General Chemistry
General Chemistry
7th Edition
ISBN: 9780073402758
Author: Chang, Raymond/ Goldsby
Publisher: McGraw-Hill College
bartleby

Concept explainers

bartleby

Videos

Question
Book Icon
Chapter 19, Problem 19.1QP

(a)

Interpretation Introduction

Interpretation: The given redox equation should be balanced by the half-reaction method.

Concept Introduction:

Half-reaction can either be oxidation or reduction reaction of a reactant component in a redox reaction.

Oxidation: Removal of electrons from a chemical species is known as oxidation.

Reduction: Addition of electrons to a chemical species is known as reduction.

Redox reaction: It is a reaction which involves both reduction and oxidation reactions.

To balance the equation, use the following steps:

The general steps involved balancing the redox reactions in both acidic and basic conditions:

  • Write the oxidation number for all of the elements involved in reaction.
  • Based on the change in the oxidation number separate the reaction into reduced and oxidized half-reactions.
  • Balance the atoms other than Oxygen and Hydrogen atom.
  • Balance the Oxygen atoms by adding water on the appropriate side.
  • Balance the hydrogen atoms by adding equal number of H+ on the opposite side.
  • Only in the basic condition, the H+ ions are needed to be balanced further. The H+ ions are balanced by the addition of equal number of OH- ions on both the sides. If there are equal number of H+ and OH- ions on any of the two sides, it will constitute the corresponding number of water molecules.
  • Balance the charge by adding appropriate number of electrons at the appropriate side.
  • Multiply the reduction and oxidation half-reactions, by the whole number such that the number of electrons released in the oxidation half-reaction should be equal to the number of electrons gained in the reduction half-reaction.
  • Add both the oxidation and reduction half-reactions and cancel out the electrons on both the sides of the reaction.
  • Write the final equation and count the number of atoms and check the charge to verify the balanced reaction.

(a)

Expert Solution
Check Mark

Answer to Problem 19.1QP

The balanced redox reaction is e+2H++H2O2+Fe2+Fe3++H2O+H2O

Explanation of Solution

The given equation in acidic condition is shown below with oxidation states:

H2O2+Fe2+Fe3++H2O

  • The hydrogen atoms and the Fe ions are balanced on both the sides.
  • Only the oxygen atoms are needed to be balanced.

Balancing oxygen:

Water molecule is needed to be added to balance the oxygen atom which is shown below:

H2O2+Fe2+Fe3++H2O+H2O

Balancing hydrogen:

Now the hydrogen atoms are needed to be balanced by the addition of protons which is shown below:

2H++H2O2+Fe2+Fe3++H2O+H2O

Balancing charge:

The charge has to be balanced by adding the required number of electrons to the more positively charged side.

e+2H++H2O2+Fe2+Fe3++H2O+H2O

So this is the overall balanced reaction.

(b)

Interpretation Introduction

Interpretation: The given redox equation should be balanced by the half-reaction method.

Concept Introduction:

Half-reaction can either be oxidation or reduction reaction of a reactant component in a redox reaction.

Oxidation: Removal of electrons from a chemical species is known as oxidation.

Reduction: Addition of electrons to a chemical species is known as reduction.

Redox reaction: It is a reaction which involves both reduction and oxidation reactions.

To balance the equation, use the following steps:

The general steps involved balancing the redox reactions in both acidic and basic conditions:

  • Write the oxidation number for all of the elements involved in reaction.
  • Based on the change in the oxidation number separate the reaction into reduced and oxidized half-reactions.
  • Balance the atoms other than Oxygen and Hydrogen atom.
  • Balance the Oxygen atoms by adding water on the appropriate side.
  • Balance the hydrogen atoms by adding equal number of H+ on the opposite side.
  • Only in the basic condition, the H+ ions are needed to be balanced further. The H+ ions are balanced by the addition of equal number of OH- ions on both the sides. If there are equal number of H+ and OH- ions on any of the two sides, it will constitute the corresponding number of water molecules.
  • Balance the charge by adding appropriate number of electrons at the appropriate side.
  • Multiply the reduction and oxidation half-reactions, by the whole number such that the number of electrons released in the oxidation half-reaction should be equal to the number of electrons gained in the reduction half-reaction.
  • Add both the oxidation and reduction half-reactions and cancel out the electrons on both the sides of the reaction.
  • Write the final equation and count the number of atoms and check the charge to verify the balanced reaction.

