Chapter 20, Problem 48A

(a)

Interpretation Introduction

Interpretation: The balanced chemical equation of the given reaction $Al\left(s\right)+Mn{O}_2\left(s\right)\to Mn\left(s\right)+A{l}_2{O}_3\left(s\right)$ (s) 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 48A

The balanced chemical equation is,

  $4Al\left(s\right)+3Mn{O}_2\left(s\right)\to 3Mn\left(s\right)+2A{l}_2{O}_3\left(s\right)$

Explanation of Solution

In the given reaction,

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

The half-reactions can be written as,

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

Reduction: $M{n}^{2+}+2e^{-}\to Mn$

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

  $2Al\left(s\right)+3Mn{O}_2\left(s\right)\to 3Mn\left(s\right)+2A{l}_2{O}_3\left(s\right)$

The count of aluminum is not equal on both sides. It can be done by multiplying 2 by it.

  $4Al\left(s\right)+3Mn{O}_2\left(s\right)\to 3Mn\left(s\right)+2A{l}_2{O}_3\left(s\right)$

(b)

Interpretation Introduction

Interpretation: The balanced chemical equation of the given reaction $K\left(s\right)+H_{2}O\left(l\right)\to KOH\left(aq\right)+H_2\left(g\right)$ (g) 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 48A

The balanced chemical equation is,

  $2K\left(s\right)+2H_{2}O\left(l\right)\to \text{ 2}KOH\left(aq\right)+H_2\left(g\right)$

Explanation of Solution

In the given reaction,

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

The half-reactions can be written as,

Oxidation: $K\to K^{+}+e^{-}$

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

The electron count is in the balanced form. But the hydrogen count is not. It can be balanced by multiplying the water, K, and $KOH$ by 2. Therefore the equation becomes,

  $2K\left(s\right)+2H_{2}O\left(l\right)\to \text{ 2}KOH\left(aq\right)+H_2\left(g\right)$

(c)

Interpretation Introduction

Interpretation: The balanced chemical equation of the given reaction $HgO\to Hg\left(l\right)+O_2\left(g\right)$ 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 48A

The balanced chemical equation is, $2HgO\to 2Hg\left(l\right)+O_2\left(g\right)$ .

Explanation of Solution

In the given reaction,

  $HgO\to Hg\left(l\right)+O_2\left(g\right)$

The half-reactions can be written as,

Oxidation: $2O^{2-}\to O_2+2e^{-}$

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

The electron count is in the balanced form. But the oxygen count is not. It can be done by multiplying $HgO$ by 2 and thereby balancing the Hg count by 2 too.

Therefore, the balanced equation becomes,

  $2HgO\to 2Hg\left(l\right)+O_2\left(g\right)$

(d)

Interpretation Introduction

Interpretation: The balanced chemical equation of the given reaction $P_4\left(s\right)+O_2\left(g\right)\to P_4{O}_{10}\left(s\right)$ 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 48A

The balanced chemical equation is, $P_4\left(s\right)+5O_2\left(g\right)\to P_4{O}_{10}\left(s\right)$

Explanation of Solution

In the given reaction,

  $P_4\left(s\right)+O_2\left(g\right)\to P_4{O}_{10}\left(s\right)$

The half-reactions can be written as,

Oxidation: $P_4\to P_4{}^{+5}+5e^{-}$

Reduction: $O_2+5e^{-}\to O_{10}{}^{-2}$

The electron count is in a balanced state but the oxygen count is not. It can be done by multiplying $O_2$ on the left-hand side by 5. Therefore the equation becomes,

  $P_4\left(s\right)+5O_2\left(g\right)\to P_4{O}_{10}\left(s\right)$