Introductory Chemistry: A Foundation
Introductory Chemistry: A Foundation
9th Edition
ISBN: 9781337399425
Author: Steven S. Zumdahl, Donald J. DeCoste
Publisher: Cengage Learning
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Chapter 18, Problem 47QAP

. Iodide ion, I- , is one of the most easily oxidized species. Balance each of the following oxidation-reduction reactions, which take place in acidic solution, by using the “half-reaction” method.

a. IO 3 ( aq ) + I ( aq ) I 2 ( aq )    b. Cr 2 O 7 2- ( aq ) + I ( aq ) Cr 3 + ( aq ) + I 2 ( aq )    c. Cu 2 + ( aq ) + I ( aq ) CuI ( s ) + I 2 ( aq )

Expert Solution
Check Mark
Interpretation Introduction

(a)

Interpretation:

The given oxidation-reduction reaction should be balanced using the half-reaction method.

Concept Introduction:

The oxidation-reduction reaction is also known as a redox reaction. In this reaction, one reactant is oxidized and other is reduced. In balancing an oxidation-reduction reaction they must be first divided into two half reactions: one is oxidation reaction and other is reduction reaction.

The balancing of redox reaction is complicated as compared to simple balancing. It is necessary to determine the half reactions of reactants undergoing oxidation and reduction. On adding the two half-reactions, net total equation can be obtained. This method of balancing redox reaction is known as half equation method.

The following rules must be followed in balancing redox reaction by half equation method:

  1. Initially, redox reaction is separated into two half equations; oxidation and reduction.
  2. Atoms other than hydrogen and oxygen are balanced first in the unbalanced half equations.
  3. Oxygen atoms are balanced by addition of water on either side of the reaction.
  4. Hydrogen ion/s is added to balance the hydrogen atom.
  5. Electrons are added to balance the charge.
  6. Half reactions are added to get the net total equation.
  7. The further addition of hydroxide ion takes place on both side of the reaction, if the solution is basic in nature to neutralise the hydrogen ion present in the solution.

Answer to Problem 47QAP

2IO3aq+12H+aq+10Iaq6H2Ol+6I2aq.

Explanation of Solution

The given reaction is as follows:

IO3aq+IaqI2aq

The above reaction can be separated into two half reactions as follows:

IO3aqI2aq...... (1)

IaqI2aq...... (2)

Here, IO3aq gets reduced to I2aq and Iaq gets oxidized to I2aq.

To balance the reaction (1), give coefficient 2 to IO3aq to balance the iodine atom.

2IO3aqI2aq

Now, oxygen atom can be balanced by adding 6 water molecules to right thus,

2IO3aqI2aq+6H2Ol

Next step is to balance hydrogen atoms by adding 12 hydrogen ions to the left:

2IO3aq+12H+aqI2aq+6H2Ol

The last step is to balance the charge thus, 4 electrons are added to left thus,

2IO3aq+12H+aq+10eI2aq+6H2Ol...... (3)

Now, balance reaction (2) by given coefficient 2 to Iaq thus,

2IaqI2aq

Now, charge can be balanced by adding 2 electrons to the right thus,

2IaqI2aq+2e...... (4)

To get the net reaction, add reaction (3) and (4) as follows:

2IO3aq+12H+aq+10eI2aq+6H2Ol5×2IaqI2aq+2e2IO3aq+12H+aq+10Iaq6H2Ol+6I2aq¯¯

Thus, the balanced chemical reaction is as follows:

2IO3aq+12H+aq+10Iaq6H2Ol+6I2aq.

Expert Solution
Check Mark
Interpretation Introduction

(b)

Interpretation:

The given oxidation-reduction reaction should be balanced using the half-reaction method.

Concept Introduction:

The oxidation-reduction reaction is also known as a redox reaction. In this reaction, one reactant is oxidized and other is reduced. In balancing an oxidation-reduction reaction they must be first divided into two half reactions: one is oxidation reaction and other is reduction reaction.

The balancing of redox reaction is complicated as compared to simple balancing. It is necessary to determine the half reactions of reactants undergoing oxidation and reduction. On adding the two half-reactions, net total equation can be obtained. This method of balancing redox reaction is known as half equation method.

The following rules must be followed in balancing redox reaction by half equation method:

  1. Initially, redox reaction is separated into two half equations; oxidation and reduction.
  2. Atoms other than hydrogen and oxygen are balanced first in the unbalanced half equations.
  3. Oxygen atoms are balanced by addition of water on either side of the reaction.
  4. Hydrogen ion/s is added to balance the hydrogen atom.
  5. Electrons are added to balance the charge.
  6. Half reactions are added to get the net total equation.
  7. The further addition of hydroxide ion takes place on both side of the reaction, if the solution is basic in nature to neutralise the hydrogen ion present in the solution.

