BuyFindarrow_forward

Chemistry & Chemical Reactivity

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

Solutions

Chapter
Section
BuyFindarrow_forward

Chemistry & Chemical Reactivity

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

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

  1. (a) Sn(s) + H+(aq) → Sn2+(aq) + H2(g)
  2. (b) Cr2O72− (aq) + Fe2+(aq) → Cr3+(aq) + Fe3+(aq)
  3. (c) MnO2(s) + Cl(aq) → Mn2+(aq) + Cl2(g)
  4. (d) CH2O(aq) + Ag+(aq) → HCO2H(aq) + Ag(s)

a)

Interpretation Introduction

Interpretation:

The redox equation in acidic solution has to be balanced.

a) Sn(s) + H+(aq) Sn2+(aq) + H2(g)

Concept introduction:

Steps for balancing redox reactions in ACIDIC 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.

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

    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 half reaction together and than simplyfy by cancelling out species that show up on both sides.
  7. 7. Confirm that the reaction is balanced in number of atoms and charge on the both sides of the arrow.
Explanation

Given reaction:

Sn(s) + H+(aq)  Sn2+(aq) + H2(g)

Steps for balancing redox reactions in ACIDIC solution:

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

    From the given reaction Sn increases their oxidation state 0 to +2. Therefore, it is an oxidation reaction.

  2. 2. Separate two half reactions.

    oxidation:      Sn(s)   Sn2+(aq)Reduction:     H+(aq) H2(g)

  3. 3.  Balance half reactions by mass

    Balance all atoms except H and O in half reaction.

    oxidation:      Sn(s)   Sn2+(aq)Reduction:     H+(aq) H2(g)

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

    oxidation:      Sn(s)   Sn2+(aq)Reduction:     H+(aq) H2(g)

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

    oxidation:      Sn(s)   Sn2+(aq)Reduction:     2H+(aq) H2(g)

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

    oxidation:      Sn(s)   Sn2+(aq)+2eReduction:     2H+(aq)+2e H2(g)

  5. 5

b)

Interpretation Introduction

Interpretation:

The redox equation in acidic solution has to be balanced.

b) Cr2O72-(aq) + Fe2+(aq) Cr3+(aq) + Fe3+(aq)

Concept introduction:

Steps for balancing redox reactions in ACIDIC 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.

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

    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 half reaction together and than simplyfy by cancelling out species that show up on both sides.
  7. 7. Confirm that the reaction is balanced in number of atoms and charge on the both sides of the arrow.

c)

Interpretation Introduction

Interpretation:

The redox equation in acidic solution has to be balanced.

c) MnO2(aq) + Cl-(aq) Mn2+(aq) + Cl2(g)

Concept introduction:

Steps for balancing redox reactions in ACIDIC 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.

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

    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 half reaction together and than simplyfy by cancelling out species that show up on both sides.
  7. 7. Confirm that the reaction is balanced in number of atoms and charge on the both sides of the arrow.

d)

Interpretation Introduction

Interpretation:

The redox equation in acidic solution has to be balanced.

d) CH2O(aq) + Ag+(aq) HCO2H(aq) + Ag(s)

Concept introduction:

Steps for balancing redox reactions in ACIDIC 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.

    Balance O atoms by adding water.

    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 half reaction together and than simplyfy by cancelling out species that show up on both sides.
  7. 7. Confirm that the reaction is balanced in number of atoms and charge on the both sides of the arrow.

Still sussing out bartleby?

Check out a sample textbook solution.

See a sample solution

The Solution to Your Study Problems

Bartleby provides explanations to thousands of textbook problems written by our experts, many with advanced degrees!

Get Started

Chapter 19 Solutions

Show all chapter solutions add
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

Additional Science Solutions

Find more solutions based on key concepts

Show solutions add

Organic nutrients include all of the following except. a. minerals b. fat c. carbohydrates d. protein

Nutrition: Concepts and Controversies - Standalone book (MindTap Course List)

Compare the lock-and-key theory with the induced-fit theory.

Chemistry for Today: General, Organic, and Biochemistry

A 2.00-g particle moving at 8.00 m/s makes a perfectly elastic head-on collision with a resting 1.00-g object. ...

Physics for Scientists and Engineers, Technology Update (No access codes included)