Chapter 20, Problem 43A

(a)

Interpretation Introduction

Interpretation: The oxidized and reduced atoms present in the given reaction, $Al\left(s\right)+Mn{O}_2\left(s\right)\to Mn\left(s\right)+A{l}_2{O}_3\left(s\right)$ is to be determined by checking the oxidation number.

Concept introduction: Zero, positive, and negative numbers that are assigned to reactants species to track the electron flow is oxidation number. When electrons are lost (oxidation) from a species, positive numbers are given while if electrons are gained(reduction), negative numbers are given. And the reactants that don't involve any change in electron count are assigned with the number zero. Certain rules must also be followed before assigning these numbers.

• The oxidation number of oxygen is -2.
• For a neutral compound, the total sum of the oxidation number is zero.
• For ionic compounds with a charge, the total sum of oxidation number is equal to its charge.

Answer to Problem 43A

The oxidized atom in the reaction is $Al$ . And the reduced atom is $Mn$ in $Mn{O}_2$ .

Explanation of Solution

In the 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 oxidation number of $Al$ increases from zero to +3 in $A{l}_2{O}_3$ due to the loss of electrons. The oxidation number of $Mn$ decreases from $+4$ to zero. So, the oxidized atom is $Al$ and the reduced atom is $M{n}^{2+}$ .

(b)

Interpretation Introduction

Interpretation: The oxidized and reduced atoms present in the given reaction, $K\left(s\right)+H_{2}O\left(l\right)\to KOH\left(aq\right)+H_2\left(g\right)$ is to be determined by checking the oxidation number.

Concept introduction: Zero, positive, and negative numbers that are assigned to reactants species to track the electron flow is oxidation number. When electrons are lost (oxidation) from a species, positive numbers are given while if electrons are gained(reduction), negative numbers are given. And the reactants that don't involve any change in electron count are assigned with the number zero. Certain rules must also be followed before assigning these numbers.

• The oxidation number of oxygen is -2.
• For a neutral compound, the total sum of the oxidation number is zero.
• For ionic compounds with a charge, the total sum of oxidation number is equal to its charge.

Answer to Problem 43A

The oxidized atom in the reaction is $K$ . And the reduced atom is $H$ in $H_{2}O$ .

Explanation of Solution

In the reaction,

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

The oxidation number of K increases from zero to +1 in $KOH$ . While the oxidation number of $H$ in $H_{2}O$ decreases from +1 to 0. So, the oxidized atom is $K$ and the reduced atom is $H$ in $H_{2}O$ .

(c)

Interpretation Introduction

Interpretation: The oxidized and reduced atoms present in the given reaction, $HgO\to Hg\left(l\right)+O_2\left(g\right)$ is to be determined by checking the oxidation number.

Concept introduction: Zero, positive, and negative numbers that are assigned to reactants species to track the electron flow is oxidation number. When electrons are lost (oxidation) from a species, positive numbers are given while if electrons are gained(reduction), negative numbers are given. And the reactants that don't involve any change in electron count are assigned with the number zero. Certain rules must also be followed before assigning these numbers.

• The oxidation number of oxygen is -2.
• For a neutral compound, the total sum of the oxidation number is zero.
• For ionic compounds with a charge, the total sum of oxidation number is equal to its charge.

Answer to Problem 43A

The oxidized atom in the reaction is $Hg$ . And the reduced atom is $O$ in $HgO$ .

Explanation of Solution

In the reaction,

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

The oxidation number of $Hg$ decreases from +2 to zero. While the oxidation number of $O$ increases from $-2$ to zero. So, the atom that undergoes oxidation is $O$ while that undergoes reduction is Hg.

(d)

Interpretation Introduction

Interpretation: The oxidized and reduced atoms present in the given reaction, $P_4\left(s\right)+O_2\left(g\right)\to P_4{O}_{10}\left(s\right)$ is to be determined by checking the oxidation number.

Concept introduction: Zero, positive, and negative numbers that are assigned to reactants species to track the electron flow is oxidation number. When electrons are lost (oxidation) from a species positive numbers are given while if electrons are gained(reduction) negative numbers are given. And the reactants that don't involve any change in electron count are assigned with the number zero. Certain rules must also be followed before assigning these numbers.

• The oxidation number of oxygen is -2.
• For a neutral compound, the total sum of the oxidation number is zero.
• For ionic compounds with a charge, the total sum of oxidation number is equal to its charge.

Answer to Problem 43A

The oxidized atom in the reaction is $P$ in $P_4$ . And the reduced atom is $P$ in $O_2$ .

Explanation of Solution

In the reaction,

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

The oxidation number of P in $P_4$ increases from zero to +5 by the loss of electrons. While the oxidation number of $O_2$ decreases from zero $-2$ by the gaining of electrons. So, the oxidized species is P and the reduced species is O.