Chapter 17, Problem 140CP

(a)

Interpretation Introduction

Interpretation: The following equations are balanced by the half-reaction method.

Concept introduction: The chemical equation is said to be balanced when the number of atoms of chemical species on the reactant side is equal to the number of chemical species on the product side. In the half reaction method, each half reaction that is oxidation and reduction reaction is separately balanced.

To determine: The given balanced chemical reaction by the half-reaction method.

Answer to Problem 140CP

Answer

The balanced redox reaction is,

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

Explanation of Solution

Explanation

Given

The given reaction is,

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

In this equation, the oxidation state of hydrogen decreased from $+1$ to 0 and the oxidation state of iron increased from 0 to $+3$. So, hydrogen undergoes reduction and iron undergoes oxidation.

Also, in this equation there are spectator ions, ${\text{H}}^{\text{+}},{\text{Cl}}^{-}{\text{in HFeCl}}_{\text{4}}\text{ and Cl in HCl}$ which are not directly involved in the redox reaction so they are eliminated and the equation becomes,

$\text{Fe}\left(s\right)+{\text{H}}^{+}\left(aq\right)\underset{}{\to }{\text{Fe}}^{3+}\left(aq\right)+{\text{H}}_{\text{2}}\left(g\right)$

The reduction half cell reaction is,

${\text{H}}^{+}\left(aq\right)\underset{}{\to }{\text{H}}_{\text{2}}\left(g\right)$

The oxidation half cell reaction is,

$\text{Fe}\left(s\right)\underset{}{\to }{\text{Fe}}^{3+}\left(aq\right)$

Now balance H atoms in the reduction half reaction by adding ${\text{H}}^{\text{+}}$ ions.

${\text{H}}^{+}\left(aq\right)+{\text{H}}^{+}\left(aq\right)\underset{}{\to }{\text{H}}_{\text{2}}\left(g\right)$

Finally, balance charge using electrons.

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

Now the charge in oxidation half reaction using electrons is to be balanced.

$\text{Fe}\left(s\right)\underset{}{\to }{\text{Fe}}^{3+}\left(aq\right)+3{\text{e}}^{-}$

The oxidation half reaction is multiplied by 2 and the reduction reaction by 3 to cancel electrons and add them.

${\text{6H}}^{+}\left(aq\right)+6{\text{e}}^{-}\underset{}{\to }{\text{3H}}_{\text{2}}\left(g\right)$

$\text{2Fe}\left(s\right)\underset{}{\to }{\text{2Fe}}^{3+}\left(aq\right)+6{\text{e}}^{-}$

$\text{2Fe}\left(s\right)+6{\text{H}}^{+}\left(aq\right)\underset{}{\to }{\text{2Fe}}^{3+}\left(aq\right)+3{\text{H}}_{\text{2}}\left(g\right)$

The ion ${\text{2Fe}}^{3+}$ is replaced by ${\text{2HFeCl}}_{\text{4}}$ and upon doing so 2 more ${\text{H}}^{\text{+}}$ ions are introduced to the left side of the reaction.

$\text{2Fe}\left(s\right)+6{\text{H}}^{+}\left(aq\right)\underset{}{\to }{\text{2HFeCl}}_4\left(aq\right)+3{\text{H}}_{\text{2}}\left(g\right)$

To balance chlorine atoms, eight moles of chloride ions were added.

$\text{2Fe}\left(s\right)+8{\text{H}}^{+}\left(aq\right)+8{\text{Cl}}^{-}\left(aq\right)\underset{}{\to }{\text{2HFeCl}}_4\left(aq\right)+3{\text{H}}_{\text{2}}\left(g\right)$

Hence, the balanced redox reaction is,

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

(b)

Interpretation Introduction

Interpretation: The following equations are balanced by the half-reaction method.

Concept introduction: The chemical equation is said to be balanced when the number of atoms of chemical species on the reactant side is equal to the number of chemical species on the product side. In the half reaction method, each half reaction that is oxidation and reduction reaction is separately balanced.

