Student Study Guide for Silberberg Chemistry: The Molecular Nature of Matter and Change
Student Study Guide for Silberberg Chemistry: The Molecular Nature of Matter and Change
7th Edition
ISBN: 9780078131615
Author: Martin Silberberg Dr.
Publisher: McGraw-Hill Education
bartleby

Concept explainers

Question
Book Icon
Chapter 21, Problem 21.12P

(a)

Interpretation Introduction

Interpretation:

The given skeleton redox reaction has to be balanced and also the oxidizing and reducing agents has to be identified.

Concept Introduction:

Oxidation: The gain of oxygen or the loss of hydrogen or the loss of an electron in a species during a redox reaction is called as oxidation.

Reduction: The loss of oxygen or the gain of hydrogen or the gain of an electron in a species during a redox reaction is called as reduction.

Oxidizing agent: The substance that is oxidized is called as a reducing agent.

Reducing agent: The substance that is reduced is called as an oxidizing agent.

Steps in balancing redox reactions:

  1. 1) Divide the overall reaction into an oxidation half-reaction and a reduction half-reaction.
  2. 2) Balance atoms other than O and H.
  3. 3) Balance O by adding H2O as needed
  4. 4) Balance H by adding H+ ion at the required side.
  5. 5) Balance charges by adding, as needed number of electrons,
  6. 6) Multiply the oxidation half-reaction with the coefficient of electrons in the reduction part.
  7. 7) Multiply the reduction half-reaction with the coefficient of electrons in the oxidation part.
  8. 8) Combine the two half-reactions, cancel out the species that appears on both side, so that number of elements that appear on both sides become equal.
  9. 9) For the reaction in acidic medium, the presence of H+ ion allowed.
  10. 10) For the reaction in basic medium, the H+ ion should be neutralized by the addition of required amount of OH ions on both the sides.

11) Cancel out the species that appears on both sides and ensure that the number of atoms on the reactant side is equal to the number of atoms on the product side.

Basic conditions to balance the half-reactions are as follows:

  • Balance the oxidation number with electrons.
  • Balance the elements except for O and H.
  • Balance O with H2O.
  • Balance H with H+ ion.
  • Balance the net charge with the electrons.
  • Add OH- on both sides to cancel all  H+ ions.

(a)

Expert Solution
Check Mark

Explanation of Solution

The reaction between ClO3(aq) and I(aq) contains two half reactions. They are divided and grouped into reactants and products with similar atoms.

ClO3(aq)Cl(aq)2I(aq)I2(s)

For above half reactions the atoms other than oxygen and hydrogen are balanced first. Then the oxygen atoms are balanced using water and the hydrogen atoms are balanced using H+. Finally, the charges are balanced using electrons.

ClO3(aq)Cl(aq)2I(aq)I2(s)

ClO3(aq)Cl(aq)+3H2O(l)2I(aq)I2(a) Oarebalanced

ClO3(aq)+6H+(aq)Cl(aq)+3H2O(l)2I(aq)I2(s) HarebalancedClO3(aq)+6H+(aq)+6eCl(aq)+3H2O(l)2I(aq)I2(s)+2e chargesarebalanced

Finally obtained above reactions are multiplied with integers which will balance the electrons lost with the electrons gained. Here 2 electrons are lost and 6 electrons are gained hence the oxidation reaction has to be multiplied with integer 3.

ClO3(aq)+6H+(aq)+6eCl(aq)+3H2O(l)6I(aq)3I2(s)+6e chargesarebalanced

The above half reactions are added as follows in which the substances appear on both reactants and products sides are gets cancelled and merge into single molecular equation.

ClO3(aq)+6H+(aq)+6eCl(aq)+3H2O(l)6I(aq)3I2(s)+6e_ClO3(aq)+6H+(aq)+6I(aq)Cl(aq)+3H2O(l)+3I2(s)

The substance that is reduced is called as an oxidizing agent. The substance that gets oxidized is termed as reducing agent.

From the balanced reaction it is clear that ClO3 is reduced to Cl and I is oxidized to I2.