(b)

Expert Solution
Check Mark

Answer to Problem 19.1QP

The balanced redox reaction is e+3H++Cu+HNO3Cu2++NO+H2O+H2O

Explanation of Solution

The given equation in acidic condition is shown below with oxidation states:

Cu+HNO3Cu2++NO+H2O

  • The nitrogen atoms are balanced on both sides.
  •  The oxygen and hydrogen atoms are needed to be balanced.

Balancing oxygen:

Water molecule is needed to be added to balance the oxygen atom which is shown below:

Cu+HNO3Cu2++NO+H2O+H2O

Balancing hydrogen:

Now the hydrogen atoms are needed to be balanced by the addition of protons which is shown below:

3H++Cu+HNO3Cu2++NO+H2O+H2O

Balancing charge:

The charge has to be balanced by adding the required number of electrons to the more positively charged side.

e+3H++Cu+HNO3Cu2++NO+H2O+H2O

So this is the overall balanced reaction.

(c)

Interpretation Introduction

Interpretation: The given redox equation should be balanced by the half-reaction method.

Concept Introduction:

Half-reaction can either be oxidation or reduction reaction of a reactant component in a redox reaction.

Oxidation: Removal of electrons from a chemical species is known as oxidation.

Reduction: Addition of electrons to a chemical species is known as reduction.

Redox reaction: It is a reaction which involves both reduction and oxidation reactions.

To balance the equation, use the following steps:

The general steps involved balancing the redox reactions in both acidic and basic conditions:

  • Write the oxidation number for all of the elements involved in reaction.
  • Based on the change in the oxidation number separate the reaction into reduced and oxidized half-reactions.
  • Balance the atoms other than Oxygen and Hydrogen atom.
  • Balance the Oxygen atoms by adding water on the appropriate side.
  • Balance the hydrogen atoms by adding equal number of H+ on the opposite side.
  • Only in the basic condition, the H+ ions are needed to be balanced further. The H+ ions are balanced by the addition of equal number of OH- ions on both the sides. If there are equal number of H+ and OH- ions on any of the two sides, it will constitute the corresponding number of water molecules.
  • Balance the charge by adding appropriate number of electrons at the appropriate side.
  • Multiply the reduction and oxidation half-reactions, by the whole number such that the number of electrons released in the oxidation half-reaction should be equal to the number of electrons gained in the reduction half-reaction.
  • Add both the oxidation and reduction half-reactions and cancel out the electrons on both the sides of the reaction.
  • Write the final equation and count the number of atoms and check the charge to verify the balanced reaction.

(c)

Expert Solution
Check Mark

Answer to Problem 19.1QP

The balanced redox reaction is 2H2O+CN-+MnO4CNO+MnO2+H2O+2OH

Explanation of Solution

The given equation in basic condition is given below:

CN-+MnO4CNO+MnO2

  • The carbon, nitrogen and manganese are being balanced on both sides.

Balancing Oxygen:

Water molecule is needed to be added to balance the oxygen atom which is shown below:

CN-+MnO4CNO+MnO2+H2O

Balancing Hydrogen:

Now the hydrogen atoms are needed to be balanced by the addition of protons which is shown below:

2H++CN-+MnO4CNO+MnO2+H2O

Balancing Protons:

The protons can be balanced by adding the equal number of hydroxide ions on both sides.

2OH+2H++CN-+MnO4CNO+MnO2+H2O+2OH

The equal number of protons and hydroxide ions on one particular side is converted into the equal number of water molecules.

2H2O+CN-+MnO4CNO+MnO2+H2O+2OH

This is the overall balanced reaction.

(d)

Interpretation Introduction

Interpretation: The given redox equation should be balanced by the half-reaction method.

Concept Introduction:

Half-reaction can either be oxidation or reduction reaction of a reactant component in a redox reaction.

Oxidation: Removal of electrons from a chemical species is known as oxidation.

Reduction: Addition of electrons to a chemical species is known as reduction.

Redox reaction: It is a reaction which involves both reduction and oxidation reactions.

