Chapter 20.3, Problem 29LC

a)

Interpretation Introduction

Interpretation : Using the oxidation number change method, the chemical reaction is to be balanced.

Concept Introduction : Redox reaction is a chemical reaction in which one reacting species undergoes oxidation while other reacting species undergo reduction.

Answer to Problem 29LC

The required balance reaction is as follows:

  $Cl{O}_3^{-}(aq)+\text{ 6}{I}^{-}{\text{(aq)+6H}}^{+}(aq)\to C{l}^{-}\text{(aq)}+\text{ 3}{I}_2(aq)+3H_{2}O(l)$

Explanation of Solution

Rules to calculate the oxidation number are as follows:

• All elements of the alkali group in a combined form with other atoms always show $\left(+1\right)$ oxidation state.
• All elements of alkaline earth metal groups in a combined form with other atoms always show $\left(+2\right)$ oxidation state.
• The charge of the oxygen atom is generally $\left(-2\right)$ .
• All elements of halogen groups in a combined form with other atoms always show $\left(-1\right)$ oxidation state.
• The charge of the hydrogen atom is always $\left(+1\right)$ .
• Atoms in their elemental state have zero oxidation state.

Given,

  $Cl{O}_3^{-}(aq)+I^{-}\text{(aq)}\to C{l}^{-}\text{(aq)}+I_2(aq)[acidicsolution]$ .

The steps of the method are as follows:

Step 1: Write the ionic form of the unbalanced chemical equation.

  $Cl{O}_3^{-}(aq)+I^{-}\text{(aq)}\to C{l}^{-}\text{(aq)}+I_2(aq)[acidicsolution]$

Step 2: Identify and write oxidation/reduction half-reactions separately.

  $\begin{array}{l}\text{+5 -1 -1 0}\\ Cl{O}_3^{-}(aq)+I^{-}\text{(aq)}\to C{l}^{-}\text{(aq)}+I_2(aq)[acidicsolution]\end{array}$

Calculation of the oxidation number of atoms is as follows:

The oxidation number of $Cl$ in $Cl{O}_{{}_3}^{-}$

is calculated as follows:

Let the oxidation number of the chlorine atom be x and the charge of oxygen atoms is always $\left(-2\right)$ .

  $\begin{array}{l}x+\left(3\times \left(-2\right)\right)+1=0\\ x-6+1=0\\ x=+5\end{array}$

Here, $\left(+1\right)$ is the negative charge.

So, let the oxidation number of the iodine atom of $I_2$ be x.

  $\begin{array}{l}2x=0\\ x=\frac{0}{2}\approx notdefiend\\ x=0\end{array}$

Step 3: The reaction is a redox reaction so, the oxidized and reduced atom is to be identified.

Now, balance the charge by adding electrons

  

The reaction after balancing the charge is as follows:

  $Cl{O}_3^{-}(aq)+6e^{-}\text{+ }{I}^{-}\text{(aq)}\to C{l}^{-}\text{(aq)}+I_2(aq)+2e^{-}$

This can be understood as follows:

  $\begin{array}{l}\\ +5-2-1\\ Cl{O}_3^{-}(aq)+6e^{-}\to C{l}^{-}\text{(aq)}\end{array}$

  $\begin{array}{l}-10\\ I^{-}\text{(aq)}\to I_2(aq)+2e^{-}\end{array}$

Step 3: Balance the atoms other than oxygen on both sides of the reaction.

  $Cl{O}_3^{-}(aq)+6e^{-}\text{+2}{I}^{-}\text{(aq)}\to C{l}^{-}\text{(aq)}+I_2(aq)+2e^{-}$

Step 4: Now, balance the oxygen atom by adding a water molecule.

  $Cl{O}_3^{-}(aq)+6e^{-}\text{+2}{I}^{-}\text{(aq)}\to C{l}^{-}\text{(aq)}+I_2(aq)+2e^{-}+3H_{2}O(l)$

Step 5: Now balance the hydrogen atom by adding ${\text{H}}^{+}$ ions as the reaction is occurring in an acidic medium.

  $Cl{O}_3^{-}(aq)+\text{ 6}{I}^{-}{\text{(aq)+6H}}^{+}(aq)\to C{l}^{-}\text{(aq)}+\text{ 3}{I}_2(aq)+3H_{2}O(l)$

The required balance reaction is as follows:

  $Cl{O}_3^{-}(aq)+\text{ 6}{I}^{-}{\text{(aq)+6H}}^{+}(aq)\to C{l}^{-}\text{(aq)}+\text{ 3}{I}_2(aq)+3H_{2}O(l)$

b)

Interpretation Introduction

Interpretation : Using the oxidation number change method, the chemical reaction is to be balanced.

