Chapter 19, Problem 6PS

a)

Interpretation Introduction

Interpretation:

The given redox equation in basic solution has to be balanced.

${\text{a) Fe(OH)}}_{\text{3}}\text{(s) + Cr(s) }\to {\text{ Cr(OH)}}_{\text{3}}{\text{(s) + Fe(OH)}}_{\text{2}}\text{(s)}$

Concept introduction:

Steps for balancing half –reactions in BASIC solution:

1. 1. Recognize the reaction as an oxidation and reduction reaction.
2. 2. Separate two half reactions.
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.

4. 4. Balance the charge by adding electrons to side with more total positive charge.
5. 5. Multiply the half reactions by appropriate factors.
6. 6. Add two half reactions and cancel the common atoms.
7. 7. Simplify by eliminating reactants and products that appears on both sides.

Answer to Problem 6PS

${\text{3Fe(OH)}}_{\text{3}}\text{(s) + Cr(s) }\to {\text{3Fe(OH)}}_{\text{2}}{\text{(s) + Cr(OH)}}_{\text{3}}\text{(s)}$

Explanation of Solution

The given reaction is s follows.

${\text{Fe(OH)}}_{\text{3}}\text{(s) + Cr(s) }\to {\text{Cr(OH)}}_3{\text{(s) + Fe(OH)}}_{\text{2}}\text{(s)}$

Oxidation states:

$\begin{array}{l}{\text{Fe(OH)}}_{\text{3}}{\text{ Cr(OH)}}_{\text{3}}{\text{ Fe(OH)}}_{\text{2}}\\ \text{x + 3(-2+1)= 0 x + 3(-2+1)= 0 x+2(-2+1)= 0}\\ \text{x + 3(-1)= 0 x +3(-1) = 0 x+2(-1)=0}\\ \text{x = +3 x = +3 x=+2}\end{array}$

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 0 to +3. Therefore, it is an oxidation state.

   Fe decrease their oxidation state +3 to +2.Therefore, it is a reduction reaction.

2. 2. Separate two half reactions.

   $\begin{array}{l}\text{Oxidation: Cr(s) }\to {\text{ Cr(OH)}}_{\text{3}}\text{(s)}\\ {\text{Reduction:Fe(OH)}}_{\text{3}}\text{(s) }\to {\text{Fe(OH)}}_{\text{2}}\text{(s)}\end{array}$

3. 3.  Balance half reactions for mass

   Balance all atoms except H and O in half reaction.

   $\begin{array}{l}\text{Oxidation: Cr(s) }\to {\text{ Cr(OH)}}_{\text{3}}\text{(s)}\\ {\text{Reduction:Fe(OH)}}_{\text{3}}\text{(s) }\to {\text{Fe(OH)}}_{\text{2}}\text{(s)}\end{array}$

   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.

   $\begin{array}{l}{\text{Oxidation: Cr(s) +3OH}}^{\text{-}}\text{(aq)}\to {\text{ Cr(OH)}}_{\text{3}}\text{(s)}\\ {\text{Reduction:Fe(OH)}}_{\text{3}}\text{(s) }\to {\text{ Fe(OH)}}_{\text{2}}{\text{(s)+OH}}^{\text{-}}\text{(aq)}\end{array}$

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

   $\begin{array}{l}{\text{Oxidation: Cr(s) +3OH}}^{\text{-}}\text{(aq)}\to {\text{ Cr(OH)}}_{\text{3}}{\text{(s)+3e}}^{-}\\ {\text{Reduction:Fe(OH)}}_{\text{3}}{\text{(s) +e}}^{-}\to {\text{ Fe(OH)}}_{\text{2}}{\text{(s)+OH}}^{\text{-}}\text{(aq)}\end{array}$

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

   $\begin{array}{l}{\text{Oxidation: Cr(s) +3OH}}^{\text{-}}\text{(aq)}\to {\text{ Cr(OH)}}_{\text{3}}{\text{(s)+3e}}^{-}\\ {\text{Reduction:3[Fe(OH)}}_{\text{3}}{\text{(s) +e}}^{-}\to {\text{ Fe(OH)}}_{\text{2}}{\text{(s)+OH}}^{\text{-}}\text{(aq)]}\end{array}$

