Chapter 20, Problem 49A

(a)

Interpretation Introduction

Interpretation: The balanced chemical equation of the given reaction $Li\left(s\right)+H_{2}O\left(l\right)\to LiOH\left(aq\right)+H_2\left(g\right)$ is to be interpreted.

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 49A

The balanced chemical equation is,

  $2Li\left(s\right)+2H_{2}O\left(l\right)\to 2LiOH\left(aq\right)+H_2\left(g\right)$

Explanation of Solution

In the given reaction,

  $Li\left(s\right)+H_{2}O\left(l\right)\to LiOH\left(aq\right)+H_2\left(g\right)$

The half-reactions can be written as,

Oxidation: $Li\to L{i}^{+}+e^{-}$

Reduction: $H_{2}O+2e^{-}\to H_2$

The electron count can be balanced by multiplying the oxidation equation by 2. Therefore, the equation becomes,

  $2Li\left(s\right)+H_{2}O\left(l\right)\to 2LiOH\left(aq\right)+H_2\left(g\right)$

The count of oxygen is not equal on both sides. It can be done by multiplying 2 by $H_{2}O$ . Thus, the balanced equation is,

  $2Li\left(s\right)+2H_{2}O\left(l\right)\to 2LiOH\left(aq\right)+H_2\left(g\right)$

(b)

Interpretation Introduction

Interpretation: The balanced chemical equation of the given reaction $K_{2}C{r}_2{O}_7\left(aq\right)+HCl\left(aq\right)\to KCl\left(aq\right)+CrC{l}_3\left(aq\right)+H_{2}O\left(l\right)$ is to be interpreted.

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 49A

The balanced chemical equation is,

  $K_{2}C{r}_2{O}_7\left(aq\right)+14HCl\left(aq\right)\to 2KCl\left(aq\right)+2CrC{l}_3\left(aq\right)+7H_{2}O\left(l\right)+3C{l}_2$

Explanation of Solution

In the given reaction,

  $K_{2}C{r}_2{O}_7\left(aq\right)+HCl\left(aq\right)\to KCl\left(aq\right)+CrC{l}_3\left(aq\right)+H_{2}O\left(l\right)$

The half-reactions can be written as,

Oxidation: $HCl\to C{l}_2+e^{-}$

Reduction: $C{r}^{6+}+3e^{-}\to C{r}^{3+}$

The electron count can be balanced by multiplying the oxidation equation by 3. Therefore the equation becomes,

  $K_{2}C{r}_2{O}_7\left(aq\right)+6HCl\left(aq\right)\to KCl\left(aq\right)+CrC{l}_3\left(aq\right)+H_{2}O\left(l\right)+3C{l}_2$

The count of potassium is not equal on both sides. It can be done by multiplying 2 by $KCl$ on the right-hand side. The chromium count is also not equal can be done by multiplying $CrC{l}_3$ by 2. Now, the count of oxygen can be balanced by the addition of 7 molecules of water on the right-hand side. Now, the chlorine and hydrogen count is not balanced it can be done by multiplying HCl by 2. Thus, the balanced equation becomes,

  $K_{2}C{r}_2{O}_7\left(aq\right)+14HCl\left(aq\right)\to 2KCl\left(aq\right)+2CrC{l}_3\left(aq\right)+7H_{2}O\left(l\right)+3C{l}_2$

(c)

Interpretation Introduction

Interpretation: The balanced chemical equation of the given reaction $Al\left(s\right)+HCl\left(aq\right)\to AlC{l}_3\left(aq\right)+H_2\left(g\right)$ is to be interpreted.

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 49A

The balanced chemical equation is,

  $2Al\left(s\right)+6HCl\left(aq\right)\to 2AlC{l}_3\left(aq\right)+3H_2\left(g\right)$

Explanation of Solution

In the given reaction,

  $Al\left(s\right)+HCl\left(aq\right)\to AlC{l}_3\left(aq\right)+H_2\left(g\right)$

The half-reactions can be written as,

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

Reduction: $2H^{+}+2e^{-}\to H_2$

The electron count can be balanced by multiplying the oxidation equation by 2 and the reduction equation by 3. Therefore, the equation becomes,

  $2Al\left(s\right)+6HCl\left(aq\right)\to 2AlC{l}_3\left(aq\right)+3H_2\left(g\right)$

The count of all the species is equal. So, this is the balanced equation.

(d)

Interpretation Introduction

Interpretation: The balanced chemical equation of the given reaction $C{l}_2\left(g\right)+H_{2}O\left(l\right)\to HCl\left(aq\right)+HClO\left(aq\right)$ is to be interpreted.

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 49A

The balanced chemical equation is,

  $C{l}_2\left(g\right)+H_{2}O\left(l\right)\to HCl\left(aq\right)+HClO\left(aq\right)$

Explanation of Solution

In the given reaction,

  $C{l}_2\left(g\right)+H_{2}O\left(l\right)\to HCl\left(aq\right)+HClO\left(aq\right)$

The half-reactions can be written as,

Oxidation: $C{l}_2\to HClO+e^{-}$

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

The electron count and atom count are all equal on both sides. Therefore, it is a balanced equation itself. So, the equation is,

  $C{l}_2\left(g\right)+H_{2}O\left(l\right)\to HCl\left(aq\right)+HClO\left(aq\right)$

(e)

Interpretation Introduction

Interpretation: The balanced chemical equation of the given reaction $I_2{O}_5\left(s\right)+CO\left(g\right)\to I_2\left(s\right)+4C{O}_2\left(g\right)$ is to be interpreted.

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 49A

The balanced chemical equation is,

  $I_2{O}_5\left(s\right)+5CO\left(g\right)\to I_2\left(s\right)+5C{O}_2\left(g\right)$

Explanation of Solution

In the given reaction,

  $I_2{O}_5\left(s\right)+CO\left(g\right)\to I_2\left(s\right)+4C{O}_2\left(g\right)$

The half-reactions are,

Oxidation: $I_7+5e^{-}\to I_2{}^{-}$

Reduction: $CO\to C{O}_2+2e^{-}$

The oxygen count can be balanced by the multiplication of 5 to $C{O}_2$ and 5 to CO. So, the equation change to,

  $I_2{O}_5\left(s\right)+5CO\left(g\right)\to I_2\left(s\right)+5C{O}_2\left(g\right)$

(f)

Interpretation Introduction

Interpretation: The balanced chemical equation of the given reaction $H_{2}O\left(l\right)+S{O}_3\left(g\right)\to H_{2}S{O}_4\left(aq\right)$ is to be interpreted.

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 49A

The balanced chemical equation is,

  $H_{2}O\left(l\right)+S{O}_3\left(g\right)\to H_{2}S{O}_4\left(aq\right)$

Explanation of Solution

In the given reaction,

  $H_{2}O\left(l\right)+S{O}_3\left(g\right)\to H_{2}S{O}_4\left(aq\right)$

The oxidation number is not changing in the given reaction. So, it is not a redox reaction. And also the number of atoms on both sides is equal. It is a balanced chemical equation itself.