Chapter 19, Problem 2QP

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

$\begin{array}{l}\left(\text{a}\right)\hfill \\ {\text{Mn}}^{\text{2+}}{\text{+ H}}_{\text{2}}{\text{O}}_{\text{2}}\underset{}{\to }{\text{ MnO}}_{\text{2}}{\text{+ H}}_{\text{2}}\text{O}\left(\text{in basic solution}\right)\hfill \\ \left(\text{b}\right)\hfill \\ \text{Bi}{\left(\text{OH}\right)}_{\text{3}}{\text{ + SnO}}_2^{2-}\underset{}{\to }{\text{ SnO}}_3^{2-}\text{ + Bi }\left(\text{in basic solution}\right)\hfill \\ \left(\text{c}\right)\hfill \\ {\text{Cr}}_{\text{2}}{\text{O}}_7^{2-}{\text{ +C}}_{\text{2}}{\text{O}}_4^2\underset{}{\to }{\text{ Cr}}^{\text{3+}}{\text{+ CO}}_{\text{2}}\left(\text{in acidic solution}\right)\hfill \\ \left(\text{d}\right)\hfill \\ {\text{ClO}}_3^{-}{\text{ + Cl}}^{\text{-}}\underset{}{\to }{\text{ Cl}}_{\text{2}}{\text{+ ClO}}_{\text{2}}\left(\text{in acidic solution}\right)\hfill \\ \left(\text{e}\right)\hfill \\ {\text{Mn}}^{\text{2+}}{\text{+ BiO}}_{\text{3}}\underset{}{\to }{\text{Bi}}^{\text{3+}}{\text{+ MnO}}_{\text{4}}\left(\text{in acidic solution}\right)\hfill \end{array}$

Interpretation Introduction

Interpretation:

The given redox equations are to be balanced.

Concept introduction:

An oxidation reaction takes place when gain of electrons takes place and a reduction reaction takes place when removal of electrons takes place.

A chemical reaction in which electrons are transferred from one species to another is known as a redox reaction.

Answer to Problem 2QP

Solution:

a) $2{\text{OH}}^{-}+{\text{Mn}}^{2+}+{\text{H}}_2{\text{O}}_2\to {\text{MnO}}_2+2{\text{H}}_2\text{O}$

b) $\text{2Bi}{\left(\text{OH}\right)}_3+3{\text{SnO}}_{{}_2}^{2-}\to \text{2Bi}+3{\text{SnO}}_{{}_3}^{2-}+3{\text{H}}_2\text{O}$

c) ${\text{14H}}^{+}+3{\text{C}}_2{\text{O}}_4^{2-}+{\text{Cr}}_2{\text{O}}_7^{2-}\to 6{\text{CO}}_2+2{\text{Cr}}^{3+}+7{\text{H}}_2\text{O}$

d) ${\text{4H}}^{+}{\text{+2Cl}}^{-}+2{\text{ClO}}_3^{-}\to {\text{Cl}}_2+{\text{2ClO}}_2+2{\text{H}}_2\text{O}$

e) ${\text{14H}}^{+}+2{\text{Mn}}^{2+}+5{\text{BiO}}_3^{-}\to 2{\text{MnO}}_4^{-}+5{\text{Bi}}^{3+}+{\text{7H}}_{\text{2}}\text{O}$

Explanation of Solution

a) ${\text{Mn}}^{2+}+{\text{H}}_2{\text{O}}_2\to {\text{MnO}}_2+{\text{H}}_2\text{O}$(in basic solution)

Step $1$: Thehalf-reactions are separated as follows:

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

Step $2$: By adding water, the oxygen atom is balanced as follows:

$\begin{array}{l}{\text{Mn}}^{2+}+2{\text{H}}_2\text{O}\to {\text{MnO}}_2+4{\text{H}}^{+}\\ {\text{H}}_2{\text{O}}_2+2{\text{H}}^{+}\to 2{\text{H}}_2\text{O}\end{array}$

Step $3$: Neutralize ${\text{H}}^{+}$ ions by adding ${\text{OH}}^{-}$ ions as follows:

$\begin{array}{l}{\text{ Mn}}^{2+}+2{\text{H}}_2\text{O}+4{\text{OH}}^{-}\to {\text{MnO}}_2+4{\text{H}}^{+}+4{\text{OH}}^{-}\\ {\text{ H}}_2{\text{O}}_2+{\text{2H}}^{+}+2{\text{OH}}^{-}\to 2{\text{OH}}^{-}+2{\text{H}}_2\text{O}\end{array}$

Step $4$: Balancing the charges as follows:

$\begin{array}{l}{\text{ Mn}}^{2+}+2{\text{H}}_2\text{O}+4{\text{OH}}^{-}\to {\text{MnO}}_2+4{\text{H}}_2\text{O}+{\text{2e}}^{-}\\ {\text{ H}}_2{\text{O}}_2+2{\text{H}}_2\text{O}+{\text{2e}}^{-}\to 2{\text{OH}}^{-}+2{\text{H}}_2\text{O}\end{array}$

Step $5$: By adding the above equations, the final equation is obtained as follows:

$\begin{array}{l}{\text{ Mn}}^{2+}+2{\text{H}}_2\text{O}+4{\text{OH}}^{-}\to {\text{MnO}}_2+4{\text{H}}_2\text{O}+{\text{2e}}^{-}\\ {\text{ H}}_2{\text{O}}_2+2{\text{H}}_2\text{O}+{\text{2e}}^{-}\to 2{\text{OH}}^{-}+2{\text{H}}_2\text{O}\\ \overline{2{\text{OH}}^{-}+{\text{Mn}}^{2+}+{\text{H}}_2{\text{O}}_2\to {\text{MnO}}_2+2{\text{H}}_2\text{O}}\end{array}$

The balanced equation is as follows:

$2{\text{OH}}^{-}+{\text{Mn}}^{2+}+{\text{H}}_2{\text{O}}_2\to {\text{MnO}}_2+2{\text{H}}_2\text{O}$

b) $\text{Bi}{\left(\text{OH}\right)}_3+{\text{SnO}}_{{}_2}^{2-}\to {\text{SnO}}_{{}_3}^{2-}+\text{Bi}$(in basic solution)

Step $1$ :Half-reactions are separated as follows:

$\begin{array}{l}{\text{Oxidation: SnO}}_{{}_2}^{2-}\to {\text{SnO}}_{{}_3}^{2-}\\ \text{Reduction: Bi}{\left(\text{OH}\right)}_3\to \text{Bi}\end{array}$

Step $2$: Electrons are added to balance the oxidation number as follows:

$\begin{array}{l}{\text{SnO}}_{{}_2}^{2-}\to {\text{SnO}}_{{}_3}^{2-}+2{\text{e}}^{-}\\ {\text{3e}}^{-}+\text{Bi}{\left(\text{OH}\right)}_3\to \text{Bi}\end{array}$

Step $3$: By adding water, the oxygen atom is balanced as follows:

$\begin{array}{l}{\text{2OH}}^{-}{\text{+SnO}}_{{}_2}^{2-}\to {\text{SnO}}_{{}_3}^{2-}+{\text{H}}_2\text{O}+2{\text{e}}^{-}\\ {\text{3e}}^{-}+\text{Bi}{\left(\text{OH}\right)}_3\to \text{Bi}+3{\text{OH}}^{-}\end{array}$

Step $4$: For getting the same number of electrons on each side of the aboveequations, the equations are multiplied by $3\text{ and 2}$, respectively, as follows:

$\begin{array}{l}{\text{6OH}}^{-}{\text{+3SnO}}_{{}_2}^{2-}\to 3{\text{SnO}}_{{}_3}^{2-}+3{\text{H}}_2\text{O}+6{\text{e}}^{-}\\ {\text{6e}}^{-}+2\text{Bi}{\left(\text{OH}\right)}_3\to 2\text{Bi}+6{\text{OH}}^{-}\end{array}$

Step $5$: Adding the aboveequations and cancelling ${\text{OH}}^{-}$ and ${\text{H}}_2\text{O}$ present on opposite sides as follows:

