Chapter 8, Problem 8.99QA

Interpretation Introduction

To find:

The balanced chemical equations for the following reactions(in slightly acidic medium)

a) $Cl{O}_3^{-}\left(aq\right)+{SO}_2\left(g\right)\to {ClO}_2\left(g\right)+{SO}_4^{2-}\left(aq\right)$

b) $Cl{O}_3^{-}\left(aq\right)+{Cl}^{-}\left(aq\right)\to {ClO}_2\left(g\right)+{Cl}_2\left(g\right)$

c) ${ClO}_3^{-}\left(aq\right)+{Cl}_2\left(g\right)\to Cl{O}_2\left(g\right)+O_2\left(g\right)$

Answer to Problem 8.99QA

Solution:

a) $2Cl{O}_3^{-}\left(aq\right)+{SO}_2\left(g\right)\to {2ClO}_2\left(g\right)+{SO}_4^{2-}\left(aq\right)$

b) $2Cl{O}_3^{-}\left(aq\right)+{2Cl}^{-}\left(aq\right)+4H^{+}\left(aq\right)\to {2ClO}_2\left(g\right)+{Cl}_2\left(g\right)+2H_{2}O\left(l\right)$

c) $2{ClO}_3^{-}\left(aq\right)+{Cl}_2\left(g\right)\to 2Cl{O}_2\left(g\right)+O_2\left(g\right)+2{Cl}^{-}\left(aq\right)$

Explanation of Solution

1) Concept:

The following steps are followed to balance a redox reaction in a basic medium:

i) Calculate the change in oxidation number values of elements which are getting reduced and oxidized.

ii) Insert the appropriate coefficient to balance the change in oxidation number values(∆O.N) before the species which are getting reduced and oxidized only. We don’t balance O atoms in this step.

iii) The ionic charges are balanced by adding appropriate number of H⁺  ions in  an acidic solution.

iv) Water molecules have to be added, if necessary, in the last step.

2) Calculation:

a)

i) A preliminary expression of the unbalanced chemical reaction involving reactant and product is:

$Cl{O}_3^{-}\left(aq\right)+{SO}_2\left(g\right)\to {ClO}_2\left(g\right)+{SO}_4^{2-}\left(aq\right)$

ii) An analysis of change in oxidation number values(∆O.N) is done for Cl and S.

O.N for Cl in $Cl{O}_3^{-}$ is +5 and that of Cl in $Cl{O}_2$ is +4.

Change in O.N (∆O.N) of Cl = -1

O.N for S in SO₂ is +4 and that of S in SO₄^2- is +6.

Change in O.N (∆O.N) of S = +2

iii) To balance these (∆O.N) values we put coefficient 2 in front of  ClO₃^-  and  1 in front of SO₂.

$2Cl{O}_3^{-}\left(aq\right)+{SO}_2\left(g\right)\to {ClO}_2\left(g\right)+{SO}_4^{2-}\left(aq\right)$

iv) To balance Cl atoms on both sides, we insert 2 before ClO₂.

$2Cl{O}_3^{-}\left(aq\right)+{SO}_2\left(g\right)\to {2ClO}_2\left(g\right)+{SO}_4^{2-}\left(aq\right)$

The above equation is a balanced chemical equation where the electrical charges on both sides are same. (-2 on both left and right side)

b)

i) A preliminary expression of the unbalanced chemical reaction involving reactant and product is:

$Cl{O}_3^{-}\left(aq\right)+{Cl}^{-}\left(aq\right)\to {ClO}_2\left(g\right)+{Cl}_2\left(g\right)$

ii) An analysis of change in oxidation number values(∆O.N) is done for Cl atoms for both reactants.

O.N for Cl in $Cl{O}_3^{-}$ is +5 and that of Cl in $Cl{O}_2$ is +4.

Change in O.N (∆O.N) of Cl = -1

O.N for Cl in Cl^- is -1 and that of Cl in Cl₂ is zero.

Change in O.N (∆O.N) of Cl (for each Cl atom) = +1

For two Cl atoms (∆O.N) = +2

iii) To balance these (∆O.N) values we put coefficient 2 in front of $Cl{O}_3^{-}$ and  1 in front of Cl^-.

$2Cl{O}_3^{-}\left(aq\right)+{Cl}^{-}\left(aq\right)\to {2ClO}_2\left(g\right)+{Cl}_2\left(g\right)$

iv) To make Cl equal in both sides, we insert coefficient 2 in front of Cl_-

$2Cl{O}_3^{-}\left(aq\right)+2{Cl}^{-}\left(aq\right)\to {2ClO}_2\left(g\right)+{Cl}_2\left(g\right)$

v) To balance O atoms (6 on left and 4 on right side) we add 2 molecules of H₂O on right side.

$2Cl{O}_3^{-}\left(aq\right)+2{Cl}^{-}\left(aq\right)\to {2ClO}_2\left(g\right)+{Cl}_2\left(g\right)+2H_{2}O\left(l\right)$

vi) To balance the electrical charges (-4 on left and zero on right) we add 4H⁺on left side.

(Since the solution is acidic)

$2Cl{O}_3^{-}\left(aq\right)+{2Cl}^{-}\left(aq\right)+4H^{+}\left(aq\right)\to {2ClO}_2\left(g\right)+{Cl}_2\left(g\right)+2H_{2}O\left(l\right)$

This above reaction is the final balanced equation.

c)

i) A preliminary expression of the unbalanced chemical reaction involving reactant and product is:

${ClO}_3^{-}\left(aq\right)+{Cl}_2\left(g\right)\to Cl{O}_2\left(g\right)+O_2\left(g\right)$

ii) To analyze the (∆O.N) for Cl₂ we  put Cl^- on the right side.

${ClO}_3^{-}\left(aq\right)+{Cl}_2\left(g\right)\to Cl{O}_2\left(g\right)+O_2\left(g\right)+{Cl}^{-}\left(aq\right)$

iii) An analysis of change in oxidation number values(∆O.N) is done for Cl atoms for both reactants.

O.N for Cl in $Cl{O}_3^{-}$ is +5 and that of Cl in $Cl{O}_2$ is +4.

Change in O.N (∆O.N) of Cl = -1

O.N for Cl in Cl₂ is -1 and that of Cl in Cl^- is zero.

Change in O.N (∆O.N) of Cl (for each Cl atom) = -1

For two Cl atoms (∆O.N) = -2

iv) To balance these (∆O.N) values we put coefficient 2 in front of ClO₃^-  and  1 in front of Cl₂.

${2ClO}_3^{-}\left(aq\right)+{Cl}_2\left(g\right)\to Cl{O}_2\left(g\right)+O_2\left(g\right)+{Cl}^{-}\left(aq\right)$

v) To balance Cl atoms, we insert coefficient 2 in front of both ClO₂ and Cl^-.

${2ClO}_3^{-}\left(aq\right)+{Cl}_2\left(g\right)\to 2Cl{O}_2\left(g\right)+O_2\left(g\right)+2{Cl}^{-}\left(aq\right)$

This is our final balanced chemical equation.(Electrical charge on both sides is -2)

Conclusion:

The given reactions are balanced in the acidic medium using the rules for balancing redox reactions.