To determine: The given balanced chemical reaction by the half-reaction method.

Answer to Problem 140CP

Answer

The balanced reaction is,

${\text{IO}}_3^{-}\left(aq\right)+6{\text{H}}^{\text{+}}\left(aq\right)+8{\text{I}}^{-}\left(aq\right)\stackrel{}{\to }{\text{3I}}_3^{-}\left(aq\right)+3{\text{H}}_{\text{2}}\text{O}\left(l\right)$.

Explanation of Solution

Explanation

Given

${\text{IO}}_3^{-}\left(aq\right)+{\text{I}}^{-}\left(aq\right)\stackrel{\text{Acid}}{\to }{\text{I}}_3^{-}\left(aq\right)$.

The reduction half reaction is,

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

The oxidation state of chlorine decreased from $-1$ to $-\frac{1}{3}$.

The oxidation half reaction is,

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

The oxidation state of iodine increased from $-{\text{1 in I}}^{-}$ to $-\frac{\text{1}}{\text{3}}{\text{ in I}}_3^{-}$.

Now all the elements except H and O in reduction half reaction are balanced and then O atoms using ${\text{H}}_{\text{2}}\text{O}$.

${\text{3IO}}_3^{-}\left(aq\right)\stackrel{}{\to }{\text{I}}_3^{-}\left(aq\right)+9{\text{H}}_{\text{2}}\text{O}\left(l\right)$

Then H atoms are balanced by adding ${\text{H}}^{+}$ ions.

${\text{3IO}}_3^{-}\left(aq\right)+18{\text{H}}^{\text{+}}\left(aq\right)\stackrel{}{\to }{\text{I}}_3^{-}\left(aq\right)+9{\text{H}}_{\text{2}}\text{O}\left(l\right)$

Finally, the charge is balanced by using electrons.

${\text{3IO}}_3^{-}\left(aq\right)+18{\text{H}}^{\text{+}}\left(aq\right)+16{\text{e}}^{-}\stackrel{}{\to }{\text{I}}_3^{-}\left(aq\right)+9{\text{H}}_{\text{2}}\text{O}\left(l\right)$

Now all elements except H and O in oxidation half reaction are balanced.

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

As there are no O and H atoms, the charge is balanced by using electrons.

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

The oxidation half reaction is multiplied by 8 and added to the other half reaction by cancelling the similar terms.

${\text{3IO}}_3^{-}\left(aq\right)+18{\text{H}}^{\text{+}}\left(aq\right)+16{\text{e}}^{-}\stackrel{}{\to }{\text{I}}_3^{-}\left(aq\right)+9{\text{H}}_{\text{2}}\text{O}\left(l\right)$

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

${\text{3IO}}_3^{-}\left(aq\right)+18{\text{H}}^{\text{+}}\left(aq\right)+24{\text{I}}^{-}\left(aq\right)\stackrel{}{\to }{\text{9I}}_3^{-}\left(aq\right)+9{\text{H}}_{\text{2}}\text{O}\left(l\right)$

This is simplified further by dividing it by 3.

${\text{IO}}_3^{-}\left(aq\right)+6{\text{H}}^{\text{+}}\left(aq\right)+8{\text{I}}^{-}\left(aq\right)\stackrel{}{\to }{\text{3I}}_3^{-}\left(aq\right)+3{\text{H}}_{\text{2}}\text{O}\left(l\right)$

Hence, the balanced reaction is,

${\text{IO}}_3^{-}\left(aq\right)+6{\text{H}}^{\text{+}}\left(aq\right)+8{\text{I}}^{-}\left(aq\right)\stackrel{}{\to }{\text{3I}}_3^{-}\left(aq\right)+3{\text{H}}_{\text{2}}\text{O}\left(l\right)$.

(c)

Interpretation Introduction

Interpretation: The following equations are balanced by the half-reaction method.

Concept introduction: The chemical equation is said to be balanced when the number of atoms of chemical species on the reactant side is equal to the number of chemical species on the product side. In the half reaction method, each half reaction that is oxidation and reduction reaction is separately balanced.

