Chapter 17, Problem 29E

Balance the following oxidation–reduction reactions that occur in acidic solution using the half–reaction method.

a. ${\text{I}}^{-}(aq)+{\text{ClO}}^{-}(aq)\to {\text{I}}_{\text{3}}{}^{-}(aq)+{\text{Cl}}^{-}(aq)$

b. ${\text{As}}_{\text{2}}{\text{O}}_{\text{3}}(s)+{\text{NO}}_{\text{3}}{}^{-}(aq)\to {\text{H}}_{\text{3}}{\text{AsO}}_{\text{4}}(aq)+\text{NO}(g)$

c. ${\text{Br}}^{-}(aq)+{\text{MnO}}_{\text{4}}{}^{-}(aq)\to {\text{Br}}_{\text{2}}(l)+{\text{Mn}}^{\text{2+}}(aq)$

d. ${\text{CH}}_{\text{3}}\text{OH}(aq)+{\text{Cr}}_{\text{2}}{\text{O}}_{\text{7}}{}^{2-}(aq)\to {\text{CH}}_{\text{2}}\text{O}(aq)+{\text{Cr}}^{3+}(aq)$

Interpretation Introduction

Interpretation:

The four oxidation-reduction reactions are given. The balancing of all the reactions in acidic media using the half-reaction method is to be done.

Concept introduction:

The reaction in which both oxidation and reduction reaction occur simultaneously is called a redox reaction. According to the law of conservation of mass, the mass of all the species in a reaction must be balanced. Therefore, balancing is necessary to conserve the mass and even the charge must be balanced to maintain the overall charge of the reaction.

To determine: The steps involved in the balancing of the given equation (a).

Answer to Problem 29E

The balanced equations are as follows,

${\text{ClO}}^{\text{-}}{}_{\left(\text{aq}\right)}{\text{+2H}}^{\text{+}}{}_{\left(\text{aq}\right)}{\text{+3I}}^{\text{-}}{}_{\left(\text{aq}\right)}\to {\text{Cl}}^{\text{-}}{}_{\left(\text{aq}\right)}{\text{+H}}_{\text{2}}{\text{O}}_{\left(\text{l}\right)}{\text{+I}}_{\text{3}}{{}^{\text{-}}}_{\left(\text{aq}\right)}$

Explanation of Solution

The balanced equation is defined as follows,

${\text{ClO}}^{\text{-}}{}_{\left(\text{aq}\right)}{\text{+2H}}^{\text{+}}{}_{\left(\text{aq}\right)}{\text{+3I}}^{\text{-}}{}_{\left(\text{aq}\right)}\to {\text{Cl}}^{\text{-}}{}_{\left(\text{aq}\right)}{\text{+H}}_{\text{2}}{\text{O}}_{\left(\text{l}\right)}{\text{+I}}_{\text{3}}{{}^{\text{-}}}_{\left(\text{aq}\right)}$

The reduction half cell reaction is,

${\text{ClO}}^{\text{-}}{}_{\left(\text{aq}\right)}\to {\text{Cl}}^{\text{-}}{}_{\left(\text{aq}\right)}$

The change in the oxidation number of chlorine is from $+1$ to $-1$.

The oxidation half reaction is,

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

The change in oxidation number of iodine is from $-1$ to $\left(-\frac{1}{3}\right)$.

As the atoms other than hydrogen and oxygen are already balanced, so directly balance the oxygen atom in the reduction half reaction by adding water to right hand side,

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

Balance the hydrogen atoms in the reduction half reaction by adding ${\text{H}}^{+}$ ions to the left hand side,

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

Balance the charge by adding appropriate number of electrons to the left hand side,

${\text{ClO}}^{\text{-}}{}_{\left(\text{aq}\right)}{\text{+2H}}^{\text{+}}{}_{\left(\text{aq}\right)}{\text{+2e}}^{\text{-}}\to {\text{Cl}}^{\text{-}}{}_{\left(\text{aq}\right)}{\text{+H}}_{\text{2}}{\text{O}}_{\left(\text{l}\right)}$ (1)

Balance the oxidation half reaction as,

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

Balance the charge by adding electrons at the appropriate side,

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

Add the oxidation and reduction half reaction to get the final equation,

$\begin{array}{c}{\text{ClO}}^{\text{-}}{}_{\left(\text{aq}\right)}{\text{+2H}}^{\text{+}}{}_{\left(\text{aq}\right)}{\text{+2e}}^{\text{-}}\to {\text{Cl}}^{\text{-}}{}_{\left(\text{aq}\right)}{\text{+H}}_{\text{2}}{\text{O}}_{\left(\text{l}\right)}\\ {\text{3I}}^{\text{-}}{}_{\left(\text{aq}\right)}\to {\text{I}}_{\text{3}}{{}^{\text{-}}}_{\left(\text{aq}\right)}{\text{+2e}}^{\text{-}}\end{array}$

