Chapter 19, Problem 58GQ

Balance the following equations.

1. (a) Zn(s) + VO²⁺(aq) → Zn²⁺(aq) + V³⁺(aq)⇄(acid solution)
2. (b) Zn(s) + VO₃⁻(aq) → V²⁺(aq) + Zn²⁺(aq)⇄(acid solution)
3. (c) Zn(s) + ClO⁻(aq) → Zn(OH)₂(s) + Cl⁻(aq)⇄(basic solution)
4. (d) ClO⁻(aq) + [Cr(OH)₄]⁻(aq) → Cl⁻(aq) + Cr₂O4²⁻(aq)⇄(basic solution)

Interpretation Introduction

Interpretation:

The following half reaction in acidic solution has to be identified.

${\text{(a) Zn(s) + VO}}^{\text{2+}}\text{(aq) }\to {\text{Zn}}^{\text{2+}}{\text{(aq) + V}}^{\text{3+}}\text{(aq) (acid solution)}$

Concept introduction:

Steps for balancing half –reactions in ACIDIC solution:

1. 1. Balance all atoms except H and O in half reaction.
2. 2. Balance O atoms by adding water to the side missing O atoms.
3. 3. Balance the H atoms by adding ${\text{H}}^{\text{+}}$ to the side missing H atoms.
4. 4. Balance the charge by adding electrons to side with more total positive charge.
5. 5. Make the number of electrons the same in both half reactions by multiplication, while avoiding fractional number of electrons.

Steps for balancing half –reactions in BASIC solution:

1. 1. Balance all atoms except H and O in half reaction.
2. 2. Balance O atoms by adding water to the side missing O atoms.
3. 3. Balance the H atoms by adding ${\text{H}}^{\text{+}}$ to the side missing H atoms.
4. 4. Balance the charge by adding electrons to side with more total positive charge.
5. 5. Make the number of electrons the same in both half reactions by multiplication, while avoiding fractional number of electrons.
6. 6. Add the same number of ${\text{OH}}^{\text{-}}$ groups as there are ${\text{H}}^{\text{+}}$ present to both sides of the equation.

Answer to Problem 58GQ

${\text{Zn(s) + 2VO}}^{\text{2+}}{\text{(aq) + 4H}}^{\text{+}}\text{(aq) }\to {\text{ Zn}}^{\text{2+}}{\text{(aq) + 2V}}^{\text{3+}}{\text{(aq) + 2H}}_{\text{2}}\text{O(l)}$

Explanation of Solution

Given reaction:

${\text{Zn(s) + VO}}^{\text{2+}}\text{(aq) }\to {\text{Zn}}^{\text{2+}}{\text{(aq) + V}}^{\text{3+}}\text{(aq) }$

Oxidation state:

$\begin{array}{l}{\text{VO}}^{\text{2+}}\\ \text{x+(-2) = +2}\\ \text{x-2 = +2}\\ \text{x = 4}\end{array}$

Steps for balancing redox reactions in ACIDIC solution:

1. 1. Recognize the reaction as an oxidation and reduction reaction.

   The oxidation state of Zn atom increases 0 to 2+.Therefore, it is a oxidation reaction.

   The oxidation state of vanadium reduces to +4 to +3. Therefore, it is a reduction reaction.

2. 2. Separate two half reactions.

   $\begin{array}{l}\text{oxidation: Zn(s) }\to {\text{ Zn}}^{\text{2+}}\text{(aq)}\\ {\text{Reduction: VO}}^{2+}\text{(aq) }\to {\text{V}}^{\text{3+}}\text{(aq)}\end{array}$

3. 3.  Balance half reactions by mass

   Balance all atoms except H and O in half reaction.

   $\begin{array}{l}\text{oxidation: Zn(s) }\to {\text{ Zn}}^{\text{2+}}\text{(aq)}\\ {\text{Reduction: VO}}^{2+}\text{(aq) }\to {\text{V}}^{\text{3+}}\text{(aq)}\end{array}$

   Balance O atoms by adding water to the side missing O atoms.

   $\begin{array}{l}\text{oxidation: Zn(s) }\to {\text{ Zn}}^{\text{2+}}\text{(aq)}\\ {\text{Reduction: VO}}^{2+}\text{(aq) }\to {\text{V}}^{\text{3+}}{\text{(aq) +2H}}_{\text{2}}\text{O(l)}\end{array}$

   Balance the H atoms by adding ${\text{H}}^{\text{+}}$ to the side missing H atoms.