To balance the equation, use the following steps:

The general steps involved balancing the redox reactions in both acidic and basic conditions:

  • Write the oxidation number for all of the elements involved in reaction.
  • Based on the change in the oxidation number separate the reaction into reduced and oxidized half-reactions.
  • Balance the atoms other than Oxygen and Hydrogen atom.
  • Balance the Oxygen atoms by adding water on the appropriate side.
  • Balance the hydrogen atoms by adding equal number of H+ on the opposite side.
  • Only in the basic condition, the H+ ions are needed to be balanced further. The H+ ions are balanced by the addition of equal number of OH- ions on both the sides. If there are equal number of H+ and OH- ions on any of the two sides, it will constitute the corresponding number of water molecules.
  • Balance the charge by adding appropriate number of electrons at the appropriate side.
  • Multiply the reduction and oxidation half-reactions, by the whole number such that the number of electrons released in the oxidation half-reaction should be equal to the number of electrons gained in the reduction half-reaction.
  • Add both the oxidation and reduction half-reactions and cancel out the electrons on both the sides of the reaction.
  • Write the final equation and count the number of atoms and check the charge to verify the balanced reaction.

(d)

Expert Solution
Check Mark

Answer to Problem 19.1QP

The balanced redox reaction is 6OH-+3H2O+Br2BrO3-+Br-+6H2O+4e

Explanation of Solution

The given equation in basic condition is given below:

Br2BrO3-+Br-

  • The bromine atoms are being balanced on both the sides.

Balancing Oxygen:

Water molecule is needed to be added to balance the oxygen atom which is shown below:

3H2O+Br2BrO3-+Br-

Balancing Hydrogen:

Now the hydrogen atoms are needed to be balanced by the addition of protons which is shown below:

3H2O+Br2BrO3-+Br-+6H+

Balancing Protons:

The protons can be balanced by adding the equal number of hydroxide ions on both sides.

6OH-+3H2O+Br2BrO3-+Br-+6H++6OH-

The equal number of protons and hydroxide ions on one particular side is converted into the equal number of water molecules.

6OH-+3H2O+Br2BrO3-+Br-+6H2O

Balancing the charge:

The charge has to be balanced by adding the required number of electrons to the more positively charged side.

6OH-+3H2O+Br2BrO3-+Br-+6H2O+4e

This is the overall balanced reaction.

(e)

Interpretation Introduction

Interpretation: The given redox equation should be balanced by the half-reaction method.

Concept Introduction:

Half-reaction can either be oxidation or reduction reaction of a reactant component in a redox reaction.

Oxidation: Removal of electrons from a chemical species is known as oxidation.

Reduction: Addition of electrons to a chemical species is known as reduction.

Redox reaction: It is a reaction which involves both reduction and oxidation reactions.

To balance the equation, use the following steps:

The general steps involved balancing the redox reactions in both acidic and basic conditions:

  • Write the oxidation number for all of the elements involved in reaction.
  • Based on the change in the oxidation number separate the reaction into reduced and oxidized half-reactions.
  • Balance the atoms other than Oxygen and Hydrogen atom.
  • Balance the Oxygen atoms by adding water on the appropriate side.
  • Balance the hydrogen atoms by adding equal number of H+ on the opposite side.
  • Only in the basic condition, the H+ ions are needed to be balanced further. The H+ ions are balanced by the addition of equal number of OH- ions on both the sides. If there are equal number of H+ and OH- ions on any of the two sides, it will constitute the corresponding number of water molecules.
  • Balance the charge by adding appropriate number of electrons at the appropriate side.
  • Multiply the reduction and oxidation half-reactions, by the whole number such that the number of electrons released in the oxidation half-reaction should be equal to the number of electrons gained in the reduction half-reaction.
  • Add both the oxidation and reduction half-reactions and cancel out the electrons on both the sides of the reaction.
  • Write the final equation and count the number of atoms and check the charge to verify the balanced reaction.

(e)

Expert Solution
Check Mark

Answer to Problem 19.1QP

The balanced redox reaction is 2I2+2S2O32-4I+S4O62-

Explanation of Solution

The given equation in acidic condition is shown below with oxidation states:

S2O32-+I2I-+S4O62-

The half reactions are:

Oxidation half-reaction:

S2O32-S4O62-

Balancing sulfur atom:

2S2O32-S4O62-

So the oxygen atoms are balanced.

