Chapter 18, Problem 27A

(a)

Interpretation Introduction

Interpretation: The given half-reaction which take place in acidic solution needs to be balanced.

${\text{IO}}_3^{-}\left(\text{aq}\right)+{\text{I}}^{-}\left(\text{aq}\right)\to {\text{I}}_{\text{2}}\left(\text{aq}\right)$

Concept Introduction: By transfer of electrons, oxidation-reduction reactions are characterized. The transfer occurs to form ions. Oxidation reaction is caused by losing electrons and reduction caused by gaining electrons.

Answer to Problem 27A

The balance reduction half-reaction is-

  ${\text{2IO}}_3^{-}\left(\text{aq}\right)+12{\text{H}}^{\text{+}}+10{\text{e}}^{-}\to {\text{I}}_2\left(\text{aq}\right)+6{\text{H}}_{\text{2}}\text{O}$

The balance oxidation half-reaction is-

${\text{2I}}^{-}\left(\text{aq}\right)\to {\text{I}}_{\text{2}}\left(\text{aq}\right)+2{\text{e}}^{-}$

Explanation of Solution

The oxidation state of the each element in this reaction is calculated as follows:

${\text{IO}}_3^{-}\left(\text{aq}\right)+{\text{I}}^{-}\left(\text{aq}\right)\to {\text{I}}_{\text{2}}\left(\text{aq}\right)$

${\text{I}}_{\text{2}}$ is in free state so, its oxidation state is zero. Oxidation state of I in ${\text{IO}}_3^{-}$ is $+5$ , O is $-2$ and ${\text{I}}^{-}$ is $-1$ .

Oxidation state decreases from $+5$ to 0 in ${\text{IO}}_3^{-}$ to ${\text{I}}_{\text{2}}$ . Hence, it is reduced. The reduction half-reaction is as follows:

${\text{IO}}_3^{-}\left(\text{aq}\right)\to {\text{I}}_2\left(\text{aq}\right)$

Balance I by multiplying 2 in reactant side.

${\text{2IO}}_3^{-}\left(\text{aq}\right)\to {\text{I}}_2\left(\text{aq}\right)$

Balance O by using ${\text{H}}_{\text{2}}\text{O}$ and H by adding ${\text{H}}^{+}$ .

${\text{2IO}}_3^{-}\left(\text{aq}\right)+12{\text{H}}^{\text{+}}\to {\text{I}}_2\left(\text{aq}\right)+6{\text{H}}_{\text{2}}\text{O}$

Now, add electrons to balance charge. To balance charge on both sides, add ten electrons on the left. Hence, the balanced equation is-

${\text{2IO}}_3^{-}\left(\text{aq}\right)+12{\text{H}}^{\text{+}}+10{\text{e}}^{-}\to {\text{I}}_2\left(\text{aq}\right)+6{\text{H}}_{\text{2}}\text{O}$ …… (1)

Oxidation state of ${\text{I}}^{-}$ is increases from $-1$ to $0$ . Hence, it is oxidized. The oxidation half-reaction is as follows:

${\text{I}}^{-}\left(\text{aq}\right)\to {\text{I}}_{\text{2}}\left(\text{aq}\right)$

To balance I in both side, multiply reactant side I by to 2.

${\text{2I}}^{-}\left(\text{aq}\right)\to {\text{I}}_{\text{2}}\left(\text{aq}\right)$

To balance charge on both sides, add $2{\text{e}}^{-}$ to the right. Hence, the balanced equation is-

${\text{2I}}^{-}\left(\text{aq}\right)\to {\text{I}}_{\text{2}}\left(\text{aq}\right)+2{\text{e}}^{-}$ …… (2)

Multiply equation (2) by 5. After that, add both the equations to get the net balance equation.

$\begin{array}{c}{\text{2IO}}_3^{-}\left(\text{aq}\right)+12{\text{H}}^{\text{+}}+10{\text{e}}^{-}\to {\text{I}}_2\left(\text{aq}\right)+6{\text{H}}_{\text{2}}\text{O}\\ {\text{10I}}^{-}\left(\text{aq}\right)\to 5{\text{I}}_{\text{2}}\left(\text{aq}\right)+10{\text{e}}^{-}\\ {\text{2IO}}_3^{-}\left(\text{aq}\right)+12{\text{H}}^{\text{+}}+{\text{10I}}^{-}\left(\text{aq}\right)\to {\text{I}}_2\left(\text{aq}\right)+5{\text{I}}_{\text{2}}\left(\text{aq}\right)+6{\text{H}}_{\text{2}}\text{O}\end{array}$

In the overall reaction, charge and elements are balanced in both sides.