   $\begin{array}{l}\text{oxidation: Zn(s) }\to {\text{ Zn}}^{\text{2+}}\text{(aq)}\\ {\text{Reduction: VO}}^{2+}{\text{(aq) + 2H}}^{+}(aq)\to {\text{V}}^{\text{3+}}{\text{(aq) +H}}_{\text{2}}\text{O(l)}\end{array}$

4. 4. Balance the charge by adding electrons to side with more total positive charge.

   $\begin{array}{l}\text{oxidation: Zn(s) }\to {\text{ Zn}}^{\text{2+}}{\text{(aq)+2e}}^{-}\\ {\text{Reduction: VO}}^{2+}{\text{(aq) + 2H}}^{+}(aq)\to {\text{V}}^{\text{3+}}{\text{(aq) +H}}_{\text{2}}{\text{O(l)+e}}^{-}\end{array}$

5. 5. Make the number of electrons the same in both half reactions by multiplication, while avoiding fractional number of electrons.

   $\begin{array}{l}\text{oxidation: Zn(s) }\to {\text{ Zn}}^{\text{2+}}{\text{(aq)+2e}}^{-}\\ {\text{Reduction: 2(VO}}^{2+}{\text{(aq) + 2H}}^{+}(aq)\to {\text{V}}^{\text{3+}}{\text{(aq) + H}}_{\text{2}}{\text{O(l)+e}}^{-})\end{array}$

6. 6.  Add the half reaction together and than simplyfy by cancelling out species that show up on both sides.

   $\begin{array}{l}\text{oxidation: Zn(s) }\to {\text{ Zn}}^{\text{2+}}{\text{(aq)+2e}}^{-}\\ {\text{Reduction: 2VO}}^{2+}{\text{(aq) + 4H}}^{+}(aq)\to {\text{V}}^{\text{3+}}{\text{(aq) + 2H}}_{\text{2}}{\text{O(l)+2e}}^{-}\\ \_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\\ {\text{Zn(s) + 2VO}}^{\text{2+}}{\text{(aq) + 4H}}^{\text{+}}\text{(aq) }\to {\text{ Zn}}^{\text{2+}}{\text{(aq) + 2V}}^{\text{3+}}{\text{(aq) + 2H}}_{\text{2}}\text{O(l)}\end{array}$

7. 7. Confirm that the reaction is balanced in number of atoms and charge on the both sides of the arrow.

             ${\text{Zn(s) + 2VO}}^{\text{2+}}{\text{(aq) + 4H}}^{\text{+}}\text{(aq) }\to {\text{ Zn}}^{\text{2+}}{\text{(aq) + 2V}}^{\text{3+}}{\text{(aq) + 2H}}_{\text{2}}\text{O(l)}$

Interpretation Introduction

Interpretation:

The following half reaction in acidic solution has to be identified.

${\text{Zn(s) + VO}}_{\text{3}}^{\text{-}}\text{(aq) }\to {\text{V}}^{\text{2+}}{\text{(aq) + Zn}}^{\text{2+}}\text{(aq)}$

Concept introduction:

Steps for balancing half –reactions in ACIDIC solution:

1. 6. Balance all atoms except H and O in half reaction.
2. 7. Balance O atoms by adding water to the side missing O atoms.
3. 8. Balance the H atoms by adding ${\text{H}}^{\text{+}}$ to the side missing H atoms.
4. 9. Balance the charge by adding electrons to side with more total positive charge.
5. 10. Make the number of electrons the same in both half reactions by multiplication, while avoiding fractional number of electrons.

Steps for balancing half –reactions in BASIC solution:

1. 7. Balance all atoms except H and O in half reaction.
2. 8. Balance O atoms by adding water to the side missing O atoms.
3. 9. Balance the H atoms by adding ${\text{H}}^{\text{+}}$ to the side missing H atoms.
4. 10. Balance the charge by adding electrons to side with more total positive charge.
5. 11. Make the number of electrons the same in both half reactions by multiplication, while avoiding fractional number of electrons.
6. 12. Add the same number of ${\text{OH}}^{\text{-}}$ groups as there are ${\text{H}}^{\text{+}}$ present to both sides of the equation.