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

   $\begin{array}{l}{\text{Oxidation: Cr(s) +3OH}}^{\text{-}}\text{(aq)}\to {\text{ Cr(OH)}}_{\text{3}}{\text{(s)+3e}}^{-}\\ {\text{Reduction:3Fe(OH)}}_{\text{3}}{\text{(s) +3e}}^{-}\to {\text{ 3Fe(OH)}}_{\text{2}}{\text{(s)+ 3OH}}^{\text{-}}\text{(aq)}\\ \text{\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_}\\ {\text{3Fe(OH)}}_{\text{3}}\text{(s) + Cr(s) }\to {\text{3Fe(OH)}}_{\text{2}}{\text{(s) + Cr(OH)}}_{\text{3}}\text{(s)}\end{array}$

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

   ${\text{3Fe(OH)}}_{\text{3}}\text{(s) + Cr(s) }\to {\text{3Fe(OH)}}_{\text{2}}{\text{(s) + Cr(OH)}}_{\text{3}}\text{(s)}$

b)

Interpretation Introduction

Interpretation:

The given redox equation in basic solution has to be balanced.

${\text{b) NiO}}_{\text{2}}\text{(s) + Zn(s) }\to {\text{Ni(OH)}}_{\text{2}}{\text{(s) + Zn(OH)}}_{\text{2}}\text{(s)}$

Concept introduction:

Steps for balancing half –reactions in BASIC solution:

1. 1. Recognize the reaction as an oxidation and reduction reaction.
2. 2. Separate two half reactions.
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.

4. 4. Balance the charge by adding electrons to side with more total positive charge.
5. 5. Multiply the half reactions by appropriate factors.
6. 6. Add two half reactions and cancel the common atoms.
7. 7. Simplify by eliminating reactants and products that appears on both sides.

Answer to Problem 6PS

${\text{NiO}}_{\text{2}}{\text{(s) + Zn(s) + 2H}}_{\text{2}}\text{O(l) }\to {\text{Ni(OH)}}_{\text{2}}{\text{(s) + Zn(OH)}}_{\text{2}}\text{(s)}$

Explanation of Solution

The given reaction is as follows.

${\text{NiO}}_{\text{2}}\text{(s) + Zn(s) }\to {\text{ Ni(OH)}}_{\text{2}}{\text{(s) + Zn(OH)}}_{\text{2}}\text{(s)}$

Oxidation states:

$\begin{array}{l}{\text{NiO}}_{\text{2}}{\text{ Ni(OH)}}_{\text{2}}{\text{ Zn(OH)}}_{\text{2}}\\ \text{x + 2(-2) = 0 x+2(-2+1)=0 x+2(-2+1)=0}\\ \text{x- 4 = 0 x+2(-1)=0 x+2(-1)=0}\\ \text{x = +4 x = +2 x=+2}\end{array}$

Steps for balancing half –reactions in BASIC solution:

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

   From the given reaction Ni atom reduces their oxidation state +4 to +2.Therefore, it is a reduction reaction.

   Zinc atom increases their oxidation state 0 to +2. Therefore, it is an oxidation reaction.

2. 2. Separate two half reactions.

   $\begin{array}{l}\text{Oxidation: Zn(s) }\to {\text{ Zn(OH)}}_{\text{2}}\text{(s)}\\ {\text{Reduction: NiO}}_{\text{2}}\text{(s) }\to {\text{ Ni(OH)}}_{\text{2}}\text{(s)}\end{array}$

3. 3.  Balance half reactions for mass

   Balance all atoms except H and O in half reaction.

   $\begin{array}{l}\text{Oxidation: Zn(s) }\to {\text{ Zn(OH)}}_{\text{2}}\text{(s)}\\ {\text{Reduction: NiO}}_{\text{2}}\text{(s) }\to {\text{ Ni(OH)}}_{\text{2}}\text{(s)}\end{array}$

   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.

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

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

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

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

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

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

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

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

            ${\text{NiO}}_{\text{2}}{\text{(s) + Zn(s) + 2H}}_{\text{2}}\text{O(l) }\to {\text{Ni(OH)}}_{\text{2}}{\text{(s) + Zn(OH)}}_{\text{2}}\text{(s)}$

c)

Interpretation Introduction

Interpretation:

The given redox equation in basic solution has to be balanced.

${\text{c) Fe(OH)}}_{\text{2}}{\text{(s) + CrO}}_{\text{4}}^{\text{2-}}\text{(aq) }\to {\text{Fe(OH)}}_{\text{3}}{\text{(s) + [Cr(OH)}}_{\text{4}}{\text{]}}^{\text{-}}\text{(aq)}$

Concept introduction:

Steps for balancing half –reactions in BASIC solution:

1. 1. Recognize the reaction as an oxidation and reduction reaction.
2. 2. Separate two half reactions.
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.

