 Zn(s) Zn2+(aq)+2e

To balance the excess O atoms on the product side we add three H2 O on the product side, and then we balance the excess H atom on the product side by adding nine H+ on the reactant side. Finally, we balance the charge by adding eight electrons to the reactant side of the half reaction.

Thus, the balanced reduction half reaction will be:

 NO3-(aq)+9H++8eNH3(aq)+3H2O

Net reaction:

Multiply the oxidation half reaction by 4, to cancel out the electrons in the net reaction. We add both the half reaction to get the net reaction.

 4Zn(s) 4Zn2+(aq)+8e

 4Zn(s)+NO3-(aq)+9H+4Zn2+(aq)+NH3(aq)+3H2O

Basic medium:

Since the reaction takes place in basic medium the excess proton on the reactant side of the net reaction is neutralized with OH-. OH- is added on both side of the equation to even out the charges. H+ and OH- combines to form H2 O, which is then cancelled out on both side of the net reaction. Thus, the balanced net reaction will be:

4Zn(s)+ NO3-(aq)+9H++9OH-4Zn2+(aq)+NH3(aq)+3H2O+ 9OH-

Or,

 4Zn(s)+NO3-(aq)+6H2O4Zn2+(aq)+NH3(aq)+ 9OH-

Interpretation Introduction

(c)

Interpretation:

The balanced equations for the following reaction in basic solution needs to be determined.

ClO-(aq) + CrO2-Cl-(aq) + CrO42-(aq)

Concept introduction:

Here are the rules to balance redox reactions in basic medium:

  1. Determine the oxidation numbers of the elements and write the oxidation and reduction half equation.
  2. Balance the atoms of elements other than O and H. Use H2 O to balance O, H+ to balance H, electrons to balance the charges. Equalize the number of electrons on both the half reactions and then add the half reactions.
  3. Add OH- to neutralize the excess protons since this is a basic medium. Add OH- to both side of the equation to balance it.

Expert Solution
Check Mark

Answer to Problem 7QAP

 2CrO2-+3ClO-(aq)+2OH- 2CrO42-(aq)+3Cl-(aq)+H2O

Explanation of Solution

ClO-(aq) + CrO2-Cl-(aq) + CrO42-(aq)

Determining the oxidation numbers:

The oxidation number of Cl in ClO- and Cl- is +1 and -1 respectively. As the oxidation number of Cl is decreasing this is the reduction half of the reaction.

The reduction half reaction: ClO-(aq) Cl-(aq)

Oxidation half reaction:  CrO2- CrO42-(aq)

To balance the excess O atoms on the product side we add two H2 O on the reactant side, and then we balance the excess H atom on the reactant side by adding four H+ on the product side. Finally, we balance the charge by adding three electrons to the reactant side of the half reaction.

Thus, the balanced oxidation half reaction will be:

 CrO2-+2H2O CrO42-(aq)+4H++3e

Similarly, to balance the excess O atoms on the reactant side we add one H2 O on the product side, and then we balance the excess H atom on the product side by adding two H+ on the reactant side. Finally, we balance the charge by adding two electrons to the reactant side of the half reaction.

Thus, the balanced reduction half reaction will be:

ClO-(aq)+2H++2eCl-(aq) + H2O

Net reaction:

Multiply the oxidation half reaction by 2 and reduction half reaction by 3, to cancel out the electrons in the net reaction. We add both the half reaction to get the net reaction.

3ClO-(aq)+6H++6e3Cl-(aq) + 3H2O

 2CrO2-+4H2O 2CrO42-(aq)+8H++6e

 2CrO2-+3ClO-(aq)+H2O 2CrO42-(aq)+3Cl-(aq)+2H+

Basic medium:

Since the reaction takes place in basic medium the excess proton on the reactant side of the net reaction is neutralized with OH-. OH- is added on both side of the equation to even out the charges. H+ and OH- combines to form H2 O, which is then cancelled out on both side of the net reaction.

 2CrO2-+3ClO-(aq)+H2O+2OH- 2CrO42-(aq)+3Cl-(aq)+2H++2OH-

Or,

 2CrO2-+3ClO-(aq)+2OH- 2CrO42-(aq)+3Cl-(aq)+H2O

Interpretation Introduction

(d)

Interpretation:

The balanced equations for the following reaction in basic solution needs to be determined.

K(s) + H2OK+(aq) + H2(g)

Concept introduction:

Here are the rules to balance redox reactions in basic medium:

  1. Determine the oxidation numbers of the elements and write the oxidation and reduction half equation.
  2. Balance the atoms of elements other than O and H. Use H2 O to balance O, H+ to balance H, electrons to balance the charges. Equalize the number of electrons on both the half reactions and then add the half reactions.
  3. Add OH- to neutralize the excess protons since this is a basic medium. Add OH- to both side of the equation to balance it.

Expert Solution
Check Mark

Answer to Problem 7QAP

2K(s)+2H2OH2(g)+2K+(aq)+2OH-

Explanation of Solution

K(s) + H2OK+(aq) + H2(g)

Determining the oxidation numbers:

The oxidation number of K in K and K+ is 0 and +1 respectively. As the oxidation number of K is increasing this is the oxidation half of the reaction.

Oxidation half reaction: K(s)K+(aq)

Reduction half reaction: H2OH2(g)

Balance the charge by adding one electron to the product side of the half reaction.

Thus, the balanced oxidation half reaction will be:

K(s)K+(aq) + e

To balance the excess O atoms on the reactant side we add one H2 O on the product side, and then we balance the excess H atom on the product side by adding two H+ on the reactant side. Finally, we balance the charge by adding two electrons to the reactant side of the half reaction.

Thus, the balanced reduction half reaction will be:

H2O+2H++2eH2(g)+H2O

Net reaction:

Multiply the oxidation half reaction by 2, to cancel out the electrons in the net reaction. We add both the half reaction to get the net reaction.

2H++2eH2(g)

2K(s)2K+(aq) + 2e

2K(s)+2H+H2(g)+2K+(aq)

Basic medium:

Since the reaction takes place in basic medium the excess proton on the reactant side of the net reaction is neutralized with OH-. OH- is added on both side of the equation to even out the charges. H+ and OH- combines to form H2 O, which is then cancelled out on both side of the net reaction.

2K(s)+2H++2OH-H2(g)+2K+(aq)+2OH-

Or,

2K(s)+2H2OH2(g)+2K+(aq)+2OH-

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