Chapter 19, Problem 1QP

Balance the following redox equations by the half-reaction method:

$\begin{array}{l}\left(\text{a}\right)\hfill \\ {\text{H}}_{\text{2}}{\text{O}}_{\text{2}}{\text{+ Fe}}^{\text{2+}}\underset{}{\to }{\text{ Fe}}^{\text{3+}}{\text{+ H}}_{\text{2}}\text{O }\left(\text{in acidic solution}\right)\hfill \\ \left(\text{b}\right)\hfill \\ {\text{Cu + HNO}}_{\text{3}}\underset{}{\to }{\text{Cu}}^{\text{2+}}{\text{+ NO + H}}_{\text{2}}\text{O }\left(\text{in acidic solution}\right)\hfill \\ \left(\text{c}\right)\hfill \\ {\text{CN}}^{\text{-}}{\text{+ MnO}}_4^{-}\underset{}{\to }{\text{ CNO}}^{\text{-}}{\text{ + MnO}}_{\text{2}}\left(\text{in basic solution}\right)\hfill \\ \left(\text{d}\right)\hfill \\ {\text{Br}}_{\text{2}}\underset{}{\to }{\text{BrO}}_3^{-}{\text{ + Br}}^{\text{-}}\left(\text{in basic solution}\right)\hfill \\ \left(\text{e}\right)\hfill \\ {\text{S}}_{\text{2}}{\text{O}}_3^2{\text{ + I}}_{\text{2}}\underset{}{\to }{\text{ I + S}}_{\text{4}}{\text{O}}_6^2\left(\text{in acidic solution}\right)\hfill \end{array}$

Interpretation Introduction

Interpretation:

The given redox equations are to be balanced by half-reaction method.

Concept introduction:

An oxidation reaction takes place when there is again of electrons and a reduction reaction occurs when removal of electrons takes place.

A chemical reaction in which electrons transfer from one species to another, such type of oxidation–reduction reaction is known as a redox reaction.

Steps for balancing equation by half-reaction method:

In step $1$, half reactions are separated. A half reaction is an oxidation and reduction reaction that is part of overall redox reaction.

Step $2$: Balance the half reactions for atoms other than oxygen and hydrogen.

Step $3$: By adding water, each half reaction will be balanced for oxygen.

Step $4$: By adding ${\text{H}}^{+}$, each half reaction will be balanced for hydrogen.

Step $5$: By the addition of electrons, both half-reactions will be balanced.

Step $\text{6}$: If the number of electrons are not same in both half cell-reaction, multiply one or both the reaction by the number required to make the number of electrons same.

Step $7$: By adding the resulting half reactions and cancelling identical species and electrons, for getting the overall balanced equation.

Step $8$: In the final equation, for each ${\text{H}}^{+}$, one ${\text{OH}}^{-}$ is added to each side, where ${\text{H}}^{+}$ and ${\text{OH}}^{-}$ combine to form water.

Step $9$: Make any necessary cancellations made necessary by the new water molecule.

Answer to Problem 1QP

Solution:

a) $2{\text{H}}^{+}+{\text{H}}_2{\text{O}}_2+{\text{2Fe}}^{2+}\to 2{\text{Fe}}^{3+}+2{\text{H}}_2\text{O}$

b) $6{\text{H}}^{+}+2{\text{HNO}}_3+3\text{Cu}\to 3{\text{Cu}}^{2+}+2\text{NO}+4{\text{H}}_2\text{O}$

c) $3{\text{CN}}^{-}+2{\text{MnO}}_4^{-}+{\text{H}}_2\text{O}\to 3{\text{CNO}}^{-}+2{\text{MnO}}_2+{\text{2OH}}^{-}$

d) ${\text{6OH}}^{-}{\text{+3Br}}_{\text{2}}\to {\text{BrO}}_3^{-}+3{\text{H}}_2\text{O}+{\text{5Br}}^{-}$

e) ${\text{2S}}_2{\text{O}}_3^{2-}+{\text{I}}_2\to 2{\text{I}}^{-}+{\text{S}}_4{\text{O}}_6^{2-}$

Explanation of Solution

a) ${\text{H}}_2{\text{O}}_2+{\text{Fe}}^{2+}\to {\text{Fe}}^{3+}+{\text{H}}_2\text{O}$ (in acidic solution)

In step $1$, half reactions are separated as follows:

$\begin{array}{l}{\text{Oxidation: Fe}}^{2+}\to {\text{Fe}}^{3+}\\ {\text{Reduction: H}}_2{\text{O}}_2\to {\text{H}}_2\text{O}\end{array}$

Step $2$: Balance the half reactions, but in this case, it is not necessary, as the half reactions are already balanced.