Answer to Problem 58GQ

${\text{3Zn(s) + 2VO}}_{\text{3}}^{\text{-}}{\text{(aq) + 12H}}^{\text{+}}\text{ (aq) }\to {\text{3Zn}}^{2+}{\text{(aq) + 2V}}^{\text{2+}}{\text{(aq) + 6H}}_{\text{2}}\text{O(l)}$

Explanation of Solution

The given redox equation is as follows.

${\text{Zn(s) + VO}}_{\text{3}}^{\text{-}}\text{(aq) }\to {\text{V}}^{\text{2+}}{\text{(aq) + Zn}}^{\text{2+}}\text{(aq)}$

Oxidation states:

$\begin{array}{l}{\text{VO}}_{\text{3}}^{\text{-}}\\ \text{x+3(-2) = 0}\\ \text{x-6 = 0}\\ \text{x = +6}\end{array}$

1. 1. Recognize the reaction as an oxidation and reduction reaction.

   The oxidation state of Zn atom increases 0 to 2+.Therefore, it is a oxidation reaction.

   The oxidation state of vanadium reduces to +6 to +2. Therefore, it is a reduction reaction.

2. 2. Separate two half reactions.

   $\begin{array}{l}\text{oxidation: Zn(s) }\to {\text{ Zn}}^{\text{2+}}\text{(aq)}\\ {\text{Reduction: VO}}_3^{-}\text{(aq) }\to {\text{V}}^{\text{2+}}\text{(aq)}\end{array}$

3. 3.  Balance half reactions by mass

   Balance all atoms except H and O in half reaction.

   $\begin{array}{l}\text{oxidation: Zn(s) }\to {\text{ Zn}}^{\text{2+}}\text{(aq)}\\ {\text{Reduction: VO}}_3^{-}\text{(aq) }\to {\text{V}}^{\text{2+}}\text{(aq)}\end{array}$

   Balance O atoms by adding water to the side missing O atoms.

   $\begin{array}{l}\text{oxidation: Zn(s) }\to {\text{ Zn}}^{\text{2+}}\text{(aq)}\\ {\text{Reduction: VO}}_3^{-}\text{(aq) }\to {\text{V}}^{\text{2+}}{\text{(aq) +3H}}_{\text{2}}\text{O(l)}\end{array}$

   Balance the H atoms by adding ${\text{H}}^{\text{+}}$ to the side missing H atoms.

   $\begin{array}{l}\text{oxidation: Zn(s) }\to {\text{ Zn}}^{\text{2+}}\text{(aq)}\\ {\text{Reduction: VO}}_3^{-}{\text{(aq) +6H}}^{+}\text{(aq)}\to {\text{V}}^{\text{2+}}{\text{(aq) +3H}}_{\text{2}}\text{O(l)}\end{array}$

4. 4. Balance the charge by adding electrons to side with more total positive charge.

   $\begin{array}{l}\text{oxidation: Zn(s) }\to {\text{ Zn}}^{\text{2+}}{\text{(aq)+2e}}^{-}\\ {\text{Reduction: VO}}_3^{-}{\text{(aq) +6H}}^{+}\text{(aq)}\to {\text{V}}^{\text{2+}}{\text{(aq) +3H}}_{\text{2}}{\text{O(l)+1e}}^{-}\end{array}$

5. 5. Make the number of electrons the same in both half reactions by multiplication, while avoiding fractional number of electrons.

   $\begin{array}{l}\text{oxidation: 3(Zn(s) }\to {\text{ Zn}}^{\text{2+}}{\text{(aq)+2e}}^{-})\\ {\text{Reduction: 2(VO}}_3^{-}{\text{(aq) +6H}}^{+}\text{(aq)}\to {\text{V}}^{\text{2+}}{\text{(aq) +3H}}_{\text{2}}{\text{O(l)+3e}}^{-})\end{array}$

6. 6.  Add the half reaction together and than simplyfy by cancelling out species that show up on both sides.

   $\begin{array}{l}\text{oxidation: 3Zn(s) }\to {\text{ 3Zn}}^{\text{2+}}{\text{(aq) +6e}}^{-}\\ {\text{Reduction: 2VO}}_3^{-}{\text{(aq) +12H}}^{+}\text{(aq)}\to 2{\text{V}}^{\text{2+}}{\text{(aq) +6H}}_{\text{2}}{\text{O(l)+6e}}^{-}\\ \_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\\ {\text{3Zn(s) + 2VO}}_{\text{3}}^{\text{-}}{\text{(aq) + 12H}}^{\text{+}}\text{ (aq) }\to {\text{3Zn}}^{2+}{\text{(aq) + 2V}}^{\text{2+}}{\text{(aq) + 6H}}_{\text{2}}\text{O(l)}\end{array}$

7. 7. Confirm that the reaction is balanced in number of atoms and charge on the both sides of the arrow.

