Chapter 4, Problem 83E

(a)

Interpretation Introduction

Interpretation:

The following equation should be balanced by the half reaction method.

  $\text{Fe(s)+HCl(aq)}\to {\text{HFeCl}}_{\text{4}}{\text{(aq)+H}}_{\text{2}}\text{(g)}$

Concept Introduction:

The chemical reaction in which both oxidation and reduction process takes place is known as redox reaction. In this reaction, transfer of electrons takes place among the elements.

A reduction or an oxidation reaction is known as half reaction.

Half reaction method:

Balance all atoms including oxygen and hydrogen atoms are carried out by addition of water molecule (to balance oxygen) and hydrogen ion (to balance hydrogen) in the half reactions. Number of electrons and charge should be balanced after that makes the number of electrons equal in both oxidation and reduction reactions by multiplying with an integer. The last step is to add both half reactions.

Answer to Problem 83E

The balanced equation is:

  $\text{2Fe(s)+8HCl(aq)}\to 4{\text{HFeCl}}_{\text{4}}{\text{(aq)+3H}}_{\text{2}}\text{(g)}$

Explanation of Solution

The given reaction is:

  $\text{Fe(s)+HCl(aq)}\to {\text{HFeCl}}_{\text{4}}{\text{(aq)+H}}_{\text{2}}\text{(g)}$

First step is to assign oxidation numbers to each element present in the given reaction.

  $\stackrel{0}{\text{Fe(s)}}\text{+}\stackrel{+1}{\text{H}}\stackrel{-1}{\text{Cl}}\text{(aq)}\to \stackrel{+1}{\text{H}}\stackrel{+3}{\text{Fe}}\stackrel{-1}{{\text{Cl}}_{\text{4}}}\text{(aq)+}\stackrel{0}{{\text{H}}_{\text{2}}}\text{(g)}$

The above reaction can be balanced by the inspection method.

Since, in the given reaction one molecule of ${\text{HFeCl}}_{\text{4}}$ is present in product side, thus in reactant, four molecules of $\text{HCl}$ must be present to balance the chlorine atoms.

The reaction is written as:

  $\text{Fe(s)+4HCl(aq)}\to {\text{HFeCl}}_{\text{4}}{\text{(aq)+H}}_{\text{2}}\text{(g)}$

Now, to balance the above reaction, 4 (coefficient) should be placed before ${\text{HFeCl}}_{\text{4}}$ , 8 (coefficient) should be placed before $\text{HCl}$ , 3 (coefficient) should be placed before ${\text{H}}_2$ and 2 (coefficient) should be placed before $\text{Fe}$ .

The balanced reaction is written as:

  $\text{2Fe(s)+8HCl(aq)}\to 4{\text{HFeCl}}_{\text{4}}{\text{(aq)+3H}}_{\text{2}}\text{(g)}$

(b)

Interpretation Introduction

Interpretation:

The following equation should be balanced by the half reaction method.

  ${\text{IO}}_3^{-}{\text{(aq)+I}}^{-}\text{(aq)}\stackrel{\text{Acidic}}{\to }{\text{I}}_3^{-}\text{(aq)}$

Concept Introduction:

The chemical reaction in which both oxidation and reduction process takes place is known as redox reaction. In this reaction, transfer of electrons takes place among the elements.

A reduction or an oxidation reaction is known as half reaction.

Half reaction method:

Balance all atoms including oxygen and hydrogen atoms are carried out by addition of water molecule (to balance oxygen) and hydrogen ion (to balance hydrogen) in the half reactions. Number of electrons and charge should be balanced after that makes the number of electrons equal in both oxidation and reduction reactions by multiplying with an integer. The last step is to add both half reactions.

Answer to Problem 83E

The balanced equation is:

  ${\text{IO}}_3^{-}{\text{(aq)+6H}}^{+}+8{\text{I}}^{-}\text{(aq)}\to 3{\text{I}}_3^{-}{\text{(aq)+3H}}_2\text{O}$

Explanation of Solution

The given reaction is:

  ${\text{IO}}_3^{-}{\text{(aq)+I}}^{-}\text{(aq)}\stackrel{\text{Acidic}}{\to }{\text{I}}_3^{-}\text{(aq)}$

First step is to assign oxidation numbers to each element present in the given reaction.

  $\stackrel{+5}{\text{I}}\stackrel{-2}{{\text{O}}_3^{-}}\text{(aq)+}\stackrel{-1}{{\text{I}}^{-}}\text{(aq)}\to \stackrel{0.33}{{\text{I}}_3^{-}}\text{(aq)}$

Now, separate the above reaction into oxidation and reduction reactions (two half-reactions).

