OWLv2 6-Months Printed Access Card for Kotz/Treichel/Townsend's Chemistry & Chemical Reactivity, 9th, 9th Edition
OWLv2 6-Months Printed Access Card for Kotz/Treichel/Townsend's Chemistry & Chemical Reactivity, 9th, 9th Edition
9th Edition
ISBN: 9781285460680
Author: Kotz, Treichel, Townsend
Publisher: Cengage Learning
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Textbook Question
Chapter 19, Problem 1PS

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

  1. (a) Cr(s) → Cr3+(aq)⇄(in acid)
  2. (b) AsH3(g) → As(s)⇄(in acid)
  3. (c) VO3(aq) → V2+(aq)⇄(in acid)
  4. (d) Ag(s) → Ag2O(s)⇄(in base)

(a)

Expert Solution
Check Mark
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) Cr(s)     Cr3+(aq)              (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.

Answer to Problem 1PS

Balanced equation: Cr(s)  Cr3+(aq) + 3e-

It is a oxidation reaction.

Explanation of Solution

The given reaction:

Cr(s) Cr3+(aq)

Steps for balancing half –reactions in ACIDIC solution:

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

    Cr(s) Cr3+(aq)

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

    Cr(s) Cr3+(aq)

    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.

    Cr(s) Cr3+(aq)

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

    Cr(s)  Cr3+(aq) + 3e-

  5. 5. Make the number of electrons the same in both half reactions by multiplication, while avoiding fractional number of electrons.

    Cr(s)  Cr3+(aq) + 3e-

    Therefore, the balanced half cell reaction is as follows.

    Cr(s)  Cr3+(aq) + 3e-

Oxidation state of chromium increases in the above reaction.

0to+3

Therefore, it is an oxidation reaction.

(b)

Expert Solution
Check Mark
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) AsH3(g) As(s)                    (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.

Answer to Problem 1PS

The balanced reaction

AsH3(g)  As(s) + 3H++3e:

It is an oxidation reaction.

Explanation of Solution

The given reaction:

AsH3(g)  As(s)

Steps for balancing half –reactions in ACIDIC solution:

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

    AsH3(g)  As(s)

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

    AsH3(g)  As(s)

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

    AsH3(g)  As(s)+3H+(aq)

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

    AsH3(g)  As(s) + 3H+(aq) + 3e-

  5. 5. Make the number of electrons the same in both half reactions by multiplication, while avoiding fractional number of electrons.

    AsH3(g)  As(s) + 3H+(aq) + 3e-

    Therefore, the balanced half cell reaction is as follows.

    AsH3(g)  As(s) + 3H+(aq) + 3e-

    From the above reaction Hydrogen atom oxidation state increases from 0 to +1.Therefore, it is an oxidation reaction.

Oxidation state of arsenic increases in the above reaction.

3to0

Therefore, it is an oxidation reaction.

 (c)

Expert Solution
Check Mark
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) VO3-(g)  V2+(aq)                (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.

Answer to Problem 1PS

The balanced half cell reaction:

VO3-(aq)  +6H+(aq)+3eV2+(aq) + 3H2O (l)

It is a reduction reaction.

Explanation of Solution

The given reaction: VO3-(aq)  V2+(aq)

Steps for balancing half –reactions in ACIDIC solution:

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

    . VO3-(aq)  V2+(aq)

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

    VO3-(aq)  V2+(aq) + 3H2O (l)

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

    VO3-(aq)+6H+  V2+(aq) + 3H2O (l)

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

    VO3-(aq)  +6H+(aq)+3eV2+(aq) + 3H2O (l)

  5. 5. Make the number of electrons the same in both half  reactions by multiplication, while avoiding fractional number of electrons.

               VO3-(aq)  +6H+(aq)+3eV2+(aq) + 3H2O (l)

Therefore, the balanced half cell reaction is as follows.

VO3-(aq)  +6H+(aq)+3eV2+(aq) + 3H2O (l)

Oxidation state of V in VO3-

VO3-x+3(-2)=-1x = -1+6x = +5

Oxidation state of chromium decreases in the above reaction.

+5to+2

Therefore, it is a reduction reaction.