${\text{6OH}}^{-}{\text{+3SnO}}_{{}_2}^{2-}+2\text{Bi}{\left(\text{OH}\right)}_3+6{\text{e}}^{-}\to 3{\text{SnO}}_{{}_3}^{2-}+2\text{Bi}+3{\text{H}}_2\text{O}+6{\text{e}}^{-}+{\text{6OH}}^{-}$

The balanced equation is as follows:

$\text{2Bi}{\left(\text{OH}\right)}_3+3{\text{SnO}}_{{}_2}^{2-}\to \text{2Bi}+3{\text{SnO}}_{{}_3}^{2-}+3{\text{H}}_2\text{O}$

c) ${\text{Cr}}_2{\text{O}}_7^{2-}+{\text{C}}_2{\text{O}}_4^{2-}+\to {\text{Cr}}^{3+}+{\text{CO}}_2$(in acidic solution)

Step $1$:Half-reactions are separated as follows:

$\begin{array}{l}{\text{Oxidation: C}}_2{\text{O}}_4^{2-}\to 2{\text{CO}}_2+2{\text{e}}^{-}\\ {\text{Reduction: Cr}}_2{\text{O}}_7^{2-}\to 2{\text{Cr}}^{3+}\end{array}$

Step $2$: Balance the oxygen by adding $7{\text{ H}}_2\text{O}$ as follows:

${\text{Cr}}_2{\text{O}}_7^{2-}\to 2{\text{Cr}}^{3+}+7{\text{ H}}_2\text{O}$

Step $3$: Balance the hydrogen by adding $14{\text{ H}}^{+}$ as follows:

${\text{Cr}}_2{\text{O}}_7^{2-}+14{\text{H}}^{+}\to 2{\text{Cr}}^{3+}+7{\text{ H}}_2\text{O}$

Step $4$: Charges are neutralized by adding electrons as follows:

${\text{Cr}}_2{\text{O}}_7^{2-}+14{\text{H}}^{+}+6{\text{e}}^{-}\to 2{\text{Cr}}^{3+}+7{\text{ H}}_2\text{O}$

Step $5$: Multiplying oxidation half-reaction with $3$ and adding to the above equation,

The equation is as follows:

$\begin{array}{l}{\text{3C}}_2{\text{O}}_4^{2-}\to 6{\text{CO}}_2+6{\text{e}}^{-}\\ {\text{Cr}}_2{\text{O}}_7^{2-}+14{\text{H}}^{+}+6{\text{e}}^{-}\to 2{\text{Cr}}^{3+}+7{\text{H}}_2\text{O}\\ \overline{{\text{Cr}}_2{\text{O}}_7^{2-}+3{\text{C}}_2{\text{O}}_4^{2-}+14{\text{H}}^{+}\to 6{\text{CO}}_2+2{\text{Cr}}^{3+}+7{\text{H}}_2\text{O}}\end{array}$

Therefore, the balanced chemical equation is as follows:

${\text{14H}}^{+}+3{\text{C}}_2{\text{O}}_4^{2-}+{\text{Cr}}_2{\text{O}}_7^{2-}\to 6{\text{CO}}_2+2{\text{Cr}}^{3+}+7{\text{H}}_2\text{O}$

d) ${\text{ClO}}_3^{-}+{\text{Cl}}^{-}\to {\text{Cl}}_2+{\text{ClO}}_2$(in acidic solution)

Step $1$:Half-reactions are separated as follows:

$\begin{array}{l}{\text{Oxidation: Cl}}^{-}\to {\text{Cl}}_2\\ {\text{Reduction: ClO}}_3^{-}\to {\text{ClO}}_2\end{array}$

Step $2$: Balance the oxidation half-reaction as follows:

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

Step $3$: Balance the reduction half-reaction by adding water as follows:

${\text{ClO}}_3^{-}\to {\text{ClO}}_2$

${\text{ClO}}_3^{-}+2{\text{H}}^{+}+{\text{e}}^{-}\to {\text{ClO}}_2+{\text{H}}_2\text{O}$

Step $4$: Multiplying the above equation by $2$, adding to the aboveequation and cancelling the same terms on opposite sides as follows:

