Chapter 21, Problem 16QAP

Interpretation Introduction

(a)

Interpretation:

To write a balanced net ionic equation for the following:

The oxidation of iodide to iodine by sulfate ion in acidic solution. Sulfur dioxide gas is also produced.

Concept introduction:

Net ionic equation is the ionic equation in which reactants are written in the form of ions if they occur as ions in a reaction medium and product form are shown as combination of ions. The charges of ions and number of each atom in the reaction are balanced.

Answer to Problem 16QAP

A balanced net ionic equation for oxidation of iodide to iodine by sulfate ion in acidic solution where Sulfur dioxide gas is also produced is,

${\text{SO}}_{\text{4}}{}^{2-}\left(\text{aq}\right)+4{\text{H}}^{+}\left(\text{aq}\right)+{\text{2I}}^{-}\left(\text{aq}\right)\rightleftharpoons {\text{I}}_2\left(\text{s}\right)+{\text{SO}}_{\text{2}}\left(\text{g}\right)+2{\text{H}}_{\text{2}}\text{O}\left(\text{l}\right)$

Explanation of Solution

The reaction of sulphate ion $\left({\text{SO}}_4{}^{2-}\right)$ to form sulfur dioxide is as follows:

${\text{SO}}_{\text{4}}{}^{2-}\left(\text{aq}\right)\rightleftharpoons {\text{SO}}_{\text{2}}\left(\text{g}\right)$

Balance the above reaction.

1. Balance all atoms except O(Oxygen)

${\text{SO}}_{\text{4}}{}^{2-}\left(\text{aq}\right)\rightleftharpoons {\text{SO}}_{\text{2}}\left(\text{g}\right)$

2. Balance oxygen atom by adding water.

${\text{SO}}_{\text{4}}{}^{2-}\left(\text{aq}\right)\rightleftharpoons {\text{SO}}_{\text{2}}\left(\text{g}\right)+2{\text{H}}_{\text{2}}\text{O}\left(\text{l}\right)$

3. Balance H by adding ${\text{H}}^{\text{+}}$ ion.

${\text{SO}}_{\text{4}}{}^{2-}\left(\text{aq}\right)+4{\text{H}}^{+}\left(\text{aq}\right)\rightleftharpoons {\text{SO}}_{\text{2}}\left(\text{g}\right)+2{\text{H}}_{\text{2}}\text{O}\left(\text{l}\right)$

4. Balance charges by adding electrons.

${\text{SO}}_{\text{4}}{}^{2-}\left(\text{aq}\right)+4{\text{H}}^{+}\left(\text{aq}\right)+2{\text{e}}^{-}\rightleftharpoons {\text{SO}}_{\text{2}}\left(\text{g}\right)+2{\text{H}}_{\text{2}}\text{O}\left(\text{l}\right)$

Hence, balanced ionic equation of reaction of sulphate with hydrogen ion is:

${\text{SO}}_{\text{4}}{}^{2-}\left(\text{aq}\right)+4{\text{H}}^{+}\left(\text{aq}\right)+2{\text{e}}^{-}\rightleftharpoons {\text{SO}}_{\text{2}}\left(\text{g}\right)+2{\text{H}}_{\text{2}}\text{O}\left(\text{l}\right)$.. .. (1)

Oxidation of iodide to iodide is,

${\text{I}}^{-}\left(\text{aq}\right)\rightleftharpoons {\text{I}}_2\left(\text{s}\right)$

1. Balance charge by adding electrons.

${\text{2I}}^{-}\left(\text{aq}\right)\rightleftharpoons {\text{I}}_2\left(\text{s}\right)+2{\text{e}}^{-}$

Hence, ionic equation of oxidation of iodide to iodide is,

${\text{2I}}^{-}\left(\text{aq}\right)\rightleftharpoons {\text{I}}_2\left(\text{s}\right)+2{\text{e}}^{-}$.. .. (2)

Addition of equation (1) and (2) gives net ionic equation.