Oxidation:

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

Reduction:

  ${\text{I}}^{-}\text{(aq)}\to {\text{I}}_3^{-}\text{(aq)}$

Balance the above two reactions except oxygen atoms and hydrogen atoms.

The reaction is written as:

Oxidation:

  ${\text{3IO}}_3^{-}\text{(aq)}\to {\text{I}}_3^{-}\text{(aq)}$

Reduction:

  ${\text{3I}}^{-}\text{(aq)}\to {\text{I}}_3^{-}\text{(aq)}$

Balance the oxygen atoms in the above two reactions.

Oxidation:

  ${\text{3IO}}_3^{-}\text{(aq)}\to {\text{I}}_3^{-}{\text{(aq)+9H}}_2\text{O}$

Reduction:

  ${\text{3I}}^{-}\text{(aq)}\to {\text{I}}_3^{-}\text{(aq)}$

Balance the hydrogen atoms in the above two reactions.

Oxidation:

  ${\text{3IO}}_3^{-}{\text{(aq)+18H}}^{+}\to {\text{I}}_3^{-}{\text{(aq)+9H}}_2\text{O}$

Reduction:

  ${\text{3I}}^{-}\text{(aq)}\to {\text{I}}_3^{-}\text{(aq)}$

Balance the charge and number of electrons in the above two reactions.

Oxidation:

  ${\text{3IO}}_3^{-}{\text{(aq)+18H}}^{+}+16e^{-}\to {\text{I}}_3^{-}{\text{(aq)+9H}}_2\text{O}$

Reduction:

  ${\text{3I}}^{-}\text{(aq)}\to {\text{I}}_3^{-}\text{(aq)+2}{e}^{-}$

Now, balance the number of gain electrons and number of lost electrons in the above two reactions.

Oxidation:

  ${\text{3IO}}_3^{-}{\text{(aq)+18H}}^{+}+16e^{-}\to {\text{I}}_3^{-}{\text{(aq)+9H}}_2\text{O}$ (1)

Reduction:

  ${\text{3I}}^{-}\text{(aq)}\to {\text{I}}_3^{-}\text{(aq)+2}{e}^{-}$ (2)

Multiply equation 2 by 8.

Oxidation:

  ${\text{3IO}}_3^{-}{\text{(aq)+18H}}^{+}+16e^{-}\to {\text{I}}_3^{-}{\text{(aq)+9H}}_2\text{O}$

Reduction:

  ${\text{24I}}^{-}\text{(aq)}\to 8{\text{I}}_3^{-}\text{(aq)+16}{e}^{-}$

Add both equations,

  ${\text{3IO}}_3^{-}{\text{(aq)+18H}}^{+}+{\text{24I}}^{-}\text{(aq)}+16e^{-}\to {\text{I}}_3^{-}{\text{(aq)+9H}}_2\text{O+}8{\text{I}}_3^{-}\text{(aq)+16}{e}^{-}$

The balanced equation is:

  ${\text{IO}}_3^{-}{\text{(aq)+6H}}^{+}+8{\text{I}}^{-}\text{(aq)}\to 3{\text{I}}_3^{-}{\text{(aq)+3H}}_2\text{O}$

(c)

Interpretation Introduction

Interpretation:

The following equation should be balanced by the half reaction method.

  ${\text{Cr(NCS)}}_6^{4-}{\text{(aq)+Ce}}^{\text{4+}}\text{(aq)}\stackrel{\text{Acidic}}{\to }{\text{Cr}}^{\text{3+}}{\text{(aq)+Ce}}^{3+}{\text{(aq)+NO}}_3^{-}({\text{aq)+CO}}_2\text{(g)}+{\text{SO}}_4^{2-}\text{(aq)}$

Concept Introduction:

The chemical reaction in which both oxidation and reduction process takes place is known as redox reaction. In this reaction, transfer of electrons takes place among the elements.

A reduction or an oxidation reaction is known as half reaction.

Half reaction method:

Balance all atoms including oxygen and hydrogen atoms are carried out by addition of water molecule (to balance oxygen) and hydrogen ion (to balance hydrogen) in the half reactions. Number of electrons and charge should be balanced after that makes the number of electrons equal in both oxidation and reduction reactions by multiplying with an integer. The last step is to add both half reactions.

