Chapter 17, Problem 17QRT

(a)

Interpretation Introduction

Interpretation:

The given redox reaction is to be balanced and the oxidizing agent and reducing agent are to be identified.

Concept Introduction:

The chemical reaction that involves the electron transfer between two species is referred to as redox reaction.  It is any chemical reaction in which change in the oxidation number of a molecule, atom or ion takes place by either gain of electrons or loss of electrons.  The species which gets oxidized in the redox reaction is referred to as reducing agent whereas the specie which gets reduced is termed as oxidizing agent.

Answer to Problem 17QRT

The balanced redox reaction is shown below.

  $\text{6FeO}\left(\text{s}\right)+{\text{O}}_{\text{3}}\left(\text{g}\right)\to 3{\text{Fe}}_2{\text{O}}_3\left(\text{s}\right)$

In this reaction, $\text{FeO}$ act as reducing agent and ${\text{O}}_3$ act as oxidizing agent.

Explanation of Solution

The given unbalanced redox reaction is shown below.

    $\text{FeO}\left(\text{s}\right)+{\text{O}}_{\text{3}}\left(\text{g}\right)\to {\text{Fe}}_2{\text{O}}_3\left(\text{s}\right)$

The iron atom and oxygen atom are balanced by putting $2$ and $\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.$ as prefix to $\text{FeO}$ and ozone respectively.

  $\text{2FeO}\left(\text{s}\right)+\frac{1}{3}{\text{O}}_{\text{3}}\left(\text{g}\right)\to {\text{Fe}}_2{\text{O}}_3\left(\text{s}\right)$

The above balanced reaction is multiplied by $3$ so as to get the balanced reaction in integers.

  $\text{6FeO}\left(\text{s}\right)+{\text{O}}_{\text{3}}\left(\text{g}\right)\to 3{\text{Fe}}_2{\text{O}}_3\left(\text{s}\right)$

Thus, the balanced chemical reaction is shown below.

  $\text{6FeO}\left(\text{s}\right)+{\text{O}}_{\text{3}}\left(\text{g}\right)\to 3{\text{Fe}}_2{\text{O}}_3\left(\text{s}\right)$

The oxidation number of iron atom in $\text{FeO}$ and ${\text{Fe}}_2{\text{O}}_3$ is $+2$ and $+3$ respectively.  Since there is increase in oxidation number, therefore, $\text{FeO}$ is oxidized to ${\text{Fe}}_2{\text{O}}_3$ and $\text{FeO}$ act as reducing agent.

The oxidation number of oxygen atom in ${\text{O}}_3$ and ${\text{Fe}}_2{\text{O}}_3$ is $0$ and $-2$ respectively.  Since there is decrease in oxidation number, therefore, ${\text{O}}_3$ is reduced to ${\text{Fe}}_2{\text{O}}_3$ and ${\text{O}}_3$ act as oxidizing agent.

(b)

Interpretation Introduction

Interpretation:

The given redox reaction is to be balanced and the oxidizing agent and reducing agent are to be identified.

Concept Introduction:

Same as part (a).

Answer to Problem 17QRT

The balanced redox reaction is shown below.

    ${\text{P}}_{\text{4}}\left(\text{s}\right)+10{\text{Br}}_2\left(l\right)\to 4{\text{PBr}}_5\left(l\right)$

In this reaction, ${\text{P}}_{\text{4}}$ act as reducing agent and ${\text{Br}}_2$ act as oxidizing agent.

Explanation of Solution

The given unbalanced redox reaction is shown below.

    ${\text{P}}_{\text{4}}\left(\text{s}\right)+{\text{Br}}_2\left(l\right)\to {\text{PBr}}_5\left(l\right)$

The phosphorous atoms and bromine atoms are balanced by putting $\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$4$}\right.$ and $\raisebox{1ex}{$5$}\!\left/ \!\raisebox{-1ex}{$2$}\right.$ as prefix to ${\text{P}}_{\text{4}}$ and ${\text{Br}}_2$ respectively.

  $\frac{1}{4}{\text{P}}_{\text{4}}\left(\text{s}\right)+\frac{5}{2}{\text{Br}}_2\left(l\right)\to {\text{PBr}}_5\left(l\right)$

The above balanced reaction is multiplied by $4$ so as to get the balanced reaction in integers.

  ${\text{P}}_{\text{4}}\left(\text{s}\right)+10{\text{Br}}_2\left(l\right)\to 4{\text{PBr}}_5\left(l\right)$

Thus, the balanced chemical reaction is shown below.

