Chapter 20, Problem 47A

(a)

Interpretation Introduction

Interpretation: The balanced chemical equation of the reaction, $Mn{O}_4{}^{-}\left(aq\right)+Cl{O}_2{}^{-}\left(aq\right)\to Mn{O}_2\left(s\right)+Cl{O}_4{}^{-}\left(aq\right)$ is to be interpreted using the half-reaction method.

Concept Introduction: The chemical reaction that involves both oxidation and reduction is the redox reaction. Redox reactions can be balanced by following the half-reaction method. That is,

• First, the given redox reaction is divided into half reactions one oxidation and the other reduction.
• Based on mass and charge the half-reactions are balanced.
• The electrons on both sides of the reaction are equalized.
• Then add the half-reactions together and the common one is eliminated to get the balanced equation.
• In the basic medium, the reaction involves the use of $O{H}^{-}$ to balance oxygen atoms. Since a base is a compound that releases $O{H}^{-}$ ions.

Answer to Problem 47A

The balanced chemical equation of the reaction is, $4Mn{O}_4{}^{-}\left(aq\right)+3Cl{O}_2{}^{-}\left(aq\right)+2H_{2}O\to 4Mn{O}_2\left(s\right)+3Cl{O}_4{}^{-}\left(aq\right)+4O{H}^{-}$

Explanation of Solution

In the given reaction,

  $Mn{O}_4{}^{-}\left(aq\right)+Cl{O}_2{}^{-}\left(aq\right)\to Mn{O}_2\left(s\right)+Cl{O}_4{}^{-}\left(aq\right)$

The half-reactions can be written as

Oxidation: $C{l}^{3+}\to C{l}^{7+}+4e^{-}$

Reduction: $M{n}^{7+}+3e^{-}\to M{n}^{4+}$

The electron count can be balanced by multiplying 3 by the oxidation and 4 by the reduction of half-reactions.

Therefore, the equation becomes,

  $4Mn{O}_4{}^{-}\left(aq\right)+3Cl{O}_2{}^{-}\left(aq\right)\to 4Mn{O}_2\left(s\right)+3Cl{O}_4{}^{-}\left(aq\right)$

Now, the oxygen count is not balanced on the left-hand side it is 22 while on the right-hand side it is 20. So it can be balanced by adding two $O{H}^{-}$ on the right-hand side. The addition of water to the left-hand side is necessary to balance the $H$ count. So, a total of two molecules of water on the left-hand side and $4O{H}^{-}$ on the right-hand side is needed.

The balanced chemical equation thus becomes,

  $4Mn{O}_4{}^{-}\left(aq\right)+3Cl{O}_2{}^{-}\left(aq\right)+2H_{2}O\to 4Mn{O}_2\left(s\right)+3Cl{O}_4{}^{-}\left(aq\right)+4O{H}^{-}$

(b)

Interpretation Introduction

Interpretation: The balanced chemical equation of the reaction, $C{r}^{3+}\left(aq\right)+Cl{O}^{-}\left(aq\right)\to Cr{O}_4{}^{2-}\left(aq\right)+C{l}^{-}\left(aq\right)$ is to be interpreted using the half-reaction method.

Concept Introduction: The chemical reaction that involves both oxidation and reduction is the redox reaction. Redox reactions can be balanced by following the half-reaction method. That is,

• First, the given redox reaction is divided into half reactions one oxidation and the other reduction.
• Based on mass and charge the half-reactions are balanced.
• The electrons on both sides of the reaction are equalized.
• Then add the half-reactions together and the common one is eliminated to get the balanced equation.
• In the basic medium, the reaction involves the use of $O{H}^{-}$ to balance oxygen atoms. Since a base is a compound that releases $O{H}^{-}$ ions.

Answer to Problem 47A

The balanced chemical equation of the reaction is, $2C{r}^{3+}\left(aq\right)+3Cl{O}^{-}\left(aq\right)+10O{H}^{-}\to 2Cr{O}_4{}^{2-}\left(aq\right)+3C{l}^{-}\left(aq\right)+5H_{2}O$ .