             ${\text{3Zn(s) + 2VO}}_{\text{3}}^{\text{-}}{\text{(aq) + 12H}}^{\text{+}}\text{ (aq) }\to {\text{3Zn}}^{2+}{\text{(aq) + 2V}}^{\text{2+}}{\text{(aq) + 6H}}_{\text{2}}\text{O(l)}$

Interpretation Introduction

Interpretation:

The following half reaction in basic solution has to be identified.

${\text{Zn(s) + ClO}}^{\text{-}}\text{(aq) }\to {\text{Zn(OH)}}_{\text{2}}{\text{(s) + Cl}}^{\text{-}}$

Concept introduction:

Steps for balancing half –reactions in ACIDIC solution:

1. 11. Balance all atoms except H and O in half reaction.
2. 12. Balance O atoms by adding water to the side missing O atoms.
3. 13. Balance the H atoms by adding ${\text{H}}^{\text{+}}$ to the side missing H atoms.
4. 14. Balance the charge by adding electrons to side with more total positive charge.
5. 15. Make the number of electrons the same in both half reactions by multiplication, while avoiding fractional number of electrons.

Steps for balancing half –reactions in BASIC solution:

1. 13. Balance all atoms except H and O in half reaction.
2. 14. Balance O atoms by adding water to the side missing O atoms.
3. 15. Balance the H atoms by adding ${\text{H}}^{\text{+}}$ to the side missing H atoms.
4. 16. Balance the charge by adding electrons to side with more total positive charge.
5. 17. Make the number of electrons the same in both half reactions by multiplication, while avoiding fractional number of electrons.
6. 18. Add the same number of ${\text{OH}}^{\text{-}}$ groups as there are ${\text{H}}^{\text{+}}$ present to both sides of the equation.

Answer to Problem 58GQ

${\text{Zn(s) + ClO}}^{\text{-}}{\text{(aq) + H}}_{\text{2}}\text{O }\to {\text{ Zn(OH)}}_{\text{2}}{\text{(s) + Cl}}^{\text{-}}\text{(aq)}$

Explanation of Solution

The given reaction is as follows.

${\text{Zn(s) + ClO}}^{\text{-}}\text{(aq) }\to {\text{Zn(OH)}}_{\text{2}}{\text{(s) + Cl}}^{\text{-}}$

Oxidation states:

$\begin{array}{l}{\text{ClO}}^{-}{\text{ Zn(OH)}}_2\\ \text{x + 1(-2) = -1 x + 2(-2}\times \text{1) =0}\\ \text{x - 2 = -1 x -4 = 0}\\ \text{x = -1+2 =+1 x = +4}\end{array}$  

Steps for balancing half –reactions in BASIC solution:

1. 1. Recognize the reaction as an oxidation and reduction reaction.

   From the given reaction Cl atom reduces their oxidation state +1 to -1.Therefore, it is a reduction reaction.

   Zinc atom increases their oxidation state 0 to +4. Therefore, it is an oxidation reaction.

2. 2. Separate two half reactions.

   $\begin{array}{l}\text{Oxidation: Zn }\to {\text{ Zn(OH)}}_2\\ {\text{Reduction ClO}}^{-}\to {\text{Cl}}^{\text{-}}\end{array}$

3. 3.  Balance half reactions for mass

   Balance all atoms except H and O in half reaction.

   $\begin{array}{l}\text{Oxidation: Zn }\to {\text{ Zn(OH)}}_2\\ {\text{Reduction ClO}}^{-}\to {\text{Cl}}^{\text{-}}\end{array}$

   Addition of ${\text{OH}}^{\text{-}}$ or ${\text{OH}}^{\text{-}}{\text{ and H}}_{\text{2}}\text{O}$ is required for mass balance in both half reactions.

   $\begin{array}{l}{\text{Oxidation: Zn +H}}_{\text{2}}{\text{O +OH}}^{-}\to {\text{ Zn(OH)}}_2\\ {\text{Reduction ClO}}^{-}\to {\text{Cl}}^{\text{-}}+{\text{OH}}^{\text{-}}\end{array}$

4. 4. Balance the charge by adding electrons to side with more total positive charge.