  ${\text{P}}_{\text{4}}\left(\text{s}\right)+10{\text{Br}}_2\left(l\right)\to 4{\text{PBr}}_5\left(l\right)$

The oxidation number of phosphorous atom in ${\text{P}}_{\text{4}}$ and ${\text{PBr}}_5$ is $0$ and $+5$ respectively.  Since there is increase in oxidation number, therefore, ${\text{P}}_{\text{4}}$ is oxidized to ${\text{PBr}}_5$ and ${\text{P}}_{\text{4}}$ act as reducing agent.

The oxidation number of bromine atom in ${\text{Br}}_2$ and ${\text{PBr}}_5$ is $0$ and $-1$ respectively.  Since there is decrease in oxidation number, therefore, ${\text{Br}}_2$ is reduced to ${\text{PBr}}_5$ and ${\text{Br}}_2$ act as oxidizing agent.

(c)

Interpretation Introduction

Interpretation:

The given redox reaction is to be balanced in acidic medium and the oxidizing agent and reducing agent are to be identified.

Concept Introduction:

Same as part (a).

Answer to Problem 17QRT

The balanced redox reaction is shown below.

  ${\text{2Co}}^{\text{2}+}\left(\text{aq}\right)+{\text{H}}_2{\text{O}}_{\text{2}}\left(\text{aq}\right)+2{\text{H}}^{+}\left(\text{aq}\right)\to 2{\text{Co}}^{\text{3}+}\left(\text{aq}\right)+2{\text{H}}_2\text{O}\left(l\right)$

In this reaction, ${\text{Co}}^{2+}$ act as reducing agent and ${\text{H}}_2{\text{O}}_{\text{2}}$ act as oxidizing agent.

Explanation of Solution

The given unbalanced redox reaction is shown below.

  ${\text{H}}_2{\text{O}}_{\text{2}}\left(\text{aq}\right)+{\text{Co}}^{\text{2}+}\left(\text{aq}\right)\to {\text{H}}_{\text{2}}\text{O}\left(l\right)+{\text{Co}}^{\text{3}+}\left(\text{aq}\right)$

The oxidation number of cobalt atom in ${\text{Co}}^{\text{2}+}$ and ${\text{Co}}^{\text{3}+}$ is $+2$ and $+3$ respectively.  Since there is increase in oxidation number, therefore, ${\text{Co}}^{\text{2}+}$ is oxidized to ${\text{Co}}^{\text{3}+}$ and ${\text{Co}}^{\text{2}+}$ act as reducing agent.

The oxidation number of oxygen atom in ${\text{H}}_2{\text{O}}_{\text{2}}$ and ${\text{H}}_2\text{O}$ is $-1$ and $-2$ respectively.  Since there is decrease in oxidation number, therefore, ${\text{H}}_2{\text{O}}_{\text{2}}$ is reduced to ${\text{H}}_2\text{O}$ and ${\text{H}}_2{\text{O}}_{\text{2}}$ act as oxidizing agent.

Therefore, the unbalanced half reactions are shown below.

Oxidation: ${\text{Co}}^{\text{2}+}\to {\text{Co}}^{\text{3}+}$

Reduction: ${\text{H}}_2{\text{O}}_{\text{2}}\to {\text{H}}_2\text{O}$

The given reaction is to be balanced in acidic medium.  Therefore, oxygen atoms are balanced by adding water molecules to the deficient side.

Oxidation: ${\text{Co}}^{\text{2}+}\to {\text{Co}}^{\text{3}+}$

Reduction: ${\text{H}}_2{\text{O}}_{\text{2}}\to 2{\text{H}}_2\text{O}$

After balancing oxygen atoms, now hydrogen atoms are balanced by adding protons $\left({\text{H}}^{+}\right)$ to the deficient side.

Oxidation: ${\text{Co}}^{\text{2}+}\to {\text{Co}}^{\text{3}+}$

Reduction: ${\text{H}}_2{\text{O}}_{\text{2}}+2{\text{H}}^{+}\to 2{\text{H}}_2\text{O}$

The charge is on both sides is balanced by adding electrons to the more positive side of the half- reaction to equal the less positive side of the half- reaction.

Oxidation: ${\text{Co}}^{\text{2}+}\to {\text{Co}}^{\text{3}+}+{\text{e}}^{-}$

Reduction: ${\text{H}}_2{\text{O}}_{\text{2}}+2{\text{H}}^{+}+2{\text{e}}^{-}\to 2{\text{H}}_2\text{O}$

Since, to make the equal loss and gain of electron, the oxidation reaction is multiplied by two.  Therefore, the overall cell reaction is obtained by addition the above two half-cell reactions.

