   Chapter 19, Problem 58GQ

Chapter
Section
Textbook Problem

Balance the following equations. (a) Zn(s) + VO2+(aq) → Zn2+(aq) + V3+(aq)⇄(acid solution) (b) Zn(s) + VO3−(aq) → V2+(aq) + Zn2+(aq)⇄(acid solution) (c) Zn(s) + ClO−(aq) → Zn(OH)2(s) + Cl−(aq)⇄(basic solution) (d) ClO−(aq) + [Cr(OH)4]−(aq) → Cl−(aq) + Cr2O42−(aq)⇄(basic solution)

(a)

Interpretation Introduction

Interpretation:

The following half reaction in acidic solution has to be identified.

(a) Zn(s) + VO2+(aq) Zn2+(aq) + V3+(aq)     (acid solution)

Concept introduction:

Steps for balancing half –reactions in ACIDIC solution:

1. 1. Balance all atoms except H and O in half reaction.
2. 2. Balance O atoms by adding water to the side missing O atoms.
3. 3. Balance the H atoms by adding H+ to the side missing H atoms.
4. 4. Balance the charge by adding electrons to side with more total positive charge.
5. 5. Make the number of electrons the same in both half reactions by multiplication, while avoiding fractional number of electrons.

Steps for balancing half –reactions in BASIC solution:

1. 1. Balance all atoms except H and O in half reaction.
2. 2. Balance O atoms by adding water to the side missing O atoms.
3. 3. Balance the H atoms by adding H+ to the side missing H atoms.
4. 4. Balance the charge by adding electrons to side with more total positive charge.
5. 5. Make the number of electrons the same in both half reactions by multiplication, while avoiding fractional number of electrons.
6. 6. Add the same number of OH- groups as there are H+ present to both sides of the equation.
Explanation

Given reaction:

Zn(s) + VO2+(aq) Zn2+(aq) + V3+(aq)

Oxidation state:

VO2+x+(-2) = +2x-2 = +2x = 4

Steps for balancing redox reactions in ACIDIC solution:

1. 1. Recognize the reaction as an oxidation and reduction reaction.

The oxidation state of Zn atom increases 0 to 2+.Therefore, it is a oxidation reaction.

The oxidation state of vanadium reduces to +4 to +3. Therefore, it is a reduction reaction.

2. 2. Separate two half reactions.

oxidation:      Zn(s)   Zn2+(aq)Reduction:     VO2+(aq) V3+(aq)

3. 3.  Balance half reactions by mass

Balance all atoms except H and O in half reaction.

oxidation:      Zn(s)   Zn2+(aq)Reduction:     VO2+(aq) V3+(aq)

Balance O atoms by adding water to the side missing O atoms.

oxidation:      Zn(s)   Zn2+(aq)Reduction:     VO2+(aq) V3+(aq) +2H2O(l)

Balance the H atoms by adding H+ to the side missing H atoms.

oxidation:      Zn(s)   Zn2+(aq)Reduction:     VO2+(aq) + 2H+(aq)V3+(aq) +H2O(l)

4. 4. Balance the charge by adding electrons to side with more total positive charge.

oxidation:      Zn(s)   Zn2+(aq)+2eReduction:     VO2+(aq) + 2H+(aq)V3+(aq) +H2O(l)+e

5. 5

(b)

Interpretation Introduction

Interpretation:

The following half reaction in acidic solution has to be identified.

Zn(s) + VO3-(aq) V2+(aq) + Zn2+(aq)

Concept introduction:

Steps for balancing half –reactions in ACIDIC solution:

1. 6. Balance all atoms except H and O in half reaction.
2. 7. Balance O atoms by adding water to the side missing O atoms.
3. 8. Balance the H atoms by adding H+ to the side missing H atoms.
4. 9. Balance the charge by adding electrons to side with more total positive charge.
5. 10. Make the number of electrons the same in both half reactions by multiplication, while avoiding fractional number of electrons.

Steps for balancing half –reactions in BASIC solution:

1. 7. Balance all atoms except H and O in half reaction.
2. 8. Balance O atoms by adding water to the side missing O atoms.
3. 9. Balance the H atoms by adding H+ to the side missing H atoms.
4. 10. Balance the charge by adding electrons to side with more total positive charge.
5. 11. Make the number of electrons the same in both half reactions by multiplication, while avoiding fractional number of electrons.
6. 12. Add the same number of OH- groups as there are H+ present to both sides of the equation.

(c)

Interpretation Introduction

Interpretation:

The following half reaction in basic solution has to be identified.

Zn(s) + ClO-(aq) Zn(OH)2(s) + Cl-

Concept introduction:

Steps for balancing half –reactions in ACIDIC solution:

1. 11. Balance all atoms except H and O in half reaction.
2. 12. Balance O atoms by adding water to the side missing O atoms.
3. 13. Balance the H atoms by adding H+ to the side missing H atoms.
4. 14. Balance the charge by adding electrons to side with more total positive charge.
5. 15. Make the number of electrons the same in both half reactions by multiplication, while avoiding fractional number of electrons.

Steps for balancing half –reactions in BASIC solution:

1. 13. Balance all atoms except H and O in half reaction.
2. 14. Balance O atoms by adding water to the side missing O atoms.
3. 15. Balance the H atoms by adding H+ to the side missing H atoms.
4. 16. Balance the charge by adding electrons to side with more total positive charge.
5. 17. Make the number of electrons the same in both half reactions by multiplication, while avoiding fractional number of electrons.
6. 18. Add the same number of OH- groups as there are H+ present to both sides of the equation.

(d)

Interpretation Introduction

Interpretation:

The following half reaction in basic solution has to be identified.

ClO-(aq) + [Cr(OH)4]-(aq)  Cl-(aq) + CrO42-

Concept introduction:

Steps for balancing half –reactions in ACIDIC solution:

1. 16. Balance all atoms except H and O in half reaction.
2. 17. Balance O atoms by adding water to the side missing O atoms.
3. 18. Balance the H atoms by adding H+ to the side missing H atoms.
4. 19. Balance the charge by adding electrons to side with more total positive charge.
5. 20. Make the number of electrons the same in both half reactions by multiplication, while avoiding fractional number of electrons.

Steps for balancing half –reactions in BASIC solution:

1. 19. Balance all atoms except H and O in half reaction.
2. 20. Balance O atoms by adding water to the side missing O atoms.
3. 21. Balance the H atoms by adding H+ to the side missing H atoms.
4. 22. Balance the charge by adding electrons to side with more total positive charge.
5. 23. Make the number of electrons the same in both half reactions by multiplication, while avoiding fractional number of electrons.
6. 24. Add the same number of OH- groups as there are H+ present to both sides of the equation.

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