The following reaction that occurs in basic solution has to be balanced. C o ( O H ) 2 + S O 3 2 − → S O 4 2 − + C o Concept introduction: Procedure for balancing the redox reaction in acidic and basic solution is given below. 1) The oxidation half reaction equation and a reduction half reaction equation of the given reaction have to be separated. 2) The half reaction equation has to be balanced with respect to all atoms other than oxygen and hydrogen. 3) Each half reaction equation has to be balanced with respect to oxygen atoms by adding the suitable number of H 2 O molecules to the side where oxygen atom is deficient. (For metal hydroxides, directly add O H − ions and skip steps 3 – 4. ) 4) In Acidic Solution: Each half reaction equation has to be balanced with respect to hydrogen atoms by adding the suitable number of H + ions to the side deficient in hydrogen atoms. In Basic Solution: Each half reaction has to be balanced with respect to hydrogen atoms by adding a number of H 2 O molecules equal to the number of excess hydrogen atoms to the side deficient in hydrogen atoms and an equal number of O H − ions to the side opposite to the added H 2 O molecules. 5) Each half reaction equation has to be balanced with respect to charge by adding the suitable number of electrons to the side with the excess positive charge. 6) Each half reaction equation has to be multiplied by an integer that makes the number of electrons given by the oxidation half reaction equation equal to the number of electrons taken by the reduction half reaction equation. 7) The complete balanced equation is obtained by adding the two half reaction equations and canceling or combining any like terms.

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Answer 1

Question

Chapter 24, Problem 24.40P

(a)

Interpretation Introduction

Interpretation:

The following reaction that occurs in basic solution has to be balanced.

Co(OH)2+ SO32−→SO42−+ Co

Concept introduction:

Procedure for balancing the redox reaction in acidic and basic solution is given below.

1) The oxidation half reaction equation and a reduction half reaction equation of the given reaction have to be separated.

2) The half reaction equation has to be balanced with respect to all atoms other than oxygen and hydrogen.

3) Each half reaction equation has to be balanced with respect to oxygen atoms by adding the suitable number of H2O molecules to the side where oxygen atom is deficient. (For metal hydroxides, directly add OH− ions and skip steps 3–4.)

4) In Acidic Solution: Each half reaction equation has to be balanced with respect to hydrogen atoms by adding the suitable number of H+ ions to the side deficient in hydrogen atoms.
In Basic Solution: Each half reaction has to be balanced with respect to hydrogen atoms by adding a number of H2O molecules equal to the number of excess hydrogen atoms to the side deficient in hydrogen atoms and an equal number of OH− ions to the side opposite to the added H2O molecules.

5) Each half reaction equation has to be balanced with respect to charge by adding the suitable number of electrons to the side with the excess positive charge.

6) Each half reaction equation has to be multiplied by an integer that makes the number of electrons given by the oxidation half reaction equation equal to the number of electrons taken by the reduction half reaction equation.

7) The complete balanced equation is obtained by adding the two half reaction equations and canceling or combining any like terms.

(a)

Expert Solution

Explanation of Solution

The given reaction can be written as two half reactions.

The oxidation half reaction is SO32−→SO42−.

The oxidation half reaction can be balanced using above procedure.

The balanced oxidation half reaction with respect to all atoms other than oxygen and hydrogen is given below.

SO32−→SO42−

The given reaction is balanced with respect to oxygen atom by adding suitable number of H2O molecules to the side where oxygen atom is deficient as follows,

SO32−+ H2O→SO42−

The above reaction is balanced by adding a number of H2O molecules equal to the number of excess hydrogen atoms to the side deficient in hydrogen atoms and an equal number of OH− ions to the side opposite to the added H2O molecules.

SO32−+ H2O + 2 OH−→SO42−+ 2 H2O

The reaction equation is balanced with respect to charge by adding the suitable number of electrons to the side with the excess positive charge as follows,

SO32−+ H2O + 2 OH−→SO42−+ 2 H2O+ 2e−

The reduction half reaction is Co(OH)2→Co.

The reduction half reaction can be balanced using above procedure.

The balanced reduction half reaction with respect to all atoms other than oxygen and hydrogen is given below.

Co(OH)2→Co

The given reaction is balanced with respect to oxygen atom by adding suitable number of H2O molecules to the side where oxygen atom is deficient as follows,

Co(OH)2→ Co+ 2 H2O

The above reaction is balanced by adding a number of H2O molecules equal to the number of excess hydrogen atoms to the side deficient in hydrogen atoms and an equal number of OH− ions to the side opposite to the added H2O molecules.

Co(OH)2+ 2 H2O→ Co+ 2 H2O+ 2 OH−

The reaction equation is balanced with respect to charge by adding the suitable number of electrons to the side with the excess positive charge as follows,

Co(OH)2+ 2 H2O+ 2 e−→ Co+ 2 H2O+ 2 OH−

The overall reaction can be balanced as follows,

SO32−+ H2O + 2 OH−→SO42−+ 2 H2O+ 2e−Co(OH)2+ 2 H2O+ 2 e−→ Co+ 2 H2O+ 2 OH−Co(OH)2+ SO32−→ SO42−+ H2O+ Co

The balanced reaction is Co(OH)2+ SO32−→ SO42−+ H2O+ Co.

(b)

Interpretation Introduction

Interpretation:

The following reaction that occurs in acidic solution has to be balanced.