    ${\text{2Co}}^{\text{2}+}+{\text{H}}_2{\text{O}}_{\text{2}}+2{\text{H}}^{+}+2{\text{e}}^{-}\to 2{\text{Co}}^{\text{3}+}+2{\text{e}}^{-}+2{\text{H}}_2\text{O}$

The simplified chemical equation after removing the chemical species of the similar kind is shown below.

    ${\text{2Co}}^{\text{2}+}\left(\text{aq}\right)+{\text{H}}_2{\text{O}}_{\text{2}}\left(\text{aq}\right)+2{\text{H}}^{+}\left(\text{aq}\right)\to 2{\text{Co}}^{\text{3}+}\left(\text{aq}\right)+2{\text{H}}_2\text{O}\left(l\right)$

Thus, the balanced chemical reaction is shown below.

  ${\text{2Co}}^{\text{2}+}\left(\text{aq}\right)+{\text{H}}_2{\text{O}}_{\text{2}}\left(\text{aq}\right)+2{\text{H}}^{+}\left(\text{aq}\right)\to 2{\text{Co}}^{\text{3}+}\left(\text{aq}\right)+2{\text{H}}_2\text{O}\left(l\right)$

(d)

Interpretation Introduction

Interpretation:

The given redox reaction is to be balanced in acidic medium and the oxidizing agent and reducing agent are to be identified.

Concept Introduction:

Same as part (a).

Answer to Problem 17QRT

The balanced redox reaction is shown below.

  ${\text{Cr}}_2{\text{O}}_{\text{7}}{}^{2-}\left(\text{aq}\right)+{\text{6Cl}}^{-}+14{\text{H}}^{+}\left(\text{aq}\right)\to 2{\text{Cr}}^{3+}\left(\text{aq}\right)+3{\text{Cl}}_2\left(\text{g}\right)+7{\text{H}}_2\text{O}\left(l\right)$

In this reaction, ${\text{Cl}}^{-}$ act as reducing agent and ${\text{Cr}}_2{\text{O}}_{\text{7}}{}^{2-}$ act as oxidizing agent.

Explanation of Solution

The given unbalanced redox reaction is shown below.

    ${\text{Cl}}^{-}\left(\text{aq}\right)+{\text{Cr}}_2{\text{O}}_{\text{7}}{}^{2-}\left(\text{aq}\right)\to {\text{Cl}}_2\left(\text{g}\right)+{\text{Cr}}^{3+}\left(\text{aq}\right)$

The oxidation number of chlorine atom in ${\text{Cl}}^{-}$ and ${\text{Cl}}_2$ is $-1$ and $0$ respectively.  Since there is increase in oxidation number, therefore, ${\text{Cl}}^{-}$ is oxidized to ${\text{Cl}}_2$ and ${\text{Cl}}^{-}$ act as reducing agent.

The oxidation number of manganese atom in ${\text{Cr}}_2{\text{O}}_{\text{7}}{}^{2-}$ and ${\text{Cr}}^{3+}$ is $+6$ and $+3$ respectively.  Since there is decrease in oxidation number, therefore, ${\text{Cr}}_2{\text{O}}_{\text{7}}{}^{2-}$ is reduced to ${\text{Cr}}^{3+}$ and ${\text{Cr}}_2{\text{O}}_{\text{7}}{}^{2-}$ act as oxidizing agent.

Therefore, the unbalanced half reactions are shown below.

Oxidation: ${\text{Cl}}^{-}\to {\text{Cl}}_2$

Reduction: ${\text{Cr}}_2{\text{O}}_{\text{7}}{}^{2-}\to {\text{Cr}}^{3+}$

The given reaction is to be balanced in acidic medium.  Therefore, oxygen atoms are balanced by adding water molecules to the deficient side.

Oxidation: ${\text{2Cl}}^{-}\to {\text{Cl}}_2$

Reduction: ${\text{Cr}}_2{\text{O}}_{\text{7}}{}^{2-}\to 2{\text{Cr}}^{3+}+7{\text{H}}_2\text{O}$

After balancing oxygen atoms, now hydrogen atoms are balanced by adding protons $\left({\text{H}}^{+}\right)$ to the deficient side.

