Chapter 21, Problem 17QAP

Interpretation Introduction

(a)

Interpretation:

The balanced net ionic equation for the following should be determined:

disproportionation reactionof iodine to give iodate and iodide ions in basic solution.

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 each atom in the reaction are balanced.

Answer to Problem 17QAP

A balanced net ionic equation for the disproportionation reactionof iodine to give iodate and iodide ion in basic solution is,

${\text{3I}}_{\text{2}}\left(\text{s}\right)+6{\text{OH}}^{-}\left(\text{aq}\right)\rightleftharpoons {\text{IO}}_{\text{3}}{}^{-}\left(\text{aq}\right)+3{\text{H}}_{\text{2}}\text{O}\left(\text{l}\right)+5{\text{I}}^{-}\left(\text{aq}\right)$

Explanation of Solution

Iodine is oxidized to iodate.

${\text{I}}_{\text{2}}\left(\text{s}\right)\rightleftharpoons {\text{IO}}_{\text{3}}{}^{-}\left(\text{aq}\right)$

Balance the above reaction.

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

${\text{I}}_{\text{2}}\left(\text{s}\right)\rightleftharpoons 2{\text{IO}}_{\text{3}}{}^{-}\left(\text{aq}\right)$

2. Balance oxygen by adding water.

${\text{I}}_{\text{2}}\left(\text{s}\right)+6{\text{H}}_{\text{2}}\text{O}\left(\text{l}\right)\rightleftharpoons 2{\text{IO}}_{\text{3}}{}^{-}\left(\text{aq}\right)$

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

${\text{I}}_{\text{2}}\left(\text{s}\right)+6{\text{H}}_{\text{2}}\text{O}\left(\text{l}\right)\rightleftharpoons 2{\text{IO}}_{\text{3}}{}^{-}\left(\text{aq}\right)+12{\text{H}}^{\text{+}}\left(\text{aq}\right)$

4. Balance charges by adding electrons.

${\text{I}}_{\text{2}}\left(\text{s}\right)+6{\text{H}}_{\text{2}}\text{O}\left(\text{l}\right)\rightleftharpoons 2{\text{IO}}_{\text{3}}{}^{-}\left(\text{aq}\right)+12{\text{H}}^{\text{+}}\left(\text{aq}\right)+10{\text{e}}^{-}$

5. Add $12{\text{OH}}^{-}$ to both side.

$\begin{array}{l}{\text{I}}_{\text{2}}\left(\text{s}\right)+6{\text{H}}_{\text{2}}\text{O}\left(\text{l}\right)+12{\text{OH}}^{-}\left(\text{aq}\right)\rightleftharpoons 2{\text{IO}}_{\text{3}}{}^{-}\left(\text{aq}\right)+12{\text{H}}^{\text{+}}\left(\text{aq}\right)+12{\text{OH}}^{-}\left(\text{aq}\right)+10{\text{e}}^{-}\\ {\text{I}}_{\text{2}}\left(\text{s}\right)+12{\text{OH}}^{-}\left(\text{aq}\right)\rightleftharpoons 2{\text{IO}}_{\text{3}}{}^{-}\left(\text{aq}\right)+6{\text{H}}_{\text{2}}\text{O}\left(\text{l}\right)+10{\text{e}}^{-}\end{array}$

Hence, ionic euation of iodine changes to iodate is,

${\text{I}}_{\text{2}}\left(\text{s}\right)+12{\text{OH}}^{-}\left(\text{aq}\right)\rightleftharpoons 2{\text{IO}}_{\text{3}}{}^{-}\left(\text{aq}\right)+6{\text{H}}_{\text{2}}\text{O}\left(\text{l}\right)+10{\text{e}}^{-}$ …… (1)

Iodine is oxidized to iodide.

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

Balance the above reaction.

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

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

2. Balance charge by adding electrons.

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

Hence, ionic equation of iodine changes to iodide is,

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

Equation (2) multiplied by 5.

${\text{5I}}_{\text{2}}\left(\text{s}\right)+10{\text{e}}^{-}\left(\text{aq}\right)\rightleftharpoons 10{\text{I}}^{-}\left(\text{aq}\right)$ …… (3)

Add equation (1) and equation (3) to get the net ionic equation.