Hence, the oxidizing agent is ClO3 and the reducing agent is I.

(b)

Interpretation Introduction

Interpretation:

The given skeleton redox reaction has to be balanced and also the oxidizing and reducing agents has to be identified.

Concept Introduction:

Oxidation: The gain of oxygen or the loss of hydrogen or the loss of an electron in a species during a redox reaction is called as oxidation.

Reduction: The loss of oxygen or the gain of hydrogen or the gain of an electron in a species during a redox reaction is called as reduction.

Oxidizing agent: The substance that is oxidized is called as a reducing agent.

Reducing agent: The substance that is reduced is called as an oxidizing agent.

Steps in balancing redox reactions:

  1. 1) Divide the overall reaction into an oxidation half-reaction and a reduction half-reaction.
  2. 2) Balance atoms other than O and H.
  3. 3) Balance O by adding H2O as needed
  4. 4) Balance H by adding H+ ion at the required side.
  5. 5) Balance charges by adding, as needed number of electrons,
  6. 6) Multiply the oxidation half-reaction with the coefficient of electrons in the reduction part.
  7. 7) Multiply the reduction half-reaction with the coefficient of electrons in the oxidation part.
  8. 8) Combine the two half-reactions, cancel out the species that appears on both side, so that number of elements that appear on both sides become equal.
  9. 9) For the reaction in acidic medium, the presence of H+ ion allowed.
  10. 10) For the reaction in basic medium, the H+ ion should be neutralized by the addition of required amount of OH ions on both the sides.

11) Cancel out the species that appears on both sides and ensure that the number of atoms on the reactant side is equal to the number of atoms on the product side.

Basic conditions to balance the half-reactions are as follows:

  • Balance the oxidation number with electrons.
  • Balance the elements except for O and H.
  • Balance O with H2O.
  • Balance H with H+ ion.
  • Balance the net charge with the electrons.
  • Add OH- on both sides to cancel all H+ ions.

(b)

Expert Solution
Check Mark

Explanation of Solution

The reaction between MnO4(aq) and SO32(aq) contains two half reactions. They are divided and grouped into reactants and products with similar atoms.

MnO4(aq)MnO2(s)SO32(aq)SO42(aq)

For above half reactions the atoms other than oxygen and hydrogen are balanced first. Then the oxygen atoms are balanced using water and the hydrogen atoms are balanced using H+. Finally, the charges are balanced using electrons.

MnO4(aq)MnO2(s)SO32(aq)SO42(aq)

MnO4(aq)MnO2(s)+2H2O(l)SO32(aq)+H2O(l)SO42(aq) Oarebalanced

MnO4(aq)+4H+(aq)MnO2(s)+2H2O(l)SO32(aq)+H2O(l)SO42(aq)+2H+(aq) HarebalancedMnO4(aq)+4H+(aq)+3eMnO2(s)+2H2O(l)SO32(aq)+H2O(l)SO42(aq)+2H+(aq)+2e chargesarebalanced

Finally obtained above reactions are multiplied with integers which will balance the electrons lost with the electrons gained. Here 2 electrons are lost and 3 electrons are gained hence the oxidation reaction has to be multiplied with integer 3 and reduction reaction has to be multiplied with 2 electrons.

2MnO4(aq)+8H+(aq)+6e2MnO2(s)+4H2O(l)3SO32(aq)+3H2O(l)3SO42(aq)+6H+(aq)+6e

The above half reactions are added as follows in which the substances appear on both reactants and products sides are gets cancelled and merge into single equation.

2MnO4(aq)+28H+(aq)+6e2MnO2(s)+4H2O(l)3SO32(aq)+3H2O(l)3SO42(aq)+6H+(aq)+6e_2MnO4(aq)+2H+(aq)+3SO32(aq)2MnO2(s)+H2O(l)+3SO42(aq)

The above reaction is balanced one in acidic solution. In basic solution the reaction is obtained by adding OH on both sides in order to neutralize H+.

