Chapter 17, Problem 27E

Interpretation Introduction

Interpretation:

The balanced equation for the redox reaction, ${\text{Cl}}_{\text{2}}\left(aq\right)\to {\text{Cl}}^{-}\left(aq\right)+\text{HOCl}\left(aq\right)$ in acidic solution is to be stated.

Concept introduction:

The oxidizer is the species whose oxidation state falls during the course of reaction and reducer is the species whose oxidation number increases. Oxidized product is the oxidation product of the reducer and reduced product is the reduction product of the oxidizer.

Answer to Problem 27E

The balanced equation for the redox reaction, ${\text{Cl}}_{\text{2}}\left(aq\right)\to {\text{Cl}}^{-}\left(aq\right)+\text{HOCl}\left(aq\right)$ in acidic solution is shown below.

${\text{Cl}}_{\text{2}}\left(aq\right)+{\text{H}}_{\text{2}}\text{O}\left(l\right)\to {\text{Cl}}^{-}\left(aq\right)+\text{HOCl}\left(aq\right)+{\text{H}}^{\text{+}}\left(aq\right)$

Explanation of Solution

The given redox reaction is shown below.

${\text{Cl}}_{\text{2}}\left(aq\right)\to {\text{Cl}}^{-}\left(aq\right)+\text{HOCl}\left(aq\right)$

The oxidation state of the central metal atom is calculated by knowing the standard oxidation states of few elements.

The oxidation state of the chlorine in $\text{HOCl}$ is calculated below.

Step-1: Write down the oxidation number of every element and for unknown take “n”.

$\begin{array}{l}\text{Cl H O}\\ \text{n }+1-\text{2}\end{array}$

Step-2: Multiply the oxidation state with their number of atoms of an element.

$\begin{array}{l}\text{Cl H O}\\ \text{n }+1-\text{2}\end{array}$

Step-3: Add the oxidation numbers and set them equal to the charge of the species.

$\begin{array}{l}\text{Cl H O}\\ \text{n }+\left(+1\right)+\left(-\text{2}\right)=0\end{array}$

Calculate the value of n by simplifying the equation as shown below.

$\begin{array}{rll}\text{n }+\left(+1\right)+\left(-\text{2}\right)& =& 0\\ \text{n}& =& 0-1+2\\ \text{n}& =& +1\end{array}$

The oxidation state of chlorine is $+1$ in $\text{HOCl}$.

The oxidation number of chlorine is $-1$ in ${\text{Cl}}^{-}$ which comes from the charge on chlorine.

The oxidation number of chlorine is zero in ${\text{Cl}}_{\text{2}}$ as this is the elemental form of chlorine.

The chlorine is reduced on going from ${\text{Cl}}_{\text{2}}\to {\text{Cl}}^{-}$ as the oxidation number decreases. Therefore it is a reduction half-reaction.

The chlorine is oxidized on going from ${\text{Cl}}_{\text{2}}\to \text{HOCl}$ as the oxidation number increases. Therefore it is an oxidation half-reaction.

The reduction half-reaction for the above reaction is shown below.

${\text{Cl}}_{\text{2}}\left(aq\right)\to {\text{Cl}}^{-}\left(aq\right)$

The balancing of the half-reactions is done by the following the steps shown below.

Step-1: Identify and balance the element getting oxidized or reduced.

The chlorine is getting reduced and its number of atoms is not balanced on both sides. Balance them by multiplying the ${\text{Cl}}^{-}$ by two on the right-hand side.

${\text{Cl}}_{\text{2}}\left(aq\right)\to 2{\text{Cl}}^{-}\left(aq\right)$

Step-2: Balance elements other than oxygen and hydrogen if any.

${\text{Cl}}_{\text{2}}\left(aq\right)\to 2{\text{Cl}}^{-}\left(aq\right)$

Step-3: Balance oxygen atoms by adding water on the appropriate side.

${\text{Cl}}_{\text{2}}\left(aq\right)\to 2{\text{Cl}}^{-}\left(aq\right)$

Step-4: Balance the hydrogen atoms by adding ${\text{H}}^{\text{+}}$ to the relevant side when necessary.

${\text{Cl}}_{\text{2}}\left(aq\right)\to 2{\text{Cl}}^{-}\left(aq\right)$

Step-5: Balance the charge by adding electrons to the appropriate side.

The charge is balanced by adding two electrons on the left-hand side of the equation.