Answer to Problem 83E

The balanced equation is:

  ${\text{Cr(NCS)}}_6^{4-}\text{(aq)}\stackrel{}{+{\text{54H}}_{\text{2}}\text{O+}\stackrel{}{{\text{97Ce}}^{\text{4+}}}\text{(aq)}\to }{\text{Cr}}^{\text{3+}}{\text{(aq)+6NO}}_3^{-}({\text{aq)+6CO}}_2\text{(g)}+6{\text{SO}}_4^{2-}{\text{(aq)+108H}}^{+}+97\stackrel{}{{\text{Ce}}^{3+}}\text{(aq)}$

Explanation of Solution

The given reaction is:

  ${\text{Cr(NCS)}}_6^{4-}{\text{(aq)+Ce}}^{\text{4+}}\text{(aq)}\stackrel{\text{Acidic}}{\to }{\text{Cr}}^{\text{3+}}{\text{(aq)+Ce}}^{3+}{\text{(aq)+NO}}_3^{-}({\text{aq)+CO}}_2\text{(g)}+{\text{SO}}_4^{2-}\text{(aq)}$

First step is to assign oxidation numbers to each element present in the given reaction.

  $\stackrel{+2}{\text{Cr}}{\text{(}\stackrel{-3}{\text{N}}\stackrel{+4}{\text{C}}\stackrel{-2}{\text{S}}\text{)}}_6^{4-}\text{(aq)+}\stackrel{+4}{{\text{Ce}}^{\text{4+}}}\text{(aq)}\stackrel{\text{Acidic}}{\to }\stackrel{+3}{{\text{Cr}}^{\text{3+}}}\text{(aq)+}\stackrel{+3}{{\text{Ce}}^{3+}}\text{(aq)+}\stackrel{+5}{\text{N}}{\stackrel{-2}{\text{O}}}_3^{-}(\text{aq)+}\stackrel{+4}{\text{C}}{\stackrel{-2}{\text{O}}}_2\text{(g)}+\stackrel{+6}{\text{S}}{\stackrel{-2}{\text{O}}}_4^{2-}\text{(aq)}$

Now, separate the above reaction into oxidation and reduction reactions (half-reactions).

Oxidation:

(Cr) $\stackrel{+2}{\text{Cr}}{\text{(}\stackrel{-3}{\text{N}}\stackrel{+4}{\text{C}}\stackrel{-2}{\text{S}}\text{)}}_6^{4-}\text{(aq)}\stackrel{\text{Acidic}}{\to }\stackrel{+3}{{\text{Cr}}^{\text{3+}}}\text{(aq)}$

(N) $\stackrel{+2}{\text{Cr}}{\text{(}\stackrel{-3}{\text{N}}\stackrel{+4}{\text{C}}\stackrel{-2}{\text{S}}\text{)}}_6^{4-}\text{(aq)}\stackrel{\text{Acidic}}{\to }\stackrel{+5}{\text{N}}{\stackrel{-2}{\text{O}}}_3^{-}(\text{aq)}$

(S) $\stackrel{+2}{\text{Cr}}{\text{(}\stackrel{-3}{\text{N}}\stackrel{+4}{\text{C}}\stackrel{-2}{\text{S}}\text{)}}_6^{4-}\text{(aq)}\stackrel{\text{Acidic}}{\to }\stackrel{+6}{\text{S}}{\stackrel{-2}{\text{O}}}_4^{2-}\text{(aq)}$

(C) $\stackrel{+2}{\text{Cr}}{\text{(}\stackrel{-3}{\text{N}}\stackrel{+4}{\text{C}}\stackrel{-2}{\text{S}}\text{)}}_6^{4-}\text{(aq)}\stackrel{\text{Acidic}}{\to }\stackrel{+4}{\text{C}}{\stackrel{-2}{\text{O}}}_2\text{(g)}$

Reduction:

  $\stackrel{+4}{{\text{Ce}}^{\text{4+}}}\text{(aq)}\stackrel{\text{Acidic}}{\to }\stackrel{+3}{{\text{Ce}}^{3+}}\text{(aq)}$

Combine all oxidation reactions:

Oxidation:

  $\stackrel{+2}{\text{Cr}}{\text{(}\stackrel{-3}{\text{N}}\stackrel{+4}{\text{C}}\stackrel{-2}{\text{S}}\text{)}}_6^{4-}\text{(aq)}\stackrel{\text{Acidic}}{\to }\stackrel{+3}{{\text{Cr}}^{\text{3+}}}\text{(aq)+}\stackrel{+5}{\text{N}}{\stackrel{-2}{\text{O}}}_3^{-}(\text{aq)+}\stackrel{+6}{\text{S}}{\stackrel{-2}{\text{O}}}_4^{2-}\text{(aq)+}\stackrel{+4}{\text{C}}{\stackrel{-2}{\text{O}}}_2\text{(g)}$

Reduction:

  $\stackrel{+4}{{\text{Ce}}^{\text{4+}}}\text{(aq)}\stackrel{\text{Acidic}}{\to }\stackrel{+3}{{\text{Ce}}^{3+}}\text{(aq)}$

Balance the above two reactions except oxygen atoms and hydrogen atoms.