IO3−+ I−→ I3−

Concept introduction:

Refer to part (a).

(b)

Expert Solution

Explanation of Solution

The given reaction can be written as two half reactions as follows,

The oxidation half reaction is I−→ I3−.

The oxidation half reaction can be balanced using above procedure.

The balanced oxidation half reaction with respect to all atoms other than oxygen and hydrogen is given below.

3 I−→ I3−

The given reaction is balanced with respect to oxygen atom by adding suitable number of H2O molecules to the side where oxygen atom is deficient as follows,

3 I−→ I3−

The above reaction is balanced with respect to hydrogen atom by adding the suitable number of H+ ions to the side deficient in hydrogen atoms as below.

3 I−→ I3−

The reaction equation is balanced with respect to charge by adding the suitable number of electrons to the side with the excess positive charge as follows,

3 I−→ I3−+ 2 e−

The reduction half reaction is IO3−→ I3−.

The reduction half reaction can be balanced using above procedure.

The balanced reduction half reaction with respect to all atoms other than oxygen and hydrogen is given below.

3 IO3−→ I3−

The given reaction is balanced with respect to oxygen atom by adding suitable number of H2O molecules to the side where oxygen atom is deficient as follows,

3 IO3−→ I3−+ 9 H2O

The above reaction is balanced with respect to hydrogen atom by adding the suitable number of H+ ions to the side deficient in hydrogen atoms as below.

3 IO3−+ 18 H+→ I3−+ 9 H2O

The reaction equation is balanced with respect to charge by adding the suitable number of electrons to the side with the excess positive charge as follows,

3 IO3−+ 18 H++ 16 e−→ I3−+ 9 H2O

The overall reaction can be balanced as follows,

3 I−→ I3−+ 2 e−]×83 IO3−+ 18 H++ 16 e−→ I3−+ 9 H2OIO3−+ 8 I−+ 6 H+→ 3 I3−+ 3 H2O

The balanced reaction is IO3−+ 8 I−+ 6 H+→ 3 I3−+ 3 H2O.

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Answer 2

Question

Chapter 24, Problem 24.40P

(a)

Interpretation Introduction

Interpretation:

The following reaction that occurs in basic solution has to be balanced.

Co(OH)2+ SO32−→SO42−+ Co

Concept introduction:

Procedure for balancing the redox reaction in acidic and basic solution is given below.

1) The oxidation half reaction equation and a reduction half reaction equation of the given reaction have to be separated.

2) The half reaction equation has to be balanced with respect to all atoms other than oxygen and hydrogen.

3) Each half reaction equation has to be balanced with respect to oxygen atoms by adding the suitable number of H2O molecules to the side where oxygen atom is deficient. (For metal hydroxides, directly add OH− ions and skip steps 3–4.)

4) In Acidic Solution: Each half reaction equation has to be balanced with respect to hydrogen atoms by adding the suitable number of H+ ions to the side deficient in hydrogen atoms.
In Basic Solution: Each half reaction has to be balanced with respect to hydrogen atoms by adding a number of H2O molecules equal to the number of excess hydrogen atoms to the side deficient in hydrogen atoms and an equal number of OH− ions to the side opposite to the added H2O molecules.

5) Each half reaction equation has to be balanced with respect to charge by adding the suitable number of electrons to the side with the excess positive charge.

6) Each half reaction equation has to be multiplied by an integer that makes the number of electrons given by the oxidation half reaction equation equal to the number of electrons taken by the reduction half reaction equation.

7) The complete balanced equation is obtained by adding the two half reaction equations and canceling or combining any like terms.

(a)

Expert Solution

Explanation of Solution

The given reaction can be written as two half reactions.

The oxidation half reaction is SO32−→SO42−.

The oxidation half reaction can be balanced using above procedure.

The balanced oxidation half reaction with respect to all atoms other than oxygen and hydrogen is given below.

SO32−→SO42−

The given reaction is balanced with respect to oxygen atom by adding suitable number of H2O molecules to the side where oxygen atom is deficient as follows,

SO32−+ H2O→SO42−

The above reaction is balanced by adding a number of H2O molecules equal to the number of excess hydrogen atoms to the side deficient in hydrogen atoms and an equal number of OH− ions to the side opposite to the added H2O molecules.

SO32−+ H2O + 2 OH−→SO42−+ 2 H2O

The reaction equation is balanced with respect to charge by adding the suitable number of electrons to the side with the excess positive charge as follows,

SO32−+ H2O + 2 OH−→SO42−+ 2 H2O+ 2e−

The reduction half reaction is Co(OH)2→Co.

The reduction half reaction can be balanced using above procedure.

The balanced reduction half reaction with respect to all atoms other than oxygen and hydrogen is given below.

Co(OH)2→Co

The given reaction is balanced with respect to oxygen atom by adding suitable number of H2O molecules to the side where oxygen atom is deficient as follows,

Co(OH)2→ Co+ 2 H2O

The above reaction is balanced by adding a number of H2O molecules equal to the number of excess hydrogen atoms to the side deficient in hydrogen atoms and an equal number of OH− ions to the side opposite to the added H2O molecules.