Oxidation: ${\text{2Cl}}^{-}\to {\text{Cl}}_2$

Reduction: ${\text{Cr}}_2{\text{O}}_{\text{7}}{}^{2-}+14{\text{H}}^{+}\to 2{\text{Cr}}^{3+}+7{\text{H}}_2\text{O}$

The charge is on both sides is balanced by adding electrons to the more positive side of the half- reaction to equal the less positive side of the half- reaction.

Oxidation: ${\text{2Cl}}^{-}\to {\text{Cl}}_2+2{\text{e}}^{-}$

Reduction: ${\text{Cr}}_2{\text{O}}_{\text{7}}{}^{2-}+14{\text{H}}^{+}+6{\text{e}}^{-}\to 2{\text{Cr}}^{3+}+7{\text{H}}_2\text{O}$

Since the electron gain is not equivalent to electron lost.  Thus, multiply the oxidation half reaction by $3$ and reduction half reaction by $1$.

Oxidation: ${\text{6Cl}}^{-}\to 3{\text{Cl}}_2+6{\text{e}}^{-}$

Reduction: ${\text{Cr}}_2{\text{O}}_{\text{7}}{}^{2-}+14{\text{H}}^{+}+6{\text{e}}^{-}\to 2{\text{Cr}}^{3+}+7{\text{H}}_2\text{O}$

Now, there is the equal loss and gain of electron in the above two half-cell reactions.  Therefore, the overall cell reaction is get by addition the above two half-cell reactions.

    ${\text{Cr}}_2{\text{O}}_{\text{7}}{}^{2-}+14{\text{H}}^{+}+6{\text{e}}^{-}+{\text{6Cl}}^{-}\to 2{\text{Cr}}^{3+}+7{\text{H}}_2\text{O}+3{\text{Cl}}_2+6{\text{e}}^{-}$

The simplified chemical equation after removing the chemical species of the similar kind is shown below.

  ${\text{Cr}}_2{\text{O}}_{\text{7}}{}^{2-}\left(\text{aq}\right)+{\text{6Cl}}^{-}+14{\text{H}}^{+}\left(\text{aq}\right)\to 2{\text{Cr}}^{3+}\left(\text{aq}\right)+3{\text{Cl}}_2\left(\text{g}\right)+7{\text{H}}_2\text{O}\left(l\right)$

Thus, the balanced chemical reaction is shown below.

  ${\text{Cr}}_2{\text{O}}_{\text{7}}{}^{2-}\left(\text{aq}\right)+{\text{6Cl}}^{-}+14{\text{H}}^{+}\left(\text{aq}\right)\to 2{\text{Cr}}^{3+}\left(\text{aq}\right)+3{\text{Cl}}_2\left(\text{g}\right)+7{\text{H}}_2\text{O}\left(l\right)$

(e)

Interpretation Introduction

Interpretation:

The given redox reaction is to be balanced and the oxidizing agent and reducing agent are to be identified.

Concept Introduction:

Same as part (a).

Answer to Problem 17QRT

The balanced redox reaction is shown below.

  $\text{3Zn}\left(\text{s}\right)+{\text{2MnO}}_{\text{4}}{}^{-}\left(\text{aq}\right)+4{\text{H}}_2\text{O}\left(l\right)\to 3\text{Zn}{\left(\text{OH}\right)}_{\text{2}}\left(\text{s}\right)+2{\text{MnO}}_{\text{2}}\left(\text{s}\right)+2{\text{OH}}^{-}\left(\text{aq}\right)$

In this reaction, $\text{Zn}$ act as reducing agent and ${\text{MnO}}_{\text{4}}$ act as oxidizing agent.

Explanation of Solution

The given unbalanced redox reaction is shown below.

  ${\text{MnO}}_{\text{4}}{}^{-}\left(\text{aq}\right)+\text{Zn}\left(\text{s}\right)\to {\text{MnO}}_{\text{2}}\left(\text{s}\right)+\text{Zn}{\left(\text{OH}\right)}_2\left(\text{s}\right)$

The oxidation number of zinc atom in $\text{Zn}$ and $\text{Zn}{\left(\text{OH}\right)}_2$ is $0$ and $+2$ respectively.  Since there is increase in oxidation number, therefore, $\text{Zn}$ is oxidized to $\text{Zn}{\left(\text{OH}\right)}_2$ and $\text{Zn}$ act as reducing agent.

The oxidation number of manganese atom in ${\text{MnO}}_{\text{4}}$ and ${\text{MnO}}_2$ is $+7$ and $+4$ respectively.  Since there is decrease in oxidation number, therefore, ${\text{MnO}}_{\text{4}}$ is reduced to ${\text{MnO}}_2$ and ${\text{MnO}}_{\text{4}}$ act as oxidizing agent.