The reaction is written as:

Oxidation:

  ${\text{Cr(NCS)}}_6^{4-}\text{(aq)}\stackrel{}{\to }{\text{Cr}}^{\text{3+}}{\text{(aq)+6NO}}_3^{-}({\text{aq)+6CO}}_2\text{(g)}+6{\text{SO}}_4^{2-}\text{(aq)}$

Reduction:

  $\stackrel{}{{\text{Ce}}^{\text{4+}}}\text{(aq)}\stackrel{}{\to }\stackrel{}{{\text{Ce}}^{3+}}\text{(aq)}$

Balance the oxygen atoms in the above two reactions.

Oxidation:

  ${\text{Cr(NCS)}}_6^{4-}\text{(aq)}\stackrel{}{+5{\text{4H}}_{\text{2}}\text{O}\to }{\text{Cr}}^{\text{3+}}{\text{(aq)+6NO}}_3^{-}({\text{aq)+6CO}}_2\text{(g)}+6{\text{SO}}_4^{2-}\text{(aq)}$

Reduction:

  $\stackrel{}{{\text{Ce}}^{\text{4+}}}\text{(aq)}\stackrel{}{\to }\stackrel{}{{\text{Ce}}^{3+}}\text{(aq)}$

Balance the hydrogen atoms in the above two reactions.

Oxidation:

  ${\text{Cr(NCS)}}_6^{4-}\text{(aq)}\stackrel{}{+5{\text{4H}}_{\text{2}}\text{O}\to }{\text{Cr}}^{\text{3+}}{\text{(aq)+6NO}}_3^{-}({\text{aq)+6CO}}_2\text{(g)}+6{\text{SO}}_4^{2-}{\text{(aq)+108H}}^{+}$

Reduction:

  $\stackrel{}{{\text{Ce}}^{\text{4+}}}\text{(aq)}\stackrel{}{\to }\stackrel{}{{\text{Ce}}^{3+}}\text{(aq)}$

Balance the charge and number of electrons in the above two reactions.

Oxidation:

  ${\text{Cr(NCS)}}_6^{4-}\text{(aq)}\stackrel{}{+5{\text{4H}}_{\text{2}}\text{O}\to }{\text{Cr}}^{\text{3+}}{\text{(aq)+6NO}}_3^{-}({\text{aq)+6CO}}_2\text{(g)}+6{\text{SO}}_4^{2-}{\text{(aq)+108H}}^{+}+97e^{-}$ (1)

Reduction:

  $\stackrel{}{{\text{Ce}}^{\text{4+}}}\text{(aq)+}{e}^{-}\stackrel{}{\to }\stackrel{}{{\text{Ce}}^{3+}}\text{(aq)}$ (2)

Now, multiply equation 2 by 97.

Oxidation:

  ${\text{Cr(NCS)}}_6^{4-}\text{(aq)}\stackrel{}{+5{\text{4H}}_{\text{2}}\text{O}\to }{\text{Cr}}^{\text{3+}}{\text{(aq)+6NO}}_3^{-}({\text{aq)+6CO}}_2\text{(g)}+6{\text{SO}}_4^{2-}{\text{(aq)+108H}}^{+}+97e^{-}$

Reduction:

  $\stackrel{}{{\text{97Ce}}^{\text{4+}}}\text{(aq)+97}{e}^{-}\stackrel{}{\to }97\stackrel{}{{\text{Ce}}^{3+}}\text{(aq)}$

Add both equations,

  ${\text{Cr(NCS)}}_6^{4-}\text{(aq)}\stackrel{}{+{\text{54H}}_{\text{2}}\text{O+}\stackrel{}{{\text{97Ce}}^{\text{4+}}}\text{(aq)+97}{e}^{-}\to }{\text{Cr}}^{\text{3+}}{\text{(aq)+6NO}}_3^{-}({\text{aq)+6CO}}_2\text{(g)}+6{\text{SO}}_4^{2-}{\text{(aq)+108H}}^{+}+97e^{-}+97\stackrel{}{{\text{Ce}}^{3+}}\text{(aq)}$

The balanced equation is:

  ${\text{Cr(NCS)}}_6^{4-}\text{(aq)}\stackrel{}{+{\text{54H}}_{\text{2}}\text{O+}\stackrel{}{{\text{97Ce}}^{\text{4+}}}\text{(aq)}\to }{\text{Cr}}^{\text{3+}}{\text{(aq)+6NO}}_3^{-}({\text{aq)+6CO}}_2\text{(g)}+6{\text{SO}}_4^{2-}{\text{(aq)+108H}}^{+}+97\stackrel{}{{\text{Ce}}^{3+}}\text{(aq)}$

(d)

Interpretation Introduction

Interpretation:

The following equation should be balanced by the half reaction method.