Co(OH)2+ 2 H2O→ Co+ 2 H2O+ 2 OH−

The reaction equation is balanced with respect to charge by adding the suitable number of electrons to the side with the excess positive charge as follows,

Co(OH)2+ 2 H2O+ 2 e−→ Co+ 2 H2O+ 2 OH−

The overall reaction can be balanced as follows,

SO32−+ H2O + 2 OH−→SO42−+ 2 H2O+ 2e−Co(OH)2+ 2 H2O+ 2 e−→ Co+ 2 H2O+ 2 OH−Co(OH)2+ SO32−→ SO42−+ H2O+ Co

The balanced reaction is Co(OH)2+ SO32−→ SO42−+ H2O+ Co.

(b)

Interpretation Introduction

Interpretation:

The following reaction that occurs in acidic solution has to be balanced.

IO3−+ I−→ I3−

Concept introduction:

Refer to part (a).

(b)

Expert Solution

Explanation of Solution

The given reaction can be written as two half reactions as follows,

The oxidation half reaction is I−→ I3−.

The oxidation half reaction can be balanced using above procedure.

The balanced oxidation half reaction with respect to all atoms other than oxygen and hydrogen is given below.

3 I−→ I3−

The given reaction is balanced with respect to oxygen atom by adding suitable number of H2O molecules to the side where oxygen atom is deficient as follows,

3 I−→ I3−

The above reaction is balanced with respect to hydrogen atom by adding the suitable number of H+ ions to the side deficient in hydrogen atoms as below.

3 I−→ I3−

The reaction equation is balanced with respect to charge by adding the suitable number of electrons to the side with the excess positive charge as follows,

3 I−→ I3−+ 2 e−

The reduction half reaction is IO3−→ I3−.

The reduction half reaction can be balanced using above procedure.

The balanced reduction half reaction with respect to all atoms other than oxygen and hydrogen is given below.

3 IO3−→ I3−

The given reaction is balanced with respect to oxygen atom by adding suitable number of H2O molecules to the side where oxygen atom is deficient as follows,

3 IO3−→ I3−+ 9 H2O

The above reaction is balanced with respect to hydrogen atom by adding the suitable number of H+ ions to the side deficient in hydrogen atoms as below.

3 IO3−+ 18 H+→ I3−+ 9 H2O

The reaction equation is balanced with respect to charge by adding the suitable number of electrons to the side with the excess positive charge as follows,

3 IO3−+ 18 H++ 16 e−→ I3−+ 9 H2O

The overall reaction can be balanced as follows,

3 I−→ I3−+ 2 e−]×83 IO3−+ 18 H++ 16 e−→ I3−+ 9 H2OIO3−+ 8 I−+ 6 H+→ 3 I3−+ 3 H2O

The balanced reaction is IO3−+ 8 I−+ 6 H+→ 3 I3−+ 3 H2O.

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Answer 3

Question

Chapter 24, Problem 24.40P

(a)

Interpretation Introduction

Interpretation:

The following reaction that occurs in basic solution has to be balanced.

Co(OH)2+ SO32−→SO42−+ Co

Concept introduction:

Procedure for balancing the redox reaction in acidic and basic solution is given below.

1) The oxidation half reaction equation and a reduction half reaction equation of the given reaction have to be separated.

2) The half reaction equation has to be balanced with respect to all atoms other than oxygen and hydrogen.

3) Each half reaction equation has to be balanced with respect to oxygen atoms by adding the suitable number of H2O molecules to the side where oxygen atom is deficient. (For metal hydroxides, directly add OH− ions and skip steps 3–4.)

4) In Acidic Solution: Each half reaction equation has to be balanced with respect to hydrogen atoms by adding the suitable number of H+ ions to the side deficient in hydrogen atoms.
In Basic Solution: Each half reaction has to be balanced with respect to hydrogen atoms by adding a number of H2O molecules equal to the number of excess hydrogen atoms to the side deficient in hydrogen atoms and an equal number of OH− ions to the side opposite to the added H2O molecules.

5) Each half reaction equation has to be balanced with respect to charge by adding the suitable number of electrons to the side with the excess positive charge.

6) Each half reaction equation has to be multiplied by an integer that makes the number of electrons given by the oxidation half reaction equation equal to the number of electrons taken by the reduction half reaction equation.

7) The complete balanced equation is obtained by adding the two half reaction equations and canceling or combining any like terms.

(a)

Expert Solution

Explanation of Solution

The given reaction can be written as two half reactions.

The oxidation half reaction is SO32−→SO42−.

The oxidation half reaction can be balanced using above procedure.

The balanced oxidation half reaction with respect to all atoms other than oxygen and hydrogen is given below.

SO32−→SO42−

The given reaction is balanced with respect to oxygen atom by adding suitable number of H2O molecules to the side where oxygen atom is deficient as follows,

SO32−+ H2O→SO42−

The above reaction is balanced by adding a number of H2O molecules equal to the number of excess hydrogen atoms to the side deficient in hydrogen atoms and an equal number of OH− ions to the side opposite to the added H2O molecules.

SO32−+ H2O + 2 OH−→SO42−+ 2 H2O

The reaction equation is balanced with respect to charge by adding the suitable number of electrons to the side with the excess positive charge as follows,

SO32−+ H2O + 2 OH−→SO42−+ 2 H2O+ 2e−

The reduction half reaction is Co(OH)2→Co.

The reduction half reaction can be balanced using above procedure.

The balanced reduction half reaction with respect to all atoms other than oxygen and hydrogen is given below.

