Chapter 17, Problem 1RQ

(a)

Interpretation Introduction

Interpretation:

Whether iodine is oxidized or reduced in reaction below has to be determined.

  ${\text{I}}_2+5{\text{Cl}}_{\text{2}}+6{\text{H}}_{\text{2}}\text{O}\to 2{\text{HIO}}_{\text{3}}+10\text{HCl}$

Concept Introduction:

Oxidation number is integer value allotted to every element. It is formal charge occupied by atom if all of its bonds are dissociated heterolytically. Below mentioned are rules to assign oxidation numbers to various elements.

1. Elements present in their free state have zero oxidation number.

2. Oxidation number of hydrogen is generally $+1$, except for metal hydrides.

3. Oxidation number of oxygen is $-2$, except for peroxides.

4. Metals have positive oxidation numbers.

5. Negative oxidation numbers are assigned to most electronegative element in covalent compounds.

6. Sum of oxidation numbers of different elements in neutral atom is zero.

7. Sum of oxidation numbers of various elements in polyatomic ion is equal to charge present on ion.

Answer to Problem 1RQ

Iodine undergoes oxidation in this reaction.

Explanation of Solution

Given chemical equation is as follows:

  ${\text{I}}_2+5{\text{Cl}}_{\text{2}}+6{\text{H}}_{\text{2}}\text{O}\to 2{\text{HIO}}_{\text{3}}+10\text{HCl}$

Oxidation of element in their elemental state is 0. Thus oxidation state of $\text{I}$ in ${\text{I}}_2$ is 0 and that of $\text{Cl}$ in ${\text{Cl}}_2$ is 0.

Expression for oxidation number in ${\text{HIO}}_3$ is as follows:

  $\left[\begin{array}{l}1\left(\text{Oxidation number of H}\right)+\\ 1\left(\text{Oxidation number of I}\right)+\\ 3\left(\text{Oxidation number of O}\right)\end{array}\right]=0$        (1)

Rearrange equation (1) for oxidation number of $\text{I}$.

  $\text{Oxidation number of I}=\left(\begin{array}{l}-\left(\text{Oxidation number of H}\right)\\ -3\left(\text{Oxidation number of O}\right)\end{array}\right)$        (2)

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

  $\begin{array}{c}\text{Oxidation number of I}=-\left(+1\right)-3\left(-2\right)\\ =+5\end{array}$

Hence, oxidation number of $\text{I}$ is $+5$.

Oxidation state of iodine in ${\text{I}}_2$ is 0 and it increases to $+5$ in ${\text{HIO}}_3$. Hence iodine undergoes oxidation in this reaction.

(b)

Interpretation Introduction

Interpretation:

Whether chlorine is oxidized or reduced in reaction below has to be determined.

  ${\text{I}}_2+5{\text{Cl}}_{\text{2}}+6{\text{H}}_{\text{2}}\text{O}\to 2{\text{HIO}}_{\text{3}}+10\text{HCl}$

Concept Introduction:

Refer to part (a).

Answer to Problem 1RQ

Chlorine undergoes reduction in this reaction.

Explanation of Solution

Given chemical equation is as follows:

  ${\text{I}}_2+5{\text{Cl}}_{\text{2}}+6{\text{H}}_{\text{2}}\text{O}\to 2{\text{HIO}}_{\text{3}}+10\text{HCl}$

Oxidation of element in their elemental state is 0. Thus oxidation state of $\text{I}$ in ${\text{I}}_2$ is 0 and that of $\text{Cl}$ in ${\text{Cl}}_2$ is 0.

Expression for oxidation number in $\text{HCl}$ is as follows:

  $\left[\begin{array}{l}1\left(\text{Oxidation number of H}\right)+\\ 1\left(\text{Oxidation number of Cl}\right)\end{array}\right]=0$        (3)

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

  $\text{Oxidation number of Cl}=-\left(\text{Oxidation number of H}\right)$        (4)

Substitute $+1$ for oxidation number of $\text{H}$ in equation (4).

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

Hence, oxidation number of $\text{Cl}$ is $-1$.

Oxidation state of chlorine in ${\text{Cl}}_2$ is 0 and it decreases to $-1$ in $\text{HCl}$. Hence chlorine undergoes reduction in this reaction.