4. 4. Balance the charge by adding electrons to side with more total positive charge.
5. 5. Multiply the half reactions by appropriate factors.
6. 6. Add two half reactions and cancel the common atoms.
7. 7. Simplify by eliminating reactants and products that appears on both sides.

Answer to Problem 6PS

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

Explanation of Solution

The given reaction is as follows.

${\text{Fe(OH)}}_{\text{2}}{\text{(s) + CrO}}_{\text{4}}^{\text{2-}}\text{(aq) }\to {\text{ Fe(OH)}}_{\text{3}}{\text{(s) + [Cr(OH)}}_{\text{4}}{\text{]}}^{\text{-}}(aq)$

Oxidation states:

$\begin{array}{l}{\text{Fe(OH)}}_{\text{2}}{\text{ CrO}}_{\text{4}}^{\text{2-}}{\text{ Fe(OH)}}_{\text{3}}{\text{ Cr(OH)}}_{\text{4}}^{\text{-}}\\ \text{x + (-2+1)= 0 x+4(-2) = -2 x+3(-2+1)=0 x+4(-2+1)=-1}\\ \text{x+(-1)= 0 x-8 = -2 x+3(-1)=0 x+4(-1)=-1}\\ \text{x = +1 x = +6 x=+3 x=+3}\end{array}$

Steps for balancing half –reactions in BASIC solution:

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

   From the given reaction, Fe increases their oxidation state +1 to +3.Therefore, it is an oxidation reaction.

   Cr atom decrease their oxidation state +6 to +3.Therefore, it is a reduction reaction.

2. 2. Separate two half reactions.

   $\begin{array}{l}{\text{Oxidation: Fe(OH)}}_{\text{2}}\text{(s) }\to {\text{Fe(OH)}}_{\text{3}}\text{(s)}\\ {\text{Reduction: CrO}}_{\text{4}}^{\text{2-}}\text{(aq) }\to {\text{[Cr(OH)}}_{\text{4}}^{\text{-}}\text{]}\end{array}$

3. 3.  Balance half reactions by mass

   Balance all atoms except H and O in half reaction.

   $\begin{array}{l}{\text{Oxidation: Fe(OH)}}_{\text{2}}\text{(s) }\to {\text{Fe(OH)}}_{\text{3}}\text{(s)}\\ {\text{Reduction: CrO}}_{\text{4}}^{\text{2-}}\text{(aq) }\to {\text{[Cr(OH)}}_{\text{4}}^{\text{-}}\text{]}\end{array}$

   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.

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

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

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

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

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

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

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

7. 7. Simplify by eliminating reactants and products that appears on both sides.${\text{3Fe(OH)}}_{\text{2}}{\text{(s) + CrO}}_{\text{4}}^{\text{2-}}{\text{(aq) + 4H}}_{\text{2}}\text{O(l) }\to {\text{ 3Fe(OH)}}_{\text{3}}{\text{(s) + [Cr(OH)}}_{\text{4}}{\text{]}}^{\text{-}}{\text{ + OH}}^{\text{-}}\text{(aq)}$

d)

Interpretation Introduction

Interpretation:

The given redox equation in basic solution has to be balanced.

${\text{d) N}}_{\text{2}}{\text{H}}_{\text{4}}{\text{(aq) + Ag}}_{\text{2}}\text{O(s) }\to {\text{N}}_{\text{2}}\text{(g) + Ag(s)}$

Concept introduction:

Steps for balancing half –reactions in BASIC solution:

1. 1. Recognize the reaction as an oxidation and reduction reaction.
2. 2. Separate two half reactions.
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.

4. 4. Balance the charge by adding electrons to side with more total positive charge.
5. 5. Multiply the half reactions by appropriate factors.
6. 6. Add two half reactions and cancel the common atoms.
7. 7. Simplify by eliminating reactants and products that appears on both sides.

Answer to Problem 6PS

${\text{2Ag}}_{\text{2}}{\text{O(s)+N}}_{\text{2}}{\text{H}}_{\text{4}}\text{(aq) }\to {\text{4Ag(s)+N}}_{\text{2}}{\text{(g)+2H}}_{\text{2}}\text{O(l)}$

Explanation of Solution

The given reaction is as follows.