$\begin{array}{l}{\text{I}}_{\text{2}}\left(\text{s}\right)+12{\text{OH}}^{-}\left(\text{aq}\right)\rightleftharpoons 2{\text{IO}}_{\text{3}}{}^{-}\left(\text{aq}\right)+6{\text{H}}_{\text{2}}\text{O}\left(\text{l}\right)+10{\text{e}}^{-}\\ {\text{5I}}_{\text{2}}\left(\text{s}\right)+10{\text{e}}^{-}\left(\text{aq}\right)\rightleftharpoons 10{\text{I}}^{-}\left(\text{aq}\right)\\ {\text{I}}_{\text{2}}\left(\text{s}\right)+12{\text{OH}}^{-}\left(\text{aq}\right)+{\text{5I}}_{\text{2}}\left(\text{s}\right)+\cancel{10{\text{e}}^{-}}\rightleftharpoons 2{\text{IO}}_{\text{3}}{}^{-}\left(\text{aq}\right)+6{\text{H}}_{\text{2}}\text{O}\left(\text{l}\right)+\cancel{10{\text{e}}^{-}}+10{\text{I}}^{-}\left(\text{aq}\right)\\ {\text{I}}_{\text{2}}\left(\text{s}\right)+12{\text{OH}}^{-}\left(\text{aq}\right)+{\text{5I}}_{\text{2}}\left(\text{s}\right)\rightleftharpoons 2{\text{IO}}_{\text{3}}{}^{-}\left(\text{aq}\right)+6{\text{H}}_{\text{2}}\text{O}\left(\text{l}\right)+10{\text{I}}^{-}\left(\text{aq}\right)\end{array}$

${\text{6I}}_{\text{2}}\left(\text{s}\right)+12{\text{OH}}^{-}\left(\text{aq}\right)\rightleftharpoons 2{\text{IO}}_{\text{3}}{}^{-}\left(\text{aq}\right)+6{\text{H}}_{\text{2}}\text{O}\left(\text{l}\right)+10{\text{I}}^{-}\left(\text{aq}\right)$

Dividing the equation by 2

Hence, net ionic equation is,

${\text{3I}}_{\text{2}}\left(\text{s}\right)+6{\text{OH}}^{-}\left(\text{aq}\right)\rightleftharpoons {\text{IO}}_{\text{3}}{}^{-}\left(\text{aq}\right)+3{\text{H}}_{\text{2}}\text{O}\left(\text{l}\right)+5{\text{I}}^{-}\left(\text{aq}\right)$

Interpretation Introduction

(b)

Interpretation:

The balanced net ionic equation for the following should be determined:

disproportionation reaction of chlorine gas to chloride and prechlorate ions in basic solution.

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 each atom in the reaction are balanced.

Answer to Problem 17QAP

A balanced net ionic equation for the disproportionation reaction of chlorine gas to chloride and prechlorate ions in basic solution is,

${\text{4Cl}}_2\left(\text{g}\right)+8{\text{OH}}^{-}\left(\text{aq}\right)\rightleftharpoons {\text{ClO}}_4{}^{-}\left(\text{aq}\right)+4{\text{H}}_{\text{2}}\text{O}\left(\text{l}\right)+7{\text{Cl}}^{-}\left(\text{aq}\right)$

Explanation of Solution

The reaction of chlorine gas to perchlorate ion is,

${\text{Cl}}_{\text{2}}\left(\text{g}\right)\rightleftharpoons {\text{ClO}}_4{}^{-}$

Balance the above reaction.

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

${\text{Cl}}_{\text{2}}\left(\text{g}\right)\rightleftharpoons 2{\text{ClO}}_4{}^{-}$

2. Balance oxygen by adding water.

${\text{Cl}}_{\text{2}}\left(\text{g}\right)+8{\text{H}}_{\text{2}}\text{O}\left(\text{l}\right)\rightleftharpoons 2{\text{ClO}}_4{}^{-}\left(\text{aq}\right)$

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

${\text{Cl}}_{\text{2}}\left(\text{g}\right)+8{\text{H}}_{\text{2}}\text{O}\left(\text{l}\right)\rightleftharpoons 2{\text{ClO}}_4{}^{-}\left(\text{aq}\right)+16{\text{H}}^{\text{+}}\left(\text{aq}\right)$

