Chemistry & Chemical Reactivity
Chemistry & Chemical Reactivity
9th Edition
ISBN: 9781133949640
Author: John C. Kotz, Paul M. Treichel, John Townsend, David Treichel
Publisher: Cengage Learning
Question
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Chapter 19, Problem 6PS

a)

Interpretation Introduction

Interpretation:

The given redox equation in basic solution has to be balanced.

a) Fe(OH)3(s) + Cr(s)  Cr(OH)3(s) + Fe(OH)2(s)

Concept introduction:

Steps for balancing half –reactions in BASIC solution:

  1. 1. Recognize the reaction as an oxidation and reduction reaction.
  2. 2. Separate two half reactions.
  3. 3.  Balance half reactions by mass

    Balance all atoms except H and O in half reaction.

    Addition of OH- or OH- and H2O is required for mass balance in both half reactions.

  4. 4. Balance the charge by adding electrons to side with more total positive charge.
  5. 5. Multiply the half reactions by appropriate factors.
  6. 6. Add two half reactions and cancel the common atoms.
  7. 7. Simplify by eliminating reactants and products that appears on both sides.

a)

Expert Solution
Check Mark

Answer to Problem 6PS

3Fe(OH)3(s) + Cr(s) 3Fe(OH)2(s) + Cr(OH)3(s)

Explanation of Solution

The given reaction is s follows.

Fe(OH)3(s) + Cr(s) Cr(OH)3(s) + Fe(OH)2(s)

Oxidation states:

Fe(OH)3              Cr(OH)3               Fe(OH)2x + 3(-2+1)= 0     x + 3(-2+1)= 0      x+2(-2+1)= 0x + 3(-1)= 0          x +3(-1) = 0          x+2(-1)=0x = +3                       x = +3                x=+2

Steps for balancing half –reactions in BASIC solution:

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

    From the given reaction, Cr increases their oxidation state 0 to +3. Therefore, it is an oxidation state.

    Fe decrease their oxidation state +3 to +2.Therefore, it is a reduction reaction.

  2. 2. Separate two half reactions.

    Oxidation: Cr(s)  Cr(OH)3(s)Reduction:Fe(OH)3(s) Fe(OH)2(s)

  3. 3.  Balance half reactions for mass

    Balance all atoms except H and O in half reaction.

    Oxidation: Cr(s)  Cr(OH)3(s)Reduction:Fe(OH)3(s) Fe(OH)2(s)

    Addition of OH- or OH- and H2O is required for mass balance in both half reactions.

    Oxidation: Cr(s) +3OH-(aq) Cr(OH)3(s)Reduction:Fe(OH)3(s)  Fe(OH)2(s)+OH-(aq)

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

    Oxidation: Cr(s) +3OH-(aq) Cr(OH)3(s)+3eReduction:Fe(OH)3(s) +e Fe(OH)2(s)+OH-(aq)

  5. 5. Multiply the half reactions by appropriate factors.

    Oxidation: Cr(s) +3OH-(aq) Cr(OH)3(s)+3eReduction:3[Fe(OH)3(s) +e Fe(OH)2(s)+OH-(aq)]

  6. 6. Add two half reactions and cancel the common atoms.

    Oxidation: Cr(s) +3OH-(aq) Cr(OH)3(s)+3eReduction:3Fe(OH)3(s) +3e 3Fe(OH)2(s)+ 3OH-(aq)______________________________________________3Fe(OH)3(s) + Cr(s) 3Fe(OH)2(s) + Cr(OH)3(s)

  7. 7. Simplify by eliminating reactants and products that appears on both sides.

    3Fe(OH)3(s) + Cr(s) 3Fe(OH)2(s) + Cr(OH)3(s)

b)

Interpretation Introduction

Interpretation:

The given redox equation in basic solution has to be balanced.

b) NiO2(s) + Zn(s) Ni(OH)2(s) + Zn(OH)2(s)

Concept introduction:

Steps for balancing half –reactions in BASIC solution:

  1. 1. Recognize the reaction as an oxidation and reduction reaction.
  2. 2. Separate two half reactions.
  3. 3.  Balance half reactions by mass

    Balance all atoms except H and O in half reaction.

    Addition of OH- or OH- and H2O is required for mass balance in both half reactions.

  4. 4. Balance the charge by adding electrons to side with more total positive charge.
  5. 5. Multiply the half reactions by appropriate factors.
  6. 6. Add two half reactions and cancel the common atoms.
  7. 7. Simplify by eliminating reactants and products that appears on both sides.

b)

Expert Solution
Check Mark

Answer to Problem 6PS

NiO2(s) + Zn(s) + 2H2O(l) Ni(OH)2(s) + Zn(OH)2(s)

Explanation of Solution

The given reaction is as follows.