Co(OH)2→Co

The given reaction is balanced with respect to oxygen atom by adding suitable number of H2O molecules to the side where oxygen atom is deficient as follows,

Co(OH)2→ Co+ 2 H2O

The above reaction is balanced by adding a number of H2O molecules equal to the number of excess hydrogen atoms to the side deficient in hydrogen atoms and an equal number of OH− ions to the side opposite to the added H2O molecules.

Co(OH)2+ 2 H2O→ Co+ 2 H2O+ 2 OH−

The reaction equation is balanced with respect to charge by adding the suitable number of electrons to the side with the excess positive charge as follows,

Co(OH)2+ 2 H2O+ 2 e−→ Co+ 2 H2O+ 2 OH−

The overall reaction can be balanced as follows,

SO32−+ H2O + 2 OH−→SO42−+ 2 H2O+ 2e−Co(OH)2+ 2 H2O+ 2 e−→ Co+ 2 H2O+ 2 OH−Co(OH)2+ SO32−→ SO42−+ H2O+ Co

The balanced reaction is Co(OH)2+ SO32−→ SO42−+ H2O+ Co.

(b)

Interpretation Introduction

Interpretation:

The following reaction that occurs in acidic solution has to be balanced.

IO3−+ I−→ I3−

Concept introduction:

Refer to part (a).

(b)

Expert Solution

Explanation of Solution

The given reaction can be written as two half reactions as follows,

The oxidation half reaction is I−→ I3−.

The oxidation half reaction can be balanced using above procedure.

The balanced oxidation half reaction with respect to all atoms other than oxygen and hydrogen is given below.

3 I−→ I3−

The given reaction is balanced with respect to oxygen atom by adding suitable number of H2O molecules to the side where oxygen atom is deficient as follows,

3 I−→ I3−

The above reaction is balanced with respect to hydrogen atom by adding the suitable number of H+ ions to the side deficient in hydrogen atoms as below.

3 I−→ I3−

The reaction equation is balanced with respect to charge by adding the suitable number of electrons to the side with the excess positive charge as follows,

3 I−→ I3−+ 2 e−

The reduction half reaction is IO3−→ I3−.

The reduction half reaction can be balanced using above procedure.

The balanced reduction half reaction with respect to all atoms other than oxygen and hydrogen is given below.

3 IO3−→ I3−

The given reaction is balanced with respect to oxygen atom by adding suitable number of H2O molecules to the side where oxygen atom is deficient as follows,

3 IO3−→ I3−+ 9 H2O

The above reaction is balanced with respect to hydrogen atom by adding the suitable number of H+ ions to the side deficient in hydrogen atoms as below.

3 IO3−+ 18 H+→ I3−+ 9 H2O

The reaction equation is balanced with respect to charge by adding the suitable number of electrons to the side with the excess positive charge as follows,

3 IO3−+ 18 H++ 16 e−→ I3−+ 9 H2O

The overall reaction can be balanced as follows,

3 I−→ I3−+ 2 e−]×83 IO3−+ 18 H++ 16 e−→ I3−+ 9 H2OIO3−+ 8 I−+ 6 H+→ 3 I3−+ 3 H2O

The balanced reaction is IO3−+ 8 I−+ 6 H+→ 3 I3−+ 3 H2O.

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Answer 4

Question

Chapter 24, Problem 24.40P

(a)

Interpretation Introduction

Interpretation:

The following reaction that occurs in basic solution has to be balanced.

Co(OH)2+ SO32−→SO42−+ Co

Concept introduction:

Procedure for balancing the redox reaction in acidic and basic solution is given below.

1) The oxidation half reaction equation and a reduction half reaction equation of the given reaction have to be separated.

2) The half reaction equation has to be balanced with respect to all atoms other than oxygen and hydrogen.

3) Each half reaction equation has to be balanced with respect to oxygen atoms by adding the suitable number of H2O molecules to the side where oxygen atom is deficient. (For metal hydroxides, directly add OH− ions and skip steps 3–4.)

4) In Acidic Solution: Each half reaction equation has to be balanced with respect to hydrogen atoms by adding the suitable number of H+ ions to the side deficient in hydrogen atoms.
In Basic Solution: Each half reaction has to be balanced with respect to hydrogen atoms by adding a number of H2O molecules equal to the number of excess hydrogen atoms to the side deficient in hydrogen atoms and an equal number of OH− ions to the side opposite to the added H2O molecules.

5) Each half reaction equation has to be balanced with respect to charge by adding the suitable number of electrons to the side with the excess positive charge.

6) Each half reaction equation has to be multiplied by an integer that makes the number of electrons given by the oxidation half reaction equation equal to the number of electrons taken by the reduction half reaction equation.

7) The complete balanced equation is obtained by adding the two half reaction equations and canceling or combining any like terms.

(a)

Expert Solution

Explanation of Solution

The given reaction can be written as two half reactions.

The oxidation half reaction is SO32−→SO42−.

The oxidation half reaction can be balanced using above procedure.

The balanced oxidation half reaction with respect to all atoms other than oxygen and hydrogen is given below.

SO32−→SO42−

The given reaction is balanced with respect to oxygen atom by adding suitable number of H2O molecules to the side where oxygen atom is deficient as follows,

SO32−+ H2O→SO42−

The above reaction is balanced by adding a number of H2O molecules equal to the number of excess hydrogen atoms to the side deficient in hydrogen atoms and an equal number of OH− ions to the side opposite to the added H2O molecules.

