Chapter 17, Problem 10PE

( a)

Interpretation Introduction

Interpretation:

Oxidizing agent, reducing agent, oxidized element and reduced element in below reaction has to be determined.

${\text{CH}}_4+{\text{O}}_2\to {\text{CO}}_2+{\text{H}}_2\text{O}$

Concept Introduction:

A redox reaction is a type of reaction that involves the change in oxidation number of a molecule, atom or ion changes due to transfer of electron from one species to another. A complete reaction involves oxidation and reduction.

Oxidation process represents loss of electrons. Reduction process represents gain of electrons. Oxidation numbers are the part of system that is formed to track electrons in reaction.

Explanation of Solution

Oxidation number of $\text{H}$ is $+1$.

Expression to calculate the oxidation number of $\text{C}$ in ${\text{CH}}_4$ is as follows:

  $\left[\left(\text{Oxidation}\text{number}\text{of}\text{C}\right)+4\left(\text{oxidation}\text{number}\text{of}\text{H}\right)\right]=0$        (1)

Rearrange equation (1).

  $\text{Oxidation}\text{number}\text{of}\text{C}=\left[-4\left(\text{oxidation}\text{number}\text{of}\text{H}\right)\right]$        (2)

Substitute of $+1$ oxidation number for $\text{H}$ in equation (2).

  $\begin{array}{c}\text{Oxidation}\text{number}\text{of}\text{C}=\left[-4\left(\text{oxidation}\text{number}\text{of}\text{H}\right)\right]\\ =\left[-4\left(+1\right)\right]\\ =-4\end{array}$

Oxidation number of $\text{O}$ is $-2$.

Expression to calculate the oxidation number of $\text{C}$ in ${\text{CO}}_2$ is as follows:

  $\left[\left(\text{Oxidation}\text{number}\text{of}\text{C}\right)+\left(\text{oxidation}\text{number}\text{of}\text{O}\right)\right]=0$        (3)

Rearrange equation (3) for oxidation number of $\text{C}$.

  $\text{Oxidation}\text{number}\text{of}\text{C}=\left[-2\left(\text{oxidation}\text{number}\text{of}\text{O}\right)\right]$        (4)

Substitute $-2$ for oxidation number of $\text{O}$ in equation (4).

  $\begin{array}{c}\text{Oxidation}\text{number}\text{of}\text{C}=\left[-2\left(\text{oxidation}\text{number}\text{of}\text{O}\right)\right]\\ =\left[-2\left(-2\right)\right]\\ =+4\end{array}$

Oxidation number of ${\text{O}}_2$ is 0.

Oxidation number of $\text{H}$ is $+1$.

Expression to calculate the oxidation number of $\text{O}$ in ${\text{H}}_2\text{O}$ is as follows:

  $\left[2\left(\text{Oxidation}\text{number}\text{of}\text{H}\right)+\left(\text{oxidation}\text{number}\text{of}\text{O}\right)\right]=0$        (5).

Rearrange equation (5) for oxidation number of $\text{O}$ in ${\text{H}}_2\text{O}$.

  $\text{Oxidation}\text{number}\text{of}\text{O}=\left[-2\left(\text{oxidation}\text{number}\text{of}\text{H}\right)\right]$        (6)

Substitute $+1$ for oxidation number of $\text{H}$ in equation (6).

  $\begin{array}{c}\text{Oxidation}\text{number}\text{of}\text{O}=\left[-2\left(\text{oxidation}\text{number}\text{of}\text{H}\right)\right]\\ =\left[-2\left(1\right)\right]\\ =-2\end{array}$

Oxidation state of $\text{C}$ in ${\text{CH}}_4$ is $-4$ and it is increased to $+4$ in ${\text{CO}}_2$. Hence, ${\text{CH}}_4$ undergoes oxidation and so it acts as reducing agent. Oxidized element in given reaction is carbon.

Oxidation state of $\text{O}$ in ${\text{O}}_2$ is 0 and it is decreased to $-2$ in ${\text{H}}_2\text{O}$. Hence, ${\text{O}}_2$ undergo reduction and so it acts as oxidizing agent. Reduced element in given reaction is oxygen.

(b)

Interpretation Introduction

Interpretation:

Oxidizing agent, reducing agent, oxidized element and reduced element in below reaction has to be determined.

  $\text{Mg}+{\text{FeCl}}_3\to \text{Fe}+{\text{MgCl}}_2$

Concept Introduction:

Refer to part (a).

Explanation of Solution

Oxidation number of $\text{Mg}$ is 0.

Oxidation number of $\text{Cl}$ is $-1$.

Expression to calculate the oxidation number of $\text{Mg}$ in ${\text{MgCl}}_2$ is as follows:

  $\left[\left(\text{Oxidation}\text{number}\text{of}\text{Mg}\right)+2\left(\text{oxidation}\text{number}\text{of}\text{Cl}\right)\right]=0$        (7)

Rearrange equation (7) for oxidation number of $\text{Mg}$ in ${\text{MgCl}}_2$.

  $\text{Oxidation}\text{number}\text{of}\text{Mg}=\left[-2\left(\text{oxidation}\text{number}\text{of}\text{Cl}\right)\right]$        (8)

Substitute $-1$ for oxidation number of $\text{Cl}$ in equation (8).

  $\begin{array}{c}\text{Oxidation}\text{number}\text{of}\text{Mg}=\left[-2\left(\text{oxidation}\text{number}\text{of}\text{Cl}\right)\right]\\ =\left[-2\left(-1\right)\right]\\ =+2\end{array}$

Expression to calculate the oxidation number of $\text{Fe}$ in ${\text{FeCl}}_3$ is as follows:

  $\left[\left(\text{Oxidation}\text{number}\text{of}\text{Fe}\right)+3\left(\text{oxidation}\text{number}\text{of}\text{Cl}\right)\right]=0$        (9)

Rearrange equation (9) for oxidation number of $\text{Fe}$ in ${\text{FeCl}}_3$.

  $\text{Oxidation}\text{number}\text{of}\text{Fe}\text{=}\left[-3\left(\text{oxidation}\text{number}\text{of}\text{Cl}\right)\right]$        (10)

Substitute $-1$ for oxidation number of $\text{Cl}$ in equation (10).

  $\begin{array}{c}\text{Oxidation}\text{number}\text{of}\text{Fe}\text{=}\left[-3\left(\text{oxidation}\text{number}\text{of}\text{Cl}\right)\right]\\ =\left[-3\left(-1\right)\right]\\ =3\end{array}$

Oxidation state of $\text{Mg}$ is 0 and it is increased to $+2$ in ${\text{MgCl}}_2$. Hence $\text{Mg}$ is oxidized and it behaves as reducing agent. Oxidized element in given reaction is magnesium.

Oxidation state of $\text{Fe}$ in ${\text{FeCl}}_3$ is $+3$ and it is increased to 0 in $\text{Fe}$. Hence ${\text{FeCl}}_3$ is reduced and it behaves as oxidizing agent. Reduced element in given reaction is iron.