To determine: The given balanced chemical reaction by the half-reaction method.

Answer to Problem 140CP

Answer

The balanced reaction is,

$\begin{array}{l}\text{Cr}{\left(\text{NCS}\right)}_6^{4-}\left(aq\right)+5{\text{4H}}_{\text{2}}\text{O}\left(l\right)+97{\text{Ce}}^{\text{4+}}\left(aq\right)\stackrel{\text{Acid}}{\to }\\ {\text{Cr}}^{3+}\left(aq\right)+6{\text{NO}}_3^{-}\left(aq\right)+6{\text{CO}}_{\text{2}}\left(g\right)+6{\text{SO}}_4^{-}\left(aq\right)+108{\text{H}}^{\text{+}}\left(aq\right)+97{\text{Ce}}^{\text{3+}}\left(aq\right)\end{array}$

Explanation of Solution

Explanation

Given

The given reaction is,

$\begin{array}{l}\text{Cr}{\left(\text{NCS}\right)}_6^{4-}\left(aq\right)+{\text{Ce}}^{4+}\left(aq\right)\stackrel{\text{Acid}}{\to }\\ {\text{Cr}}^{3+}\left(aq\right)+{\text{Ce}}^{3+}\left(aq\right)+{\text{NO}}_3^{-}\left(aq\right)+{\text{CO}}_{\text{2}}\left(g\right)+{\text{SO}}_4^{-}\left(aq\right)\end{array}$.

The reduction half reaction is,

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

The oxidation half reaction is,

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

Now all elements except H and O in oxidation half reaction are balanced.

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

And balance oxygen atoms using ${\text{H}}_{\text{2}}\text{O}$.

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

Then H atoms are balanced by adding ${\text{H}}^{\text{+}}$ ions.

$\begin{array}{l}\text{Cr}{\left(\text{NCS}\right)}_6^{4-}\left(aq\right)+5{\text{4H}}_{\text{2}}\text{O}\left(l\right)\stackrel{\text{Acid}}{\to }\\ {\text{Cr}}^{3+}\left(aq\right)+6{\text{NO}}_3^{-}\left(aq\right)+6{\text{CO}}_{\text{2}}\left(g\right)+6{\text{SO}}_4^{-}\left(aq\right)+108{\text{H}}^{\text{+}}\left(aq\right)\end{array}$

Finally, the charge is balanced by using electrons.

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

Now balance all elements except H and O in reduction half reaction.

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

As there are no O and H atoms the charge is balanced by using electrons.

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

The reduction half reaction is multiplied by 97 and is added to the other half reaction by cancelling the similar terms.

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

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

The overall balanced reaction is,

$\begin{array}{l}\text{Cr}{\left(\text{NCS}\right)}_6^{4-}\left(aq\right)+5{\text{4H}}_{\text{2}}\text{O}\left(l\right)+97{\text{Ce}}^{\text{4+}}\left(aq\right)\stackrel{\text{Acid}}{\to }\\ {\text{Cr}}^{3+}\left(aq\right)+6{\text{NO}}_3^{-}\left(aq\right)+6{\text{CO}}_{\text{2}}\left(g\right)+6{\text{SO}}_4^{-}\left(aq\right)+108{\text{H}}^{\text{+}}\left(aq\right)+97{\text{Ce}}^{\text{3+}}\left(aq\right)\end{array}$

(d)

Interpretation Introduction

Interpretation: The following equations are balanced by the half-reaction method.

Concept introduction: The chemical equation is said to be balanced when the number of atoms of chemical species on the reactant side is equal to the number of chemical species on the product side. In the half reaction method, each half reaction that is oxidation and reduction reaction is separately balanced.

To determine: The given balanced chemical reaction by the half-reaction method.