  ${\text{CrI}}_{\text{3}}{\text{(aq)+Cl}}_2\text{(g)}\stackrel{\text{Basic}}{\to }{\text{CrO}}_4^{2-}{\text{(aq)+IO}}_4^{-}{\text{(aq)+Cl}}^{-}(\text{aq)}$

Concept Introduction:

The chemical reaction in which both oxidation and reduction process takes place is known as redox reaction. In this reaction, transfer of electrons takes place among the elements.

A reduction or an oxidation reaction is known as half reaction.

Half reaction method:

Balance all atoms including oxygen and hydrogen atoms are carried out by addition of water molecule (to balance oxygen) and hydrogen ion (to balance hydrogen) in the half reactions. Number of electrons and charge should be balanced after that makes the number of electrons equal in both oxidation and reduction reactions by multiplying with an integer. The last step is to add both half reactions.

Answer to Problem 83E

The balanced equation is:

  ${\text{2CrI}}_{\text{3}}{\text{(aq)+27Cl}}_2\text{(g)}\stackrel{}{+64{\text{OH}}^{-}\to }{\text{2CrO}}_4^{2-}{\text{(aq)+6IO}}_4^{-}{\text{(aq)+32H}}_{\text{2}}\text{O(l)}+{\text{54Cl}}^{-}(\text{aq)}$

Explanation of Solution

The given reaction is:

  ${\text{CrI}}_{\text{3}}{\text{(aq)+Cl}}_2\text{(g)}\stackrel{\text{Basic}}{\to }{\text{CrO}}_4^{2-}{\text{(aq)+IO}}_4^{-}{\text{(aq)+Cl}}^{-}(\text{aq)}$

First step is to assign oxidation numbers to each element present in the given reaction.

  $\stackrel{+3}{\text{Cr}}\stackrel{-1}{{\text{I}}_{\text{3}}}\text{(aq)+}\stackrel{0}{{\text{Cl}}_2}\text{(g)}\stackrel{\text{Basic}}{\to }\stackrel{+6}{\text{Cr}}\stackrel{-2}{{\text{O}}_4^{2-}}\text{(aq)+}\stackrel{+7}{\text{I}}\stackrel{-2}{{\text{O}}_4^{-}}\text{(aq)+}\stackrel{-1}{{\text{Cl}}^{-}}(\text{aq)}$

Now, separate the above reaction into oxidation and reduction reactions (half-reactions).

Oxidation:

(Cr) $\stackrel{+3}{\text{Cr}}\stackrel{-1}{{\text{I}}_{\text{3}}}\text{(aq)}\stackrel{}{\to }\stackrel{+6}{\text{Cr}}\stackrel{-2}{{\text{O}}_4^{2-}}\text{(aq)}$

(I) $\stackrel{+3}{\text{Cr}}\stackrel{-1}{{\text{I}}_{\text{3}}}\text{(aq)}\stackrel{}{\to }\stackrel{+7}{\text{I}}\stackrel{-2}{{\text{O}}_4^{-}}\text{(aq)}$

Reduction:

  $\stackrel{0}{{\text{Cl}}_2}\text{(g)}\stackrel{}{\to }\stackrel{-1}{{\text{Cl}}^{-}}(\text{aq)}$

Combine all oxidation reactions:

Oxidation:

  ${\text{CrI}}_{\text{3}}\text{(aq)}\stackrel{}{\to }{\text{CrO}}_4^{2-}{\text{(aq)+IO}}_4^{-}\text{(aq)}$

Reduction:

  ${\text{Cl}}_2\text{(g)}\stackrel{}{\to }{\text{Cl}}^{-}(\text{aq)}$

Balance the above two reactions except oxygen atoms and hydrogen atoms.

The reaction is written as:

Oxidation:

  ${\text{CrI}}_{\text{3}}\text{(aq)}\stackrel{}{\to }{\text{CrO}}_4^{2-}{\text{(aq)+3IO}}_4^{-}\text{(aq)}$

Reduction:

  ${\text{Cl}}_2\text{(g)}\stackrel{}{\to }{\text{2Cl}}^{-}(\text{aq)}$

Balance the oxygen atoms in the above two reactions.

