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

A common laboratory analysis for iron is to titrate aqueous iron(II) ion with a solution of potassium permanganate of precisely known concentration. Use the half-reaction method to write the balanced net ionic equation for the reaction in acid solution.

MnO4(aq) + Fe2+(aq) → Mn2+(aq) + Fe3+(aq)

Identify the oxidizing agent, the reducing agent, the substance oxidized, and the substance reduced (Figure 19.4).

Interpretation Introduction

Interpretation:

The balanced net ionic equation for the given reaction has to be given and also identify the oxidizing and reducing agent in the reaction.

Concept introduction:

Oxidation reaction:

The loss of electrons or the gain of oxigen atoms.And also increase their oxidation number.

Ag  Ag++ e-

In the above reaction , Ag atom lose one electron and change their oxidation state 0 to 1.

Reduction reaction:

Gaining electrons or adding hydrogen atoms. And also decrease their oxidation number.

Fe2++2e-  Fe

In the above reaction Fe2+ ion gaining 2 electrons and reduce their oxidation number +2 to 0.

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.

Oxidizing agent:  A reagent which increases the oxidation number of an element of a given substance.

Reducing agent: A reagent ht lowers the oxidation number of a given element.

Explanation

The given reaction:

MnO4-(aq) + Fe2+(aq) Mn2+(aq) + Fe3+(aq)

Oxidation states:

MnO4-x+4(-2)= -1x-8= -1x= +7

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

    Mn reduces their oxidation state +7 to +2 therefore, it is a reduction reaction.

    And Iron atom increases their oxidation state +2 to +3 therefore, it is a oxidation reaction.

  2. 2. Separate two half reactions.

    Oxidation:     Fe2+  (aq)   Fe3+(aq)Reduction:     MnO4(aq)  Mn2+(aq)

  3. 3.  Balance half reactions by mass.

    Balance all atoms except H and O in half reaction.

    Oxidation:     Fe2+  (aq)   Fe3+(aq)Reduction:     MnO4(aq)  Mn2+(aq)

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

    Oxidation:     Fe2+  (aq)   Fe3+(aq)Reduction:     MnO4(aq)  Mn2+(aq)+ 4H2O(l)

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

    Oxidation:     Fe2+  (aq)   Fe3+(aq)Reduction:     MnO4(aq) +8H+ Mn2+(aq)+ 4H2O(l)

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

    Oxidation:     Fe2+(aq)   Fe3+(aq)+ eReduction:     MnO4(aq) +8H++5e Mn2+(aq)+ 4H2O(l)

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

    Oxidation:     5(Fe2+(aq)   Fe3+(aq)+ e)Reduction:     MnO4(aq) +8H++5e Mn2+(aq)+ 4H2O(l)

  6. 6

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

Diets with sufficient protein may provide more satiety than diets that are low in protein.

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

How can errors in the cell cycle lead to cancer in humans?

Human Heredity: Principles and Issues (MindTap Course List)

How can star clusters confirm astronomers’ theories of stellar evolution?

Horizons: Exploring the Universe (MindTap Course List)

Three long, parallel conductors each carry a current of I = 2.00 A. Figure P29.7 is an end view of the conducto...

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