Answer to Problem 47QAP

Cr2O72aq+14H+aq+6Iaq2Cr3+aq+7H2Ol+I2aq.

Explanation of Solution

The given reaction is as follows:

Cr2O72aq+IaqCr3+aq+I2aq

The two reactions can be separated into half reactions as follows:

Cr2O72aqCr3+aq...... (1)

And,

IaqI2aq...... (2)

Reaction (1) can be balanced by giving coefficient 2 to Cr3+aq thus,

Cr2O72aq2Cr3+aq

Add 7 water molecules to the right to balance oxygen atoms thus,

Cr2O72aq2Cr3+aq+7H2Ol

Hydrogen atom can be balanced by adding 14 hydrogen ions to the left

Cr2O72aq+14H+aq2Cr3+aq+7H2Ol

Last step is to balance the charge, the net charge on left side is + 12 and on right side is + 6 thus, 6 electrons are added to left to balance the charge thus,

Cr2O72aq+14H+aq+6e2Cr3+aq+7H2Ol...... (3)

In reaction (2), iodine atom is balanced by giving coefficient 2 to Iaq thus,

2IaqI2aq

Next step is to balance the charge; two electrons are added to the right to balance the two negative charges on left side:

2IaqI2aq+2e...... (4)

To get the net reaction, add reaction (3) and (4) as follows:

Cr2O72aq+14H+aq+6e2Cr3+aq+7H2Ol3×2IaqI2aq+2eCr2O72aq+14H+aq+6Iaq2Cr3+aq+7H2Ol+I2aq¯¯

Thus, the balanced chemical reaction is as follows:

Cr2O72aq+14H+aq+6Iaq2Cr3+aq+7H2Ol+I2aq.

Expert Solution
Check Mark
Interpretation Introduction

(c)

Interpretation:

The given oxidation-reduction reaction should be balanced using the half-reaction method.

Concept Introduction:

The oxidation-reduction reaction is also known as a redox reaction. In this reaction, one reactant is oxidized and other is reduced. In balancing an oxidation-reduction reaction they must be first divided into two half reactions: one is oxidation reaction and other is reduction reaction.

The balancing of redox reaction is complicated as compared to simple balancing. It is necessary to determine the half reactions of reactants undergoing oxidation and reduction. On adding the two half-reactions, net total equation can be obtained. This method of balancing redox reaction is known as half equation method.

The following rules must be followed in balancing redox reaction by half equation method:

  1. Initially, redox reaction is separated into two half equations; oxidation and reduction.
  2. Atoms other than hydrogen and oxygen are balanced first in the unbalanced half equations.
  3. Oxygen atoms are balanced by addition of water on either side of the reaction.
  4. Hydrogen ion/s is added to balance the hydrogen atom.
  5. Electrons are added to balance the charge.
  6. Half reactions are added to get the net total equation.
  7. The further addition of hydroxide ion takes place on both side of the reaction, if the solution is basic in nature to neutralise the hydrogen ion present in the solution.

Answer to Problem 47QAP

2Cu2+aq+4Iaq2CuIs+I2aq.

Explanation of Solution

The given reaction is as follows:

Cu2+aq+IaqCuIs+I2aq

The above reaction can be separated into two half reactions as follows:

Cu2+aq+IaqCuIs..... (1)

And,

IaqI2aq..... (2)

In reaction (1), charge can be balanced by adding electrons. Since, there is + 1 charge on left thus, 1 electron is added to left:

Cu2+aq+Iaq+eCuIs..... (3)

Now, in reaction (2), iodine atom can be balanced by giving coefficient 2 to Iaq thus,

2IaqI2aq

The charge can be balanced by adding 2 electrons to the right thus,

2IaqI2aq+2e..... (4)

To get the net reaction, add reaction (3) and (4) as follows:

2Cu2+aq+Iaq+eCuIs2IaqI2aq+2e2Cu2+aq+4Iaq2CuIs+I2aq¯¯

Thus, the balanced chemical reaction is as follows:

2Cu2+aq+4Iaq2CuIs+I2aq.