Answer to Problem 140CP

Answer

The balanced reaction is,

$\begin{array}{l}{\text{2CrI}}_{\text{3}}\left(s\right)+27{\text{Cl}}_{\text{2}}\left(g\right)+64{\text{OH}}^{-}\left(aq\right)\stackrel{}{\to }\\ {\text{2CrO}}_4^{-}\left(aq\right)+6{\text{IO}}_4^{-}\left(aq\right)+5{\text{4Cl}}^{-}\left(aq\right)+32{\text{H}}_{\text{2}}\text{O}\left(l\right)\end{array}$

Explanation of Solution

Explanation

Given

The given reaction is,

${\text{CrI}}_{\text{3}}\left(s\right)+{\text{Cl}}_{\text{2}}\left(g\right)\stackrel{\text{Base}}{\to }{\text{CrO}}_4^{-}\left(aq\right)+{\text{IO}}_4^{-}\left(aq\right)+{\text{Cl}}^{-}\left(aq\right)$.

Chromium is oxidized from $+3$ to $+6$, iodine is oxidized from $-1$ to $+7$ and chlorine is reduced from 0 to $-1$.

The reduction half reaction is,

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

The oxidation half reaction is,

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

Now all elements except H and O in oxidation half reaction are balanced

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

And then oxygen atoms are balanced by using ${\text{H}}_{\text{2}}\text{O}$.

${\text{CrI}}_{\text{3}}\left(s\right)+1{\text{6H}}_{\text{2}}\text{O}\left(l\right)\stackrel{}{\to }{\text{CrO}}_4^{-}\left(aq\right)+3{\text{IO}}_4^{-}\left(aq\right)$

Then hydrogen atoms are balanced by adding ${\text{H}}^{\text{+}}$ ions.

${\text{CrI}}_{\text{3}}\left(s\right)+1{\text{6H}}_{\text{2}}\text{O}\left(l\right)\stackrel{}{\to }{\text{CrO}}_4^{-}\left(aq\right)+3{\text{IO}}_4^{-}\left(aq\right)+32{\text{H}}^{\text{+}}\left(aq\right)$

Finally, the charge is balanced by using electrons.

${\text{CrI}}_{\text{3}}\left(s\right)+1{\text{6H}}_{\text{2}}\text{O}\left(l\right)\stackrel{}{\to }{\text{CrO}}_4^{-}\left(aq\right)+3{\text{IO}}_4^{-}\left(aq\right)+32{\text{H}}^{\text{+}}\left(aq\right)+27{\text{e}}^{-}$

Now all elements except H and O in oxidation half reaction are balanced

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

As there are no O and H atoms, the charge is balanced using electrons.

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

The reduction half reactions are multiplied by 27 and oxidation half reaction by 2 and were added by cancelling similar terms.

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

${\text{27Cl}}_{\text{2}}\left(g\right)+54{\text{e}}^{-}\stackrel{}{\to }{\text{54Cl}}^{-}\left(aq\right)$

$\begin{array}{l}{\text{2CrI}}_{\text{3}}\left(s\right)+32{\text{H}}_{\text{2}}\text{O}\left(l\right)+27{\text{Cl}}_{\text{2}}\left(g\right)\stackrel{}{\to }\\ {\text{2CrO}}_4^{-}\left(aq\right)+6{\text{IO}}_4^{-}\left(aq\right)+64{\text{H}}^{\text{+}}\left(aq\right)+5{\text{4Cl}}^{-}\left(aq\right)\end{array}$

In basic medium, ${\text{OH}}^{-}$ ions are added on both sides to neutralize ${\text{H}}^{\text{+}}$ ions.