(b)

Interpretation Introduction

Interpretation: The given half-reaction which take place in acidic solution needs to be balanced.

${\text{Cr}}_2{\text{O}}_7^{2-}\left(\text{aq}\right)+{\text{I}}^{-}\left(\text{aq}\right)\to {\text{Cr}}^{3+}\left(\text{aq}\right){\text{+I}}_{\text{2}}\left(\text{aq}\right)$

Concept Introduction: By transfer of electrons, oxidation-reduction reactions are characterized. The transfer occurs to form ions. Oxidation reaction is caused by losing electrons and reduction caused by gaining electrons.

Answer to Problem 27A

The balance reduction half-reaction is-

  ${\text{Cr}}_{\text{2}}{\text{O}}_7^{2-}\left(\text{aq}\right)+14{\text{H}}^{\text{+}}+6{\text{e}}^{-}\to 2{\text{Cr}}^{3+}\left(\text{aq}\right)+7{\text{H}}_{\text{2}}\text{O}$

The balance oxidation half-reaction is-

${\text{2I}}^{-}\left(\text{aq}\right)\to {\text{I}}_{\text{2}}\left(\text{aq}\right)+2{\text{e}}^{-}$

Explanation of Solution

The oxidation state of the each element in this reaction is calculated as follows:

${\text{Cr}}_2{\text{O}}_7^{2-}\left(\text{aq}\right)+{\text{I}}^{-}\left(\text{aq}\right)\to {\text{Cr}}^{3+}\left(\text{aq}\right){\text{+I}}_{\text{2}}\left(\text{aq}\right)$

${\text{I}}_{\text{2}}$ is in free state so, its oxidation state is zero. Oxidation state of ${\text{Cr}}^{3+}$ is 3. Oxidation state of Cr in ${\text{Cr}}_{\text{2}}{\text{O}}_7^{2-}$ is $+6$ , O is $-2$ . And ${\text{I}}^{-}$ is $-1$ .

Oxidation state decreases from $+5$ to 0 in ${\text{Cr}}_{\text{2}}{\text{O}}_7^{2-}$ to ${\text{Cr}}^{3+}$ . Hence, it is reduced. The reduction half-reaction is as follows:

${\text{Cr}}_{\text{2}}{\text{O}}_7^{2-}\to {\text{Cr}}^{3+}$

Balance Cr by multiplying 2 in product side.

${\text{Cr}}_{\text{2}}{\text{O}}_7^{2-}\to 2{\text{Cr}}^{3+}$

Balance O by using ${\text{H}}_{\text{2}}\text{O}$ and H by adding ${\text{H}}^{+}$ .

${\text{Cr}}_{\text{2}}{\text{O}}_7^{2-}\left(\text{aq}\right)+14{\text{H}}^{\text{+}}\to 2{\text{Cr}}^{3+}\left(\text{aq}\right)+7{\text{H}}_{\text{2}}\text{O}$

Now, add electrons to balance charge. To balance charge on both sides, add six electrons on the left. Hence, the balanced equation is-

${\text{Cr}}_{\text{2}}{\text{O}}_7^{2-}\left(\text{aq}\right)+14{\text{H}}^{\text{+}}+6{\text{e}}^{-}\to 2{\text{Cr}}^{3+}\left(\text{aq}\right)+7{\text{H}}_{\text{2}}\text{O}$ …… (1)

Oxidation state of ${\text{I}}^{-}$ is increases from $-1$ to $0$ . Hence, it is oxidized. The oxidation half-reaction is as follows:

${\text{I}}^{-}\left(\text{aq}\right)\to {\text{I}}_{\text{2}}\left(\text{aq}\right)$

To balance I in both side, multiply reactant side I by to 2.

${\text{2I}}^{-}\left(\text{aq}\right)\to {\text{I}}_{\text{2}}\left(\text{aq}\right)$

To balance charge on both sides, add $2{\text{e}}^{-}$ to the right. Hence, the balanced equation is-

${\text{2I}}^{-}\left(\text{aq}\right)\to {\text{I}}_{\text{2}}\left(\text{aq}\right)+2{\text{e}}^{-}$ …… (2)

Multiply equation (2) by 3. After that, add both the equation to get the net balance equation.