4. Balance charges by adding electrons.

${\text{Cl}}_{\text{2}}\left(\text{g}\right)+8{\text{H}}_{\text{2}}\text{O}\left(\text{l}\right)\rightleftharpoons 2{\text{ClO}}_4{}^{-}\left(\text{aq}\right)+16{\text{H}}^{\text{+}}\left(\text{aq}\right)+14{\text{e}}^{-}$

5. Add $16{\text{OH}}^{-}$ to both side.

$\begin{array}{l}{\text{Cl}}_{\text{2}}\left(\text{g}\right)+8{\text{H}}_{\text{2}}\text{O}\left(\text{l}\right)+16{\text{OH}}^{-}\left(\text{aq}\right)\rightleftharpoons 2{\text{ClO}}_4{}^{-}\left(\text{aq}\right)+16{\text{H}}^{\text{+}}\left(\text{aq}\right)+16{\text{OH}}^{-}\left(\text{aq}\right)+14e^{-}\\ {\text{Cl}}_{\text{2}}\left(\text{g}\right)+8{\text{H}}_{\text{2}}\text{O}\left(\text{l}\right)+16{\text{OH}}^{-}\left(\text{aq}\right)\rightleftharpoons 2{\text{ClO}}_4{}^{-}\left(\text{aq}\right)+16{\text{H}}_{\text{2}}\text{O}\left(\text{l}\right)+14e^{-}\\ {\text{Cl}}_{\text{2}}\left(\text{g}\right)+16{\text{OH}}^{-}\left(\text{aq}\right)\rightleftharpoons 2{\text{ClO}}_4{}^{-}\left(\text{aq}\right)+8{\text{H}}_{\text{2}}\text{O}\left(\text{l}\right)+14e^{-}\end{array}$

Hence, ionic equation of chlorine gas changes to perchlorate ion is,

${\text{Cl}}_{\text{2}}\left(\text{g}\right)+16{\text{OH}}^{-}\left(\text{aq}\right)\rightleftharpoons 2{\text{ClO}}_4{}^{-}\left(\text{aq}\right)+8{\text{H}}_{\text{2}}\text{O}\left(\text{l}\right)+14e^{-}$ …… (3)

Chlorine is oxidized to iodide.

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

Balance the above reaction.

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

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

2. Balance charge by adding electrons.

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

Equation (3) is multiplied by 7.

${\text{7Cl}}_{\text{2}}\left(\text{s}\right)+14{\text{e}}^{-}\rightleftharpoons 14{\text{Cl}}^{-}\left(\text{aq}\right)$ ……. (5)

Add equation (3) and (4) to get the net ionic equation.

$\begin{array}{l}{\text{Cl}}_{\text{2}}\left(\text{g}\right)+16{\text{OH}}^{-}\left(\text{aq}\right)+{\text{7Cl}}_{\text{2}}\left(\text{s}\right)+\cancel{14{\text{e}}^{-}}\rightleftharpoons 2{\text{ClO}}_4{}^{-}\left(\text{aq}\right)+8{\text{H}}_{\text{2}}\text{O}\left(\text{l}\right)+14{\text{Cl}}^{-}\left(\text{aq}\right)+\cancel{14e^{-}}\\ 8{\text{Cl}}_2\left(\text{g}\right)+16{\text{OH}}^{-}\left(\text{aq}\right)\rightleftharpoons 2{\text{ClO}}_4{}^{-}\left(\text{aq}\right)+8{\text{H}}_{\text{2}}\text{O}\left(\text{l}\right)+14{\text{Cl}}^{-}\left(\text{aq}\right)\end{array}$

Dividing the equation by 2:

Hence, net ionic equation is,

${\text{4Cl}}_2\left(\text{g}\right)+8{\text{OH}}^{-}\left(\text{aq}\right)\rightleftharpoons {\text{ClO}}_4{}^{-}\left(\text{aq}\right)+4{\text{H}}_{\text{2}}\text{O}\left(\text{l}\right)+7{\text{Cl}}^{-}\left(\text{aq}\right)$