Oxidation:

  ${\text{CrI}}_{\text{3}}{\text{(aq)+16H}}_2\text{O}\stackrel{}{\to }{\text{CrO}}_4^{2-}{\text{(aq)+3IO}}_4^{-}\text{(aq)}$

Reduction:

  ${\text{Cl}}_2\text{(g)}\stackrel{}{\to }{\text{2Cl}}^{-}(\text{aq)}$

Balance the hydrogen atoms in the above two reactions.

Oxidation:

  ${\text{CrI}}_{\text{3}}{\text{(aq)+16H}}_2\text{O}\stackrel{}{\to }{\text{CrO}}_4^{2-}{\text{(aq)+3IO}}_4^{-}{\text{(aq)+32H}}^{+}$

Reduction:

  ${\text{Cl}}_2\text{(g)}\stackrel{}{\to }{\text{2Cl}}^{-}(\text{aq)}$

Balance the charge and number of electrons in the above two reactions.

Oxidation:

  ${\text{CrI}}_{\text{3}}{\text{(aq)+16H}}_2\text{O}\stackrel{}{\to }{\text{CrO}}_4^{2-}{\text{(aq)+3IO}}_4^{-}{\text{(aq)+32H}}^{+}+27e^{-}$ (1)

Reduction:

  ${\text{Cl}}_2\text{(g)+2}{e}^{-}\stackrel{}{\to }{\text{2Cl}}^{-}(\text{aq)}$ (2)

Now, multiply equation 1 by 2 and equation 2 by 27.

Oxidation:

  ${\text{2CrI}}_{\text{3}}{\text{(aq)+32H}}_2\text{O}\stackrel{}{\to }{\text{2CrO}}_4^{2-}{\text{(aq)+6IO}}_4^{-}{\text{(aq)+64H}}^{+}+54e^{-}$

Reduction:

  ${\text{27Cl}}_2\text{(g)+54}{e}^{-}\stackrel{}{\to }{\text{54Cl}}^{-}(\text{aq)}$

Add both equations,

  ${\text{2CrI}}_{\text{3}}{\text{(aq)+32H}}_2{\text{O+27Cl}}_2\text{(g)+54}{e}^{-}\stackrel{}{\to }{\text{2CrO}}_4^{2-}{\text{(aq)+6IO}}_4^{-}{\text{(aq)+64H}}^{+}+54e^{-}+{\text{54Cl}}^{-}(\text{aq)}$

The equation is written as:

  ${\text{2CrI}}_{\text{3}}{\text{(aq)+32H}}_2{\text{O+27Cl}}_2\text{(g)}\stackrel{}{\to }{\text{2CrO}}_4^{2-}{\text{(aq)+6IO}}_4^{-}{\text{(aq)+64H}}^{+}+{\text{54Cl}}^{-}(\text{aq)}$

For basic medium: Now, add hydroxide ion on both sides of the reaction to balance the hydrogen ion.

The equation is written as:

  ${\text{2CrI}}_{\text{3}}{\text{(aq)+32H}}_2{\text{O+27Cl}}_2\text{(g)}\stackrel{}{+64{\text{OH}}^{-}\to }{\text{2CrO}}_4^{2-}{\text{(aq)+6IO}}_4^{-}{\text{(aq)+64H}}^{+}+{\text{54Cl}}^{-}({\text{aq)+64 OH}}^{-}$

The balance reaction is:

  ${\text{2CrI}}_{\text{3}}{\text{(aq)+27Cl}}_2\text{(g)}\stackrel{}{+64{\text{OH}}^{-}\to }{\text{2CrO}}_4^{2-}{\text{(aq)+6IO}}_4^{-}{\text{(aq)+32H}}_{\text{2}}\text{O(l)}+{\text{54Cl}}^{-}(\text{aq)}$

(e)

Interpretation Introduction

Interpretation:

The following equation should be balanced by the half reaction method.

  ${\text{Fe(CN)}}_6^{4-}{\text{(aq)+Ce}}^{4+}\text{(aq)}\stackrel{\text{Basic}}{\to }{\text{Ce(OH)}}_{\text{3}}(\text{s})+{\text{Fe(OH)}}_{\text{3}}(\text{s})+{\text{CO}}_3^{2-}({\text{aq)+NO}}_3^{-}(\text{aq)}$

Concept Introduction:

The chemical reaction in which both oxidation and reduction process takes place is known as redox reaction. In this reaction, transfer of electrons takes place among the elements.