 (d)

Expert Solution
Check Mark
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) Ag(s)      Ag2O(s)               (in base)

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.

Answer to Problem 1PS

The balanced half cell reaction:

2Ag(s) + 2OH-(aq) Ag2O(s) + H2O(l) + 2e-

It is a oxidation reaction.

Explanation of Solution

The given reaction:

Ag(s)  Ag2O(s)

Steps for balancing half –reactions in BASIC solution:

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

    2Ag(s)  Ag2O(s)

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

    2Ag(s) + 2OH(aq)  Ag2O(s)+H2O(l)

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

    2Ag(s) + 2OH(aq)  Ag2O(s)+H2O(l)

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

    2Ag(s) + OH(aq)  Ag2O(s)+H2O(l)+2e

  5. 5. Make the number of electrons the same in both half  reactions by multiplication, while avoiding fractional number of electrons.

    2Ag(s) + OH(aq)  Ag2O(s)+H2O(l)+2e

  6. 6. Add the same number of OH- groups as there are H+ present to both sides of the equation.

             2Ag(s) + OH(aq)  Ag2O(s)+H2O(l)+2e

Therefore, the balanced half reaction is as follows.

2Ag(s) + OH(aq)  Ag2O(s)+H2O(l)+2e

Oxidation state of silver decreases in the above reaction:

0to1

Therefore, it is an oxidation reaction.

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Chapter 19 Solutions

OWLv2 6-Months Printed Access Card for Kotz/Treichel/Townsend's Chemistry & Chemical Reactivity, 9th, 9th Edition