Cancel similar terms on both the sides. The final equation is,

${\text{ClO}}^{\text{-}}{}_{\left(\text{aq}\right)}{\text{+2H}}^{\text{+}}{}_{\left(\text{aq}\right)}{\text{+3I}}^{\text{-}}{}_{\left(\text{aq}\right)}\to {\text{Cl}}^{\text{-}}{}_{\left(\text{aq}\right)}{\text{+H}}_{\text{2}}{\text{O}}_{\left(\text{l}\right)}{\text{+I}}_{\text{3}}{{}^{\text{-}}}_{\left(\text{aq}\right)}$

Interpretation Introduction

Interpretation:

The four oxidation-reduction reactions are given. The balancing of all the reactions in acidic media using the half-reaction method is to be done.

Concept introduction:

The reaction in which both oxidation and reduction reaction occur simultaneously is called a redox reaction. According to the law of conservation of mass, the mass of all the species in a reaction must be balanced. Therefore, balancing is necessary to conserve the mass and even the charge must be balanced to maintain the overall charge of the reaction.

To determine: The steps involved in the balancing of the given equation (b).

Answer to Problem 29E

The balanced equations are as follows,

${\text{3As}}_{\text{2}}{\text{O}}_{\text{3}}{}_{\left(\text{s}\right)}\text{+}{\text{4H}}^{\text{+}}{}_{\left(\text{aq}\right)}\text{+}{\text{4NO}}_{\text{3}}{{}^{\text{-}}}_{\left(\text{aq}\right)}\text{+}{\text{7H}}_{\text{2}}{\text{O}}_{\left(\text{l}\right)}\to {\text{6H}}_{\text{3}}{\text{AsO}}_{\text{4}}{}_{\left(\text{aq}\right)}\text{+}{\text{4NO}}_{\left(\text{g}\right)}$

Explanation of Solution

The balanced equation is defined as follows,

${\text{3As}}_{\text{2}}{\text{O}}_{\text{3}}{}_{\left(\text{s}\right)}{\text{+4H}}^{\text{+}}{}_{\left(\text{aq}\right)}{\text{+4NO}}_{\text{3}}{{}^{\text{-}}}_{\left(\text{aq}\right)}{\text{+7H}}_{\text{2}}{\text{O}}_{\left(\text{l}\right)}\to {\text{6H}}_{\text{3}}{\text{AsO}}_{\text{4}}{}_{\left(\text{aq}\right)}{\text{+4NO}}_{\left(\text{g}\right)}$

The reduction half cell reaction is,

${\text{NO}}_{\text{3}}{}^{\text{-}}\left(\text{aq}\right)\to \text{NO}\left(\text{g}\right)$

The change in the oxidation number of nitrogen is from $+5$ to $+2$.

The oxidation half reaction is,

${\text{As}}_{\text{2}}{\text{O}}_{\text{3}}{}_{\left(\text{s}\right)}\to {\text{H}}_{\text{3}}{\text{AsO}}_{\text{4}}{}_{\left(\text{aq}\right)}$

The change in oxidation number of arsenic is from $+3$ to $+5$.

Elements except hydrogen and oxygen are already balanced, therefore, directly balance the oxygen in the reduction half reaction by adding water to right hand side,

${\text{NO}}_{\text{3}}{{}^{\text{-}}}_{\left(\text{aq}\right)}\to {\text{NO}}_{\left(\text{g}\right)}{\text{+2H}}_{\text{2}}{\text{O}}_{\left(\text{l}\right)}$

Balance the hydrogen atoms in the reduction half reaction by adding ${\text{H}}^{+}$ ions to the left hand side,

${\text{NO}}_{\text{3}}{{}^{\text{-}}}_{\left(\text{aq}\right)}{\text{+4H}}^{\text{+}}{}_{\left(\text{aq}\right)}\to {\text{NO}}_{\left(\text{g}\right)}{\text{+2H}}_{\text{2}}{\text{O}}_{\left(\text{l}\right)}$

Balance the charge by adding appropriate number of electrons to the left hand side,