Therefore, the unbalanced half reactions are shown below.

Oxidation: $\text{Zn}\to \text{Zn}{\left(\text{OH}\right)}_{\text{2}}$

Reduction: ${\text{MnO}}_{\text{4}}{}^{-}\to {\text{MnO}}_{\text{2}}$

The given reaction is to be balanced in basic medium.  Therefore, oxygen atoms are balanced by adding water molecules to the deficient side.

Oxidation: $\text{Zn}+2{\text{H}}_2\text{O}\to \text{Zn}{\left(\text{OH}\right)}_{\text{2}}$

Reduction: ${\text{MnO}}_{\text{4}}{}^{-}\to {\text{MnO}}_{\text{2}}+2{\text{H}}_2\text{O}$

After balancing oxygen atoms, hydrogen atoms are balanced by adding water molecules $\left({\text{H}}_{\text{2}}\text{O}\right)$ to the deficient side and equal number of hydroxide ions $\left({\text{OH}}^{-}\right)$ to the other side.

Oxidation: $\text{Zn}+2{\text{H}}_2\text{O}+2{\text{OH}}^{-}\to \text{Zn}{\left(\text{OH}\right)}_{\text{2}}+2{\text{H}}_{\text{2}}\text{O}$

Reduction: ${\text{MnO}}_{\text{4}}{}^{-}+2{\text{H}}_2\text{O}\to {\text{MnO}}_{\text{2}}+4{\text{OH}}^{-}$

The charge is on both sides is balanced by adding electrons to the more positive side of the half- reaction to equal the less positive side of the half- reaction.

Oxidation: $\text{Zn}+2{\text{H}}_2\text{O}+2{\text{OH}}^{-}\to \text{Zn}{\left(\text{OH}\right)}_{\text{2}}+2{\text{H}}_{\text{2}}\text{O}+2{\text{e}}^{-}$

Reduction: ${\text{MnO}}_{\text{4}}{}^{-}+3{\text{e}}^{-}+2{\text{H}}_2\text{O}\to {\text{MnO}}_{\text{2}}+4{\text{OH}}^{-}$

Since the electron gain is not equivalent to electron lost.  Thus, multiply the oxidation half reaction by $3$ and reduction half reaction by $2$.

Oxidation: $\text{3Zn}+6{\text{H}}_2\text{O}+6{\text{OH}}^{-}\to 3\text{Zn}{\left(\text{OH}\right)}_{\text{2}}+6{\text{H}}_{\text{2}}\text{O}+6{\text{e}}^{-}$

Reduction: ${\text{2MnO}}_{\text{4}}{}^{-}+6{\text{e}}^{-}+4{\text{H}}_2\text{O}\to 2{\text{MnO}}_{\text{2}}+8{\text{OH}}^{-}$

Now, there is the equal loss and gain of electron in the above two half-cell reactions.  Therefore, the overall cell reaction is get by addition the above two half-cell reactions.

    $\begin{array}{l}\text{3Zn}+6{\text{H}}_2\text{O}+6{\text{OH}}^{-}+{\text{2MnO}}_{\text{4}}{}^{-}+6{\text{e}}^{-}+4{\text{H}}_2\text{O}\to \\ 3\text{Zn}{\left(\text{OH}\right)}_{\text{2}}+6{\text{H}}_{\text{2}}\text{O}+6{\text{e}}^{-}+2{\text{MnO}}_{\text{2}}+8{\text{OH}}^{-}\end{array}$

The simplified chemical equation after removing the chemical species of the similar kind is shown below.

  $\text{3Zn}\left(\text{s}\right)+{\text{2MnO}}_{\text{4}}{}^{-}\left(\text{aq}\right)+4{\text{H}}_2\text{O}\left(l\right)\to 3\text{Zn}{\left(\text{OH}\right)}_{\text{2}}\left(\text{s}\right)+2{\text{MnO}}_{\text{2}}\left(\text{s}\right)+2{\text{OH}}^{-}\left(\text{aq}\right)$

Thus, the balanced chemical reaction is shown below.