NiO2(s) + Zn(s)  Ni(OH)2(s) + Zn(OH)2(s)

Oxidation states:

NiO2                 Ni(OH)2               Zn(OH)2x + 2(-2) = 0     x+2(-2+1)=0         x+2(-2+1)=0x- 4 = 0             x+2(-1)=0              x+2(-1)=0x = +4                 x = +2                  x=+2

Steps for balancing half –reactions in BASIC solution:

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

    From the given reaction Ni atom reduces their oxidation state +4 to +2.Therefore, it is a reduction reaction.

    Zinc atom increases their oxidation state 0 to +2. Therefore, it is an oxidation reaction.

  2. 2. Separate two half reactions.

    Oxidation:  Zn(s)  Zn(OH)2(s)Reduction: NiO2(s)  Ni(OH)2(s)

  3. 3.  Balance half reactions for mass

    Balance all atoms except H and O in half reaction.

    Oxidation:  Zn(s)  Zn(OH)2(s)Reduction: NiO2(s)  Ni(OH)2(s)

    Addition of OH- or OH- and H2O is required for mass balance in both half reactions.

    Oxidation:  Zn(s) +2OH-(aq) Zn(OH)2(s)Reduction: NiO2(s) +2H2O Ni(OH)2(s)+2OH-(aq)

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

    Oxidation:  Zn(s) +2OH-(aq) Zn(OH)2(s) + 2eReduction: NiO2(s) +2H2O + 2e Ni(OH)2(s)+2OH-(aq)

  5. 5. Multiply the half reactions by appropriate factors.

    Oxidation:  Zn(s) +2OH-(aq) Zn(OH)2(s) + 2eReduction: NiO2(s) +2H2O + 2e Ni(OH)2(s)+2OH-(aq)

  6. 6. Add two half reactions and cancel the common atoms.

    Oxidation:  Zn(s) +2OH-(aq)  Zn(OH)2(s) + 2e-Reduction: NiO2(s) +2H2O + 2e- Ni(OH)2(s)+2OH-(aq)__________________________________________________NiO2(s) + Zn(s) + 2H2O(l)  Ni(OH)2(s) + Zn(OH)2(s)

  7. 7. Simplify by eliminating reactants and products that appears on both sides.

            NiO2(s) + Zn(s) + 2H2O(l) Ni(OH)2(s) + Zn(OH)2(s)

c)

Interpretation Introduction

Interpretation:

The given redox equation in basic solution has to be balanced.

c) Fe(OH)2(s) + CrO42-(aq) Fe(OH)3(s) + [Cr(OH)4]-(aq)

Concept introduction:

Steps for balancing half –reactions in BASIC solution:

  1. 1. Recognize the reaction as an oxidation and reduction reaction.
  2. 2. Separate two half reactions.
  3. 3.  Balance half reactions by mass

    Balance all atoms except H and O in half reaction.

    Addition of OH- or OH- and H2O is required for mass balance in both half reactions.

  4. 4. Balance the charge by adding electrons to side with more total positive charge.
  5. 5. Multiply the half reactions by appropriate factors.
  6. 6. Add two half reactions and cancel the common atoms.
  7. 7. Simplify by eliminating reactants and products that appears on both sides.

c)

Expert Solution
Check Mark

Answer to Problem 6PS

3Fe(OH)2(s) + CrO42-(aq) + 4H2O(l)  3Fe(OH)3(s) + [Cr(OH)4]- + OH-(aq)

Explanation of Solution

The given reaction is as follows.

Fe(OH)2(s) + CrO42-(aq)   Fe(OH)3(s) + [Cr(OH)4]-(aq)

Oxidation states:

Fe(OH)2                CrO42-                 Fe(OH)3         Cr(OH)4-x + (-2+1)= 0         x+4(-2) = -2      x+3(-2+1)=0    x+4(-2+1)=-1x+(-1)= 0               x-8 = -2             x+3(-1)=0         x+4(-1)=-1x = +1                    x = +6               x=+3                 x=+3

Steps for balancing half –reactions in BASIC solution:

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

    From the given reaction, Fe increases their oxidation state +1 to +3.Therefore, it is an oxidation reaction.

    Cr atom decrease their oxidation state +6 to +3.Therefore, it is a reduction reaction.

  2. 2. Separate two half reactions.

    Oxidation: Fe(OH)2(s) Fe(OH)3(s)Reduction: CrO42-(aq) [Cr(OH)4-]

  3. 3.  Balance half reactions by mass

    Balance all atoms except H and O in half reaction.

    Oxidation: Fe(OH)2(s) Fe(OH)3(s)Reduction: CrO42-(aq) [Cr(OH)4-]

    Addition of OH- or OH- and H2O is required for mass balance in both half reactions.

