Chapter 10, Problem 10.121AP

Interpretation Introduction

Interpretation: The balance form of given chemical reaction occurring in basic solution is to be stated.

Concept introduction: In balanced form of chemical reaction number of atoms and total charge of reactants and products are same.

To determine: The balance form of given chemical reaction occurring in basic solution.

Answer to Problem 10.121AP

The balanced reaction is,

${\text{IO}}_{\text{3}}{}^{-}\left(aq\right)+3{\text{HSO}}_3{}^{-}\left(aq\right)+3{\text{OH}}^{-}\left(aq\right)\to {\text{I}}^{-}\left(aq\right)+3{\text{SO}}_4{}^{2-}\left(aq\right)+3{\text{H}}_{\text{2}}\text{O}\left(l\right)$

Explanation of Solution

The given unbalanced chemical reaction is,

${\text{IO}}_{\text{3}}{}^{-}\left(aq\right)+{\text{HSO}}_3{}^{-}\left(aq\right)\to {\text{I}}^{-}\left(aq\right)+{\text{SO}}_4{}^{2-}\left(aq\right)$ (1)

In balanced form of chemical reaction number of atoms and total charge of reactants and products are same.

To balance the above equation (1) following steps are followed.

Step 1:

Calculate the oxidation state of each atom of given charge compounds.

${\text{IO}}_{\text{3}}{}^{-}$

The oxidation state of oxygen is $-2$.

The oxidation state of iodine is assumed to be $x$. Thus, the oxidation state of iodine in ${\text{IO}}_{\text{3}}{}^{-}$ is calculated as,

$\begin{array}{c}1\times x+3\times \left(-2\right)=-1\\ x-6=-1\\ x=+5\end{array}$

${\text{HSO}}_3{}^{-}$

The oxidation state of $\text{H}$ is $+1$. The oxidation state of oxygen is $-2$. The oxidation state of sulfur is assumed to be $x$. Thus, the oxidation state of sulfur in ${\text{HSO}}_3{}^{-}$ is calculated as,

$\begin{array}{c}1\times \left(+1\right)+1\times x+3\times \left(-2\right)=-1\\ x-5=-1\\ x=+4\end{array}$

${\text{SO}}_4{}^{2-}$

The oxidation state of oxygen is $-2$.

The oxidation state of sulfur is assumed to be $x$. Thus, the oxidation state of sulfur in ${\text{SO}}_4{}^{2-}$ is calculated as,

$\begin{array}{c}1\times x+4\times \left(-2\right)=-2\\ x-8=-2\\ x=+6\end{array}$

${\text{I}}^{-}$

The oxidation state of ${\text{I}}^{-}$ is $-1$.

Step 2:

Identify the oxidation and reduction reaction and separate them as half reactions.

In oxidation process, oxidation state of atoms increased by some number.

In reduction process, oxidation state of atoms decreased by some number.

$\begin{array}{c}\begin{array}{cccc}\text{Oxidation:}& {\text{HSO}}_3{}^{-}\left(aq\right)& \to & {\text{SO}}_4{}^{2-}\left(aq\right)\\ & +4& & +6\end{array}\\ \begin{array}{cccc}\text{Reduction:}& {\text{IO}}_{\text{3}}{}^{-}\left(aq\right)& \to & {\text{I}}^{-}\left(aq\right)\\ & +5& & -1\end{array}\end{array}$

Step 3:

Balance the atoms of reactant and product.

Balance $\text{O}$ by the addition of one ${\text{H}}_{\text{2}}\text{O}$ in oxidation half reaction on the right side.

Balance $\text{H}$ by the addition of one ${\text{H}}^{+}$ in oxidation half reaction on the left side. To neutralize ${\text{H}}^{+}$ add same number of ${\text{OH}}^{-}$ on both sides.

Balance $\text{O}$ by the addition of three ${\text{H}}_{\text{2}}\text{O}$ in reduction half reaction on the right side.

Balance $\text{H}$ by the addition of six ${\text{H}}^{+}$ in reduction half reaction on the left side. To neutralize ${\text{H}}^{+}$ add same number of ${\text{OH}}^{-}$ on both sides.

$\begin{array}{l}\text{Oxidation:}{\text{HSO}}_3{}^{-}\left(aq\right)+{\text{H}}_2\text{O}\left(l\right)\to {\text{SO}}_4{}^{2-}\left(aq\right)\\ {\text{HSO}}_3{}^{-}\left(aq\right)+{\text{H}}_2\text{O}\left(l\right)+3{\text{OH}}^{-}\left(aq\right)\to {\text{SO}}_4{}^{2-}\left(aq\right)+3{\text{H}}^{+}\left(aq\right)+3{\text{OH}}^{-}\left(aq\right)\\ \text{Reduction:}{\text{IO}}_{\text{3}}{}^{-}\left(aq\right)\to {\text{I}}^{-}\left(aq\right)+3{\text{H}}_2\text{O}\left(l\right)\\ {\text{IO}}_{\text{3}}{}^{-}\left(aq\right)+6{\text{H}}^{+}\left(aq\right)+6{\text{OH}}^{-}\left(aq\right)\to {\text{I}}^{-}\left(aq\right)+3{\text{H}}_2\text{O}\left(l\right)+6{\text{OH}}^{-}\left(aq\right)\end{array}$

Step 4:

Balance each half reaction with respect to charge.