$\begin{array}{l}{\text{SO}}_{\text{4}}{}^{2-}\left(\text{aq}\right)+4{\text{H}}^{+}\left(\text{aq}\right)+\cancel{2{\text{e}}^{-}}+{\text{2I}}^{-}\left(\text{aq}\right)\rightleftharpoons {\text{I}}_2\left(\text{s}\right)+{\text{SO}}_{\text{2}}\left(\text{g}\right)+\cancel{2{\text{e}}^{-}}+2{\text{H}}_{\text{2}}\text{O}\left(\text{l}\right)\\ {\text{SO}}_{\text{4}}{}^{2-}\left(\text{aq}\right)+4{\text{H}}^{+}\left(\text{aq}\right)+{\text{2I}}^{-}\left(\text{aq}\right)\rightleftharpoons {\text{I}}_2\left(\text{s}\right)+{\text{SO}}_{\text{2}}\left(\text{g}\right)+2{\text{H}}_{\text{2}}\text{O}\left(\text{l}\right)\end{array}$

Hence, net ionic equation is,

${\text{SO}}_{\text{4}}{}^{2-}\left(\text{aq}\right)+4{\text{H}}^{+}\left(\text{aq}\right)+{\text{2I}}^{-}\left(\text{aq}\right)\rightleftharpoons {\text{I}}_2\left(\text{s}\right)+{\text{SO}}_{\text{2}}\left(\text{g}\right)+2{\text{H}}_{\text{2}}\text{O}\left(\text{l}\right)$

Interpretation Introduction

(b)

Interpretation:

To write balanced net ionic equation for the following:

The preparation of iodine from an iodide salt and chlorine gas.

Concept introduction:

Net ionic equation is the ionic equation in which reactants are written in the form of ions if they occur as ions in a reaction medium and product form are shown as combination of ions. The charges of ions and number of each atom in the reaction are balanced.

Answer to Problem 16QAP

${\text{2I}}^{-}\left(\text{aq}\right){\text{+Cl}}_{\text{2}}\left(\text{g}\right)\rightleftharpoons {\text{I}}_2\left(\text{s}\right)+2{\text{Cl}}^{-}\left(\text{aq}\right)$

Explanation of Solution

Iodide ion is oxidized as below:

${\text{I}}^{-}\left(\text{aq}\right)\rightleftharpoons {\text{I}}_2\left(\text{s}\right)$

Balance the above reaction.

1. Balance all atoms except O(Oxygen) and H(Hydrogen)

${\text{2I}}^{-}\left(\text{aq}\right)\rightleftharpoons {\text{I}}_2\left(\text{s}\right)$

2. Balance charge by adding electrons.

${\text{2I}}^{-}\left(\text{aq}\right)\rightleftharpoons {\text{I}}_2\left(\text{s}\right)+2{\text{e}}^{-}$

Hence, ionic equation of oxidation of iodide to iodide is,

${\text{2I}}^{-}\left(\text{aq}\right)\rightleftharpoons {\text{I}}_2\left(\text{s}\right)+2{\text{e}}^{-}$.. .. (3)

Chlorine gas is reduced as follows:

${\text{Cl}}_{\text{2}}\left(\text{g}\right)\rightleftharpoons {\text{Cl}}^{-}\left(\text{aq}\right)$

Balance the above reaction.

1. Balance all atoms except O(Oxygen) and H(Hydrogen)

${\text{Cl}}_{\text{2}}\left(\text{g}\right)\rightleftharpoons 2{\text{Cl}}^{-}\left(\text{aq}\right)$

2. Balance charge by adding electrons.

${\text{Cl}}_{\text{2}}\left(\text{g}\right)+2{\text{e}}^{-}\rightleftharpoons 2{\text{Cl}}^{-}\left(\text{aq}\right)$

Hence, ionic equation of oxidation of iodide to iodide is,

${\text{Cl}}_{\text{2}}\left(\text{g}\right)+2{\text{e}}^{-}\rightleftharpoons 2{\text{Cl}}^{-}\left(\text{aq}\right)$.. .. (4)

Net ionic equation is the summation equation (3) and (4).

$\begin{array}{l}{\text{2I}}^{-}\left(\text{aq}\right){\text{+Cl}}_{\text{2}}\left(\text{g}\right)+\cancel{2{\text{e}}^{-}}\rightleftharpoons {\text{I}}_2\left(\text{s}\right)+\cancel{2{\text{e}}^{-}}+2{\text{Cl}}^{-}\left(\text{aq}\right)\\ {\text{2I}}^{-}\left(\text{aq}\right){\text{+Cl}}_{\text{2}}\left(\text{g}\right)\rightleftharpoons {\text{I}}_2\left(\text{s}\right)+2{\text{Cl}}^{-}\left(\text{aq}\right)\end{array}$

Hence, net ionic equation is,

${\text{2I}}^{-}\left(\text{aq}\right){\text{+Cl}}_{\text{2}}\left(\text{g}\right)\rightleftharpoons {\text{I}}_2\left(\text{s}\right)+2{\text{Cl}}^{-}\left(\text{aq}\right)$