Chapter 20, Problem 52A

(a)

Interpretation Introduction

Interpretation: The elements that undergo oxidation and reduction and the reducing and oxidizing agent in the given reaction, $Mn{O}_2\left(s\right)+HCl\left(aq\right)\to MnC{l}_2\left(aq\right)+C{l}_2\left(g\right)+H_{2}O\left(l\right)$ to be determined.

Concept Introduction: The process that involves the gain of electrons is reduction while the one which involves the loss of electrons is oxidation. The chemical species that undergo oxidation by the removal of electrons is the reducing agent while the one that undergoes reduction by the gain of electrons is the oxidizing agent.

Answer to Problem 52A

The elements in the reaction, $Mn{O}_2\left(s\right)+HCl\left(aq\right)\to MnC{l}_2\left(aq\right)+C{l}_2\left(g\right)+H_{2}O\left(l\right)$ ,

Reduced is $M{n}^{4+}$

Oxidised is $C{l}^{-}$

Oxidizing agent is $Mn{O}_2$

Reducing agent is $HCl$

Explanation of Solution

The given reaction, $Mn{O}_2\left(s\right)+HCl\left(aq\right)\to MnC{l}_2\left(aq\right)+C{l}_2\left(g\right)+H_{2}O\left(l\right)$ .

It involves the reduction of $M{n}^{4+}$ in the $Mn{O}_2$ to the $M{n}^{2+}$ in $MnC{l}_2$ . The oxidation of $C{l}^{-}$ from $HCl$ to zero in $C{l}_2$ . So, the species that undergo reduction is $M{n}^{4+}$ , while, those that undergo oxidation is $C{l}^{-}$ . The oxidizing agent is $Mn{O}_2$ and the reducing agent is $HCl$ .

(b)

Interpretation Introduction

Interpretation: The elements that undergo oxidation and reduction and the reducing and oxidizing agent in the given reaction, $Cu\left(s\right)+HN{O}_3\left(aq\right)\to Cu{\left(N{O}_3\right)}_2\left(aq\right)+N{O}_2\left(g\right)+H_{2}O\left(l\right)$ is to be determined.

Concept Introduction: The process that involves the gain of electrons is reduction while the one which involves the loss of electrons is oxidation. The chemical species that undergo oxidation by the removal of electrons is the reducing agent while the one that undergoes reduction by the gain of electrons is the oxidizing agent.

Answer to Problem 52A

The elements in the reaction, $Cu\left(s\right)+HN{O}_3\left(aq\right)\to Cu{\left(N{O}_3\right)}_2\left(aq\right)+N{O}_2\left(g\right)+H_{2}O\left(l\right)$ ,

Reduced is $N$

Oxidised is $Cu$

Oxidizing agent is $HN{O}_3$

Reducing agent is $Cu$

Explanation of Solution

The given reaction,

  $Cu\left(s\right)+HN{O}_3\left(aq\right)\to Cu{\left(N{O}_3\right)}_2\left(aq\right)+N{O}_2\left(g\right)+H_{2}O\left(l\right)$

It involves the removal of electrons from $Cu$ to $C{u}^{2+}$ , oxidation. And the oxidation number of N changes from +5 to +4 by the gain of electrons. So, the species that undergo oxidation is Cu and that undergoes a reduction in $N$ .

And, the reducing agent is $Cu$ and oxidizing agent is $HN{O}_3$ .

(c)

Interpretation Introduction

Interpretation: The elements that undergo oxidation and reduction and the reducing and oxidizing agent in the given reaction, a $P\left(s\right)+HN{O}_3\left(aq\right)+H_{2}O\to NO\left(g\right)+H_{3}P{O}_4\left(aq\right)$ are to be determined.

Concept Introduction: The process that involves the gain of electrons is reduction while the one which involves the loss of electrons is oxidation. The chemical species that undergo oxidation by the removal of electrons is the reducing agent while the one that undergoes reduction by the gain of electrons is the oxidizing agent.

Answer to Problem 52A

The elements in the reaction, $P\left(s\right)+HN{O}_3\left(aq\right)+H_{2}O\to NO\left(g\right)+H_{3}P{O}_4\left(aq\right)$ ,

Reduced is $N$

Oxidised is $P$

Oxidizing agent is $HN{O}_3$

Reducing agent is $P$

Explanation of Solution

The given reaction,

  $P\left(s\right)+HN{O}_3\left(aq\right)+H_{2}O\to NO\left(g\right)+H_{3}P{O}_4\left(aq\right)$

The oxidation number of $P$ changes from zero to +5 in $H_{3}P{O}_4$ . And the oxidation number of N changes from +5 in $HN{O}_3$ to +2 in $NO$ . So the species that undergo oxidation is $P$ and that undergoes reduction is N. And the reducing agent is $P$ and oxidizing agent is $HN{O}_3$ .

(d)

Interpretation Introduction

Interpretation: The elements that undergo oxidation and reduction and the reducing and oxidizing agent in the given reaction, $Bi{\left(OH\right)}_3\left(s\right)+N{a}_{2}Sn{O}_2\left(aq\right)\to Bi\left(s\right)+N{a}_{2}Sn{O}_3\left(aq\right)+H_{2}O\left(l\right)$ is to be determined.

Concept Introduction: The process that involves the gain of electrons is reduction while the one which involves the loss of electrons is oxidation. The chemical species that undergo oxidation by the removal of electrons is the reducing agent while the one that undergoes reduction by the gain of electrons is the oxidizing agent.

Answer to Problem 52A

The elements in the reaction, $Bi{\left(OH\right)}_3\left(s\right)+N{a}_{2}Sn{O}_2\left(aq\right)\to Bi\left(s\right)+N{a}_{2}Sn{O}_3\left(aq\right)+H_{2}O\left(l\right)$ ,

Reduced is $B{i}^{3+}$

Oxidised is $S{n}^{2+}$

Oxidizing agent is $Bi{\left(OH\right)}_3$

Reducing agent is $N{a}_{2}Sn{O}_3$

Explanation of Solution

The given reaction, $Bi{\left(OH\right)}_3\left(s\right)+N{a}_{2}Sn{O}_2\left(aq\right)\to Bi\left(s\right)+N{a}_{2}Sn{O}_3\left(aq\right)+H_{2}O\left(l\right)$

It involves the reduction of $B{i}^{3+}$ in $Bi{\left(OH\right)}_3$ to zero in Bismuth by the gain of electrons. The oxidation of $S{n}^{2+}$ in $N{a}_{2}Sn{O}_2$ changes to $S{n}^{3+}$ in $N{a}_{2}Sn{O}_3$ by the loss of electrons. So, the species that undergo oxidation is $S{n}^{2+}$ and those that undergo reduction is $B{i}^{3+}$ .

And the oxidizing agent is $Bi{\left(OH\right)}_3$ and the reducing agent is $N{a}_{2}Sn{O}_2$ .