$\begin{array}{l}{\text{2CrI}}_{\text{3}}\left(s\right)+32{\text{H}}_{\text{2}}\text{O}\left(l\right)+27{\text{Cl}}_{\text{2}}\left(g\right)+64{\text{OH}}^{-}\left(aq\right)\stackrel{}{\to }\\ {\text{2CrO}}_4^{-}\left(aq\right)+6{\text{IO}}_4^{-}\left(aq\right)+64{\text{H}}^{\text{+}}\left(aq\right)+5{\text{4Cl}}^{-}\left(aq\right)+64{\text{OH}}^{-}\left(aq\right)\end{array}$

Hence, the balanced reaction is,

$\begin{array}{l}{\text{2CrI}}_{\text{3}}\left(s\right)+27{\text{Cl}}_{\text{2}}\left(g\right)+64{\text{OH}}^{-}\left(aq\right)\stackrel{}{\to }\\ {\text{2CrO}}_4^{-}\left(aq\right)+6{\text{IO}}_4^{-}\left(aq\right)+5{\text{4Cl}}^{-}\left(aq\right)+32{\text{H}}_{\text{2}}\text{O}\left(l\right)\end{array}$

(e)

Interpretation Introduction

Interpretation: The following equations are balanced by the half-reaction method.

Concept introduction: The chemical equation is said to be balanced when the number of atoms of chemical species on the reactant side is equal to the number of chemical species on the product side. In the half reaction method, each half reaction that is oxidation and reduction reaction is separately balanced.

To determine: The given balanced chemical reaction by the half-reaction method.

Answer to Problem 140CP

Answer

The balanced reaction is,

$\begin{array}{l}\text{Fe}{\left(\text{CN}\right)}_6^{4-}\left(aq\right)+49{\text{Ce}}^{\text{4+}}\left(aq\right)+21{\text{6OH}}^{-}\left(aq\right)\stackrel{}{\to }\\ \text{Fe}{\left(\text{OH}\right)}_{\text{3}}\left(s\right)+6{\text{CO}}_3^{2-}\left(aq\right)+6{\text{NO}}_3^{-}\left(aq\right)+3{\text{3H}}_{\text{2}}\text{O}\left(l\right)+\\ 4\text{9Ce}{\left(\text{OH}\right)}_{\text{3}}\left(aq\right)\end{array}$

Explanation of Solution

Explanation

Given

$\begin{array}{l}\text{Fe}{\left(\text{CN}\right)}_6^{4-}\left(aq\right)+{\text{Ce}}^{4+}\left(aq\right)\stackrel{\text{Base}}{\to }\\ \text{Ce}{\left(\text{OH}\right)}_{\text{3}}\left(aq\right)+\text{Fe}{\left(\text{OH}\right)}_{\text{3}}\left(s\right)+{\text{CO}}_3^{2-}\left(aq\right)+{\text{NO}}_3^{-}\left(aq\right)\end{array}$.

The reduction half reaction is.

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

The oxidation half reaction is,

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

Now all elements except H and O in oxidation half reaction are balanced

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

And then balance O atoms using ${\text{H}}_{\text{2}}\text{O}$.

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

Then H atoms are balanced by adding ${\text{H}}^{\text{+}}$ ions,

$\begin{array}{l}\text{Fe}{\left(\text{CN}\right)}_6^{4-}\left(aq\right)+3{\text{6H}}_{\text{2}}\text{O}\left(l\right)\stackrel{}{\to }\\ \text{Fe}{\left(\text{OH}\right)}_{\text{3}}\left(s\right)+6{\text{CO}}_3^{2-}\left(aq\right)+6{\text{NO}}_3^{-}\left(aq\right)+6{\text{9H}}^{\text{+}}\left(aq\right)\end{array}$

Finally, the charge is balanced by using electrons.

$\begin{array}{l}\text{Fe}{\left(\text{CN}\right)}_6^{4-}\left(aq\right)+3{\text{6H}}_{\text{2}}\text{O}\left(l\right)\stackrel{}{\to }\\ \text{Fe}{\left(\text{OH}\right)}_{\text{3}}\left(s\right)+6{\text{CO}}_3^{2-}\left(aq\right)+6{\text{NO}}_3^{-}\left(aq\right)+6{\text{9H}}^{\text{+}}\left(aq\right)+49{\text{e}}^{-}\end{array}$

Now the O atoms are balanced by using ${\text{H}}_{\text{2}}\text{O}$ in reduction half reaction.