A reduction or an oxidation reaction is known as half reaction.

Half reaction method:

Balance all atoms including oxygen and hydrogen atoms are carried out by addition of water molecule (to balance oxygen) and hydrogen ion (to balance hydrogen) in the half reactions. Number of electrons and charge should be balanced after that makes the number of electrons equal in both oxidation and reduction reactions by multiplying with an integer. The last step is to add both half reactions.

Answer to Problem 83E

The balanced equation is:

  $\begin{array}{l}{\text{Fe(CN)}}_6^{4-}{\text{(aq)+61Ce}}^{4+}{\text{(aq)+258OH}}^{-}\\ \to {\text{Fe(OH)}}_{\text{3}}(\text{s})+6{\text{CO}}_3^{2-}({\text{aq)+6NO}}_3^{-}(\text{aq)}+61{\text{Ce(OH)}}_{\text{3}}(\text{s})+{\text{36H}}_{\text{2}}\text{O}\end{array}$

Explanation of Solution

The given reaction is:

  ${\text{Fe(CN)}}_6^{4-}{\text{(aq)+Ce}}^{4+}\text{(aq)}\stackrel{\text{Basic}}{\to }{\text{Ce(OH)}}_{\text{3}}(\text{s})+{\text{Fe(OH)}}_{\text{3}}(\text{s})+{\text{CO}}_3^{2-}({\text{aq)+NO}}_3^{-}(\text{aq)}$

First step is to assign oxidation numbers to each element present in the given reaction.

  $\stackrel{+2}{\text{Fe}}{\text{(}\stackrel{+2}{\text{C}}\stackrel{-3}{\text{N}}\text{)}}_6^{4-}\text{(aq)+}{\stackrel{+4}{\text{Ce}}}^{4+}\text{(aq)}\stackrel{\text{Basic}}{\to }\stackrel{+3}{\text{Ce}}{\text{(}\stackrel{-2}{\text{O}}\stackrel{+1}{\text{H}}\text{)}}_{\text{3}}(\text{s})+\stackrel{+3}{\text{Fe}}{\text{(}\stackrel{-2}{\text{O}}\stackrel{+1}{\text{H}}\text{)}}_{\text{3}}(\text{s})+\stackrel{+4}{\text{C}}\stackrel{-2}{{\text{O}}_3^{2-}}(\text{aq)+}\stackrel{+5}{\text{N}}\stackrel{-2}{{\text{O}}_3^{-}}(\text{aq)}$

Now, separate the above reaction into oxidation and reduction reactions (half-reactions).

Oxidation:

(Fe) ${\text{Fe(CN)}}_6^{4-}\text{(aq)}\to {\text{Fe(OH)}}_{\text{3}}(\text{s})$

(C) ${\text{Fe(CN)}}_6^{4-}\text{(aq)}\to {\text{CO}}_3^{2-}(\text{aq))}$

(N) ${\text{Fe(CN)}}_6^{4-}\text{(aq)}\to {\text{NO}}_3^{-}(\text{aq)}$

Reduction:

  ${\text{Ce}}^{4+}\text{(aq)}\to {\text{Ce(OH)}}_{\text{3}}(\text{s})$

Combine all oxidation reactions:

Oxidation:

  ${\text{Fe(CN)}}_6^{4-}\text{(aq)}\to {\text{Fe(OH)}}_{\text{3}}(\text{s})+{\text{CO}}_3^{2-}({\text{aq)+NO}}_3^{-}(\text{aq)}$

Reduction:

  ${\text{Ce}}^{4+}\text{(aq)}\to {\text{Ce(OH)}}_{\text{3}}(\text{s})$

Balance the above two reactions except oxygen atoms and hydrogen atoms.

The reaction is written as:

Oxidation:

  ${\text{Fe(CN)}}_6^{4-}\text{(aq)}\to {\text{Fe(OH)}}_{\text{3}}(\text{s})+6{\text{CO}}_3^{2-}({\text{aq)+6NO}}_3^{-}(\text{aq)}$

Reduction:

  ${\text{Ce}}^{4+}\text{(aq)}\to {\text{Ce(OH)}}_{\text{3}}(\text{s})$

Balance the oxygen atoms in the above two reactions.

Oxidation:

  ${\text{Fe(CN)}}_6^{4-}{\text{(aq)+39H}}_{\text{2}}\text{O}\to {\text{Fe(OH)}}_{\text{3}}(\text{s})+6{\text{CO}}_3^{2-}({\text{aq)+6NO}}_3^{-}(\text{aq)}$

Reduction:

  ${\text{Ce}}^{4+}{\text{(aq)+3H}}_{\text{2}}\text{O}\to {\text{Ce(OH)}}_{\text{3}}(\text{s})$

Balance the hydrogen atoms in the above two reactions.