Concept Introduction : Oxidation reaction is a part of redox reaction in which the removal of electrons or addition of oxygen takes place.

Answer to Problem 29LC

The required balance reaction is as follows:

  $5C_2{O}_4^{2-}(aq)+16H^{+}+2Mn{O}_4{}^{-}\text{(aq)}\to 2M{n}^{2+}\text{(aq)}+10C{O}_2(aq)+8H_{2}O(l)$

Explanation of Solution

Rules to calculate the oxidation number are as follows:

• All elements of the alkali group in a combined form with other atoms always show $\left(+1\right)$ oxidation state.
• All elements of alkaline earth metal groups in a combined form with other atoms always show $\left(+2\right)$ oxidation state.
• The charge of the oxygen atom is generally $\left(-2\right)$ .
• All elements of halogen groups in a combined form with other atoms always show $\left(-1\right)$ oxidation state.
• The charge of the hydrogen atom is always $\left(+1\right)$ .
• Atoms in their elemental state have zero oxidation state.

Given,

  $C_2{O}_4^{2-}(aq)+Mn{O}_4{}^{-}\text{(aq)}\to M{n}^{2+}\text{(aq)}+C{O}_2(aq)[acidicsolution]$ .

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The steps of the method are as follows:

Step 1: Write the ionic form of the unbalanced chemical equation.

  $C_2{O}_4^{2-}(aq)+Mn{O}_4{}^{-}\text{(aq)}\to M{n}^{2+}\text{(aq)}+C{O}_2(aq)[acidicsolution]$

Step 2: Identify and write oxidation/reduction half-reactions separately.

  $\begin{array}{l}\text{+3 +7 +2 +4}\\ C_2{O}_4^{2-}(aq)+Mn{O}_4{}^{-}\text{(aq)}\to M{n}^{2+}\text{(aq)}+C{O}_2(aq)[acidicsolution]\end{array}$

Calculation of the oxidation number of atoms is as follows:

The oxidation number of $C$ in $C_2{O}_{{}_4}^{2-}$

is calculated as follows:

Let the oxidation number of the carbon atom be x and the charge of oxygen atoms is always $\left(-2\right)$ .

  $\begin{array}{l}2x+\left(4\times \left(-2\right)\right)+2=0\\ 2x-8+2=0\\ 2x=6\\ x=\frac{6}{2}=+3\end{array}$

Here, $\left(+2\right)$ is the negative charge.

So, let the oxidation number of the iodine atom of $Mn$ be x in $Mn{O}_{{}_4}^{-}$ .

  $\begin{array}{l}x+(4\times (-2))+1=0\\ x-8+1=0\\ x=+7\end{array}$

Step 3: The reaction is a redox reaction so, the oxidized and reduced atom is to be identified.

Now, balance the charge by adding electrons

  

The reaction after balancing the charge is as follows:

  $C_2{O}_4^{2-}(aq)+Mn{O}_4{}^{-}{\text{(aq)+10e}}^{-}\to M{n}^{2+}\text{(aq)}+C{O}_2(aq)+10e^{-}$

This can be understood as follows:

  $\begin{array}{l}+3\text{ +4}\\ C_2{O}_4^{2-}(aq)\to +C{O}_2(aq)+10e^{-}\end{array}$

  $\begin{array}{l}+7\text{ +2}\\ 2Mn{O}_4{}^{-}{\text{(aq)+10e}}^{-}\to 2M{n}^{2+}\text{(aq)}\end{array}$

Step 3: Balance the atoms other than oxygen on both sides of the reaction.

  $5C_2{O}_4^{2-}(aq)+2Mn{O}_4{}^{-}\text{(aq)}\to 2M{n}^{2+}\text{(aq)}+10C{O}_2(aq)$

Step 4: Now, balance the oxygen atom by adding a water molecule.

  $5C_2{O}_4^{2-}(aq)+2Mn{O}_4{}^{-}\text{(aq)}\to 2M{n}^{2+}\text{(aq)}+10C{O}_2(aq)+8H_{2}O(l)$

Step 5: Now balance the hydrogen atom by adding ${\text{H}}^{+}$ ions as the reaction is occurring in an acidic medium.

  $5C_2{O}_4^{2-}(aq)+16H^{+}+2Mn{O}_4{}^{-}\text{(aq)}\to 2M{n}^{2+}\text{(aq)}+10C{O}_2(aq)+8H_{2}O(l)$

The required balance reaction is as follows:

  $5C_2{O}_4^{2-}(aq)+16H^{+}+2Mn{O}_4{}^{-}\text{(aq)}\to 2M{n}^{2+}\text{(aq)}+10C{O}_2(aq)+8H_{2}O(l)$

c)

Interpretation Introduction

Interpretation : Using the oxidation number change method, the chemical reaction is to be balanced.