SO32−+ H2O + 2 OH−→SO42−+ 2 H2O

The reaction equation is balanced with respect to charge by adding the suitable number of electrons to the side with the excess positive charge as follows,

SO32−+ H2O + 2 OH−→SO42−+ 2 H2O+ 2e−

The reduction half reaction is Co(OH)2→Co.

The reduction half reaction can be balanced using above procedure.

The balanced reduction half reaction with respect to all atoms other than oxygen and hydrogen is given below.

Co(OH)2→Co

The given reaction is balanced with respect to oxygen atom by adding suitable number of H2O molecules to the side where oxygen atom is deficient as follows,

Co(OH)2→ Co+ 2 H2O

The above reaction is balanced by adding a number of H2O molecules equal to the number of excess hydrogen atoms to the side deficient in hydrogen atoms and an equal number of OH− ions to the side opposite to the added H2O molecules.

Co(OH)2+ 2 H2O→ Co+ 2 H2O+ 2 OH−

The reaction equation is balanced with respect to charge by adding the suitable number of electrons to the side with the excess positive charge as follows,

Co(OH)2+ 2 H2O+ 2 e−→ Co+ 2 H2O+ 2 OH−

The overall reaction can be balanced as follows,

SO32−+ H2O + 2 OH−→SO42−+ 2 H2O+ 2e−Co(OH)2+ 2 H2O+ 2 e−→ Co+ 2 H2O+ 2 OH−Co(OH)2+ SO32−→ SO42−+ H2O+ Co

The balanced reaction is Co(OH)2+ SO32−→ SO42−+ H2O+ Co.

(b)

Interpretation Introduction

Interpretation:

The following reaction that occurs in acidic solution has to be balanced.

IO3−+ I−→ I3−

Concept introduction:

Refer to part (a).

(b)

Expert Solution

Explanation of Solution

The given reaction can be written as two half reactions as follows,

The oxidation half reaction is I−→ I3−.

The oxidation half reaction can be balanced using above procedure.

The balanced oxidation half reaction with respect to all atoms other than oxygen and hydrogen is given below.

3 I−→ I3−

The given reaction is balanced with respect to oxygen atom by adding suitable number of H2O molecules to the side where oxygen atom is deficient as follows,

3 I−→ I3−

The above reaction is balanced with respect to hydrogen atom by adding the suitable number of H+ ions to the side deficient in hydrogen atoms as below.

3 I−→ I3−

The reaction equation is balanced with respect to charge by adding the suitable number of electrons to the side with the excess positive charge as follows,

3 I−→ I3−+ 2 e−

The reduction half reaction is IO3−→ I3−.

The reduction half reaction can be balanced using above procedure.

The balanced reduction half reaction with respect to all atoms other than oxygen and hydrogen is given below.

3 IO3−→ I3−

The given reaction is balanced with respect to oxygen atom by adding suitable number of H2O molecules to the side where oxygen atom is deficient as follows,

3 IO3−→ I3−+ 9 H2O

The above reaction is balanced with respect to hydrogen atom by adding the suitable number of H+ ions to the side deficient in hydrogen atoms as below.

3 IO3−+ 18 H+→ I3−+ 9 H2O

The reaction equation is balanced with respect to charge by adding the suitable number of electrons to the side with the excess positive charge as follows,

3 IO3−+ 18 H++ 16 e−→ I3−+ 9 H2O

The overall reaction can be balanced as follows,

3 I−→ I3−+ 2 e−]×83 IO3−+ 18 H++ 16 e−→ I3−+ 9 H2OIO3−+ 8 I−+ 6 H+→ 3 I3−+ 3 H2O

The balanced reaction is IO3−+ 8 I−+ 6 H+→ 3 I3−+ 3 H2O.

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Answer 5

Chapter 24, Problem 24.40P

(a)

Interpretation Introduction

Interpretation:

The following reaction that occurs in basic solution has to be balanced.

Co(OH)2+ SO32−→SO42−+ Co

Concept introduction:

Procedure for balancing the redox reaction in acidic and basic solution is given below.

1) The oxidation half reaction equation and a reduction half reaction equation of the given reaction have to be separated.

2) The half reaction equation has to be balanced with respect to all atoms other than oxygen and hydrogen.

3) Each half reaction equation has to be balanced with respect to oxygen atoms by adding the suitable number of H2O molecules to the side where oxygen atom is deficient. (For metal hydroxides, directly add OH− ions and skip steps 3–4.)

4) In Acidic Solution: Each half reaction equation has to be balanced with respect to hydrogen atoms by adding the suitable number of H+ ions to the side deficient in hydrogen atoms.
In Basic Solution: Each half reaction has to be balanced with respect to hydrogen atoms by adding a number of H2O molecules equal to the number of excess hydrogen atoms to the side deficient in hydrogen atoms and an equal number of OH− ions to the side opposite to the added H2O molecules.

5) Each half reaction equation has to be balanced with respect to charge by adding the suitable number of electrons to the side with the excess positive charge.

6) Each half reaction equation has to be multiplied by an integer that makes the number of electrons given by the oxidation half reaction equation equal to the number of electrons taken by the reduction half reaction equation.

7) The complete balanced equation is obtained by adding the two half reaction equations and canceling or combining any like terms.