2MnO4(aq)+2H+(aq)+3SO32(aq)+2OH2MnO2(s)+H2O(l)+3SO42(aq)+2OH2MnO4(aq)+12H2O(l)+3SO32(aq)2MnO2(s)+H2O(l)+3SO42(aq)+2OHBalancedreaction:__2MnO4(aq)+H2O(l)+3SO32(aq)2MnO2(s)+3SO42(aq)+2OH(aq)

The substance that is reduced is called as an oxidizing agent. The substance that gets oxidized is termed as reducing agent.

From the balanced reaction it is clear that MnO4 is reduced to MnO2 and SO32 is oxidized to SO42.

Hence, the oxidizing agent is MnO4 and the reducing agent is SO32.

(c)

Interpretation Introduction

Interpretation:

The given skeleton redox reaction has to be balanced and also the oxidizing and reducing agents has to be identified.

Concept Introduction:

Oxidation: The gain of oxygen or the loss of hydrogen or the loss of an electron in a species during a redox reaction is called as oxidation.

Reduction: The loss of oxygen or the gain of hydrogen or the gain of an electron in a species during a redox reaction is called as reduction.

Oxidizing agent: The substance that is oxidized is called as a reducing agent.

Reducing agent: The substance that is reduced is called as an oxidizing agent.

Steps in balancing redox reactions:

  1. 1) Divide the overall reaction into an oxidation half-reaction and a reduction half-reaction.
  2. 2) Balance atoms other than O and H.
  3. 3) Balance O by adding H2O as needed
  4. 4) Balance H by adding H+ ion at the required side.
  5. 5) Balance charges by adding, as needed number of electrons,
  6. 6) Multiply the oxidation half-reaction with the coefficient of electrons in the reduction part.
  7. 7) Multiply the reduction half-reaction with the coefficient of electrons in the oxidation part.
  8. 8) Combine the two half-reactions, cancel out the species that appears on both side, so that number of elements that appear on both sides become equal.
  9. 9) For the reaction in acidic medium, the presence of H+ ion allowed.
  10. 10) For the reaction in basic medium, the H+ ion should be neutralized by the addition of required amount of OH ions on both the sides.

11) Cancel out the species that appears on both sides and ensure that the number of atoms on the reactant side is equal to the number of atoms on the product side.

Basic conditions to balance the half-reactions are as follows:

  • Balance the oxidation number with electrons.
  • Balance the elements except for O and H.
  • Balance O with H2O.
  • Balance H with H+ ion.
  • Balance the net charge with the electrons.
  • Add OH- on both sides to cancel all  H+ ions.

(c)

Expert Solution
Check Mark

Explanation of Solution

The reaction between MnO4 and H2O2(aq) contains two half reactions. They are divided and grouped into reactants and products with similar atoms.

MnO4(aq)Mn2+(aq)H2O2(aq)O2(g)

For above half reactions the atoms other than oxygen and hydrogen are balanced first. Then the oxygen atoms are balanced using water and the hydrogen atoms are balanced using H+. Finally, the charges are balanced using electrons.

MnO4(aq)Mn2+(aq)H2O2(aq)O2(g)

MnO4(aq)Mn2+(aq)+4H2O(l)H2O2(aq)O2(g) Oarebalanced

MnO4(aq)+8H+(aq)Mn2+(aq)+4H2O(l)H2O2(aq)O2(g)+2H+(aq) HarebalancedMnO4(aq)+8H+(aq)+5eMn2+(aq)+4H2O(l)H2O2(aq)O2(g)+2H+(aq)+2e chargesarebalanced

Finally obtained above reactions are multiplied with integers which will balance the electrons lost with the electrons gained. Here 2 electrons are lost and 5 electrons are gained hence the oxidation reaction has to be multiplied with integer 5 and the reduction reaction is multiplied with 2.

2MnO4(aq)+16H+(aq)+10e2Mn2+(aq)+8H2O(l)5H2O2(aq)5O2(g)+10H+(aq)+10e

The above half reactions are added as follows in which the substances appear on both reactants and products sides are gets cancelled and merge into single molecular equation.