  $\text{3Zn}\left(\text{s}\right)+{\text{2MnO}}_{\text{4}}{}^{-}\left(\text{aq}\right)+4{\text{H}}_2\text{O}\left(l\right)\to 3\text{Zn}{\left(\text{OH}\right)}_{\text{2}}\left(\text{s}\right)+2{\text{MnO}}_{\text{2}}\left(\text{s}\right)+2{\text{OH}}^{-}\left(\text{aq}\right)$

(f)

Interpretation Introduction

Interpretation:

The given redox reaction is to be balanced and the oxidizing agent and reducing agent are to be identified.

Concept Introduction:

Same as part (a).

Answer to Problem 17QRT

The balanced redox reaction is shown below.

  ${\text{O}}_{\text{3}}\text{(g)}+\text{NO(g)}\to {\text{O}}_{\text{2}}\text{(g)}+{\text{NO}}_{\text{2}}\text{(g)}$

In this reaction, $\text{NO}$ act as reducing agent and ${\text{O}}_3$ act as oxidizing agent.

Explanation of Solution

The given redox reaction is shown below.

  ${\text{O}}_{\text{3}}\text{(g)}+\text{NO(g)}\to {\text{O}}_{\text{2}}\text{(g)}+{\text{NO}}_{\text{2}}\text{(g)}$

The given redox reaction is already balanced and there are equal numbers of each atom both the sides.

The oxidation number of nitrogen atom in $\text{NO}$ and ${\text{NO}}_2$ is $+2$ and $+4$ respectively.  Since there is increase in oxidation number, therefore, $\text{NO}$ is oxidized to ${\text{NO}}_2$ and $\text{NO}$ act as reducing agent.

The oxidation number of oxygen atom in ${\text{O}}_3$ and ${\text{NO}}_2$ is $0$ and $-2$ respectively.  Since there is decrease in oxidation number, therefore, ${\text{O}}_3$ is reduced to ${\text{NO}}_2$ and ${\text{O}}_3$ act as oxidizing agent.

(g)

Interpretation Introduction

Interpretation:

The given redox reaction is to be balanced in basic medium and the oxidizing agent and reducing agent are to be identified.

Concept Introduction:

Same as part (a).

Answer to Problem 17QRT

The balanced redox reaction is shown below.

  ${\text{C}}_{\text{3}}{\text{H}}_{\text{8}}\left(\text{g}\right)+5{\text{O}}_2\left(\text{g}\right)\to 3{\text{CO}}_2\left(\text{g}\right)+4{\text{H}}_2\text{O}\left(l\right)$

In this reaction, ${\text{C}}_{\text{3}}{\text{H}}_{\text{8}}$ act as reducing agent and ${\text{O}}_2$ act as oxidizing agent.

Explanation of Solution

The given unbalanced redox reaction is shown below.

    ${\text{C}}_{\text{3}}{\text{H}}_{\text{8}}\left(\text{g}\right)+{\text{O}}_2\left(\text{g}\right)\to {\text{CO}}_2\left(\text{g}\right)+{\text{H}}_2\text{O}\left(l\right)$

The phosphorous atoms and bromine atoms are balanced by putting $5$, $3$ and $4$ as prefix to ${\text{O}}_{\text{2}}$, ${\text{CO}}_2$ and ${\text{H}}_2\text{O}$ respectively.

  ${\text{C}}_{\text{3}}{\text{H}}_{\text{8}}\left(\text{g}\right)+5{\text{O}}_2\left(\text{g}\right)\to 3{\text{CO}}_2\left(\text{g}\right)+4{\text{H}}_2\text{O}\left(l\right)$

Thus, the balanced chemical reaction is shown below.

  ${\text{C}}_{\text{3}}{\text{H}}_{\text{8}}\left(\text{g}\right)+5{\text{O}}_2\left(\text{g}\right)\to 3{\text{CO}}_2\left(\text{g}\right)+4{\text{H}}_2\text{O}\left(l\right)$

The oxidation number of carbon atom in ${\text{C}}_{\text{3}}{\text{H}}_{\text{8}}$ and ${\text{CO}}_2$ is $+2,+3$ and $+4$ respectively.  Since there is increase in oxidation number, therefore, ${\text{C}}_{\text{3}}{\text{H}}_{\text{8}}$ is oxidized to ${\text{CO}}_2$ and ${\text{C}}_{\text{3}}{\text{H}}_{\text{8}}$ act as reducing agent.

The oxidation number of oxygen atom in ${\text{O}}_2$ and ${\text{H}}_2\text{O}$ is $0$ and $-2$ respectively.  Since there is decrease in oxidation number, therefore, ${\text{O}}_2$ is reduced to ${\text{H}}_2\text{O}$ and ${\text{O}}_2$ act as oxidizing agent.