Chapter 8, Problem 8.114QA

Interpretation Introduction

To find:

a) Assign oxidation numbers to the elements in each compound and balance the redox reaction

b) Write a net ionic equation describing the formation of chlorine

c) Balance the equation for conversion of $C{l}_2$ to $HCl$ and $HOCl$

Answer to Problem 8.114QA

Solution:

a) The balanced redox reaction is as follows:

$4NaCl\left(aq\right)+2H_{2}S{O}_4\left(aq\right)+Mn{O}_2\left(s\right)\to 2N{a}_{2}S{O}_4\left(aq\right)+MnC{l}_2\left(aq\right)+2H_{2}O\left(l\right)+C{l}_2\left(g\right)$

The oxidation numbers (O.N.) of the elements in each compound are as follows:

Compound	Element	O.N.
$NaCl\left(aq\right)$	$Na$ $Cl$	$+1$ $-1$
$H_{2}S{O}_4\left(aq\right)$	$H$ $S$ $O$	$+1$ $+6$ $-2$
$Mn{O}_2\left(s\right)$	$Mn$ $O$	$+4$ $-2$
$N{a}_{2}S{O}_4\left(aq\right)$	$Na$ $S$ $O$	$+1$ $+6$ $-2$
$MnC{l}_2\left(aq\right)$	$Mn$ $Cl$	$+2$ $-1$
$H_{2}O\left(l\right)$	$H$ $O$	$+1$ $-1$
$C{l}_2\left(g\right)$	$Cl$	$-1$

b) The net ionic equation for the formation of chlorine is

$2C{l}^{-}\left(aq\right)+4H^{+}\left(aq\right)+Mn{O}_2\left(s\right)\to M{n}^{2+}\left(aq\right)+2H_{2}O\left(l\right)+C{l}_2\left(g\right)$

c) The balanced equation for conversion of $C{l}_2$  to $HCl$ and $HOCl$ is

$C{l}_2\left(g\right)+H_{2}O\left(l\right)\to HCl\left(aq\right)+HOCl\left(aq\right)$

Explanation of Solution

We are asked to assign oxidation numbers to all the elements in the given balanced equation. Oxidation number of an atom in a molecule or ion is a measure of the number of electrons it has as compared to the number it would have if it were a free atom. The O.N. values of monoatomic ions are the same as their charges. In molecules and polyatomic ions, atoms have O.N values that are related to the number of covalent bonds they form.

For balancing redox reactions, first, the change in oxidation numbers ($∆O.N.$) is calculated. The $∆O.N.$ is adjusted by assigning coefficients to the molecules containing the unbalanced atoms. The ionic charges are adjusted by adding $H^{+}$ for the acidic medium or $O{H}^{-}$ for the basic medium. Finally, the $H$ and $O$ atoms are balanced by the addition of $H_{2}O$ to the appropriate sides.

a) The reaction to be balanced is as follows:

$NaCl\left(aq\right)+H_{2}S{O}_4\left(aq\right)+Mn{O}_2\left(s\right)\to N{a}_{2}S{O}_4\left(aq\right)+MnC{l}_2\left(aq\right)+H_{2}O\left(l\right)+C{l}_2\left(g\right)$

We first adjust the $Cl$ atoms by assigning a coefficient of 4 to $NaCl$.

$4NaCl\left(aq\right)+H_{2}S{O}_4\left(aq\right)+Mn{O}_2\left(s\right)\to N{a}_{2}S{O}_4\left(aq\right)+MnC{l}_2\left(aq\right)+H_{2}O\left(l\right)+C{l}_2\left(g\right)$

To adjust the number of $Na$ atoms, we assign a coefficient of $2$ to $N{a}_{2}S{O}_4$.

$4NaCl\left(aq\right)+H_{2}S{O}_4\left(aq\right)+Mn{O}_2\left(s\right)\to 2N{a}_{2}S{O}_4\left(aq\right)+MnC{l}_2\left(aq\right)+H_{2}O\left(l\right)+C{l}_2\left(g\right)$

Lastly, in order to adjust the $O$ and $S$ atoms, we assign coefficients of $2$ and $2$ to $H_{2}S{O}_4$ and $H_{2}O$ respectively.

$4NaCl\left(aq\right)+{2H}_{2}S{O}_4\left(aq\right)+Mn{O}_2\left(s\right)\to 2N{a}_{2}S{O}_4\left(aq\right)+MnC{l}_2\left(aq\right)+2H_{2}O\left(l\right)+C{l}_2\left(g\right)$

This is the balanced form of the equation.

b) In order to write the net ionic equation, we split the aqueous compounds into their corresponding ions:

$4N{a}^{+}\left(aq\right)+4C{l}^{-}\left(aq\right)+4H^{+}\left(aq\right)+2S{O}_4^{2-}\left(aq\right)+Mn{O}_2\left(s\right)\to 4N{a}^{+}\left(aq\right)+2S{O}_4^{2-}\left(aq\right)+M{n}^{2+}\left(aq\right)+2C{l}^{-}\left(aq\right)+2H_{2}O\left(l\right)+C{l}_2\left(g\right)$

We now cancel the $4N{a}^{+}$ ions, the $2C{l}^{-}$ ions, and the $2S{O}_4^{2-}$ ions from both the sides to get the net ionic reaction.

$2C{l}^{-}\left(aq\right)+4H^{+}\left(aq\right)+Mn{O}_2\left(s\right)\to M{n}^{2+}\left(aq\right)+2H_{2}O\left(l\right)+C{l}_2\left(g\right)$

c) Chlorine gas reacts with water to form hydrochloric acid and hypochlorous acid. The balanced reaction is as follows:

$C{l}_2\left(g\right)+H_{2}O\left(l\right)\to HCl\left(aq\right)+HOCl\left(aq\right)$

It can be seen that the reaction is already balanced.

Conclusion:

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