(a)

Expert Solution

Explanation of Solution

The given reaction can be written as two half reactions.

The oxidation half reaction is SO32−→SO42−.

The oxidation half reaction can be balanced using above procedure.

The balanced oxidation half reaction with respect to all atoms other than oxygen and hydrogen is given below.

SO32−→SO42−

The given reaction is balanced with respect to oxygen atom by adding suitable number of H2O molecules to the side where oxygen atom is deficient as follows,

SO32−+ H2O→SO42−

The above reaction is balanced by adding a number of H2O molecules equal to the number of excess hydrogen atoms to the side deficient in hydrogen atoms and an equal number of OH− ions to the side opposite to the added H2O molecules.

SO32−+ H2O + 2 OH−→SO42−+ 2 H2O

The reaction equation is balanced with respect to charge by adding the suitable number of electrons to the side with the excess positive charge as follows,

SO32−+ H2O + 2 OH−→SO42−+ 2 H2O+ 2e−

The reduction half reaction is Co(OH)2→Co.

The reduction half reaction can be balanced using above procedure.

The balanced reduction half reaction with respect to all atoms other than oxygen and hydrogen is given below.

Co(OH)2→Co

The given reaction is balanced with respect to oxygen atom by adding suitable number of H2O molecules to the side where oxygen atom is deficient as follows,

Co(OH)2→ Co+ 2 H2O

The above reaction is balanced by adding a number of H2O molecules equal to the number of excess hydrogen atoms to the side deficient in hydrogen atoms and an equal number of OH− ions to the side opposite to the added H2O molecules.

Co(OH)2+ 2 H2O→ Co+ 2 H2O+ 2 OH−

The reaction equation is balanced with respect to charge by adding the suitable number of electrons to the side with the excess positive charge as follows,

Co(OH)2+ 2 H2O+ 2 e−→ Co+ 2 H2O+ 2 OH−

The overall reaction can be balanced as follows,

SO32−+ H2O + 2 OH−→SO42−+ 2 H2O+ 2e−Co(OH)2+ 2 H2O+ 2 e−→ Co+ 2 H2O+ 2 OH−Co(OH)2+ SO32−→ SO42−+ H2O+ Co

The balanced reaction is Co(OH)2+ SO32−→ SO42−+ H2O+ Co.

(b)

Interpretation Introduction

Interpretation:

The following reaction that occurs in acidic solution has to be balanced.

IO3−+ I−→ I3−

Concept introduction:

Refer to part (a).

(b)

Expert Solution

Explanation of Solution

The given reaction can be written as two half reactions as follows,

The oxidation half reaction is I−→ I3−.

The oxidation half reaction can be balanced using above procedure.

The balanced oxidation half reaction with respect to all atoms other than oxygen and hydrogen is given below.

3 I−→ I3−

The given reaction is balanced with respect to oxygen atom by adding suitable number of H2O molecules to the side where oxygen atom is deficient as follows,

3 I−→ I3−

The above reaction is balanced with respect to hydrogen atom by adding the suitable number of H+ ions to the side deficient in hydrogen atoms as below.

3 I−→ I3−

The reaction equation is balanced with respect to charge by adding the suitable number of electrons to the side with the excess positive charge as follows,

3 I−→ I3−+ 2 e−

The reduction half reaction is IO3−→ I3−.

The reduction half reaction can be balanced using above procedure.

The balanced reduction half reaction with respect to all atoms other than oxygen and hydrogen is given below.

3 IO3−→ I3−

The given reaction is balanced with respect to oxygen atom by adding suitable number of H2O molecules to the side where oxygen atom is deficient as follows,

3 IO3−→ I3−+ 9 H2O

The above reaction is balanced with respect to hydrogen atom by adding the suitable number of H+ ions to the side deficient in hydrogen atoms as below.

3 IO3−+ 18 H+→ I3−+ 9 H2O

The reaction equation is balanced with respect to charge by adding the suitable number of electrons to the side with the excess positive charge as follows,

3 IO3−+ 18 H++ 16 e−→ I3−+ 9 H2O

The overall reaction can be balanced as follows,

3 I−→ I3−+ 2 e−]×83 IO3−+ 18 H++ 16 e−→ I3−+ 9 H2OIO3−+ 8 I−+ 6 H+→ 3 I3−+ 3 H2O

The balanced reaction is IO3−+ 8 I−+ 6 H+→ 3 I3−+ 3 H2O.

Answer 6

Chapter 24, Problem 24.40P

(a)

Interpretation Introduction

Interpretation:

The following reaction that occurs in basic solution has to be balanced.

Co(OH)2+ SO32−→SO42−+ Co

Concept introduction:

Procedure for balancing the redox reaction in acidic and basic solution is given below.

1) The oxidation half reaction equation and a reduction half reaction equation of the given reaction have to be separated.

2) The half reaction equation has to be balanced with respect to all atoms other than oxygen and hydrogen.

3) Each half reaction equation has to be balanced with respect to oxygen atoms by adding the suitable number of H2O molecules to the side where oxygen atom is deficient. (For metal hydroxides, directly add OH− ions and skip steps 3–4.)

4) In Acidic Solution: Each half reaction equation has to be balanced with respect to hydrogen atoms by adding the suitable number of H+ ions to the side deficient in hydrogen atoms.
In Basic Solution: Each half reaction has to be balanced with respect to hydrogen atoms by adding a number of H2O molecules equal to the number of excess hydrogen atoms to the side deficient in hydrogen atoms and an equal number of OH− ions to the side opposite to the added H2O molecules.