   $\begin{array}{l}{\text{Oxidation: Zn +H}}_{\text{2}}{\text{O +OH}}^{-}\to {\text{ Zn(OH)}}_2\\ {\text{Reduction ClO}}^{-}\to {\text{Cl}}^{\text{-}}+{\text{OH}}^{\text{-}}\end{array}$

5. 5. Multiply the half reactions by appropriate factors.

   $\begin{array}{l}{\text{Oxidation: Zn +H}}_{\text{2}}{\text{O +OH}}^{-}\to {\text{ Zn(OH)}}_2\\ {\text{Reduction ClO}}^{-}\to {\text{Cl}}^{\text{-}}+{\text{OH}}^{\text{-}}\end{array}$

6. 6. Add two half reactions and cancel the common atoms.

   $\begin{array}{l}{\text{Oxidation: Zn +H}}_{\text{2}}{\text{O +OH}}^{-}\to {\text{ Zn(OH)}}_2\\ {\text{Reduction ClO}}^{-}\to {\text{Cl}}^{\text{-}}+{\text{OH}}^{\text{-}}\\ \_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\\ {\text{Zn(s) + ClO}}^{\text{-}}\text{(aq) }\to {\text{Zn(OH)}}_{\text{2}}{\text{(s) + Cl}}^{\text{-}}\end{array}$

7. 7. Simplify by eliminating reactants and products that appears on both sides.

            ${\text{Zn(s) + ClO}}^{\text{-}}{\text{(aq) + H}}_{\text{2}}\text{O }\to {\text{ Zn(OH)}}_{\text{2}}{\text{(s) + Cl}}^{\text{-}}\text{(aq)}$

Interpretation Introduction

Interpretation:

The following half reaction in basic solution has to be identified.

${\text{ClO}}^{\text{-}}{\text{(aq) + [Cr(OH)}}_{\text{4}}{\text{]}}^{\text{-}}\text{(aq) }\to {\text{Cl}}^{\text{-}}{\text{(aq) + CrO}}_{\text{4}}^{\text{2-}}$

Concept introduction:

Steps for balancing half –reactions in ACIDIC solution:

1. 16. Balance all atoms except H and O in half reaction.
2. 17. Balance O atoms by adding water to the side missing O atoms.
3. 18. Balance the H atoms by adding ${\text{H}}^{\text{+}}$ to the side missing H atoms.
4. 19. Balance the charge by adding electrons to side with more total positive charge.
5. 20. Make the number of electrons the same in both half reactions by multiplication, while avoiding fractional number of electrons.

Steps for balancing half –reactions in BASIC solution:

1. 19. Balance all atoms except H and O in half reaction.
2. 20. Balance O atoms by adding water to the side missing O atoms.
3. 21. Balance the H atoms by adding ${\text{H}}^{\text{+}}$ to the side missing H atoms.
4. 22. Balance the charge by adding electrons to side with more total positive charge.
5. 23. Make the number of electrons the same in both half reactions by multiplication, while avoiding fractional number of electrons.
6. 24. Add the same number of ${\text{OH}}^{\text{-}}$ groups as there are ${\text{H}}^{\text{+}}$ present to both sides of the equation.

Answer to Problem 58GQ

${\text{3ClO}}^{\text{-}}{\text{(aq) + 2[Cr(OH)}}_{\text{4}}{\text{]}}^{\text{-}}{\text{(aq) + 2OH}}^{\text{-}}\text{(aq) }\to {\text{3Cl}}^{\text{-}}{\text{(aq)+2CrO}}_{\text{4}}^{\text{2-}}{\text{(aq)+5H}}_{\text{2}}\text{O(l)}$

Explanation of Solution

The given reaction is as follows.

${\text{ClO}}^{\text{-}}{\text{(aq) + [Cr(OH)}}_{\text{4}}{\text{]}}^{\text{-}}\text{(aq) }\to {\text{Cl}}^{\text{-}}{\text{(aq) + CrO}}_{\text{4}}^{\text{2-}}$

Oxidation states:

   $\begin{array}{l}{\text{ClO}}^{\text{-}}\\ \text{x+1(-2)= -1}\\ \text{x-2=-1}\\ \text{x= -1+2= +1}\end{array}$         $\begin{array}{l}{\text{Cr(OH)}}_{\text{4}}^{\text{-}}\\ \text{x+4(-1)=-1}\\ \text{x-4= -1}\\ \text{x= +3}\end{array}$       $\begin{array}{l}{\text{CrO}}_{\text{4}}^{\text{2-}}\\ x+4(-2)=-2\\ x-8=-2\\ x=+6\end{array}$

Steps for balancing half –reactions in BASIC solution:

1. 1. Recognize the reaction as an oxidation and reduction reaction.