${\text{Ce}}^{4+}\left(aq\right)+{\text{3H}}_{\text{2}}\text{O}\left(l\right)\stackrel{}{\to }\text{Ce}{\left(\text{OH}\right)}_{\text{3}}\left(aq\right)$

Then H atoms are balanced by adding ${\text{H}}^{\text{+}}$ ions.

${\text{Ce}}^{4+}\left(aq\right)+{\text{3H}}_{\text{2}}\text{O}\left(l\right)\stackrel{}{\to }\text{Ce}{\left(\text{OH}\right)}_{\text{3}}\left(aq\right)+{\text{3H}}^{+}\left(aq\right)$

The charge is balanced by using electrons.

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

The reduction half reaction is multiplied by 49 and is added to the other half reaction by cancelling the similar terms.

$\begin{array}{l}\text{Fe}{\left(\text{CN}\right)}_6^{4-}\left(aq\right)+3{\text{6H}}_{\text{2}}\text{O}\left(l\right)\stackrel{}{\to }\\ \text{Fe}{\left(\text{OH}\right)}_{\text{3}}\left(s\right)+6{\text{CO}}_3^{2-}\left(aq\right)+6{\text{NO}}_3^{-}\left(aq\right)+6{\text{9H}}^{\text{+}}\left(aq\right)+49{\text{e}}^{-}\end{array}$

${\text{49Ce}}^{4+}\left(aq\right)+147{\text{H}}_{\text{2}}\text{O}\left(l\right)+49{\text{e}}^{-}\stackrel{}{\to }\text{49Ce}{\left(\text{OH}\right)}_{\text{3}}\left(aq\right)+147{\text{H}}^{+}\left(aq\right)$

$\begin{array}{l}\text{Fe}{\left(\text{CN}\right)}_6^{4-}\left(aq\right)+183{\text{H}}_{\text{2}}\text{O}\left(l\right)+49{\text{Ce}}^{\text{4+}}\left(aq\right)\stackrel{}{\to }\\ \text{Fe}{\left(\text{OH}\right)}_{\text{3}}\left(s\right)+6{\text{CO}}_3^{2-}\left(aq\right)+6{\text{NO}}_3^{-}\left(aq\right)+216{\text{H}}^{\text{+}}\left(aq\right)+4\text{9Ce}{\left(\text{OH}\right)}_{\text{3}}\left(aq\right)\end{array}$

In basic medium, ${\text{OH}}^{-}$ ions are added on both sides to neutralize ${\text{H}}^{\text{+}}$ ions.

$\begin{array}{l}\text{Fe}{\left(\text{CN}\right)}_6^{4-}\left(aq\right)+183{\text{H}}_{\text{2}}\text{O}\left(l\right)+49{\text{Ce}}^{\text{4+}}\left(aq\right)+21{\text{6OH}}^{-}\left(aq\right)\stackrel{}{\to }\\ \text{Fe}{\left(\text{OH}\right)}_{\text{3}}\left(s\right)+6{\text{CO}}_3^{2-}\left(aq\right)+6{\text{NO}}_3^{-}\left(aq\right)+216{\text{H}}^{\text{+}}\left(aq\right)+\\ 4\text{9Ce}{\left(\text{OH}\right)}_{\text{3}}\left(aq\right)+21{\text{6OH}}^{-}\left(aq\right)\end{array}$

Hence, the balanced reaction is,

$\begin{array}{l}\text{Fe}{\left(\text{CN}\right)}_6^{4-}\left(aq\right)+49{\text{Ce}}^{\text{4+}}\left(aq\right)+21{\text{6OH}}^{-}\left(aq\right)\stackrel{}{\to }\\ \text{Fe}{\left(\text{OH}\right)}_{\text{3}}\left(s\right)+6{\text{CO}}_3^{2-}\left(aq\right)+6{\text{NO}}_3^{-}\left(aq\right)+3{\text{3H}}_{\text{2}}\text{O}\left(l\right)+\\ 4\text{9Ce}{\left(\text{OH}\right)}_{\text{3}}\left(aq\right)\end{array}$