Step $3$: By adding water, each half reaction will be balanced for oxygen. In the oxidationhalf-reaction, oxygen is not present and it is already balanced. The reduction half-reactionneeds one molecule of water on the product side. The reaction will be as follows:

${\text{H}}_2{\text{O}}_2\to {\text{H}}_2\text{O}+{\text{H}}_2\text{O}$ or ${\text{H}}_2{\text{O}}_2\to 2{\text{H}}_2\text{O}$

Step $4$: By adding ${\text{H}}^{+}$, each half reaction will be balanced for hydrogen. In the oxidation, half-reaction, hydrogen is not present and it is already balanced. The reductionhalf-reactionneeds two ${\text{H}}^{+}$ on the reactant side. The reaction will be as follows:

${\text{H}}_2{\text{O}}_2+2{\text{H}}^{+}\to 2{\text{H}}_2\text{O}$

Step $5$: By the addition of electrons, both half-reactions will be balanced. The total charge on the left is $+2$, and on the right, the total charge is $+3$, in the oxidation-half reaction. If one electron is added to the side of the products, on each side, the total charge will become $+2$. The reaction is as follows:

${\text{Fe}}^{2+}\to {\text{Fe}}^{3+}+{\text{e}}^{-}$

The total charge on the left is $+2$, and on the right, the total charge is $0$, in the reduction-half reaction. If two electrons are added to the side of reactants, on each side, the total charge will become $0$. The reaction is as follows:

${\text{H}}_2{\text{O}}_2+2{\text{H}}^{+}+2{\text{e}}^{-}\to 2{\text{H}}_2\text{O}$

Step $6$: The oxidation half-reaction is multiplied by $2$ so that the number of electrons in both half-reactions becomes the same. The reaction is as follows:

$2\left({\text{Fe}}^{2+}\to {\text{Fe}}^{3+}+{\text{e}}^{-}\right)={\text{2Fe}}^{2+}\to 2{\text{Fe}}^{3+}+2{\text{e}}^{-}$

Step $7$: By adding the resulting half reactions, and cancelling identical species and electrons for getting the overall balanced equation as follows:

$\begin{array}{l}\underset{\_}{\begin{array}{l}{\text{ 2Fe}}^{2+}\to 2{\text{Fe}}^{3+}+\cancel{2{\text{e}}^{-}}\\ {\text{ H}}_2{\text{O}}_2+2{\text{H}}^{+}+\cancel{2{\text{e}}^{-}}\to 2{\text{H}}_2\text{O}\end{array}}\\ 2{\text{H}}^{+}+{\text{H}}_2{\text{O}}_2+{\text{2Fe}}^{2+}\to 2{\text{Fe}}^{3+}+2{\text{H}}_2\text{O}\end{array}$

Therefore, the balanced chemical equation is $2{\text{H}}^{+}+{\text{H}}_2{\text{O}}_2+{\text{2Fe}}^{2+}\to 2{\text{Fe}}^{3+}+2{\text{H}}_2\text{O}$.

b) $\text{Cu}+{\text{HNO}}_3\to {\text{Cu}}^{2+}+\text{NO}+{\text{H}}_2\text{O}$ (in acidic solution)

In step $1$, half reactions are separated as follows:

$\begin{array}{l}\text{Oxidation: Cu}\to {\text{Cu}}^{2+}\\ {\text{Reduction: HNO}}_3\to \text{NO}\end{array}$

Step $2$: Balance the half reactions, but in this case, it is not necessary as the half reactions are already balanced.

Step $3$: By adding water, each half reaction will be balanced for oxygen. In the oxidationhalf-reaction, oxygen is not present and it is already balanced. The reductionhalf-reactionneeds two molecules of water on side of the products. The reaction will be as follows:

${\text{HNO}}_3\to \text{NO}+2{\text{H}}_2\text{O}$

Step $4$: By adding ${\text{H}}^{+}$, each half reaction will be balanced for hydrogen. In the oxidationhalf-reaction, hydrogen is not present and it is already balanced. The reduction half-reactionneedsthree ${\text{H}}^{+}$ on side of reactants. The reaction will be as follows:

${\text{3H}}^{+}+{\text{HNO}}_3\to \text{NO}+2{\text{H}}_2\text{O}$

Step $5$: By the addition of electrons, both half-reactions will be balanced. The total charge on the left is $0$, and on the right, total charge is $+2$, in the oxidation-half reaction. If two electrons are added to the side of the products, on each side, the total charge will become $0$. The reaction is as follows:

$\text{Cu}\to {\text{Cu}}^{2+}+2{\text{e}}^{-}$

The total charge on the left is $+3$, and on the right, total charge is $0$, in the reduction-half reaction. If three electrons are added to the side of reactants, on each side, the total charge will become $0$. The reaction is as follows:

${\text{3H}}^{+}+{\text{HNO}}_3+3{\text{e}}^{-}\to \text{NO}+2{\text{H}}_2\text{O}$

Step $6$: The oxidation half-reaction is multiplied by $3$ and the reduction half-reaction is multiplied by $2$, so that the number of electrons in both half reactions becomes the same. The reaction is as follows:

$\begin{array}{l}3\left(\text{Cu}\to {\text{Cu}}^{2+}+2{\text{e}}^{-}\right)=3\text{Cu}\to 3{\text{Cu}}^{2+}+6{\text{e}}^{-}\\ 2\left({\text{3H}}^{+}+{\text{HNO}}_3+3{\text{e}}^{-}\to \text{NO}+2{\text{H}}_2\text{O}\right)=6{\text{H}}^{+}+2{\text{HNO}}_3+6{\text{e}}^{-}\to 2\text{NO}+4{\text{H}}_2\text{O}\end{array}$

Step $7$: By adding the resulting half reactions and cancelling identical species and electrons, for getting the overall balanced equation as follows:

$\begin{array}{l}\underset{\_}{\begin{array}{l}3\text{Cu}\to 3{\text{Cu}}^{2+}+\cancel{6{\text{e}}^{-}}\\ 6{\text{H}}^{+}+2{\text{HNO}}_3+\cancel{6{\text{e}}^{-}}\to 2\text{NO}+4{\text{H}}_2\text{O}\end{array}}\\ 6{\text{H}}^{+}+2{\text{HNO}}_3+3\text{Cu}\to 3{\text{Cu}}^{2+}+2\text{NO}+4{\text{H}}_2\text{O}\end{array}$

Therefore, the balanced chemical equation is $6{\text{H}}^{+}+2{\text{HNO}}_3+3\text{Cu}\to 3{\text{Cu}}^{2+}+2\text{NO}+4{\text{H}}_2\text{O}$.

c) ${\text{CN}}^{-}{\text{+MnO}}_4^{-}\to {\text{CNO}}^{-}+{\text{MnO}}_2$ (in basic solution)

In step $1$, half reactions are separated as:

$\begin{array}{l}{\text{Oxidation: CN}}^{-}\to {\text{CNO}}^{-}\\ {\text{Reduction: MnO}}_4^{-}\to {\text{MnO}}_2\end{array}$

Step $2$: Balance the half reactions, but in this case, it is not necessary, as the half reactions are already balanced.

Step $3$: By adding water, each half reaction will be balanced for oxygen. The oxidation half-reactionneeds one molecule of water on side of reactants. The reduction half-reactionneeds two molecules of water on side of the products. The reaction will be as:

$\begin{array}{l}{\text{H}}_2\text{O}+{\text{CN}}^{-}\to {\text{CNO}}^{-}\\ {\text{MnO}}_4^{-}\to {\text{MnO}}_2+2{\text{H}}_2\text{O}\end{array}$

Step $4$: By adding ${\text{H}}^{+}$, each half reaction will be balanced for hydrogen. The oxidationhalf-reaction needs two ${\text{H}}^{+}$ on side of the products. The reductionhalf-reactionneedsfour ${\text{H}}^{+}$ on side of reactants. The reaction will be as:

$\begin{array}{l}{\text{ H}}_2\text{O}+{\text{CN}}^{-}\to {\text{CNO}}^{-}+2{\text{H}}^{+}\\ {\text{4H}}^{+}+{\text{MnO}}_4^{-}\to {\text{MnO}}_2+2{\text{H}}_2\text{O}\end{array}$

Step $5$: By the addition of electrons, both half-reactions will be balanced. The total charge on the left is $-1$ and on the right, total charge is $+1$, in the oxidation-half reaction. If two electrons are added to the side of the products, on each side, the total charge will become $-1$. The reaction is as:

${\text{H}}_2\text{O}+{\text{CN}}^{-}\to {\text{CNO}}^{-}+2{\text{H}}^{+}+2{\text{e}}^{-}$

The total charge on the left is $+3$ and on the right, total charge is $0$, in the reduction-half reaction. If three electrons are added to the side of reactants, on each side, the total charge will become $0$. The reaction is as:

${\text{3e}}^{-}+{\text{4H}}^{+}+{\text{MnO}}_4^{-}\to {\text{MnO}}_2+2{\text{H}}_2\text{O}$