${\text{N}}_{\text{2}}{\text{H}}_4{\text{(aq) + Ag}}_{\text{2}}\text{O(s) }\to {\text{N}}_{\text{2}}\text{(g) + Ag(s)}$

Oxidation states:

$\begin{array}{l}{\text{N}}_{\text{2}}{\text{H}}_4{\text{ Ag}}_{\text{2}}\text{O}\\ \text{2x + 4(1)=0 2x + (-2)=0}\\ \text{2x +4=0 2x-2=0}\\ \text{x = -2 x = +1}\end{array}$

Steps for balancing half –reactions in BASIC solution:

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

   N atoms increases their oxidation state -2 to 0.Therefroe, it is an oxidation reaction.

   Ag atom decreases their oxidation state+1 to 0.Therefore, it is a reduction reaction.

2. 2. Separate two half reactions.

   $\begin{array}{l}{\text{Oxidation: N}}_{\text{2}}{\text{H}}_4\text{(aq) }\to {\text{ N}}_{\text{2}}\text{(g)}\\ {\text{Reduction: Ag}}_{\text{2}}\text{O(aq) }\to \text{Ag(aq)}\end{array}$

3. 3.  Balance half reactions for mass.

   Balance all atoms except H and O in half reaction.

   $\begin{array}{l}{\text{Oxidation: N}}_{\text{2}}{\text{H}}_4\text{(aq) }\to {\text{ N}}_{\text{2}}\text{(g)}\\ {\text{Reduction: Ag}}_{\text{2}}\text{O(aq) }\to 2\text{Ag(aq)}\end{array}$

   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.

   $\begin{array}{l}{\text{Oxidation: N}}_{\text{2}}{\text{H}}_{\text{4}}{\text{(aq) + 4OH}}^{\text{-}}\to {\text{ N}}_{\text{2}}{\text{(g) +4H}}_{\text{2}}\text{O(l)}\\ {\text{Reduction: Ag}}_{\text{2}}{\text{O(aq) +H}}_{\text{2}}\text{O }\to {\text{2Ag(aq) + 2OH}}^{\text{-}}(aq)\end{array}$

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

   $\begin{array}{l}{\text{Oxidation: N}}_{\text{2}}{\text{H}}_{\text{4}}{\text{(aq) + 4OH}}^{\text{-}}\to {\text{ N}}_{\text{2}}{\text{(g) +4H}}_{\text{2}}{\text{O(l) + 4e}}^{-}\\ {\text{Reduction: Ag}}_{\text{2}}{\text{O(aq) +H}}_{\text{2}}{\text{O(l)+2e}}^{-}\to {\text{2Ag(aq) + 2OH}}^{\text{-}}\end{array}$

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

   $\begin{array}{l}{\text{Oxidation: N}}_{\text{2}}{\text{H}}_{\text{4}}{\text{(aq) + 4OH}}^{\text{-}}\to {\text{ N}}_{\text{2}}{\text{(g) +4H}}_{\text{2}}{\text{O(l) + 4e}}^{-}\\ {\text{Reduction: 2[Ag}}_{\text{2}}{\text{O(aq) +H}}_{\text{2}}{\text{O(l)+2e}}^{-}\to {\text{2Ag(aq) + 2OH}}^{\text{-}}]\end{array}$

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

   $\begin{array}{l}{\text{Oxidation: N}}_{\text{2}}{\text{H}}_{\text{4}}{\text{(aq) + 4OH}}^{\text{-}}\to {\text{N}}_{\text{2}}{\text{(g) +4H}}_{\text{2}}{\text{O(l) + 4e}}^{\text{-}}\\ {\text{Reduction: 2Ag}}_{\text{2}}{\text{O(aq) + 2H}}_{\text{2}}{\text{O(l)+4e}}^{\text{-}}\to {\text{4Ag(aq) + 4OH}}^{\text{-}}\\ \text{\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_}\\ {\text{2Ag}}_{\text{2}}{\text{O(s)+N}}_{\text{2}}{\text{H}}_{\text{4}}\text{(aq) }\to {\text{4Ag(s)+N}}_{\text{2}}{\text{(g)+2H}}_{\text{2}}\text{O(l)}\end{array}$

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

   ${\text{2Ag}}_{\text{2}}{\text{O(s)+N}}_{\text{2}}{\text{H}}_{\text{4}}\text{(aq) }\to {\text{4Ag(s)+N}}_{\text{2}}{\text{(g)+2H}}_{\text{2}}\text{O(l)}$