Chapter 20, Problem 44A

(a)

Interpretation Introduction

Interpretation: The given redox reaction $Al\left(s\right)+C{l}_2\left(g\right)\to AlC{l}_3\left(s\right)$ is to be balanced.

Concept Introduction: The chemical reaction that involves both oxidation and reduction is the redox reaction. Redox reactions can be balanced by following the half-reaction method. That is,

• First, the given redox reaction is divided into half reactions one oxidation and the other reduction.
• Based on mass and charge the half-reactions are balanced.
• The electrons on both sides of the reaction are equalized.
• Then add the half-reactions together and the common one is eliminated to get the balanced equation.

Answer to Problem 44A

The balanced equation is $2Al\left(s\right)+3C{l}_2\left(g\right)\to 2AlC{l}_3\left(s\right)$ .

Explanation of Solution

In the reaction,

  $Al\left(s\right)+C{l}_2\left(g\right)\to AlC{l}_3\left(s\right)$

The half-reaction involved in the reaction can be given as,

Oxidation: $Al\to A{l}^{3+}+3e^{-}$

Reduction: $C{l}_2+3e^{-}\to 2C{l}^{-}$

The half-reaction can be balanced by 2 to the oxidation and 3 to the reduction reaction.

The overall reaction after the cancellation of common species is,

  $2Al\left(s\right)+3C{l}_2\left(g\right)\to 2AlC{l}_3\left(s\right)$

(b)

Interpretation Introduction

Interpretation: The given redox reaction $Al\left(s\right)+F{e}_2{O}_3\left(s\right)\to A{l}_2{O}_3\left(s\right)+Fe\left(s\right)$ is to be balanced.

Concept Introduction: The chemical reaction that involves both oxidation and reduction is the redox reaction. Redox reactions can be balanced by following the half-reaction method. That is,

• First, the given redox reaction is divided into half reactions one oxidation and the other reduction.
• Based on mass and charge the half-reactions are balanced.
• The electrons on both sides of the reaction are equalized.
• Then add the half-reactions together and the common one is eliminated to get the balanced equation.

Answer to Problem 44A

The balanced equation is $2Al\left(s\right)+F{e}_2{O}_3\left(s\right)\to A{l}_2{O}_3\left(s\right)+2Fe\left(s\right)$ .

Explanation of Solution

In the reaction,

  $Al\left(s\right)+F{e}_2{O}_3\left(s\right)\to A{l}_2{O}_3\left(s\right)+Fe\left(s\right)$

The half-reactions can be given as,

  $Al\to A{l}^{3+}+3e^{-}$

  $F{e}_3+3e^{-}\to Fe$

The reaction can be balanced by adding 2 to Al on the reactant side and 2 to Fe on the product side.

Therefore, the balanced equation after cancellation is,

  $2Al\left(s\right)+F{e}_2{O}_3\left(s\right)\to A{l}_2{O}_3\left(s\right)+2Fe\left(s\right)$

(c)

Interpretation Introduction

Interpretation: The given redox reaction $KOH\left(aq\right)+C{l}_2\left(g\right)\to KCl{O}_3\left(aq\right)+KCl\left(aq\right)+H_{2}O$ is to be balanced.

Concept Introduction: The chemical reaction that involves both oxidation and reduction is the redox reaction. Redox reactions can be balanced by following the half-reaction method. That is,

• First, the given redox reaction is divided into half reactions one oxidation and the other reduction.
• Based on mass and charge the half-reactions are balanced.
• The electrons on both sides of the reaction are equalized.
• Then add the half-reactions together and the common one is eliminated to get the balanced equation.

Answer to Problem 44A

The balanced equation is $6KOH\left(aq\right)+3C{l}_2\left(g\right)\to KCl{O}_3\left(aq\right)+5KCl\left(aq\right)+3H_{2}O$ .

Explanation of Solution

In the reaction,

  $KOH\left(aq\right)+C{l}_2\left(g\right)\to KCl{O}_3\left(aq\right)+KCl\left(aq\right)+H_{2}O$

The elements in the reactant and product side must be balanced to get a balanced chemical equation.

