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 101AP
Interpretation Introduction

(a)

Interpretation:

The given oxidation-reduction reaction should be balanced

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.

Expert Solution
Check Mark

Answer to Problem 101AP

2MnO4aq+16H+aq+10Iaq2Mn2+aq+8H2Ol+5I2aq.

Explanation of Solution

The given reaction is as follows:

Iaq+MnO4aqI2aq+Mn2+aq

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

IaqI2aq...... (1)

And,

MnO4aqMn2+aq...... (2)

In reaction (1), give coefficient 2 to I to balance the number of iodine atoms.

2IaqI2aq

Add two electrons to the right to balance the charge thus,

2IaqI2aq+2e...... (3)

In reaction (2), add 4 water molecules to right side of the reaction to balance the number of oxygen atom,

MnO4aqMn2+aq+4H2Ol

Now, add 8 hydrogen ions to the left, to balance the hydrogen atoms.

MnO4aq+8H+aqMn2+aq+4H2Ol

Last step is to balance the charge, the net charge on left side is + 7 and that on right side is + 2 thus, add 5 electrons to the left to balance the charge,

MnO4aq+8H+aq+5eMn2+aq+4H2Ol...... (4)

To get the overall reaction, add reaction (3) and (4)

5×2IaqI2aq+2e2×MnO4aq+8H+aq+5eMn2+aq+4H2Ol2MnO4aq+16H+aq+10Iaq2Mn2+aq+8H2Ol+5I2aq¯¯

Thus, the balance oxidation-reduction reaction is as follows:

2MnO4aq+16H+aq+10Iaq2Mn2+aq+8H2Ol+5I2aq.

Interpretation Introduction

(b)

Interpretation:

The given oxidation-reduction reaction should be balanced

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.

Expert Solution
Check Mark

Answer to Problem 101AP

4Cr3+aq+14HO2l+3S2O82aq2Cr2O27aq+28H+aq+6SO43.

Explanation of Solution

The given reaction is as follows:

S2O82aq+Cr3+aqSO43aq+Cr2O27aq

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

S2O82aqSO43aq...... (1)

And,

Cr3+aqCr2O27aq...... (2)

In reaction (1), give coefficient 2 to SO43aq to balance the number of sulfur atoms.

S2O82aq2SO43aq

Add 4 electrons to the left to balance the charge thus,

S2O82aq+4e2SO43aq...... (3)

In reaction (2), give coefficient 2 to Cr3+ to balance the chromium atom thus,

2Cr3+aqCr2O27aq

Add 7 water molecules to left side of the reaction to balance the number of oxygen atom,

2Cr3+aq+7HO2lCr2O27aq

Balance the hydrogen atoms to add 14 hydrogen ions to the right:

2Cr3+aq+7HO2lCr2O27aq+14H+aq

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

2Cr3+aq+7HO2lCr2O27aq+14H+aq+6e...... (4)

To get the overall reaction, add reaction (3) and (4)

3×S2O82aq+4e2SO43aq2×2Cr3+aq+7HO2lCr2O27aq+14H+aq+6e4Cr3+aq+14HO2l+3S2O82aq2Cr2O27aq+28H+aq+6SO43¯¯

Thus, the balance oxidation-reduction reaction is as follows:

4Cr3+aq+14HO2l+3S2O82aq2Cr2O27aq+28H+aq+6SO43.

Interpretation Introduction

(c)

Interpretation:

The given oxidation-reduction reaction should be balanced

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.

Expert Solution
Check Mark

Answer to Problem 101AP

5BiO3aq+14H+aq+2Mn2+aq5Bi3+aq+7H2Ol+2MnO4aq.

Explanation of Solution

The given reaction is as follows:

BiO3aq+Mn2+aqBi3+aq+MnO4aq

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

BiO3aqBi3+aq...... (1)

And,

Mn2+aqMnO4aq...... (2)

In reaction (1), add 3 water molecules on right side to balance the number of oxygen atoms,

BiO3aqBi3+aq+3H2Ol

To balance the hydrogen atom, add 6 hydrogen ions to the left thus,

BiO3aq+6H+aqBi3+aq+3H2Ol

Now, balance the charge by adding 2 electrons to the left

BiO3aq+6H+aq+2eBi3+aq+3H2Ol...... (3)

In reaction (2), add 4 water molecules on left to balance the oxygen atoms.

Mn2+aq+4H2OlMnO4aq

Hydrogen atoms can be balanced by adding 8 hydrogen ions on the right:

Mn2+aq+4H2OlMnO4aq+8H+aq

Last step is to balance the charge, add 5 electrons to the right:

Mn2+aq+4H2OlMnO4aq+8H+aq+5e..... (4)

To get the overall reaction, add reaction (3) and (4)

5×BiO3aq+6H+aq+2eBi3+aq+3H2Ol2×Mn2+aq+4H2OlMnO4aq+8H+aq+5e5BiO3aq+14H+aq+2Mn2+aq5Bi3+aq+7H2Ol+2MnO4aq¯¯

Thus, the balance oxidation-reduction reaction is as follows:

5BiO3aq+14H+aq+2Mn2+aq5Bi3+aq+7H2Ol+2MnO4aq.

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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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