${\text{NO}}_{\text{3}}{{}^{\text{-}}}_{\left(\text{aq}\right)}{\text{+4H}}^{\text{+}}{}_{\left(\text{aq}\right)}{\text{+3e}}^{\text{-}}\to {\text{NO}}_{\left(\text{g}\right)}{\text{+2H}}_{\text{2}}{\text{O}}_{\left(\text{l}\right)}$ (3)

Balance the oxidation half reaction as,

${\text{As}}_{\text{2}}{\text{O}}_{\text{3}}{}_{\left(\text{s}\right)}\to {\text{2H}}_{\text{3}}{\text{AsO}}_{\text{4}}{}_{\left(\text{aq}\right)}$

Balance the oxygen in the reduction half reaction by adding water to left hand side,

${\text{As}}_{\text{2}}{\text{O}}_{\text{3}}{}_{\left(\text{s}\right)}{\text{+5H}}_{\text{2}}{\text{O}}_{\left(\text{l}\right)}\to {\text{2H}}_{\text{3}}{\text{AsO}}_{\text{4}}{}_{\left(\text{aq}\right)}$

Balance the hydrogen atoms in the reduction half reaction by adding ${\text{H}}^{+}$ ions to the right hand side,

${\text{As}}_{\text{2}}{\text{O}}_{\text{3}}{}_{\left(\text{s}\right)}{\text{+5H}}_{\text{2}}{\text{O}}_{\left(\text{l}\right)}\to {\text{2H}}_{\text{3}}{\text{AsO}}_{\text{4}}{}_{\left(\text{aq}\right)}{\text{+4H}}^{\text{+}}$

Balance the charge by adding electrons at the appropriate side,

${\text{As}}_{\text{2}}{\text{O}}_{\text{3}}{}_{\left(\text{s}\right)}{\text{+5H}}_{\text{2}}{\text{O}}_{\left(\text{l}\right)}\to {\text{2H}}_{\text{3}}{\text{AsO}}_{\text{4}}{}_{\left(\text{aq}\right)}{\text{+4H}}^{\text{+}}{\text{+4e}}^{\text{-}}$ (4)

Multiply equation (3) by $4$ and equation (4) by $3$ to cancel out the electrons and then add the oxidation and reduction half reaction to get the final equation,

$\begin{array}{c}{\text{3As}}_{\text{2}}{\text{O}}_{\text{3}}{}_{\left(\text{s}\right)}{\text{+15H}}_{\text{2}}{\text{O}}_{\left(\text{l}\right)}\to {\text{6H}}_{\text{3}}{\text{AsO}}_{\text{4}}{}_{\left(\text{aq}\right)}{\text{+12H}}^{\text{+}}{\text{+12e}}^{\text{-}}\\ {\text{4NO}}_{\text{3}}{{}^{\text{-}}}_{\left(\text{aq}\right)}{\text{+16H}}^{\text{+}}{}_{\left(\text{aq}\right)}{\text{+12e}}^{\text{-}}\to {\text{4NO}}_{\left(\text{g}\right)}{\text{+8H}}_{\text{2}}{\text{O}}_{\left(\text{l}\right)}\end{array}$

Cancel similar terms on both the sides to get the final equation as,

${\text{3As}}_{\text{2}}{\text{O}}_{\text{3}}{}_{\left(\text{s}\right)}{\text{+4H}}^{\text{+}}{}_{\left(\text{aq}\right)}{\text{+4NO}}_{\text{3}}{{}^{\text{-}}}_{\left(\text{aq}\right)}{\text{+7H}}_{\text{2}}{\text{O}}_{\left(\text{l}\right)}\to {\text{6H}}_{\text{3}}{\text{AsO}}_{\text{4}}{}_{\left(\text{aq}\right)}{\text{+4NO}}_{\left(\text{g}\right)}$

Interpretation Introduction

Interpretation:

The four oxidation-reduction reactions are given. The balancing of all the reactions in acidic media using the half-reaction method is to be done.

Concept introduction:

The reaction in which both oxidation and reduction reaction occur simultaneously is called a redox reaction. According to the law of conservation of mass, the mass of all the species in a reaction must be balanced. Therefore, balancing is necessary to conserve the mass and even the charge must be balanced to maintain the overall charge of the reaction.

To determine: The steps involved in the balancing of the given equation (c).

