Chapter 20.2, Problem 19LC

(a)

Interpretation Introduction

Interpretation: The oxidized and reduced species in the reaction, $2Na\left(s\right)\text{+}C{l}_2\left(l\right)\to 2NaCl\left(s\right)$ is to be identified using a change in oxidation number.

Concept Introduction: The zero, positive or negative numbers that are assigned to an atom for indicating its degree of oxidation and reduction is the oxidation number. This number indicates the number of electrons that are either gained or lost during the bond formation in a chemical compound.

Answer to Problem 19LC

The chemical species that undergo oxidation is $Na$ and that undergo reduction is $C{l}_2$ in the given reaction.

Explanation of Solution

The completion of chemical reactions results in a change in the oxidation number of chemical compounds. From the change in oxidation number values, the chemical species that have undergone oxidation and reduction can be easily found.

The one in which the oxidation number is found to be decreased is said to be reduced. Since reduction involves the gain of electrons. While the one in which the oxidation number is found to be increased is said to be oxidized. Since oxidation involves the loss of electrons.

In the reaction,

  $2Na\left(s\right)\text{+}C{l}_2\left(l\right)\to 2NaCl\left(s\right)$

The oxidation number of sodium, $Na$ increases from zero to one by the loss of an electron. So, here $Na$ is said to be oxidized.

The oxidation number of chlorine, $C{l}_2$ decreased from zero to $-1$ by the gain of electrons. So $C{l}_2$ is said to be reduced.

(b)

Interpretation Introduction

Interpretation: The oxidized and reduced species in the reaction, $2HN{O}_3\left(aq\right)+6HI\left(aq\right)\to 2NO\left(g\right)+3I_2\left(s\right)+4H_{2}O\left(l\right)$ is to be identified using a change in oxidation number.

Concept Introduction: The zero, positive or negative numbers that are assigned to an atom for indicating its degree of oxidation and reduction is the oxidation number. This number indicates the number of electrons that are either gained or lost during the bond formation in a chemical compound.

Answer to Problem 19LC

The chemical species that undergo oxidation is $I^{-}$ and that undergoes reduction is nitrogen in the given reaction.

Explanation of Solution

The completion of chemical reactions results in a change in the oxidation number of chemical compounds. From the change in oxidation number values, the chemical species that have undergone oxidation and reduction can be easily found.

The one in which the oxidation number is found to be decreased is said to be reduced. Since reduction involves the gain of electrons. While the one in which the oxidation number is found to be increased is said to be oxidized. Since oxidation involves the loss of electrons.

In the reaction,

  $2HN{O}_3\left(aq\right)+6HI\left(aq\right)\to 2NO\left(g\right)+3I_2\left(s\right)+4H_{2}O\left(l\right)$

The oxidation number of Iodine, $I^{-}$ present in $HI$ increases from $-1$ to zero by the loss of an electron. So, $I^{-}$ is said to be oxidized.

The oxidation number of nitrogen in $HN{O}_3$ decreases from +5 to +2 by the gain of electrons. So, nitrogen is said to be reduced.

(c)

Interpretation Introduction

Interpretation: The oxidized and reduced species in the reaction, $2H_{2}S\left(g\right)+2HN{O}_3\left(aq\right)\to 3S\left(s\right)+2NO\left(g\right)+4H_{2}O\left(l\right)$ is to be identified using a change in oxidation number.

Concept Introduction: The zero, positive or negative numbers that are assigned to an atom for indicating its degree of oxidation and reduction is the oxidation number. This number indicates the number of electrons that are either gained or lost during the bond formation in a chemical compound.

Answer to Problem 19LC

The chemical species that undergo oxidation is sulfur and that undergoes reduction is nitrogen in the given reaction.

Explanation of Solution

The completion of chemical reactions results in a change in the oxidation number of chemical compounds. From the change in oxidation number values, the chemical species that have undergone oxidation and reduction can be easily found.

The one in which the oxidation number is found to be decreased is said to be reduced. Since reduction involves the gain of electrons. While the one in which the oxidation number is found to be increased is said to be oxidized. Since oxidation involves the loss of electrons.

In the reaction,

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

The oxidation number of sulfur presents in $H_{2}S$ increases from $-2$ to zero by the loss of electrons. So it is said to be oxidized.

The oxidation number of nitrogen in $HN{O}_3$ decreases from +5 to $-2$ by the gain of electrons. So, nitrogen is found to be reduced.

(d)

Interpretation Introduction

Interpretation: The oxidized and reduced species in the reaction, $2PbS{O}_4\left(s\right)+2H_{2}O\left(l\right)\to 2Pb\left(s\right)+Pb{O}_2\left(s\right)+2H_{2}S{O}_4\left(aq\right)$ is to be identified using a change in oxidation number.

Concept Introduction: The zero, positive or negative numbers that are assigned to an atom for indicating its degree of oxidation and reduction is the oxidation number. This number indicates the number of electrons that are either gained or lost during the bond formation in a chemical compound.

Answer to Problem 19LC

The chemical species that undergo oxidation is $P{b}^{2+}$ and that undergo reduction is $P{b}^{2+}$ in the given reaction. The same chemical species is oxidized and reduced in this reaction.

Explanation of Solution

The completion of chemical reactions results in a change in the oxidation number of chemical compounds. From the change in oxidation number values, the chemical species that have undergone oxidation and reduction can be easily found.

The one in which the oxidation number is found to be decreased is said to be reduced. Since reduction involves the gain of electrons. While the one in which the oxidation number is found to be increased is said to be oxidized. Since oxidation involves the loss of electrons.

In the reaction,

  $2PbS{O}_4\left(s\right)+2H_{2}O\left(l\right)\to 2Pb\left(s\right)+Pb{O}_2\left(s\right)+2H_{2}S{O}_4\left(aq\right)$

The oxidation number of lead, $P{b}^{2+}$ in $PbS{O}_4$ decreases from +2 to zero by the gain of electrons and forms $Pb$ . So it is said to be reduced.

The oxidation number of lead, $P{b}^{2+}$ in $PbS{O}_4$ increased from +2 to +4 in $Pb{O}_2$ by the loss of electrons. So, it is said to be oxidized.

Therefore, the oxidized and reduced species are the same in this case. The reaction is disproportionate.