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Chemistry & Chemical Reactivity

10th Edition
John C. Kotz + 3 others
ISBN: 9781337399074

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BuyFindarrow_forward

Chemistry & Chemical Reactivity

10th Edition
John C. Kotz + 3 others
ISBN: 9781337399074
Textbook Problem

Write balanced equations for the following half-reactions. Specify whether each is an oxidation or reduction.

  1. (a) H2O2(aq) → O2(g)⇄(in acid)
  2. (b) H2C2O4(aq) → CO2(g)⇄(in acid)
  3. (c) NO3(aq) → NO(g)⇄(in acid)
  4. (d) MnO4(aq) → MnO2(s)⇄(in base)

(a)

Interpretation Introduction

Interpretation:    

The balanced equation for the following half reaction has to be identified and also specify whether the reaction is an oxidation or reduction reaction.

a) H2O2(aq)      O2(g)    (in acid)

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

The given reaction:

H2O2(aq)  O2(g)

Steps for balancing half –reactions in ACIDIC solution:

  1. 1. Balance all atoms except H and O in half reaction.

    H2O2(aq)  O2(g)

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

    H2O2(aq)  O2(g)

    No need to add water because oxygen atoms are already balanced.

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

    H2O2(aq)   O2(g) + 2H+(aq)

  4. 4

(b)

Interpretation Introduction

Interpretation:    

The balanced equation for the following half reaction has to be identified and also specify whether the reaction is an oxidation or reduction reaction.

b) H2C2O4(aq)  CO2(g) (in acid)

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 balanced equation for the following half reaction has to be identified and also specify whether the reaction is an oxidation or reduction reaction.

c) NO3-(aq)        NO(g)   (in acid)

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 balanced equation for the following half reaction has to be identified and also specify whether the reaction is an oxidation or reduction reaction.

d) MnO4-(aq)    MnO2(s) (in acid)

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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Sect-19.8 P-19.11CYUSect-19.9 P-1.1ACPSect-19.9 P-1.2ACPSect-19.9 P-1.3ACPSect-19.9 P-2.1ACPSect-19.9 P-2.2ACPSect-19.9 P-2.3ACPSect-19.9 P-2.4ACPSect-19.9 P-2.5ACPCh-19 P-1PSCh-19 P-2PSCh-19 P-3PSCh-19 P-4PSCh-19 P-5PSCh-19 P-6PSCh-19 P-7PSCh-19 P-8PSCh-19 P-9PSCh-19 P-10PSCh-19 P-11PSCh-19 P-12PSCh-19 P-13PSCh-19 P-14PSCh-19 P-15PSCh-19 P-16PSCh-19 P-17PSCh-19 P-18PSCh-19 P-19PSCh-19 P-20PSCh-19 P-21PSCh-19 P-22PSCh-19 P-23PSCh-19 P-24PSCh-19 P-25PSCh-19 P-26PSCh-19 P-27PSCh-19 P-28PSCh-19 P-29PSCh-19 P-30PSCh-19 P-31PSCh-19 P-32PSCh-19 P-33PSCh-19 P-34PSCh-19 P-35PSCh-19 P-36PSCh-19 P-37PSCh-19 P-38PSCh-19 P-39PSCh-19 P-40PSCh-19 P-41PSCh-19 P-42PSCh-19 P-43PSCh-19 P-44PSCh-19 P-45PSCh-19 P-46PSCh-19 P-47PSCh-19 P-48PSCh-19 P-49PSCh-19 P-50PSCh-19 P-51PSCh-19 P-52PSCh-19 P-53PSCh-19 P-54PSCh-19 P-55PSCh-19 P-56PSCh-19 P-57GQCh-19 P-58GQCh-19 P-59GQCh-19 P-60GQCh-19 P-61GQCh-19 P-62GQCh-19 P-63GQCh-19 P-64GQCh-19 P-65GQCh-19 P-66GQCh-19 P-67GQCh-19 P-68GQCh-19 P-69GQCh-19 P-70GQCh-19 P-71GQCh-19 P-72GQCh-19 P-73GQCh-19 P-74GQCh-19 P-75GQCh-19 P-76GQCh-19 P-77GQCh-19 P-78GQCh-19 P-79GQCh-19 P-80GQCh-19 P-81GQCh-19 P-82GQCh-19 P-83GQCh-19 P-84GQCh-19 P-85GQCh-19 P-86GQCh-19 P-87GQCh-19 P-88GQCh-19 P-89GQCh-19 P-90GQCh-19 P-91GQCh-19 P-92GQCh-19 P-93GQCh-19 P-94GQCh-19 P-95GQCh-19 P-96GQCh-19 P-97GQCh-19 P-98GQCh-19 P-99GQCh-19 P-100GQCh-19 P-101GQCh-19 P-102GQCh-19 P-103GQCh-19 P-104GQCh-19 P-105ILCh-19 P-106ILCh-19 P-107ILCh-19 P-108ILCh-19 P-109ILCh-19 P-110ILCh-19 P-111SCQCh-19 P-112SCQCh-19 P-113SCQCh-19 P-114SCQCh-19 P-115SCQ

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