Concept Introduction : Reduction reaction is a part of redox reaction which involves the addition of electrons and the species that undergo reduction act as an oxidizing agent.

Answer to Problem 29LC

The required balance reaction is as follows:

  $Cl{O}_3^{-}(aq)+\text{ 6}{I}^{-}{\text{(aq)+6H}}^{+}(aq)\to C{l}^{-}\text{(aq)}+\text{ 3}{I}_2(aq)+3H_{2}O(l)$

Explanation of Solution

Rules to calculate the oxidation number are as follows:

• All elements of the alkali group in a combined form with other atoms always show $\left(+1\right)$ oxidation state.
• All elements of alkaline earth metal groups in a combined form with other atoms always show $\left(+2\right)$ oxidation state.
• The charge of the oxygen atom is generally $\left(-2\right)$ .
• All elements of halogen groups in a combined form with other atoms always show $\left(-1\right)$ oxidation state.
• The charge of the hydrogen atom is always $\left(+1\right)$ .
• Atoms in their elemental state have zero oxidation state.

Given,

  $B{r}_2(l)+S{O}_2\text{(g)}\to B{r}^{-}\text{(aq)}+S{O}_4{}^{2-}(aq)[acidicsolution]$ .

The steps of the method are as follows:

Step 1: Write the ionic form of the unbalanced chemical equation.

  $Cl{O}_3^{-}(aq)+I^{-}\text{(aq)}\to C{l}^{-}\text{(aq)}+I_2(aq)[acidicsolution]$

Step 2: Identify and write oxidation/reduction reactions.

  $\begin{array}{l}\text{0 +4 -2 -1 +6 -2}\\ B{r}_2(l)+S{O}_2\text{(g)}\to B{r}^{-}\text{(aq)}+S{O}_4{}^{2-}(aq)[acidicsolution]\end{array}$

Calculation of the oxidation number of atoms is as follows:

The oxidation number of $S$ in $S{O}_2$

is calculated as follows:

Let the oxidation number of the chlorine atom be x and the charge of oxygen atoms is always $\left(-2\right)$ .

  $\begin{array}{l}x+\left(2\times \left(-2\right)\right)=0\\ x-4=0\\ x=+4\end{array}$

So, let the oxidation number of the iodine atom of $B{r}_2$ be x.

  $\begin{array}{l}2x=0\\ x=\frac{0}{2}\approx notdefiend\\ x=0\end{array}$

Step 3: The reaction is a redox reaction so, the oxidized and reduced atom is to be identified.

Now, balance the charge by adding electrons

  

The reaction after balancing the charge is as follows:

  $B{r}_2(l)+S{O}_2(g)\to {\text{ SO}}_4{}^{2-}\text{(aq)}+{\text{ Br}}^{-}(aq)$

This can be understood as follows:

  $\begin{array}{l}\text{0 }-2-1\\ B{r}_2(l)+2e^{-}\to {\text{ 2Br}}^{-}\text{(aq)}\end{array}$

  $\begin{array}{l}\text{+4 -2 +6 -2 }\\ {\text{ SO}}_2\text{(g)}\to S{O}_4^{2-}(aq)+2e^{-}\end{array}$

Step 4: Balance the atoms other than oxygen on both sides of the reaction.

  $B{r}_2(l)+S{O}_2(g)\to {\text{ SO}}_4{}^{2-}\text{(aq)}+{\text{ Br}}^{-}(aq)$

Step 5: Now, balance the oxygen atom by adding a water molecule.

  $B{r}_2(l)+S{O}_2(g)+2H_{2}O\to {\text{ SO}}_4{}^{2-}\text{(aq)}+{\text{ Br}}^{-}(aq)$

Step 6: Now balance the hydrogen atom by adding ${\text{H}}^{+}$ ions as the reaction is occurring in an acidic medium.

  $B{r}_2(l)+S{O}_2(g)+2H_{2}O\to {\text{ SO}}_4{}^{2-}\text{(aq)}+{\text{ Br}}^{-}(aq)$

The required balance reaction is as follows:

  $B{r}_2(l)+{\text{ SO}}_2{\text{(g)+2H}}_2\text{O}(l)\to {\text{ 2Br}}^{-}\text{(aq)}+{\text{ SO}}_4^{2-}(aq)+4H^{+}(aq)$