Answer to Problem 29E

The balanced equations are as follows,

${\text{2MnO}}_{\text{4}}{{}^{\text{-}}}_{\left(\text{aq}\right)}\text{+}{\text{10Br}}^{\text{-}}{}_{\left(\text{aq}\right)}\text{+}{\text{16H}}^{\text{+}}{}_{\left(\text{aq}\right)}\to {\text{2Mn}}^{\text{2+}}{}_{\left(\text{aq}\right)}\text{+}{\text{5Br}}_{\text{2}}{}_{\left(\text{l}\right)}\text{+}{\text{8H}}_{\text{2}}{\text{O}}_{\left(\text{l}\right)}$

Explanation of Solution

The balanced equation is defined as follows,

${\text{2MnO}}_{\text{4}}{{}^{\text{-}}}_{\left(\text{aq}\right)}{\text{+10Br}}^{\text{-}}{}_{\left(\text{aq}\right)}{\text{+16H}}^{\text{+}}{}_{\left(\text{aq}\right)}\to {\text{2Mn}}^{\text{2+}}{}_{\left(\text{aq}\right)}{\text{+5Br}}_{\text{2}}{}_{\left(\text{l}\right)}{\text{+8H}}_{\text{2}}{\text{O}}_{\left(\text{l}\right)}$

The reduction half cell reaction is,

${\text{MnO}}_{\text{4}}{{}^{\text{-}}}_{\left(\text{aq}\right)}\to {\text{Mn}}^{\text{2+}}{}_{\left(\text{aq}\right)}$

The change in the oxidation number of manganese is from $+7$ to $+2$.

The oxidation half reaction is,

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

The change in oxidation number of bromine is from $-1$ to $0$.

All the elements except hydrogen and oxygen are already balanced so directly balance the oxygen in the reduction half reaction by adding water to right hand side,

${\text{MnO}}_{\text{4}}{{}^{\text{-}}}_{\left(\text{aq}\right)}\to {\text{Mn}}^{\text{2+}}{}_{\left(\text{aq}\right)}{\text{+4H}}_{\text{2}}{\text{O}}_{\left(\text{l}\right)}$

Balance the hydrogen atoms in the reduction half reaction by adding ${\text{H}}^{+}$ ions to the left hand side,

${\text{MnO}}_{\text{4}}{{}^{\text{-}}}_{\left(\text{aq}\right)}{\text{+8H}}^{\text{+}}{}_{\left(\text{aq}\right)}\to {\text{Mn}}^{\text{2+}}{}_{\left(\text{aq}\right)}{\text{+4H}}_{\text{2}}{\text{O}}_{\left(\text{l}\right)}$

Balance the charge by adding appropriate number of electrons to the left hand side,

${\text{MnO}}_{\text{4}}{{}^{\text{-}}}_{\left(\text{aq}\right)}{\text{+8H}}^{\text{+}}{}_{\left(\text{aq}\right)}{\text{+5e}}^{\text{-}}\to {\text{Mn}}^{\text{2+}}{}_{\left(\text{aq}\right)}{\text{+4H}}_{\text{2}}{\text{O}}_{\left(\text{l}\right)}$ (5)

Balance the oxidation half reaction as,

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

As there are no atoms of hydrogen or oxygen, so directly balance the charge by adding electrons at the appropriate side,

${\text{2Br}}^{\text{-}}{}_{\left(\text{aq}\right)}\to {\text{Br}}_{\text{2}}{}_{\left(\text{l}\right)}{\text{+2e}}^{\text{-}}$ (6)

Multiply equation $\left(5\right)$ by $2$ and equation $\left(6\right)$ by $5$ to cancel out the electrons and then add the oxidation and reduction half reaction to get the final equation,

$\begin{array}{c}{\text{2MnO}}_{\text{4}}{{}^{\text{-}}}_{\left(\text{aq}\right)}{\text{+16H}}^{\text{+}}{}_{\left(\text{aq}\right)}{\text{+10e}}^{\text{-}}\to {\text{2Mn}}^{\text{2+}}{}_{\left(\text{aq}\right)}{\text{+8H}}_{\text{2}}{\text{O}}_{\left(\text{l}\right)}\\ {\text{10Br}}^{\text{-}}{}_{\left(\text{aq}\right)}\to {\text{5Br}}_{\text{2}}{}_{\left(\text{l}\right)}{\text{+10e}}^{\text{-}}\end{array}$

Cancel similar terms on both the sides. The final equation is,

${\text{2MnO}}_{\text{4}}{{}^{\text{-}}}_{\left(\text{aq}\right)}{\text{+10Br}}^{\text{-}}{}_{\left(\text{aq}\right)}{\text{+16H}}^{\text{+}}{}_{\left(\text{aq}\right)}\to {\text{2Mn}}^{\text{2+}}{}_{\left(\text{aq}\right)}{\text{+5Br}}_{\text{2}}{}_{\left(\text{l}\right)}{\text{+8H}}_{\text{2}}{\text{O}}_{\left(\text{l}\right)}$

Interpretation Introduction

Interpretation:

The four oxidation-reduction reactions are given. The balancing of all the reactions in acidic media using the half-reaction method is to be done.