$\begin{array}{l}2{\text{Cl}}^{-}\to {\text{Cl}}_2+2{\text{e}}^{-}\\ {\text{2ClO}}_3^{-}+4{\text{H}}^{+}+2{\text{e}}^{-}\to 2{\text{ClO}}_2+2{\text{H}}_2\text{O}\\ \overline{2{\text{Cl}}^{-}+{\text{2ClO}}_3^{-}+4{\text{H}}^{+}\to {\text{Cl}}_2+2{\text{ClO}}_2+2{\text{H}}_2\text{O}}\end{array}$

The balanced equation for this reaction will be as follows:

${\text{4H}}^{+}{\text{+2Cl}}^{-}+2{\text{ClO}}_3^{-}\to {\text{Cl}}_2+{\text{2ClO}}_2+2{\text{H}}_2\text{O}$

e) ${\text{Mn}}^{2+}+{\text{BiO}}_3^{-}\to {\text{Bi}}^{3+}+{\text{MnO}}_4^{-}$(in acidic solution)

Step $1$:Half reactions are separated as follows:

$\begin{array}{l}{\text{Oxidation: BiO}}_3^{-}\to {\text{Bi}}^{3+}\\ {\text{Reduction: Mn}}^{\text{2+}}\to {\text{MnO}}_4^{-}\end{array}$

Step $2$: Water is added to balance the oxygen on each side as follows:

$\begin{array}{l}{\text{BiO}}_3^{-}\to {\text{Bi}}^{3+}+3{\text{H}}_2\text{O}\\ {\text{Mn}}^{\text{2+}}+4{\text{H}}_2\text{O}\to {\text{MnO}}_4^{-}\end{array}$

Step $3$: By adding ${\text{H}}^{+}$ on each side, hydrogen is balanced as follows:

$\begin{array}{l}{\text{BiO}}_3^{-}+6{\text{H}}^{+}\to {\text{Bi}}^{3+}+3{\text{H}}_2\text{O}\\ {\text{Mn}}^{\text{2+}}+4{\text{H}}_2\text{O}\to {\text{MnO}}_4^{-}+8{\text{H}}^{+}\end{array}$

Step $4$: Now electrons are added to balance the charges as follows:

$\begin{array}{l}{\text{BiO}}_3^{-}+6{\text{H}}^{+}+2{\text{e}}^{-}\to {\text{Bi}}^{3+}+3{\text{H}}_2\text{O}\\ {\text{Mn}}^{\text{2+}}+4{\text{H}}_2\text{O}\to {\text{MnO}}_4^{-}+8{\text{H}}^{+}+5{\text{e}}^{-}\end{array}$

Step $5$: The equations are multiplied by $\text{5 and 2}$ respectively as follows:

$\begin{array}{l}{\text{5BiO}}_3^{-}+30{\text{H}}^{+}+10{\text{e}}^{-}\to 5{\text{Bi}}^{3+}+15{\text{H}}_2\text{O}\\ {\text{2Mn}}^{\text{2+}}+8{\text{H}}_2\text{O}\to 2{\text{MnO}}_4^{-}+16{\text{H}}^{+}+10{\text{e}}^{-}\end{array}$

Step $6$: Adding the equations and cancelling the same terms on opposite sides as follows:

$\begin{array}{l}{\text{5BiO}}_3^{-}+30{\text{H}}^{+}+10{\text{e}}^{-}\to 5{\text{Bi}}^{3+}+15{\text{H}}_2\text{O}\\ {\text{2Mn}}^{\text{2+}}+8{\text{H}}_2\text{O}\to 2{\text{MnO}}_4^{-}+16{\text{H}}^{+}+10{\text{e}}^{-}\\ \overline{{\text{14H}}^{+}+2{\text{Mn}}^{2+}+5{\text{BiO}}_3^{-}\to 2{\text{MnO}}_4^{-}+5{\text{Bi}}^{3+}+{\text{7H}}_{\text{2}}\text{O}}\end{array}$

The balanced equation for this reaction will be as follows:

${\text{14H}}^{+}+2{\text{Mn}}^{2+}+5{\text{BiO}}_3^{-}\to 2{\text{MnO}}_4^{-}+5{\text{Bi}}^{3+}+{\text{7H}}_{\text{2}}\text{O}$