Ch. 19.8 - Prob. 19.11CYUCh. 19.9 - Prob. 1.1ACPCh. 19.9 - Prob. 1.2ACPCh. 19.9 - Prob. 1.3ACPCh. 19.9 - Prob. 2.1ACPCh. 19.9 - Use standard reduction potentials to determine...Ch. 19.9 - Prob. 2.3ACPCh. 19.9 - The overall reaction for the production of Cu(OH)2...Ch. 19.9 - Assume the following electrochemical cell...Ch. 19 - Write balanced equations for the following...Ch. 19 - Write balanced equations for the following...Ch. 19 - Balance the following redox equations. All occur...Ch. 19 - Balance the following redox equations. All occur...Ch. 19 - Balance the following redox equations. All occur...Ch. 19 - Prob. 6PSCh. 19 - A voltaic cell is constructed using the reaction...Ch. 19 - A voltaic cell is constructed using the reaction...Ch. 19 - The half-cells Fe2+(aq) | Fe(s) and O2(g) | H2O...Ch. 19 - The half cells Sn2+(aq) |Sn(s) and Cl2(g) |Cl(aq)...Ch. 19 - For each of the following electrochemical cells,...Ch. 19 - For each of the following electrochemical cells,...Ch. 19 - Use cell notation to depict an electrochemical...Ch. 19 - Use cell notation to depict an electrochemical...Ch. 19 - What are the similarities and differences between...Ch. 19 - What reactions occur when a lead storage battery...Ch. 19 - Calculate the value of E for each of the following...Ch. 19 - Calculate the value of E for each of the following...Ch. 19 - Balance each of the following unbalanced...Ch. 19 - Balance each of the following unbalanced...Ch. 19 - Consider the following half-reactions: (a) Based...Ch. 19 - Prob. 22PSCh. 19 - Which of the following elements is the best...Ch. 19 - Prob. 24PSCh. 19 - Which of the following ions is most easily...Ch. 19 - From the following list, identify the ions that...Ch. 19 - (a) Which halogen is most easily reduced in acidic...Ch. 19 - Prob. 28PSCh. 19 - Calculate the potential delivered by a voltaic...Ch. 19 - Calculate the potential developed by a voltaic...Ch. 19 - One half-cell in a voltaic cell is constructed...Ch. 19 - One half-cell in a voltaic cell is constructed...Ch. 19 - One half-cell in a voltaic cell is constructed...Ch. 19 - One half-cell in a voltaic cell is constructed...Ch. 19 - Calculate rG and the equilibrium constant for the...Ch. 19 - Prob. 36PSCh. 19 - Use standard reduction potentials (Appendix M) for...Ch. 19 - Use the standard reduction potentials (Appendix M)...Ch. 19 - Use the standard reduction potentials (Appendix M)...Ch. 19 - Use the standard reduction potentials (Appendix M)...Ch. 19 - Prob. 41PSCh. 19 - Prob. 42PSCh. 19 - Which product, O2 or F2, is more likely to form at...Ch. 19 - Which product, Ca or H2, is more likely to form at...Ch. 19 - An aqueous solution of KBr is placed in a beaker...Ch. 19 - An aqueous solution of Na2S is placed in a beaker...Ch. 19 - In the electrolysis of a solution containing...Ch. 19 - In the electrolysis of a solution containing...Ch. 19 - Electrolysis of a solution of CuSO4(aq) to give...Ch. 19 - Electrolysis of a solution of Zn(NO3)2(aq) to give...Ch. 19 - A voltaic cell can be built using the reaction...Ch. 19 - Assume the specifications of a Ni-Cd voltaic cell...Ch. 19 - Use E values to predict which of the following...Ch. 19 - Prob. 54PSCh. 19 - Prob. 55PSCh. 19 - Prob. 56PSCh. 19 - Prob. 57GQCh. 19 - Balance the following equations. (a) Zn(s) +...Ch. 19 - Magnesium metal is oxidized, and silver ions are...Ch. 19 - You want to set up a series of voltaic cells with...Ch. 19 - Prob. 61GQCh. 19 - Prob. 62GQCh. 19 - In the table of standard reduction potentials,...Ch. 19 - Prob. 64GQCh. 19 - Four voltaic cells are set up. In each, one...Ch. 19 - The following half-cells are available: (i)...Ch. 19 - Prob. 67GQCh. 19 - Prob. 68GQCh. 19 - A potential of 0.142 V is recorded (under standard...Ch. 19 - Prob. 70GQCh. 19 - The standard potential, E, for the reaction of...Ch. 19 - An electrolysis cell for aluminum production...Ch. 19 - Electrolysis of molten NaCl is done in cells...Ch. 19 - A current of 0.0100 A is passed through a solution...Ch. 19 - A current of 0.44 A is passed through a solution...Ch. 19 - Prob. 76GQCh. 19 - Prob. 77GQCh. 19 - Prob. 78GQCh. 19 - The products formed in the electrolysis of aqueous...Ch. 19 - Predict the products formed in the electrolysis of...Ch. 19 - Prob. 81GQCh. 19 - The metallurgy of aluminum involves electrolysis...Ch. 19 - Prob. 83GQCh. 19 - Prob. 84GQCh. 19 - Prob. 85GQCh. 19 - Prob. 86GQCh. 19 - Two Ag+(aq) | Ag(s) half-cells are constructed....Ch. 19 - Calculate equilibrium constants for the following...Ch. 19 - Prob. 89GQCh. 19 - Use the table of standard reduction potentials...Ch. 19 - Prob. 91GQCh. 19 - Prob. 92GQCh. 19 - Prob. 93GQCh. 19 - A voltaic cell is constructed in which one...Ch. 19 - An expensive but lighter alternative to the lead...Ch. 19 - The specifications for a lead storage battery...Ch. 19 - Manganese may play an important role in chemical...Ch. 19 - Prob. 98GQCh. 19 - Iron(II) ion undergoes a disproportionation...Ch. 19 - Copper(I) ion disproportionates to copper metal...Ch. 19 - Prob. 101GQCh. 19 - Prob. 102GQCh. 19 - Can either sodium or potassium metal be used as a...Ch. 19 - Galvanized steel pipes are used in the plumbing of...Ch. 19 - Consider an electrochemical cell based on the...Ch. 19 - Prob. 106ILCh. 19 - A silver coulometer (Study Question 106) was used...Ch. 19 - Four metals, A, B, C, and D, exhibit the following...Ch. 19 - Prob. 109ILCh. 19 - The amount of oxygen, O2, dissolved in a water...Ch. 19 - Prob. 111SCQCh. 19 - The free energy change for a reaction, rG, is the...Ch. 19 - Prob. 113SCQCh. 19 - (a) Is it easier to reduce water in acid or base?...Ch. 19 - Prob. 115SCQ
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