Explanation of Solution

In the given reaction,

  $C{r}^{3+}\left(aq\right)+Cl{O}^{-}\left(aq\right)\to Cr{O}_4{}^{2-}\left(aq\right)+C{l}^{-}\left(aq\right)$

The half-reactions can be written as,

Oxidation: $C{r}^{3+}\to C{r}^{6+}+3e^{-}$

Reduction: $C{l}^{-}+2e^{-}\to C{l}^{-}$

The electron count can be balanced by multiplying the oxidation equation by 2 and reduction half reactions by 3.

Now the equation becomes,

  $2C{r}^{3+}\left(aq\right)+3Cl{O}^{-}\left(aq\right)\to 2Cr{O}_4{}^{2-}\left(aq\right)+3C{l}^{-}\left(aq\right)$

The oxygen count is not equal on both sides. It can be balanced by the proper addition of $O{H}^{-}$ ions since the basic medium is used. The addition of $10O{H}^{-}$ on the left-hand side and 5 water molecules on the right-hand side makes the equation balanced.

So, the balanced equation is,

  $2C{r}^{3+}\left(aq\right)+3Cl{O}^{-}\left(aq\right)+10O{H}^{-}\to 2Cr{O}_4{}^{2-}\left(aq\right)+3C{l}^{-}\left(aq\right)+5H_{2}O$

(c)

Interpretation Introduction

Interpretation: The balanced chemical equation of the reaction, $M{n}^{3+}\left(aq\right)+I^{-}\left(aq\right)\to M{n}^{2+}\left(aq\right)+I{O}_3{}^{-}\left(aq\right)$ is to be interpreted using the half-reaction method.

Concept Introduction: The chemical reaction that involves both oxidation and reduction is the redox reaction. Redox reactions can be balanced by following the half-reaction method. That is,

• First, the given redox reaction is divided into half reactions one oxidation and the other reduction.
• Based on mass and charge the half-reactions are balanced.
• The electrons on both sides of the reaction are equalized.
• Then add the half-reactions together and the common one is eliminated to get the balanced equation.
• In the basic medium, the reaction involves the use of $O{H}^{-}$ to balance oxygen atoms. Since a base is a compound that releases $O{H}^{-}$ ions.

Answer to Problem 47A

The balanced chemical equation of the reaction is, $6M{n}^{3+}\left(aq\right)+I^{-}\left(aq\right)+6O{H}^{-}\to 6M{n}^{2+}\left(aq\right)+I{O}_3{}^{-}\left(aq\right)+3H_{2}O$

Explanation of Solution

In the given reaction,

  $M{n}^{3+}\left(aq\right)+I^{-}\left(aq\right)\to M{n}^{2+}\left(aq\right)+I{O}_3{}^{-}\left(aq\right)$

The half-reactions of this reaction are,

Oxidation: $M{n}^{3+}\to M{n}^{2+}+1e^{-}$

Reduction: $I^{-}+6e^{-}\to I^{5+}$

The electron count can be balanced by multiplying the oxidation reaction by 6. Therefore the equation becomes,

  $6M{n}^{3+}\left(aq\right)+I^{-}\left(aq\right)\to 6M{n}^{2+}\left(aq\right)+I{O}_3{}^{-}\left(aq\right)$

Now the oxygen count is not equal on both sides it can be balanced by the proper addition of $O{H}^{-}$ and water molecules. The addition of $6O{H}^{-}$ on the left-hand side and 3 water molecules on the right-hand side make the equation balanced.

The balanced equation is,

  $6M{n}^{3+}\left(aq\right)+I^{-}\left(aq\right)+6O{H}^{-}\to 6M{n}^{2+}\left(aq\right)+I{O}_3{}^{-}\left(aq\right)+3H_{2}O$