$\begin{array}{c}{\text{Cr}}_{\text{2}}{\text{O}}_7^{2-}\left(\text{aq}\right)+14{\text{H}}^{\text{+}}+6{\text{e}}^{-}\to 2{\text{Cr}}^{3+}\left(\text{aq}\right)+7{\text{H}}_{\text{2}}\text{O}\\ {\text{6I}}^{-}\left(\text{aq}\right)\to 3{\text{I}}_{\text{2}}\left(\text{aq}\right)+6{\text{e}}^{-}\\ {\text{Cr}}_{\text{2}}{\text{O}}_7^{2-}\left(\text{aq}\right)+14{\text{H}}^{\text{+}}+{\text{6I}}^{-}\left(\text{aq}\right)\to 2{\text{Cr}}^{3+}\left(\text{aq}\right)+7{\text{H}}_{\text{2}}\text{O}+3{\text{I}}_{\text{2}}\left(\text{aq}\right)\end{array}$

In the overall reaction, charge and elements are balanced in both sides.

(c)

Interpretation Introduction

Interpretation: The given half-reaction which take place in acidic solution needs to be balanced.

${\text{Cu}}^{\text{2+}}\left(\text{aq}\right)+{\text{I}}^{-}\left(\text{aq}\right)\to \text{Cu}\left(\text{s}\right)+{\text{I}}_{\text{2}}\left(\text{aq}\right)$

Concept Introduction: By transfer of electrons, oxidation-reduction reactions are characterized. The transfer occurs to form ions. Oxidation reaction is caused by losing electrons and reduction caused by gaining electrons.

Answer to Problem 27A

The balance reduction half-reaction is-

  ${\text{2Cu}}^{2+}\left(\text{aq}\right)+2{\text{e}}^{-}\to 2\text{Cu}\left(\text{aq}\right)$

The balance oxidation half-reaction is-

  ${\text{2I}}^{-}\left(\text{aq}\right)\to {\text{I}}_{\text{2}}\left(\text{aq}\right)+2{\text{e}}^{-}$

Explanation of Solution

The oxidation state of the each element in this reaction is calculated as follows:

${\text{Cu}}^{\text{2+}}\left(\text{aq}\right)+{\text{I}}^{-}\left(\text{aq}\right)\to \text{Cu}\left(\text{s}\right)+{\text{I}}_{\text{2}}\left(\text{aq}\right)$

${\text{I}}_{\text{2}}$ and Cu is in free state so, its oxidation state is zero. Oxidation state of ${\text{Cu}}^{\text{2+}}$ is $+2$ , ${\text{I}}^{-}$ is $-1$ . Oxidation state decreases from $+2$ to +1 in ${\text{Cu}}^{\text{2+}}$ to $\text{CuI}$ . Hence, it is reduced. The reduction half-reaction is as follows:

${\text{Cu}}^{2+}\left(\text{aq}\right)\to \text{Cu}\left(\text{aq}\right)$

Now, add electrons to balance charge. To balance charge on both sides, add ten electrons on the left. Hence, the balanced equation is-

${\text{Cu}}^{2+}\left(\text{aq}\right)+2{\text{e}}^{-}\to \text{Cu}\left(\text{aq}\right)$ …… (1)

Oxidation state of ${\text{I}}^{-}$ is increases from $-1$ to $0$ . Hence, it is oxidized. The oxidation half-reaction is as follows:

${\text{I}}^{-}\left(\text{aq}\right)\to {\text{I}}_{\text{2}}\left(\text{aq}\right)$

To balance I in both side, multiply reactant side I by to 2.

${\text{2I}}^{-}\left(\text{aq}\right)\to {\text{I}}_{\text{2}}\left(\text{aq}\right)$

To balance charge on both side add $2{\text{e}}^{-}$ to the right. Hence, the balanced equation is-

${\text{2I}}^{-}\left(\text{aq}\right)\to {\text{I}}_{\text{2}}\left(\text{aq}\right)+2{\text{e}}^{-}$ …… (2)

Multiply equation (2) by 5. After that, add both the equation to get the net balance equation.

$\begin{array}{c}{\text{Cu}}^{2+}\left(\text{aq}\right)+2{\text{e}}^{-}\to \text{Cu}\left(\text{aq}\right)\\ {\text{2I}}^{-}\left(\text{aq}\right)\to {\text{I}}_{\text{2}}\left(\text{aq}\right)+2{\text{e}}^{-}\\ {\text{Cu}}^{2+}\left(\text{aq}\right){\text{+2I}}^{-}\left(\text{aq}\right)\to \text{Cu}\left(\text{aq}\right){\text{+I}}_{\text{2}}\left(\text{aq}\right)\end{array}$

In the overall reaction, charge and elements are balanced in both sides.