5) Each half reaction equation has to be balanced with respect to charge by adding the suitable number of electrons to the side with the excess positive charge.

6) Each half reaction equation has to be multiplied by an integer that makes the number of electrons given by the oxidation half reaction equation equal to the number of electrons taken by the reduction half reaction equation.

7) The complete balanced equation is obtained by adding the two half reaction equations and canceling or combining any like terms.

(a)

Expert Solution

Explanation of Solution

The given reaction can be written as two half reactions.

The oxidation half reaction is SO32−→SO42−.

The oxidation half reaction can be balanced using above procedure.

The balanced oxidation half reaction with respect to all atoms other than oxygen and hydrogen is given below.

SO32−→SO42−

The given reaction is balanced with respect to oxygen atom by adding suitable number of H2O molecules to the side where oxygen atom is deficient as follows,

SO32−+ H2O→SO42−

The above reaction is balanced by adding a number of H2O molecules equal to the number of excess hydrogen atoms to the side deficient in hydrogen atoms and an equal number of OH− ions to the side opposite to the added H2O molecules.

SO32−+ H2O + 2 OH−→SO42−+ 2 H2O

The reaction equation is balanced with respect to charge by adding the suitable number of electrons to the side with the excess positive charge as follows,

SO32−+ H2O + 2 OH−→SO42−+ 2 H2O+ 2e−

The reduction half reaction is Co(OH)2→Co.

The reduction half reaction can be balanced using above procedure.

The balanced reduction half reaction with respect to all atoms other than oxygen and hydrogen is given below.

Co(OH)2→Co

The given reaction is balanced with respect to oxygen atom by adding suitable number of H2O molecules to the side where oxygen atom is deficient as follows,

Co(OH)2→ Co+ 2 H2O

The above reaction is balanced by adding a number of H2O molecules equal to the number of excess hydrogen atoms to the side deficient in hydrogen atoms and an equal number of OH− ions to the side opposite to the added H2O molecules.

Co(OH)2+ 2 H2O→ Co+ 2 H2O+ 2 OH−

The reaction equation is balanced with respect to charge by adding the suitable number of electrons to the side with the excess positive charge as follows,

Co(OH)2+ 2 H2O+ 2 e−→ Co+ 2 H2O+ 2 OH−

The overall reaction can be balanced as follows,

SO32−+ H2O + 2 OH−→SO42−+ 2 H2O+ 2e−Co(OH)2+ 2 H2O+ 2 e−→ Co+ 2 H2O+ 2 OH−Co(OH)2+ SO32−→ SO42−+ H2O+ Co

The balanced reaction is Co(OH)2+ SO32−→ SO42−+ H2O+ Co.

Answer 7

Chapter 24, Problem 24.40P

(a)

Interpretation Introduction

Interpretation:

The following reaction that occurs in basic solution has to be balanced.

Co(OH)2+ SO32−→SO42−+ Co

Concept introduction:

Procedure for balancing the redox reaction in acidic and basic solution is given below.

1) The oxidation half reaction equation and a reduction half reaction equation of the given reaction have to be separated.

2) The half reaction equation has to be balanced with respect to all atoms other than oxygen and hydrogen.

3) Each half reaction equation has to be balanced with respect to oxygen atoms by adding the suitable number of H2O molecules to the side where oxygen atom is deficient. (For metal hydroxides, directly add OH− ions and skip steps 3–4.)

4) In Acidic Solution: Each half reaction equation has to be balanced with respect to hydrogen atoms by adding the suitable number of H+ ions to the side deficient in hydrogen atoms.
In Basic Solution: Each half reaction has to be balanced with respect to hydrogen atoms by adding a number of H2O molecules equal to the number of excess hydrogen atoms to the side deficient in hydrogen atoms and an equal number of OH− ions to the side opposite to the added H2O molecules.

5) Each half reaction equation has to be balanced with respect to charge by adding the suitable number of electrons to the side with the excess positive charge.

6) Each half reaction equation has to be multiplied by an integer that makes the number of electrons given by the oxidation half reaction equation equal to the number of electrons taken by the reduction half reaction equation.

7) The complete balanced equation is obtained by adding the two half reaction equations and canceling or combining any like terms.

Answer 8

(a)

Expert Solution

Explanation of Solution

The given reaction can be written as two half reactions.

The oxidation half reaction is SO32−→SO42−.

The oxidation half reaction can be balanced using above procedure.

The balanced oxidation half reaction with respect to all atoms other than oxygen and hydrogen is given below.

SO32−→SO42−

The given reaction is balanced with respect to oxygen atom by adding suitable number of H2O molecules to the side where oxygen atom is deficient as follows,

SO32−+ H2O→SO42−

The above reaction is balanced by adding a number of H2O molecules equal to the number of excess hydrogen atoms to the side deficient in hydrogen atoms and an equal number of OH− ions to the side opposite to the added H2O molecules.

SO32−+ H2O + 2 OH−→SO42−+ 2 H2O

The reaction equation is balanced with respect to charge by adding the suitable number of electrons to the side with the excess positive charge as follows,

SO32−+ H2O + 2 OH−→SO42−+ 2 H2O+ 2e−

The reduction half reaction is Co(OH)2→Co.