Oxidation:

  ${\text{Fe(CN)}}_6^{4-}{\text{(aq)+39H}}_{\text{2}}\text{O}\to {\text{Fe(OH)}}_{\text{3}}(\text{s})+6{\text{CO}}_3^{2-}({\text{aq)+6NO}}_3^{-}({\text{aq)+75H}}^{+}$

Reduction:

  ${\text{Ce}}^{4+}{\text{(aq)+3H}}_{\text{2}}\text{O}\to {\text{Ce(OH)}}_{\text{3}}(\text{s})+3{\text{H}}^{\text{+}}$

Balance the charge and number of electrons in the above two reactions.

Oxidation:

  ${\text{Fe(CN)}}_6^{4-}{\text{(aq)+39H}}_{\text{2}}\text{O}\to {\text{Fe(OH)}}_{\text{3}}(\text{s})+6{\text{CO}}_3^{2-}({\text{aq)+6NO}}_3^{-}({\text{aq)+75H}}^{+}+61e^{-}$ (1)

Reduction:

  ${\text{Ce}}^{4+}{\text{(aq)+3H}}_{\text{2}}\text{O+}{e}^{-}\to {\text{Ce(OH)}}_{\text{3}}(\text{s})+3{\text{H}}^{\text{+}}$ (2)

Now, multiply equation 2 by 61.

Oxidation:

  ${\text{Fe(CN)}}_6^{4-}{\text{(aq)+39H}}_{\text{2}}\text{O}\to {\text{Fe(OH)}}_{\text{3}}(\text{s})+6{\text{CO}}_3^{2-}({\text{aq)+6NO}}_3^{-}({\text{aq)+75H}}^{+}+61e^{-}$

Reduction:

  ${\text{61Ce}}^{4+}{\text{(aq)+183H}}_{\text{2}}\text{O+61}{e}^{-}\to 61{\text{Ce(OH)}}_{\text{3}}(\text{s})+183{\text{H}}^{\text{+}}$

Add both equations,

  $\begin{array}{l}{\text{Fe(CN)}}_6^{4-}{\text{(aq)+39H}}_{\text{2}}{\text{O+61Ce}}^{4+}{\text{(aq)+183H}}_{\text{2}}\text{O+61}{e}^{-}\\ \to {\text{Fe(OH)}}_{\text{3}}(\text{s})+6{\text{CO}}_3^{2-}({\text{aq)+6NO}}_3^{-}({\text{aq)+75H}}^{+}+61e^{-}+61{\text{Ce(OH)}}_{\text{3}}(\text{s})+183{\text{H}}^{\text{+}}\end{array}$

The equation is written as:

  $\begin{array}{l}{\text{Fe(CN)}}_6^{4-}{\text{(aq)+222H}}_{\text{2}}{\text{O+61Ce}}^{4+}\text{(aq)}\\ \to {\text{Fe(OH)}}_{\text{3}}(\text{s})+6{\text{CO}}_3^{2-}({\text{aq)+6NO}}_3^{-}({\text{aq)+258H}}^{+}+61{\text{Ce(OH)}}_{\text{3}}(\text{s})\end{array}$

For basic medium: Now, add hydroxide ion on both sides of the reaction to balance the hydrogen ion.

The equation is written as:

  $\begin{array}{l}{\text{Fe(CN)}}_6^{4-}{\text{(aq)+222H}}_{\text{2}}{\text{O+61Ce}}^{4+}{\text{(aq)+258OH}}^{-}\\ \to {\text{Fe(OH)}}_{\text{3}}(\text{s})+6{\text{CO}}_3^{2-}({\text{aq)+6NO}}_3^{-}({\text{aq)+258H}}^{+}+61{\text{Ce(OH)}}_{\text{3}}(\text{s})+{\text{258OH}}^{-}\end{array}$

The balance reaction is:

  $\begin{array}{l}{\text{Fe(CN)}}_6^{4-}{\text{(aq)+61Ce}}^{4+}{\text{(aq)+258OH}}^{-}\\ \to {\text{Fe(OH)}}_{\text{3}}(\text{s})+6{\text{CO}}_3^{2-}({\text{aq)+6NO}}_3^{-}(\text{aq)}+61{\text{Ce(OH)}}_{\text{3}}(\text{s})+{\text{36H}}_{\text{2}}\text{O}\end{array}$