2MnO4(aq)+616H+(aq)+10e2Mn2+(aq)+8H2O(l)5H2O2(aq)5O2(g)+10H+(aq)+10e_2MnO4(aq)+6H+(aq)+5H2O2(aq)2Mn2+(aq)+8H2O(l)+5O2(g)

The substance that is reduced is called as an oxidizing agent. The substance that gets oxidized is termed as reducing agent.

From the balanced reaction it is clear that MnO4 is reduced to Mn2+ and H2O2 is oxidized to H2O.

Hence, the oxidizing agent is MnO4 and the reducing agent is H2O2.

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 21 Solutions

Student Study Guide for Silberberg Chemistry: The Molecular Nature of Matter and Change

Ch. 21.4 - Prob. 21.6AFPCh. 21.4 - Prob. 21.6BFPCh. 21.4 - Prob. 21.7AFPCh. 21.4 - Prob. 21.7BFPCh. 21.4 - Prob. 21.8AFPCh. 21.4 - Prob. 21.8BFPCh. 21.7 - The most ionic and least ionic of the common...Ch. 21.7 - Prob. 21.9BFPCh. 21.7 - Prob. 21.10AFPCh. 21.7 - Prob. 21.10BFPCh. 21.7 - Prob. 21.11AFPCh. 21.7 - Prob. 21.11BFPCh. 21.7 - In the final steps of the ETC, iron and copper...Ch. 21.7 - Prob. B21.2PCh. 21 - Prob. 21.1PCh. 21 - Prob. 21.2PCh. 21 - Prob. 21.3PCh. 21 - Water is used to balance O atoms in the...Ch. 21 - Prob. 21.5PCh. 21 - Prob. 21.6PCh. 21 - Prob. 21.7PCh. 21 - Prob. 21.8PCh. 21 - Prob. 21.9PCh. 21 - Prob. 21.10PCh. 21 - Prob. 21.11PCh. 21 - Prob. 21.12PCh. 21 - Prob. 21.13PCh. 21 - Prob. 21.14PCh. 21 - Prob. 21.15PCh. 21 - Prob. 21.16PCh. 21 - Prob. 21.17PCh. 21 - Prob. 21.18PCh. 21 - Prob. 21.19PCh. 21 - Prob. 21.20PCh. 21 - Aqua regia, a mixture of concentrated HNO3 and...Ch. 21 - Consider the following general voltaic...Ch. 21 - Why does a voltaic cell not operate unless the two...Ch. 21 - Prob. 21.24PCh. 21 - Prob. 21.25PCh. 21 - Prob. 21.26PCh. 21 - Consider the following voltaic cell: In which...Ch. 21 - Consider the following voltaic cell: In which...Ch. 21 - Prob. 21.29PCh. 21 - Prob. 21.30PCh. 21 - A voltaic cell is constructed with an Fe/Fe2+...Ch. 21 - Prob. 21.32PCh. 21 - Prob. 21.33PCh. 21 - Prob. 21.34PCh. 21 - Prob. 21.35PCh. 21 - What does a negative indicate about a redox...Ch. 21 - Prob. 21.37PCh. 21 - In basic solution, Se2− and ions react...Ch. 21 - Prob. 21.39PCh. 21 - Prob. 21.40PCh. 21 - Use the emf series (Appendix D) to arrange each...Ch. 21 - Prob. 21.42PCh. 21 - Prob. 21.43PCh. 21 - Prob. 21.44PCh. 21 - Prob. 21.45PCh. 21 - Prob. 21.46PCh. 21 - Prob. 21.47PCh. 21 - Prob. 21.48PCh. 21 - Prob. 21.49PCh. 21 - Prob. 21.50PCh. 21 - Prob. 21.51PCh. 21 - Prob. 21.52PCh. 21 - Prob. 21.53PCh. 21 - Prob. 21.54PCh. 21 - Prob. 