The reduction half reaction can be balanced using above procedure.

The balanced reduction half reaction with respect to all atoms other than oxygen and hydrogen is given below.

Co(OH)2→Co

The given reaction is balanced with respect to oxygen atom by adding suitable number of H2O molecules to the side where oxygen atom is deficient as follows,

Co(OH)2→ Co+ 2 H2O

The above reaction is balanced by adding a number of H2O molecules equal to the number of excess hydrogen atoms to the side deficient in hydrogen atoms and an equal number of OH− ions to the side opposite to the added H2O molecules.

Co(OH)2+ 2 H2O→ Co+ 2 H2O+ 2 OH−

The reaction equation is balanced with respect to charge by adding the suitable number of electrons to the side with the excess positive charge as follows,

Co(OH)2+ 2 H2O+ 2 e−→ Co+ 2 H2O+ 2 OH−

The overall reaction can be balanced as follows,

SO32−+ H2O + 2 OH−→SO42−+ 2 H2O+ 2e−Co(OH)2+ 2 H2O+ 2 e−→ Co+ 2 H2O+ 2 OH−Co(OH)2+ SO32−→ SO42−+ H2O+ Co

The balanced reaction is Co(OH)2+ SO32−→ SO42−+ H2O+ Co.

Answer 9

(a)

Expert Solution

Answer 10

(a)

Expert Solution

Answer 11

Expert Solution

Answer 12

(b)

Interpretation Introduction

Interpretation:

The following reaction that occurs in acidic solution has to be balanced.

IO3−+ I−→ I3−

Concept introduction:

Refer to part (a).

(b)

Expert Solution

Explanation of Solution

The given reaction can be written as two half reactions as follows,

The oxidation half reaction is I−→ I3−.

The oxidation half reaction can be balanced using above procedure.

The balanced oxidation half reaction with respect to all atoms other than oxygen and hydrogen is given below.

3 I−→ I3−

The given reaction is balanced with respect to oxygen atom by adding suitable number of H2O molecules to the side where oxygen atom is deficient as follows,

3 I−→ I3−

The above reaction is balanced with respect to hydrogen atom by adding the suitable number of H+ ions to the side deficient in hydrogen atoms as below.

3 I−→ I3−

The reaction equation is balanced with respect to charge by adding the suitable number of electrons to the side with the excess positive charge as follows,

3 I−→ I3−+ 2 e−

The reduction half reaction is IO3−→ I3−.

The reduction half reaction can be balanced using above procedure.

The balanced reduction half reaction with respect to all atoms other than oxygen and hydrogen is given below.

3 IO3−→ I3−

The given reaction is balanced with respect to oxygen atom by adding suitable number of H2O molecules to the side where oxygen atom is deficient as follows,

3 IO3−→ I3−+ 9 H2O

The above reaction is balanced with respect to hydrogen atom by adding the suitable number of H+ ions to the side deficient in hydrogen atoms as below.

3 IO3−+ 18 H+→ I3−+ 9 H2O

The reaction equation is balanced with respect to charge by adding the suitable number of electrons to the side with the excess positive charge as follows,

3 IO3−+ 18 H++ 16 e−→ I3−+ 9 H2O

The overall reaction can be balanced as follows,

3 I−→ I3−+ 2 e−]×83 IO3−+ 18 H++ 16 e−→ I3−+ 9 H2OIO3−+ 8 I−+ 6 H+→ 3 I3−+ 3 H2O

The balanced reaction is IO3−+ 8 I−+ 6 H+→ 3 I3−+ 3 H2O.

Answer 13

(b)

Interpretation Introduction

Interpretation:

The following reaction that occurs in acidic solution has to be balanced.

IO3−+ I−→ I3−

Concept introduction:

Refer to part (a).

Answer 14

(b)

Expert Solution

Explanation of Solution

The given reaction can be written as two half reactions as follows,

The oxidation half reaction is I−→ I3−.

The oxidation half reaction can be balanced using above procedure.

The balanced oxidation half reaction with respect to all atoms other than oxygen and hydrogen is given below.

3 I−→ I3−

The given reaction is balanced with respect to oxygen atom by adding suitable number of H2O molecules to the side where oxygen atom is deficient as follows,

3 I−→ I3−

The above reaction is balanced with respect to hydrogen atom by adding the suitable number of H+ ions to the side deficient in hydrogen atoms as below.

3 I−→ I3−

The reaction equation is balanced with respect to charge by adding the suitable number of electrons to the side with the excess positive charge as follows,

3 I−→ I3−+ 2 e−

The reduction half reaction is IO3−→ I3−.

The reduction half reaction can be balanced using above procedure.

The balanced reduction half reaction with respect to all atoms other than oxygen and hydrogen is given below.

3 IO3−→ I3−

The given reaction is balanced with respect to oxygen atom by adding suitable number of H2O molecules to the side where oxygen atom is deficient as follows,

3 IO3−→ I3−+ 9 H2O

The above reaction is balanced with respect to hydrogen atom by adding the suitable number of H+ ions to the side deficient in hydrogen atoms as below.

3 IO3−+ 18 H+→ I3−+ 9 H2O

The reaction equation is balanced with respect to charge by adding the suitable number of electrons to the side with the excess positive charge as follows,

3 IO3−+ 18 H++ 16 e−→ I3−+ 9 H2O