The oxygen in the product and the reactant side are not equal. To balance them 6 is added to $KOH$ , 3 is added to $H_{2}O$ , 3 is added to $C{l}_2$ , and 5 is added to $KCl$ on the reactant side.

So, the balanced chemical equation is,

  $6KOH\left(aq\right)+3C{l}_2\left(g\right)\to KCl{O}_3\left(aq\right)+5KCl\left(aq\right)+3H_{2}O$

(d)

Interpretation Introduction

Interpretation: The given redox reaction $HN{O}_3\left(aq\right)+H_{2}S\left(g\right)\to S\left(s\right)+NO\left(g\right)+H_{2}O\left(s\right)$ is to be balanced.

Concept Introduction: The chemical reaction that involves both oxidation and reduction is the redox reaction. Redox reactions can be balanced by following the half-reaction method. That is,

• First, the given redox reaction is divided into half reactions one oxidation and the other reduction.
• Based on mass and charge the half-reactions are balanced.
• The electrons on both sides of the reaction are equalized.
• Then add the half-reactions together and the common one is eliminated to get the balanced equation.

Answer to Problem 44A

The balanced equation is $2HN{O}_3\left(aq\right)+3H_{2}S\left(g\right)\to 3S\left(s\right)+2NO\left(g\right)+4H_{2}O\left(s\right)$ .

Explanation of Solution

In the reaction,

  $HN{O}_3\left(aq\right)+H_{2}S\left(g\right)\to S\left(s\right)+NO\left(g\right)+H_{2}O\left(s\right)$

The half-reactions can be represented as

  $S^{2-}\to S+2e^{-}$

  $N^{5+}+3e^{-}\to N^{2+}$

The above half reaction can be balanced by multiplying the oxidation reaction by 2 and the reduction reaction by 3.

The overall reaction after cancellation is,

  $2HN{O}_3\left(aq\right)+3H_{2}S\left(g\right)\to 3S\left(s\right)+2NO\left(g\right)+4H_{2}O\left(s\right)$

(e)

Interpretation Introduction

Interpretation: The given redox reaction $KI{O}_4\left(aq\right)+KI\left(aq\right)+HCl\left(aq\right)\to KCl\left(aq\right)+I_2\left(s\right)+H_{2}O\left(l\right)$ is to be balanced.

Concept Introduction: The chemical reaction that involves both oxidation and reduction is the redox reaction. Redox reactions can be balanced by following the half-reaction method. That is,

• First, the given redox reaction is divided into half reactions one oxidation and the other reduction.
• Based on mass and charge the half-reactions are balanced.
• The electrons on both sides of the reaction are equalized.
• Then add the half-reactions together and the common one is eliminated to get the balanced equation.

Answer to Problem 44A

The balanced equation is $KI{O}_4\left(aq\right)+7KI\left(aq\right)+8HCl\left(aq\right)\to 8KCl\left(aq\right)+4I_2\left(s\right)+4H_{2}O\left(l\right)$

Explanation of Solution

In the reaction,

  $KI{O}_4\left(aq\right)+KI\left(aq\right)+HCl\left(aq\right)\to KCl\left(aq\right)+I_2\left(s\right)+H_{2}O\left(l\right)$

The number of oxygen in reactants and products is not equal. It can be balanced by adding 4 to $H_{2}O$ on the product side. Now the number of H is not balanced adding 8 to $HCl$ in the reactant will make it balanced. The Cl in the product must also be balanced by adding 8 to $KCl$ on the product side. Now the number of potassium can be balanced by adding 7 to KI.

Then, I can be balanced by adding 4 to $I_2$ on the product side. So, the balanced equation becomes,

  $KI{O}_4\left(aq\right)+7KI\left(aq\right)+8HCl\left(aq\right)\to 8KCl\left(aq\right)+4I_2\left(s\right)+4H_{2}O\left(l\right)$