21.55PCh. 21 - Prob. 21.56PCh. 21 - Prob. 21.57PCh. 21 - Prob. 21.58PCh. 21 - Prob. 21.59PCh. 21 - Prob. 21.60PCh. 21 - Prob. 21.61PCh. 21 - Prob. 21.62PCh. 21 - Prob. 21.63PCh. 21 - Prob. 21.64PCh. 21 - Prob. 21.65PCh. 21 - Prob. 21.66PCh. 21 - Prob. 21.67PCh. 21 - Prob. 21.68PCh. 21 - Prob. 21.69PCh. 21 - Prob. 21.70PCh. 21 - Prob. 21.71PCh. 21 - Prob. 21.72PCh. 21 - Prob. 21.73PCh. 21 - Prob. 21.74PCh. 21 - Prob. 21.75PCh. 21 - Prob. 21.76PCh. 21 - Prob. 21.77PCh. 21 - Prob. 21.78PCh. 21 - Prob. 21.79PCh. 21 - Prob. 21.80PCh. 21 - Prob. 21.81PCh. 21 - Consider the following general electrolytic...Ch. 21 - Prob. 21.83PCh. 21 - Prob. 21.84PCh. 21 - Prob. 21.85PCh. 21 - Prob. 21.86PCh. 21 - In the electrolysis of molten NaBr: What product...Ch. 21 - Prob. 21.88PCh. 21 - Prob. 21.89PCh. 21 - Prob. 21.90PCh. 21 - Prob. 21.91PCh. 21 - Prob. 21.92PCh. 21 - Prob. 21.93PCh. 21 - Prob. 21.94PCh. 21 - Prob. 21.95PCh. 21 - Prob. 21.96PCh. 21 - Prob. 21.97PCh. 21 - Write a balanced half-reaction for the product...Ch. 21 - Prob. 21.99PCh. 21 - Prob. 21.100PCh. 21 - Prob. 21.101PCh. 21 - Prob. 21.102PCh. 21 - Prob. 21.103PCh. 21 - Prob. 21.104PCh. 21 - Prob. 21.105PCh. 21 - Prob. 21.106PCh. 21 - Prob. 21.107PCh. 21 - Prob. 21.108PCh. 21 - Prob. 21.109PCh. 21 - Prob. 21.110PCh. 21 - Prob. 21.111PCh. 21 - Prob. 21.112PCh. 21 - Prob. 21.113PCh. 21 - Prob. 21.114PCh. 21 - Prob. 21.115PCh. 21 - Prob. 21.116PCh. 21 - Prob. 21.117PCh. 21 - Prob. 21.118PCh. 21 - Prob. 21.119PCh. 21 - Prob. 21.120PCh. 21 - To examine the effect of ion removal on cell...Ch. 21 - Prob. 21.122PCh. 21 - Prob. 21.123PCh. 21 - Prob. 21.124PCh. 21 - Prob. 21.125PCh. 21 - Prob. 21.126PCh. 21 - Commercial electrolytic cells for producing...Ch. 21 - Prob. 21.129PCh. 21 - Prob. 21.130PCh. 21 - The following reactions are used in...Ch. 21 - Prob. 21.132PCh. 21 - Prob. 21.133PCh. 21 - Prob. 21.134PCh. 21 - Prob. 21.135PCh. 21 - If the Ecell of the following cell is 0.915 V,...Ch. 21 - Prob. 21.137PCh. 21 - Prob. 21.138PCh. 21 - Prob. 21.139PCh. 21 - Prob. 21.140PCh. 21 - Prob. 21.141PCh. 21 - Prob. 21.142PCh. 21 - Prob. 21.143PCh. 21 - Prob. 21.144PCh. 21 - Prob. 21.145PCh. 21 - Prob. 21.146PCh. 21 - Prob. 21.147PCh. 21 - Both Ti and V are reactive enough to displace H2...Ch. 21 - For the reaction ∆G° = 87.8 kJ/mol Identity the...Ch. 21 - Two voltaic cells are to be joined so that one...Ch. 21 - Prob. 21.152PCh. 21 - Prob. 21.153P
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