Chemistry: Matter and Change
Chemistry: Matter and Change
1st Edition
ISBN: 9780078746376
Author: Dinah Zike, Laurel Dingrando, Nicholas Hainen, Cheryl Wistrom
Publisher: Glencoe/McGraw-Hill School Pub Co
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Chapter 19, Problem 72A
Interpretation Introduction

(a)

Interpretation:

The given redox reaction is to be balanced using half reaction method.

Concept introduction:

Half reaction method to balance ionic equation in acidic solution involve:-

  • Assign the oxidation number to each element.
  • Check which element is oxidized and which element is reduced. Write oxidation and reduction half reactions.
  • Check the change in oxidation number for the elements which are oxidized and reduced.
  • Then add enough hydrogen ions and water molecules to the equation to balance hydrogen atoms on both sides.
  • Adjust the coefficients so that number of electrons lost in oxidation is equal to electrons gained in reduction.
  • Add the balanced half reactions.

Expert Solution
Check Mark

Answer to Problem 72A

3Cl+2NO3+2H+2NO+3ClO+H2O

Explanation of Solution

The given redox reaction is:

Cl+NO3NO+ClO

Assigning the oxidation number-

Cl1+N+5O3N+2O+Cl+1O

The oxidation numbers of atoms shows that

Cl is oxidized and N is reduced.

Change in oxidation number of Cl=+2

Change in oxidation number of N=3

ClClO+2e

NO3+3eNO

Balancing the atoms and charges:

Cl+H2OClO+2e+2H+

NO3+3e+4H+NO+2H2O

Now adding both reactions after multiplying oxidation half cell reaction with 3 and reduction half reaction with 2,

3Cl+2NO3+2H+2NO+3ClO+H2O

Interpretation Introduction

(b)

Interpretation:

The given redox reaction is to be balanced using half reaction method.

Concept introduction:

Half reaction method to balance ionic equation in acidic solution involve:-

  • Assign the oxidation number to each element.
  • Check which element is oxidized and which element is reduced. Write oxidation and reduction half reactions.
  • Check the change in oxidation number for the elements which are oxidized and reduced.
  • Then add enough hydrogen ions and water molecules to the equation to balance hydrogen atoms on both sides.
  • Adjust the coefficients so that number of electrons lost in oxidation is equal to electrons gained in reduction.
  • Add the balanced half reactions.

Expert Solution
Check Mark

Answer to Problem 72A

5Br+IO3+6H+2Br2+IBr+3H2O

Explanation of Solution

The given redox reaction is:

Br+IO3Br2+IBr

Assigning the oxidation number-

Br1+I+5O3Br20+I+1Br

The oxidation numbers of atoms shows that

Br is oxidized and I is reduced.

Change in oxidation number of Br=+1

Change in oxidation number of I=4

BrBr2+e

Br+IO3+4eIBr

Balancing the atoms and charges:

2BrBr2+2e

Br+IO3+4e+6H+IBr+3H2O

Now adding both reactions after multiplying oxidation half cell reaction with 2

5Br+IO3+6H+2Br2+IBr+3H2O

Interpretation Introduction

(c)

Interpretation:

The given redox reaction is to be balanced using half reaction method.

Concept introduction:

Half reaction method to balance ionic equation in acidic solution involve:-

  • Assign the oxidation number to each element.
  • Check which element is oxidized and which element is reduced. Write oxidation and reduction half reactions.
  • Check the change in oxidation number for the elements which are oxidized and reduced.
  • Then add enough hydrogen ions and water molecules to the equation to balance hydrogen atoms on both sides.
  • Adjust the coefficients so that number of electrons lost in oxidation is equal to electrons gained in reduction.
  • Add the balanced half reactions.

Expert Solution
Check Mark

Answer to Problem 72A

I2+S2O32+H2OS2O42+2I+2H+

Explanation of Solution

The given redox reaction is:

I2+Na2S2O3Na2S2O4+NaI

Reaction in ionic form:

I2+S2O32S2O42+I

Assigning the oxidation number-

I20+S+22O32S+32O42+I1

The oxidation numbers of atoms shows that

S is oxidized and I is reduced.