Step $6$: The oxidation half-reaction is multiplied by $3$ and reduction half-reaction is multiplied by $2$ so that the number of electrons in both half reactions becomes the same. The reaction is as:

$\begin{array}{l}3\left({\text{H}}_2\text{O}+{\text{CN}}^{-}\to {\text{CNO}}^{-}+2{\text{H}}^{+}+2{\text{e}}^{-}\right)=3{\text{H}}_2\text{O}+3{\text{CN}}^{-}\to 3{\text{CNO}}^{-}+6{\text{H}}^{+}+6{\text{e}}^{-}\\ 2\left({\text{3e}}^{-}+{\text{4H}}^{+}+{\text{MnO}}_4^{-}\to {\text{MnO}}_2+2{\text{H}}_2\text{O}\right)=6{\text{e}}^{-}+8{\text{H}}^{+}+2{\text{MnO}}_4^{-}\to 2{\text{MNO}}_2+4{\text{H}}_2\text{O}\end{array}$

Step $7$: By adding the resulting half reactions and cancelling identical species and electrons for getting the overall balanced equation as:

$\begin{array}{l}\underset{\_}{\begin{array}{l}3{\text{H}}_2\text{O}+3{\text{CN}}^{-}\to 3{\text{CNO}}^{-}+6{\text{H}}^{+}+\cancel{6{\text{e}}^{-}}\\ \cancel{6{\text{e}}^{-}}+8{\text{H}}^{+}+2{\text{MnO}}_4^{-}\to 2{\text{MnO}}_2+4{\text{H}}_2\text{O}\end{array}}\\ 3{\text{CN}}^{-}+2{\text{MnO}}_4^{-}+2{\text{H}}^{+}\to 3{\text{CNO}}^{-}+2{\text{MnO}}_2+{\text{H}}_2\text{O}\end{array}$

Step $8$: In the final equation, for each ${\text{H}}^{+}$, one ${\text{OH}}^{-}$ is added to each side, where ${\text{H}}^{+}$ and ${\text{OH}}^{-}$ combine to form water. The reaction is as:

$\begin{array}{l}\underset{\_}{\begin{array}{l}3{\text{CN}}^{-}+2{\text{MnO}}_4^{-}+2{\text{H}}^{+}\to 3{\text{CNO}}^{-}+2{\text{MnO}}_2+{\text{H}}_2\text{O}\\ +{\text{2OH}}^{-}+{\text{2OH}}^{-}\end{array}}\\ {\text{2H}}_2\text{O}+2{\text{MnO}}_4^{-}+3{\text{CN}}^{-}\to 2{\text{MnO}}_2+3{\text{CNO}}^{-}+{\text{H}}_2\text{O}+{\text{2OH}}^{-}\end{array}$

Step $9$: The molecules of water will cancel in the result of step $8$ and reaction is as:

$3{\text{CN}}^{-}+2{\text{MnO}}_4^{-}+{\text{H}}_2\text{O}\to 3{\text{CNO}}^{-}+2{\text{MnO}}_2+{\text{2OH}}^{-}$

Therefore, the balanced chemical equation is $3{\text{CN}}^{-}+2{\text{MnO}}_4^{-}+{\text{H}}_2\text{O}\to 3{\text{CNO}}^{-}+2{\text{MnO}}_2+{\text{2OH}}^{-}$

d) ${\text{Br}}_{\text{2}}\to {\text{BrO}}_3^{-}+{\text{Br}}^{-}$ (in basic solution)

The reaction is balanced by following the steps mentioned in subparts a, b, and c which include separation of half-cell reactions and balancing them. By balancing oxygen by addition of water and then adding electrons and cancelling the identical terms as in the previous subparts, the balanced equation for this reaction will be as:

${\text{6OH}}^{-}{\text{+3Br}}_{\text{2}}\to {\text{BrO}}_3^{-}+3{\text{H}}_2\text{O}+{\text{5Br}}^{-}$

e) ${\text{S}}_2{\text{O}}_3^{2-}+{\text{I}}_2\to {\text{I}}^{-}+{\text{S}}_4{\text{O}}_6^{2-}$ (in acidic solution)

The reaction is balanced by following the steps mentioned in subparts a, b, and c which include separation of half-cell reactions and balancing them. By balancing oxygen by addition of water and then adding electrons and cancelling the identical terms as in the previous subparts, the balanced equation for this reaction will be as ${\text{2S}}_2{\text{O}}_3^{2-}+{\text{I}}_2\to 2{\text{I}}^{-}+{\text{S}}_4{\text{O}}_6^{2-}$