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Chapter 18 Solutions

Introductory Chemistry: A Foundation

Ch. 18 - Prob. 5ALQCh. 18 - Prob. 6ALQCh. 18 - In balancing oxidation-reduction equations, why is...Ch. 18 - What does it mean for a substance to be oxidized?...Ch. 18 - Label the following parts of the galvanic cell....Ch. 18 - Prob. 1QAPCh. 18 - Prob. 2QAPCh. 18 - For each of the following oxidation-reduction...Ch. 18 - For each of the following oxidation-reduction...Ch. 18 - For each of the following oxidation-reduction...Ch. 18 - Prob. 6QAPCh. 18 - Prob. 7QAPCh. 18 - Prob. 8QAPCh. 18 - Explain why, although it is not an ionic compound,...Ch. 18 - Prob. 10QAPCh. 18 - Prob. 11QAPCh. 18 - Prob. 12QAPCh. 18 - Prob. 13QAPCh. 18 - . Assign oxidation states to all of the atoms in...Ch. 18 - Prob. 15QAPCh. 18 - Prob. 16QAPCh. 18 - . What is the oxidation state of chlorine in each...Ch. 18 - . What is the oxidation state of manganese in each...Ch. 18 - Prob. 19QAPCh. 18 - Prob. 20QAPCh. 18 - Prob. 21QAPCh. 18 - Prob. 22QAPCh. 18 - Prob. 23QAPCh. 18 - Prob. 24QAPCh. 18 - Prob. 25QAPCh. 18 - Prob. 26QAPCh. 18 - . Does an oxidizing agent donate or accept...Ch. 18 - Prob. 28QAPCh. 18 - Prob. 29QAPCh. 18 - Prob. 30QAPCh. 18 - Prob. 31QAPCh. 18 - Prob. 32QAPCh. 18 - Prob. 33QAPCh. 18 - Prob. 34QAPCh. 18 - Prob. 35QAPCh. 18 - Prob. 36QAPCh. 18 - Prob. 37QAPCh. 18 - Prob. 38QAPCh. 18 - Prob. 39QAPCh. 18 - Prob. 40QAPCh. 18 - Prob. 41QAPCh. 18 - Prob. 42QAPCh. 18 - Prob. 43QAPCh. 18 - Prob. 44QAPCh. 18 - . Balance each of the following...Ch. 18 - Prob. 46QAPCh. 18 - . Iodide ion, I- , is one of the most easily...Ch. 18 - Prob. 48QAPCh. 18 - Prob. 49QAPCh. 18 - Prob. 50QAPCh. 18 - . In which direction do electrons flow in a...Ch. 18 - Prob. 52QAPCh. 18 - . Consider the oxidation-reduction reaction...Ch. 18 - . Consider the oxidation—reduction reaction...Ch. 18 - Prob. 55QAPCh. 18 - Prob. 56QAPCh. 18 - Prob. 57QAPCh. 18 - Prob. 58QAPCh. 18 - Prob. 59QAPCh. 18 - Prob. 60QAPCh. 18 - Prob. 61QAPCh. 18 - . What are some important uses of electrolysis?Ch. 18 - . Although aluminum is one of the most abundant...Ch. 18 - . The “Chemistry in Focus” segment Water-Powered...Ch. 18 - Prob. 65APCh. 18 - Prob. 66APCh. 18 - Prob. 67APCh. 18 - Prob. 68APCh. 18 - Prob. 69APCh. 18 - Prob. 70APCh. 18 - Prob. 71APCh. 18 - Prob. 72APCh. 18 - Prob. 73APCh. 18 - . To obtain useful electrical energy from an...Ch. 18 - Prob. 75APCh. 18 - Prob. 76APCh. 18 - Prob. 77APCh. 18 - Prob. 78APCh. 18 - . The “pressure” on electrons to flow from one...Ch. 18 - Prob. 80APCh. 18 - Prob. 81APCh. 18 - Prob. 82APCh. 18 - Prob. 83APCh. 18 - . For each of the following unbalanced...Ch. 18 - Prob. 85APCh. 18 - Prob. 86APCh. 18 - Prob. 87APCh. 18 - . Balance each of the following...Ch. 18 - . Balance each of the following...Ch. 18 - . For each of the following oxidation-reduction...Ch. 18 - . For each of the following oxidation-reduction...Ch. 18 - . Assign oxidation sates to all of the atoms in...Ch. 18 - . Assign oxidation states to all of the atoms in...Ch. 18 - Prob. 94APCh. 18 - Prob. 95APCh. 18 - . Assign oxidation states to all of the atoms in...Ch. 18 - Prob. 97APCh. 18 - . In each of the following reactions, identify...Ch. 18 - . Balance each of the following half-reactions....Ch. 18 - Prob. 100APCh. 18 - Prob. 101APCh. 18 - Prob. 102APCh. 18 - . Consider the oxidation—reduction reaction...Ch. 18 - Prob. 104APCh. 18 - Prob. 105CP
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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