Change in oxidation number of Speratom=+2

Change in oxidation number of Iperatom=2

S2O32S2O42+2e

I2+2eI

Balancing the atoms and charges:

S2O32+H2OS2O42+2e+2H+

I2+2e2I

Now adding both reactions

I2+S2O32+H2OS2O42+2I+2H+

Chapter 19 Solutions

Chemistry: Matter and Change

Ch. 19.1 - Prob. 11SSCCh. 19.1 - Prob. 12SSCCh. 19.1 - Prob. 13SSCCh. 19.1 - Prob. 14SSCCh. 19.2 - Prob. 15PPCh. 19.2 - Prob. 16PPCh. 19.2 - Prob. 17PPCh. 19.2 - Prob. 18PPCh. 19.2 - Prob. 19PPCh. 19.2 - Prob. 20PPCh. 19.2 - Prob. 21PPCh. 19.2 - Prob. 22PPCh. 19.2 - Prob. 23PPCh. 19.2 - Prob. 24PPCh. 19.2 - Prob. 25PPCh. 19.2 - Prob. 26SSCCh. 19.2 - Prob. 27SSCCh. 19.2 - Prob. 28SSCCh. 19.2 - Prob. 29SSCCh. 19.2 - Prob. 30SSCCh. 19.2 - Prob. 31SSCCh. 19.2 - Prob. 32SSCCh. 19 - Prob. 33ACh. 19 - Prob. 34ACh. 19 - Prob. 35ACh. 19 - Prob. 36ACh. 19 - Prob. 37ACh. 19 - Prob. 38ACh. 19 - Prob. 39ACh. 19 - Prob. 40ACh. 19 - Prob. 41ACh. 19 - Prob. 42ACh. 19 - Prob. 43ACh. 19 - Prob. 44ACh. 19 - Prob. 45ACh. 19 - Prob. 46ACh. 19 - Prob. 47ACh. 19 - Prob. 48ACh. 19 - Prob. 49ACh. 19 - Prob. 50ACh. 19 - Prob. 51ACh. 19 - Prob. 52ACh. 19 - Prob. 53ACh. 19 - Prob. 54ACh. 19 - Prob. 55ACh. 19 - Prob. 56ACh. 19 - Prob. 57ACh. 19 - Prob. 58ACh. 19 - Prob. 59ACh. 19 - Prob. 60ACh. 19 - Prob. 61ACh. 19 - Prob. 62ACh. 19 - Prob. 63ACh. 19 - Prob. 64ACh. 19 - Prob. 65ACh. 19 - Prob. 66ACh. 19 - Prob. 67ACh. 19 - Prob. 68ACh. 19 - Prob. 69ACh. 19 - Prob. 70ACh. 19 - Prob. 71ACh. 19 - Prob. 72ACh. 19 - Prob. 73ACh. 19 - Prob. 74ACh. 19 - Prob. 75ACh. 19 - Prob. 76ACh. 19 - Prob. 77ACh. 19 - Prob. 78ACh. 19 - Prob. 79ACh. 19 - Prob. 80ACh. 19 - Prob. 81ACh. 19 - Prob. 82ACh. 19 - Prob. 83ACh. 19 - Prob. 84ACh. 19 - Prob. 85ACh. 19 - Prob. 86ACh. 19 - Prob. 87ACh. 19 - Prob. 88ACh. 19 - Prob. 89ACh. 19 - Prob. 90ACh. 19 - Prob. 91ACh. 19 - Prob. 92ACh. 19 - Prob. 93ACh. 19 - Prob. 94ACh. 19 - Prob. 95ACh. 19 - Prob. 96ACh. 19 - Prob. 1STPCh. 19 - Prob. 2STPCh. 19 - Prob. 3STPCh. 19 - Prob. 4STPCh. 19 - Prob. 5STPCh. 19 - Prob. 6STPCh. 19 - Prob. 7STPCh. 19 - Prob. 8STPCh. 19 - Prob. 9STPCh. 19 - Prob. 10STPCh. 19 - Prob. 11STPCh. 19 - Prob. 12STPCh. 19 - Prob. 13STPCh. 19 - Prob. 14STPCh. 19 - Prob. 15STPCh. 19 - Prob. 16STPCh. 19 - Prob. 17STPCh. 19 - Prob. 18STP
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Balancing Redox Reactions in Acidic and Basic Conditions; Author: Professor Dave Explains;https://www.youtube.com/watch?v=N6ivvu6xlog;License: Standard YouTube License, CC-BY