   From the given reaction, Cr increases their oxidation state +3 to +6.Therefore, it is an oxidation reaction.

   Chlorine atom decrease their oxidation state +1 to -1.Therefore, it is a reduction reaction.

2. 2. Separate two half reactions.

   $\begin{array}{l}{\text{Oxidation: [Cr(OH)}}_4{]}^{-}\to {\text{CrO}}_{\text{4}}^{\text{2-}}(aq)\\ {\text{Reduction: ClO}}^{-}\text{(aq) }\to {\text{ Cl}}^{-}\end{array}$

3. 3.  Balance half reactions by mass

   Balance all atoms except H and O in half reaction.

   Addition of ${\text{OH}}^{\text{-}}$ or ${\text{OH}}^{\text{-}}{\text{ and H}}_{\text{2}}\text{O}$ is required for mass balance in both half reactions.

   $\begin{array}{l}{\text{Oxidation: [Cr(OH)}}_4{]}^{-}+{\text{OH}}^{\text{-}}\to {\text{CrO}}_{\text{4}}^{\text{2-}}(aq)+{\text{4H}}_{\text{2}}\text{O}\\ {\text{Reduction: ClO}}^{-}\text{(aq) }\to {\text{ Cl}}^{-}{\text{ +H}}_{\text{2}}\text{O(l)}\end{array}$

4. 4. Balance the charge by adding electrons to side with more total positive charge.

   $\begin{array}{l}{\text{Oxidation: [Cr(OH)}}_4{]}^{-}+{\text{OH}}^{\text{-}}\to {\text{CrO}}_{\text{4}}^{\text{2-}}(aq)+{\text{4H}}_{\text{2}}\text{O}\\ {\text{Reduction: ClO}}^{-}\text{(aq) }\to {\text{ Cl}}^{-}{\text{ +H}}_{\text{2}}\text{O(l)}\end{array}$

5. 5. Multiply the half reactions by appropriate factors.

   $\begin{array}{l}{\text{Oxidation: 2([Cr(OH)}}_4{]}^{-}+{\text{OH}}^{\text{-}}\to {\text{CrO}}_{\text{4}}^{\text{2-}}(aq)+4H_{2}O)\\ {\text{Reduction: 3(ClO}}^{-}\text{(aq) }\to {\text{ Cl}}^{-}{\text{ +H}}_{\text{2}}\text{O(l))}\end{array}$

6. 6. Add two half reactions and cancel the common atoms.

   $\begin{array}{l}{\text{Oxidation: 2[Cr(OH)}}_4{]}^{-}+2{\text{OH}}^{\text{-}}\to 2{\text{CrO}}_{\text{4}}^{\text{2-}}(aq)+8H_{2}O\\ {\text{Reduction: 3ClO}}^{-}\text{(aq) }\to {\text{ 3Cl}}^{-}{\text{ +3H}}_{\text{2}}\text{O(l)}\\ \text{\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_}\\ {\text{3ClO}}^{\text{-}}{\text{(aq) + 2[Cr(OH)}}_{\text{4}}{\text{]}}^{\text{-}}{\text{(aq) + 2OH}}^{\text{-}}\text{(aq) }\to {\text{3Cl}}^{\text{-}}{\text{(aq)+2CrO}}_{\text{4}}^{\text{2-}}{\text{(aq)+5H}}_{\text{2}}\text{O(l)}\end{array}$

7. 7. Simplify by eliminating reactants and products that appears on both sides.

           ${\text{3ClO}}^{\text{-}}{\text{(aq) + 2[Cr(OH)}}_{\text{4}}{\text{]}}^{\text{-}}{\text{(aq) + 2OH}}^{\text{-}}\text{(aq) }\to {\text{3Cl}}^{\text{-}}{\text{(aq)+2CrO}}_{\text{4}}^{\text{2-}}{\text{(aq)+5H}}_{\text{2}}\text{O(l)}$