${\text{Cl}}_{\text{2}}\left(aq\right)+2{\text{e}}^{-}\to 2{\text{Cl}}^{-}\left(aq\right)$

Step-6: Recheck the equation to be sure that it is perfectly balanced.

The equation is completely balanced and is shown below.

${\text{Cl}}_{\text{2}}\left(aq\right)+2{\text{e}}^{-}\to 2{\text{Cl}}^{-}\left(aq\right)$ …(1)

The oxidation half-reaction for the above reaction is shown below.

${\text{Cl}}_{\text{2}}\left(aq\right)\to \text{HOCl}\left(aq\right)$

The balancing of the half-reactions is done by the following the steps shown below.

Step-1: Identify and balance the element getting oxidized or reduced.

The chlorine is getting oxidized and its number of atoms is not balanced on both sides. Balance them by multiplying $\text{HOCl}\left(aq\right)$ by two on the right-hand side of the equation.

${\text{Cl}}_{\text{2}}\left(aq\right)\to 2\text{HOCl}\left(aq\right)$

Step-2: Balance elements other than oxygen and hydrogen if any.

${\text{Cl}}_{\text{2}}\left(aq\right)\to 2\text{HOCl}\left(aq\right)$

Step-3: Balance oxygen atoms by adding water on the appropriate side.

Add two water molecules on the left-hand side of the equation.

${\text{Cl}}_{\text{2}}\left(aq\right)+2{\text{H}}_{\text{2}}\text{O}\left(l\right)\to 2\text{HOCl}\left(aq\right)$

Step-4: Balance the hydrogen atoms by adding ${\text{H}}^{\text{+}}$ to the relevant side when necessary.

To balance hydrogen atoms add two ${\text{H}}^{\text{+}}$ on the right-hand side of the equation.

${\text{Cl}}_{\text{2}}\left(aq\right)+2{\text{H}}_{\text{2}}\text{O}\left(l\right)\to 2\text{HOCl}\left(aq\right)+2{\text{H}}^{\text{+}}\left(aq\right)$

Step-5: Balance the charge by adding electrons to the appropriate side.

The charge is balanced by adding two electrons on the right-hand side of the equation.

${\text{Cl}}_{\text{2}}\left(aq\right)+2{\text{H}}_{\text{2}}\text{O}\left(l\right)\to 2\text{HOCl}\left(aq\right)+2{\text{H}}^{\text{+}}\left(aq\right)+2{\text{e}}^{-}$

Step-6: Recheck the equation to be sure that it is perfectly balanced.

The equation is completely balanced and is shown below.

${\text{Cl}}_{\text{2}}\left(aq\right)+2{\text{H}}_{\text{2}}\text{O}\left(l\right)\to 2\text{HOCl}\left(aq\right)+2{\text{H}}^{\text{+}}\left(aq\right)+2{\text{e}}^{-}$ …(2)

The balanced redox equation is obtained by adding equation (1) and (2) in such a way that electrons are canceled out.

Add both equations (1) and (2).

$\begin{array}{c}{\text{Cl}}_{\text{2}}\left(aq\right)+2{\text{e}}^{-}\to 2{\text{Cl}}^{-}\left(aq\right)\\ +\\ {\text{Cl}}_{\text{2}}\left(aq\right)+2{\text{H}}_{\text{2}}\text{O}\left(l\right)\to 2\text{HOCl}\left(aq\right)+2{\text{H}}^{\text{+}}\left(aq\right)+2{\text{e}}^{-}\end{array}$

The equation obtained after adding these equations is shown below.

${\text{2Cl}}_{\text{2}}\left(aq\right)+2{\text{H}}_{\text{2}}\text{O}\left(l\right)\to 2{\text{Cl}}^{-}\left(aq\right)+2\text{HOCl}\left(aq\right)+2{\text{H}}^{\text{+}}\left(aq\right)$

Divide whole equation by two to eliminate the common factor.

${\text{Cl}}_{\text{2}}\left(aq\right)+{\text{H}}_{\text{2}}\text{O}\left(l\right)\to {\text{Cl}}^{-}\left(aq\right)+\text{HOCl}\left(aq\right)+{\text{H}}^{\text{+}}\left(aq\right)$

The equation is now completely balanced equation.

Conclusion

The balanced equation of the redox reaction is shown below.

${\text{Cl}}_{\text{2}}\left(aq\right)+{\text{H}}_{\text{2}}\text{O}\left(l\right)\to {\text{Cl}}^{-}\left(aq\right)+\text{HOCl}\left(aq\right)+{\text{H}}^{\text{+}}\left(aq\right)$