Concept introduction:

The reaction in which both oxidation and reduction reaction occur simultaneously is called a redox reaction. According to the law of conservation of mass, the mass of all the species in a reaction must be balanced. Therefore, balancing is necessary to conserve the mass and even the charge must be balanced to maintain the overall charge of the reaction.

To determine: The steps involved in the balancing of the given equation (d).

Answer to Problem 29E

The balanced equations are as follows,

${\text{Cr}}_{\text{2}}{\text{O}}_{\text{7}}{{}^{\text{2-}}}_{\left(\text{aq}\right)}{\text{+8H}}^{\text{+}}{}_{\left(\text{aq}\right)}{\text{+3CH}}_{\text{3}}{\text{OH}}_{\left(\text{aq}\right)}\to {\text{2Cr}}^{\text{3+}}{}_{\left(\text{aq}\right)}{\text{+3CH}}_{\text{2}}{\text{O}}_{\left(\text{aq}\right)}{\text{+7H}}_{\text{2}}{\text{O}}_{\left(\text{l}\right)}$

Explanation of Solution

The balanced equation is defined as follows,

${\text{Cr}}_{\text{2}}{\text{O}}_{\text{7}}{{}^{\text{2-}}}_{\left(\text{aq}\right)}{\text{+8H}}^{\text{+}}{}_{\left(\text{aq}\right)}{\text{+3CH}}_{\text{3}}{\text{OH}}_{\left(\text{aq}\right)}\to {\text{2Cr}}^{\text{3+}}{}_{\left(\text{aq}\right)}{\text{+3CH}}_{\text{2}}{\text{O}}_{\left(\text{aq}\right)}{\text{+7H}}_{\text{2}}{\text{O}}_{\left(\text{l}\right)}$

The reduction half cell reaction is,

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

The change in the oxidation number of chromium is from $+7$ to $+2$.

The oxidation half reaction is,

${\text{CH}}_{\text{3}}{\text{OH}}_{\left(\text{aq}\right)}\to {\text{CH}}_{\text{2}}{\text{O}}_{\left(\text{aq}\right)}$

The change in oxidation number of carbon is from $-2$ to $0$.

Balance all the atoms except hydrogen and oxygen in the reduction half cell,

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

Balance the oxygen in the reduction half reaction by adding water to right hand side,

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

Balance the hydrogen atoms in the reduction half reaction by adding ${\text{H}}^{+}$ ions to the left hand side,

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

Balance the charge by adding appropriate number of electrons to the left hand side,

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

As all the atoms except hydrogen are already balanced so directly balance the oxidation half reaction as,

${\text{CH}}_{\text{3}}{\text{OH}}_{\left(\text{aq}\right)}\to {\text{CH}}_{\text{2}}{\text{O}}_{\left(\text{aq}\right)}{\text{+2H}}^{\text{+}}{}_{\left(\text{aq}\right)}$

Balance the charge by adding electrons at the appropriate side,

${\text{CH}}_{\text{3}}{\text{OH}}_{\left(\text{aq}\right)}\to {\text{CH}}_{\text{2}}{\text{O}}_{\left(\text{aq}\right)}{\text{+2H}}^{\text{+}}{}_{\left(\text{aq}\right)}{\text{+2e}}^{\text{-}}$ (8)

Multiply equation (8) by $3$ to cancel out the electrons and then add the oxidation and reduction half reaction to get the final equation,

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

Cancel similar terms on both the sides. The final equation is,

${\text{Cr}}_{\text{2}}{\text{O}}_{\text{7}}{{}^{\text{2-}}}_{\left(\text{aq}\right)}{\text{+8H}}^{\text{+}}{}_{\left(\text{aq}\right)}{\text{+3CH}}_{\text{3}}{\text{OH}}_{\left(\text{aq}\right)}\to {\text{2Cr}}^{\text{3+}}{}_{\left(\text{aq}\right)}{\text{+3CH}}_{\text{2}}{\text{O}}_{\left(\text{aq}\right)}{\text{+7H}}_{\text{2}}{\text{O}}_{\left(\text{l}\right)}$