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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

Balance the following redox equations. All occur in basic solution.

  1. (a) Fe(OH)3(s) + Cr(s) → Cr(OH)3(s) + Fe(OH)2(s)
  2. (b) NiO2(s) + Zn(s) → Ni(OH)2(s) + Zn(OH)2(s)
  3. (c) Fe(OH)2(s) + CrO42−(aq) → Fe(OH)3(s) + [Cl(OH)4](aq)
  4. (d) N2H4(aq) + Ag2O(s) → N2(g) + Ag(s)

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.
Explanation

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

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.

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.

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.

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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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