Chapter 18, Problem 46A

(a)

Interpretation Introduction

Interpretation: The given oxidation-reduction reaction needs to be balanced and the species undergoing oxidation and reduction needs to be identified.

$\text{Fe}\left(\text{s}\right)+{\text{O}}_{\text{2}}\left(\text{g}\right)\to {\text{Fe}}_{\text{2}}{\text{O}}_{\text{3}}\left(\text{s}\right)$

Concept Introduction: By transfer of electrons, oxidation-reduction reactions are characterized. The transfer occurs to form ions. Oxidation reaction is caused by losing electrons and reduction is caused by gaining electrons.

Answer to Problem 46A

Balanced reaction is-

$\text{4Fe}\left(\text{s}\right)+3{\text{O}}_{\text{2}}\left(\text{g}\right)\to 2{\text{Fe}}_{\text{2}}{\text{O}}_{\text{3}}\left(\text{s}\right)$

In this reaction, Fe is oxidized and O is reduced.

Explanation of Solution

The given reaction is as follows:

$\text{Fe}\left(\text{s}\right)+{\text{O}}_{\text{2}}\left(\text{g}\right)\to {\text{Fe}}_{\text{2}}{\text{O}}_{\text{3}}\left(\text{s}\right)$

As number of Fe is one and O is 2 in reactant side and Fe is $2$ and O is $\text{3}$ in product side, multiply reactant side Fe by $4$ and O by $3$ also, product side by 2. Hence, balance equation is-

$\text{4Fe}\left(\text{s}\right)+3{\text{O}}_{\text{2}}\left(\text{g}\right)\to 2{\text{Fe}}_{\text{2}}{\text{O}}_{\text{3}}\left(\text{s}\right)$

Fe and ${\text{O}}_{\text{2}}$ are in free state so, oxidation state of both the element is $0$ . ${\text{Fe}}_{\text{2}}{\text{O}}_{\text{3}}$ contains ${\text{Fe}}^{\text{3+}}$ and ${\text{O}}^{2-}$ ions. Oxidation state of Fe is $3+$ and O is $2-$ . Each Fe changes its oxidation state from $0$ to $+3$ . An increase in oxidation state takes place. Hence, Fe is oxidized. Oxidation state of O is decreases from $0$ to $-2$ . Hence, it is reduced.

(b)

Interpretation Introduction

Interpretation: The given oxidation-reduction reaction needs to be balanced and the species undergoing oxidation and reduction needs to be identified.

$\text{Al}\left(\text{s}\right)+{\text{Cl}}_{\text{2}}\left(\text{g}\right)\to {\text{AlCl}}_{\text{3}}\left(\text{s}\right)$

Concept Introduction: By transfer of electrons, oxidation-reduction reactions are characterized. The transfer occurs to form ions. Oxidation reaction is caused by losing electrons and reduction is caused by gaining electrons.

Answer to Problem 46A

Balanced reaction is-

$\text{2Al}\left(\text{s}\right)+3{\text{Cl}}_{\text{2}}\left(\text{g}\right)\to 2{\text{AlCl}}_{\text{3}}\left(\text{s}\right)$

In this reaction, Al is oxidized and Cl is reduced.

Explanation of Solution

The given reaction is as follows:

$\text{Al}\left(\text{s}\right)+{\text{Cl}}_{\text{2}}\left(\text{g}\right)\to {\text{AlCl}}_{\text{3}}\left(\text{s}\right)$

As number of Al is one and Cl is 2 in reactant side and Al is $1$ and Cl is $\text{3}$ in product side, multiply reactant side Al by $2$ and Cl by $3$ also, product side by $2$ . Hence, balance equation is-

$\text{2Al}\left(\text{s}\right)+3{\text{Cl}}_{\text{2}}\left(\text{g}\right)\to 2{\text{AlCl}}_{\text{3}}\left(\text{s}\right)$

Al and ${\text{Cl}}_{\text{2}}$ are in free state so, oxidation state of both the element is $0$ . ${\text{AlCl}}_{\text{3}}$ contains ${\text{Al}}^{\text{3+}}$ and ${\text{Cl}}^{-}$ ions. Oxidation state of Al is $3+$ and O is $1-$ . Each Al changes its oxidation state from $0$ to $+3$ . An increase in oxidation state takes place. Hence, Al is oxidized. Oxidation state of Cl is decreases from $0$ to $-1$ . Hence, it is reduced.

(c)

Interpretation Introduction

Interpretation: The given oxidation-reduction reaction needs to be balanced and the species undergoing oxidation and reduction needs to be identified.

$\text{Mg}\left(\text{s}\right)+{\text{P}}_{\text{4}}\left(\text{s}\right)\to {\text{Mg}}_3{\text{P}}_2\left(\text{s}\right)$

Concept Introduction: By transfer of electrons, oxidation-reduction reactions are characterized. The transfer occurs to form ions. Oxidation reaction is caused by losing electrons and reduction is caused by gaining electrons.

Answer to Problem 46A

Balanced reaction is-

$\text{6Mg}\left(\text{s}\right)+{\text{P}}_{\text{4}}\left(\text{s}\right)\to 2{\text{Mg}}_3{\text{P}}_2\left(\text{s}\right)$

In this reaction, Al is oxidized and P is reduced.

Explanation of Solution

The given reaction is as follows:

$\text{Mg}\left(\text{s}\right)+{\text{P}}_{\text{4}}\left(\text{s}\right)\to {\text{Mg}}_3{\text{P}}_2\left(\text{s}\right)$

Number of Mg is one and P is 4 in reactant side and Mg is $3$ and P is $2$ in product side thus, multiply reactant side Mg by 6 and product side ${\text{Mg}}_3{\text{P}}_2$ by 2. Hence, balance equation is-

$\text{6Mg}\left(\text{s}\right)+{\text{P}}_{\text{4}}\left(\text{s}\right)\to 2{\text{Mg}}_3{\text{P}}_2\left(\text{s}\right)$

Mg and ${\text{P}}_{\text{2}}$ are in free state so, oxidation state of both the element is $0$ . ${\text{Mg}}_3{\text{P}}_2$ contains ${\text{Mg}}^{\text{2+}}$ and ${\text{P}}^{3-}$ ions. Oxidation state of Mg is $2+$ and P is $3-$ . Each Mg changes its oxidation state from $0$ to $+2$ . An increase in oxidation state takes place. Hence, Mg is oxidized. Oxidation state of P is decreases from $0$ to $-3$ . Hence, it is reduced.