Balancing charge:

The charge has to be balanced by adding the required number of electrons to the more positively charged side.

2S2O32-S4O62-+2e-

Reduction half-reaction:

I2I

Balancing iodine atom:

I22I

Balancing charge:

The charge has to be balanced by adding the required number of electrons to the more positively charged side.

e+I22I

Determining the overall balanced equation:

The balanced reduction half-reaction is e+I22I

The balanced oxidation half-reaction is 2S2O32-S4O62-+2e-

2(e+I22I)(2S2O32-S4O62-+2e-)

2e-+2I2+2S2O32-4I+S4O62-+2e-

2I2+2S2O32-4I+S4O62-Thisistheoverallbalancedequation in the acidic condition.

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Chapter 19 Solutions

General Chemistry

Ch. 19.5 - Prob. 2PECh. 19.5 - Prob. 1RCCh. 19.7 - Prob. 1RCCh. 19.8 - Prob. 1RCCh. 19.8 - An aqueous solution of Mg(NO3)2 is electrolyzed....Ch. 19.8 - Prob. 2RCCh. 19.8 - Prob. 2PECh. 19.8 - Prob. 3RCCh. 19 - Prob. 19.1QPCh. 19 - Prob. 19.2QPCh. 19 - Prob. 19.3QPCh. 19 - Prob. 19.4QPCh. 19 - Prob. 19.5QPCh. 19 - Prob. 19.6QPCh. 19 - 19.7 What is the difference between the...Ch. 19 - Prob. 19.8QPCh. 19 - Prob. 19.9QPCh. 19 - Prob. 19.10QPCh. 19 - Prob. 19.11QPCh. 19 - Prob. 19.12QPCh. 19 - Prob. 19.13QPCh. 19 - 19.14 Which of the following reagents can oxidize...Ch. 19 - 19.15 Consider the following half-reactions: (aq)...Ch. 19 - 19.16 Predict whether the following reactions...Ch. 19 - Prob. 19.17QPCh. 19 - Prob. 19.18QPCh. 19 - Prob. 19.19QPCh. 19 - Prob. 19.20QPCh. 19 - Prob. 19.21QPCh. 19 - Prob. 19.22QPCh. 19 - Prob. 19.23QPCh. 19 - Prob. 19.24QPCh. 19 - Prob. 19.25QPCh. 19 - Prob. 19.26QPCh. 19 - Prob. 19.27QPCh. 19 - Prob. 19.28QPCh. 19 - Prob. 19.29QPCh. 19 - Prob. 19.30QPCh. 19 - Prob. 19.31QPCh. 19 - Prob. 19.33QPCh. 19 - Prob. 19.34QPCh. 19 - 19.35 Explain the differences between a primary...Ch. 19 - Prob. 19.36QPCh. 19 - Prob. 19.37QPCh. 19 - Prob. 19.38QPCh. 19 - Prob. 19.39QPCh. 19 - Prob. 19.40QPCh. 19 - Prob. 19.41QPCh. 19 - Prob. 19.42QPCh. 19 - 19.43 What is the difference between a galvanic...Ch. 19 - Prob. 19.44QPCh. 19 - Prob. 19.45QPCh. 19 - Prob. 19.46QPCh. 19 - Prob. 19.47QPCh. 19 - Prob. 19.48QPCh. 19 - Prob. 19.49QPCh. 19 - Prob. 19.50QPCh. 19 - 19.51 Calculate the amounts of Cu and Br2 produced...Ch. 19 - Prob. 19.52QPCh. 19 - Prob. 19.53QPCh. 19 - Prob. 19.54QPCh. 19 - 19.55 What is the hourly production rate of...Ch. 19 - Prob. 19.56QPCh. 19 - Prob. 19.57QPCh. 19 - Prob. 19.58QPCh. 19 - Prob. 19.59QPCh. 19 - Prob. 19.60QPCh. 19 - Prob. 19.61QPCh. 19 - Prob. 19.62QPCh. 19 - Prob. 19.63QPCh. 19 - Prob. 19.64QPCh. 19 - Prob. 19.65QPCh. 19 - 19.66 A sample of iron ore weighing 0.2792 g was...Ch. 19 - Prob. 19.67QPCh. 19 - Prob. 19.68QPCh. 19 - Prob. 19.69QPCh. 19 - Prob. 19.70QPCh. 19 - Prob. 19.71QPCh. 19 - Prob. 19.72QPCh. 19 - Prob. 19.73QPCh. 19 - Prob. 19.74QPCh. 19 - Prob. 19.75QPCh. 19 - Prob. 19.76QPCh. 19 - Prob. 19.77QPCh. 19 - Prob. 19.78QPCh. 19 - Prob. 19.79QPCh. 19 - Prob. 19.80QPCh. 19 - Prob. 19.81QPCh. 19 - Prob. 19.82QPCh. 19 - Prob. 19.83QPCh. 19 - Prob. 19.84QPCh. 19 - Prob. 19.86QPCh. 19 - Prob. 19.87QPCh. 19 - Prob. 19.88QPCh. 19 - Prob. 19.89QPCh. 19 - Prob. 19.90QPCh. 19 - Prob. 19.91QPCh. 19 - Prob. 19.92QPCh. 19 - Prob. 19.93QPCh. 19 - Prob. 19.94QPCh. 19 - Prob. 19.95QPCh. 19 - Prob. 19.96QPCh. 19 - Prob. 19.97QPCh. 19 - Prob. 19.98QPCh. 19 - Prob. 19.99QPCh. 19 - Prob. 19.100QPCh. 19 - Prob. 19.101QPCh. 19 - 19.102 The magnitudes (but not the signs) of the...Ch. 19 - Prob. 19.103QPCh. 19 - Prob. 19.104QPCh. 19 - Prob. 19.105QPCh. 19 - Prob. 19.106QPCh. 19 - Prob. 19.107QPCh. 19 - Prob. 19.108QPCh. 19 - Prob. 19.109QPCh. 19 - Prob. 19.110QPCh. 19 - 19.111 A spoon was silver-plated electro lyrically...Ch. 19 - Prob. 19.112QPCh. 19 - Prob. 19.113QPCh. 19 - Prob. 19.114QPCh. 19 - Prob. 19.115QPCh. 19 - Prob. 19.116QPCh. 19 - Prob. 19.117QPCh. 19 - Prob. 19.118QPCh. 19 - Prob. 19.119QPCh. 19 - Prob. 19.120QPCh. 19 - Prob. 19.121SPCh. 19 - Prob. 19.122SPCh. 19 - Prob. 19.123SPCh. 19 - Prob. 19.124SPCh. 19 - Prob. 19.125SPCh. 19 - Prob. 19.126SPCh. 19 - Prob. 19.128SPCh. 19 - Prob. 19.129SPCh. 19 - Prob. 19.130SP
Knowledge Booster
Background pattern image
Chemistry
Learn more about
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, chemistry and related others by exploring similar questions and additional content below.
Recommended textbooks for you
Text book image
Chemistry
Chemistry
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Cengage Learning
Text book image
Chemistry
Chemistry
ISBN:9781259911156
Author:Raymond Chang Dr., Jason Overby Professor
Publisher:McGraw-Hill Education
Text book image
Principles of Instrumental Analysis
Chemistry
ISBN:9781305577213
Author:Douglas A. Skoog, F. James Holler, Stanley R. Crouch
Publisher:Cengage Learning
Text book image
Organic Chemistry
Chemistry
ISBN:9780078021558
Author:Janice Gorzynski Smith Dr.
Publisher:McGraw-Hill Education
Text book image
Chemistry: Principles and Reactions
Chemistry
ISBN:9781305079373
Author:William L. Masterton, Cecile N. Hurley
Publisher:Cengage Learning
Text book image
Elementary Principles of Chemical Processes, Bind...
Chemistry
ISBN:9781118431221
Author:Richard M. Felder, Ronald W. Rousseau, Lisa G. Bullard
Publisher:WILEY
Electrolysis; Author: Tyler DeWitt;https://www.youtube.com/watch?v=dRtSjJCKkIo;License: Standard YouTube License, CC-BY