    Oxidation: Fe(OH)2(s)+OH-(aq) Fe(OH)3(s)Reduction: CrO42-(aq)+4H2[Cr(OH)4-](aq)+4OH-(aq)

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

    Oxidation: Fe(OH)2(s)+OH-(aq) Fe(OH)3(s) + eReduction: CrO42-(aq)+4H2O + 3e [Cr(OH)4-](aq)+4OH-(aq)

  5. 5. Multiply the half reactions by appropriate factors.

    Oxidation: 3[Fe(OH)2(s)+OH-(aq) Fe(OH)3(s) +e]Reduction: CrO42-(aq)+4H2O + 3e [Cr(OH)4-](aq)+4OH-(aq)

  6. 6. Add two half reactions and cancel the common atoms.

    Oxidation: 3Fe(OH)2(s)+ 3OH-(aq) 3Fe(OH)3(s) +3e-]Reduction: CrO42-(aq)+4H2O + 3e- [Cr(OH)4-](aq)+4OH-(aq)_____________________________________________________3Fe(OH)2(s)+CrO42-(aq) + 4H2O(l)  3Fe(OH)3(s) + [Cr(OH)4]- + OH-(aq)

  7. 7. Simplify by eliminating reactants and products that appears on both sides.3Fe(OH)2(s) + CrO42-(aq) + 4H2O(l)  3Fe(OH)3(s) + [Cr(OH)4]- + OH-(aq)

d)

Interpretation Introduction

Interpretation:

The given redox equation in basic solution has to be balanced.

d) N2H4(aq) + Ag2O(s) N2(g) + Ag(s)

Concept introduction:

Steps for balancing half –reactions in BASIC solution:

  1. 1. Recognize the reaction as an oxidation and reduction reaction.
  2. 2. Separate two half reactions.
  3. 3.  Balance half reactions by mass

    Balance all atoms except H and O in half reaction.

    Addition of OH- or OH- and H2O is required for mass balance in both half reactions.

  4. 4. Balance the charge by adding electrons to side with more total positive charge.
  5. 5. Multiply the half reactions by appropriate factors.
  6. 6. Add two half reactions and cancel the common atoms.
  7. 7. Simplify by eliminating reactants and products that appears on both sides.

d)

Expert Solution
Check Mark

Answer to Problem 6PS

2Ag2O(s)+N2H4(aq)  4Ag(s)+N2(g)+2H2O(l)

Explanation of Solution

The given reaction is as follows.

N2H4(aq) + Ag2O(s)  N2(g) + Ag(s)

Oxidation states:

N2H4                   Ag2O2x + 4(1)=0          2x + (-2)=02x +4=0                2x-2=0x = -2                    x = +1

Steps for balancing half –reactions in BASIC solution:

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

    N atoms increases their oxidation state -2 to 0.Therefroe, it is an oxidation reaction.

    Ag atom decreases their oxidation state+1 to 0.Therefore, it is a reduction reaction.

  2. 2. Separate two half reactions.

    Oxidation: N2H4(aq)  N2(g)Reduction: Ag2O(aq) Ag(aq)

  3. 3.  Balance half reactions for mass.

    Balance all atoms except H and O in half reaction.

    Oxidation: N2H4(aq)  N2(g)Reduction: Ag2O(aq) 2Ag(aq)

    Addition of OH- or OH- and H2O is required for mass balance in both half reactions.

    Oxidation: N2H4(aq) + 4OH- N2(g)  +4H2O(l)Reduction: Ag2O(aq) +H22Ag(aq) + 2OH-(aq)

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

    Oxidation: N2H4(aq) + 4OH- N2(g)  +4H2O(l) + 4eReduction: Ag2O(aq) +H2O(l)+2e 2Ag(aq) + 2OH-

  5. 5. Multiply the half reactions by appropriate factors.

    Oxidation: N2H4(aq) + 4OH- N2(g)  +4H2O(l) + 4eReduction: 2[Ag2O(aq) +H2O(l)+2e 2Ag(aq) + 2OH-]

  6. 6. Add two half reactions and cancel the common atoms.

    Oxidation: N2H4(aq) + 4OH- N2(g)  +4H2O(l) + 4e-Reduction: 2Ag2O(aq) + 2H2O(l)+4e- 4Ag(aq) + 4OH-_______________________________________________2Ag2O(s)+N2H4(aq)  4Ag(s)+N2(g)+2H2O(l)

  7. 7. Simplify by eliminating reactants and products that appears on both sides.

    2Ag2O(s)+N2H4(aq)  4Ag(s)+N2(g)+2H2O(l)

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Chapter 19 Solutions

Chemistry & Chemical Reactivity

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(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. 57GQCh. 19 - Prob. 58GQCh. 19 - In the table of standard reduction potentials,...Ch. 19 - Prob. 60GQCh. 19 - Four voltaic cells are set up. 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