Balance the charge of oxidation half reaction by the addition of $2e^{-}$ on right side.

Balance the charge of reduction half reaction by the addition of $6e^{-}$ on left side.

$\begin{array}{l}\text{Oxidation:}{\text{HSO}}_3{}^{-}\left(aq\right)+{\text{H}}_2\text{O}\left(l\right)+3{\text{OH}}^{-}\left(aq\right)\to {\text{SO}}_4{}^{2-}\left(aq\right)+3{\text{H}}^{+}\left(aq\right)+3{\text{OH}}^{-}\left(aq\right)+2e^{-}\\ \text{Reduction:}{\text{IO}}_{\text{3}}{}^{-}\left(aq\right)+6{\text{H}}^{+}\left(aq\right)+6{\text{OH}}^{-}\left(aq\right)+6e^{-}\to {\text{I}}^{-}\left(aq\right)+3{\text{H}}_2\text{O}\left(l\right)+6{\text{OH}}^{-}\left(aq\right)\\ {\text{IO}}_{\text{3}}{}^{-}\left(aq\right)+6{\text{H}}_2\text{O}\left(l\right)+6e^{-}\to {\text{I}}^{-}\left(aq\right)+3{\text{H}}_2\text{O}\left(l\right)+6{\text{OH}}^{-}\left(aq\right)\end{array}$

Step 5:

Make the number of electrons equal on both sides of oxidation and reduction half reaction.

Multiply oxidation half reaction with three on both sides.

$\begin{array}{l}\text{Oxidation:}3\times \left[{\text{HSO}}_3{}^{-}\left(aq\right)+{\text{H}}_2\text{O}\left(l\right)+3{\text{OH}}^{-}\left(aq\right)\to {\text{SO}}_4{}^{2-}\left(aq\right)+3{\text{H}}^{+}\left(aq\right)+3{\text{OH}}^{-}\left(aq\right)+2e^{-}\right]\\ {\text{3HSO}}_3{}^{-}\left(aq\right)+3{\text{H}}_2\text{O}\left(l\right)+9{\text{OH}}^{-}\left(aq\right)\to 3{\text{SO}}_4{}^{2-}\left(aq\right)+9{\text{H}}^{+}\left(aq\right)+9{\text{OH}}^{-}\left(aq\right)+6e^{-}\\ {\text{3HSO}}_3{}^{-}\left(aq\right)+3{\text{H}}_2\text{O}\left(l\right)+9{\text{OH}}^{-}\left(aq\right)\to 3{\text{SO}}_4{}^{2-}\left(aq\right)+9{\text{H}}_2\text{O}\left(l\right)+6e^{-}\\ \text{Reduction:}{\text{IO}}_{\text{3}}{}^{-}\left(aq\right)+6{\text{H}}_2\text{O}\left(l\right)+6e^{-}\to {\text{I}}^{-}\left(aq\right)+3{\text{H}}_2\text{O}\left(l\right)+6{\text{OH}}^{-}\left(aq\right)\end{array}$

Step 6:

To get overall reaction add the half reactions together and remove common terms.

$\begin{array}{l}\text{Oxidation:}{\text{3HSO}}_3{}^{-}\left(aq\right)+3{\text{H}}_2\text{O}\left(l\right)+9{\text{OH}}^{-}\left(aq\right)\to 3{\text{SO}}_4{}^{2-}\left(aq\right)+9{\text{H}}_2\text{O}\left(l\right)+6e^{-}\\ \text{Reduction:}{\text{IO}}_{\text{3}}{}^{-}\left(aq\right)+6{\text{H}}_2\text{O}\left(l\right)+6e^{-}\to {\text{I}}^{-}\left(aq\right)+3{\text{H}}_2\text{O}\left(l\right)+6{\text{OH}}^{-}\left(aq\right)\\ \overline{\underset{\_}{\text{Overall:}{\text{IO}}_{\text{3}}{}^{-}\left(aq\right)+3{\text{HSO}}_3{}^{-}\left(aq\right)+3{\text{OH}}^{-}\left(aq\right)\to {\text{I}}^{-}\left(aq\right)+3{\text{SO}}_4{}^{2-}\left(aq\right)+3{\text{H}}_{\text{2}}\text{O}\left(l\right)}}\end{array}$

Thus, the balanced reaction is,

${\text{IO}}_{\text{3}}{}^{-}\left(aq\right)+3{\text{HSO}}_3{}^{-}\left(aq\right)+3{\text{OH}}^{-}\left(aq\right)\to {\text{I}}^{-}\left(aq\right)+3{\text{SO}}_4{}^{2-}\left(aq\right)+3{\text{H}}_{\text{2}}\text{O}\left(l\right)$

Conclusion

The balanced reaction is,

${\text{IO}}_{\text{3}}{}^{-}\left(aq\right)+3{\text{HSO}}_3{}^{-}\left(aq\right)+3{\text{OH}}^{-}\left(aq\right)\to {\text{I}}^{-}\left(aq\right)+3{\text{SO}}_4{}^{2-}\left(aq\right)+3{\text{H}}_{\text{2}}\text{O}\left(l\right)$