Question

Balance the following equation in basic medium by ion electron method and oxidation number methods and identify the oxidising agent and the reducing agent: $N_{2}H_{4}(l)+ClO_{3}^{-}(aq)\longrightarrow NO(g)+Cl^{-}(g)$

Answer 1

$\begin{matrix} { N }_{ 2 }{ H }_{ 4(l) } \\ -2 \end{matrix}\quad +\quad \begin{matrix} { Cl{ O }_{ 3 }^{ - } }_{ (aq) } \\ +5 \end{matrix}\quad \rightarrow \quad \begin{matrix} { NO }_{ (g) } \\ +2 \end{matrix}\quad +\quad \begin{matrix} { Cl }_{ (aq) } \\ -1 \end{matrix}$

The oxidation number of N increases from -2 in ${ N }_{ 2 }{ H }_{ 4 }$ to +2 in NO and the oxidation number of Cl decreases from + 5 in ${ Cl{ O }_{ 3 }^{ - } }$ to -1 in ${ Cl }^{ - }$.

Hence, in this reaction, ${ N }_{ 2 }{ H }_{ 4 }$ is the reducing agent and ${ Cl{ O }_{ 3 }^{ - } }$ is the oxidizing agent.

i) Ion-electron method:

The oxidation half equation is:

${ N }_{ 2 }^{ -2 }{ H }_{ 4(l) }\quad \rightarrow \quad { N }^{ +2 }{ O }_{ (g) }$

The N atoms are balanced as:

${ N }_{ 2 }{ H }_{ 4(l) }\quad \rightarrow \quad { 2NO }_{ (g) }$

The oxidation number is balanced by adding 8 electrons as:

${ N }_{ 2 }{ H }_{ 4(l) }\quad \rightarrow \quad { 2NO }_{ (g) }\quad +\quad { 8e }^{ - }$

The charge is balanced by adding 8 OH-ions as:

${ N }_{ 2 }{ H }_{ 4(l) }\quad +\quad { 8OH }_{ (aq) }^{ - }\quad \rightarrow \quad { 2NO }_{ (g) }\quad +\quad { 8e }^{ - }$

The O atoms are balanced by adding 6H2O as:

${ N }_{ 2 }{ H }_{ 4(l) }\quad +\quad { 8OH }_{ (aq) }^{ - }\quad \rightarrow \quad { 2NO }_{ (g) }\quad +\quad { 6H }_{ 2 }{ O }_{ (l) }\quad +\quad { 8e }^{ - }\quad .....\quad (i)$

The reduction half equation is:

${ C }^{ +5 }{ IO }_{ 3(aq) }^{ - }\quad \rightarrow \quad { C }^{ -1 }{ I }_{ (aq) }^{ - }$

The oxidation number is balanced by adding 6 electrons as:

${ ClO }_{ 3(aq) }^{ - }\quad +\quad { 6e }^{ - }\quad \rightarrow \quad { Cl }_{ (aq) }^{ - }$

The charge is balanced by adding 6OH- ions as:

${ ClO }_{ 3(aq) }^{ - }\quad +\quad { 6e }^{ - }\quad \rightarrow \quad { Cl }_{ (aq) }\quad +\quad { 6OH }_{ (aq) }^{ - }$

The O atoms are balanced by adding 3H2O as:

${ ClO }_{ 3(aq) }^{ - }\quad +\quad 3{ H }_{ 2 }{ O }_{ (l) }\quad +\quad { 6e }^{ - }\quad \rightarrow \quad { Cl }_{ (aq) }^{ - }\quad +\quad { 6OH }_{ (aq) }^{ - }\quad .....\quad (ii)$

The balanced equation can be obtained by multiplying equation (i) with 3 and equation (ii) with 4 and then adding them as:

$3{ N }_{ 2 }{ H }_{ 4(l) }\quad +\quad 4Cl{ O }_{ 3(aq) }^{ - }\quad \rightarrow \quad 6{ NO }_{ (g) }\quad +\quad 4{ Cl }_{ (aq) }^{ - }\quad +\quad 6{ H }_{ 2 }{ O }_{ (l) }$

ii) Oxidation number method:

Total decrease in oxidation number of $N = 2 \times 4 = 8$

Total increase in oxidation number of $Cl = 1 \times 6 = 6$

On multiplying ${ N }_{ 2 }{ H }_{ 4 }$ with 3 and $Cl{ O }_{ 3 }^{ - }$ with 4 to balance the increase and decrease in O and N, we get:

$3{ N }_{ 2 }{ H }_{ 4(l) }\quad +\quad 4Cl{ O }_{ 3(aq) }^{ - }\quad \rightarrow \quad { NO }_{ (g) }\quad +\quad { Cl }_{ (aq) }^{ - }$

The N and Cl atoms are balanced as:

$3{ N }_{ 2 }{ H }_{ 4(l) }\quad +\quad 4Cl{ O }_{ 3(aq) }^{ - }\quad \rightarrow \quad 6{ NO }_{ (g) }\quad +\quad 4{ Cl }_{ (aq) }^{ - }$

The O atoms are balanced by adding 6H2O as:

$3{ N }_{ 2 }{ H }_{ 4(l) }\quad +\quad 4Cl{ O }_{ 3(aq) }^{ - }\quad \rightarrow \quad 6{ NO }_{ (g) }\quad +\quad 4{ Cl }_{ (aq) }^{ - }\quad +\quad 6{ H }_{ 2 }{ O }_{ (l) }$

This is the required balanced equation.

iii) In the above reaction ${ N }_{ 2 }{ H }_{ 4 }$ is the reducing agent and $Cl{ O }_{ 3 }^{ - }$ is the oxidizing agent.