Chapter 20.3, Problem 23SSC

Interpretation Introduction

Interpretation:

Different product formation on electrolysis of brine solution and molten sodium chloride is to be explained.

Concept introduction:

Electrochemistry is study of relation between electrical energy and chemical changes that takes place in redox reaction. There are three main aspects of electrochemistry branch, those are:

1. Electrolysis
2. Electrochemical cell
3. Electrolytic conduction.

Electrolysis is a process in which decomposition of electrolyte occurs on flow of electricity through its aqueous solution or molten state. The instrument used to carry out electrolysis is electrolytic cell. Example of electrolytic cell is when a secondary battery is recharged.

Answer to Problem 23SSC

When molten $\text{NaCl}$ undergo electrolysis, only ${\text{Na}}^{+}$ and ${\text{Cl}}^{-}$ ions are generated whereas when brine is electrolyzed, apart from ${\text{Na}}^{+}$ and ${\text{Cl}}^{-}$ ions, there is an additional possibility of generation of ${\text{H}}^{+}$ and ${\text{OH}}^{-}$ ions. Hence as per the ease of redox reaction or lower discharge potential of ions, product is formed. Therefore, different product formation is observed on electrolysis of brine solution and molten sodium chloride.

Explanation of Solution

When molten $\text{NaCl}$ undergo electrolysis, ${\text{Na}}^{+}$ and ${\text{Cl}}^{-}$ ions are generated. ${\text{Na}}^{+}$ moves to cathode and undergo reduction to $\text{Na}$ and ${\text{Cl}}^{-}$ move to anode and oxidizes to ${\text{Cl}}_2$ gas.

The half cell reactions at electrodes are as follows:

At anode: ${\text{Na}}^{+}\left(\text{l}\right)+{\text{e}}^{-}\to \text{Na}\left(\text{l}\right)$

At cathode: ${\text{2Cl}}^{-}\left(\text{l}\right)\to {\text{Cl}}_2\left(\text{g}\right)+2{\text{e}}^{-}$

And the net cell reaction is as follows:

${\text{2Cl}}^{-}\left(\text{l}\right)+2{\text{Na}}^{+}\left(\text{l}\right)\to {\text{Cl}}_2\left(\text{g}\right)+2\text{Na}\left(\text{l}\right)$

When brine or aqueous $\text{NaCl}$ undergo electrolysis, apart from ${\text{Na}}^{+}$ and ${\text{Cl}}^{-}$ ions there is possibility of ${\text{H}}^{+}$ and ${\text{OH}}^{-}$ ions to be generated due to presence of large amount of water. Therefore, at cathode 2 reactions are possible which are as follows:

$\begin{array}{l}{\text{Na}}^{+}\left(\text{aq}\right)+{\text{e}}^{-}\to \text{Na}\left(\text{s}\right)\\ {\text{2H}}_{\text{2}}\text{O}\left(\text{l}\right)+{\text{2e}}^{-}\to {\text{H}}_{\text{2}}\left(\text{g}\right){\text{+2OH}}^{-}\left(\text{aq}\right)\end{array}$

Since water is easy to reduce than ${\text{Na}}^{+}$, so ${\text{H}}_{\text{2}}\left(\text{g}\right)$ is evolved at cathode.

And at anode, similarly, 2 reactions are possible which are as follows:

$\begin{array}{l}{\text{2Cl}}^{-}\left(\text{aq}\right)\to {\text{Cl}}_2\left(\text{g}\right)+2{\text{e}}^{-}\\ {\text{2H}}_{\text{2}}\text{O}\left(\text{l}\right)\to {\text{O}}_{\text{2}}\left(\text{g}\right){\text{+4H}}^{+}\left(\text{aq}\right)+4{\text{e}}^{-}\end{array}$

Since ${\text{Cl}}_2$ is desired product so its quantity is kept large at anode hence, oxidation of ${\text{Cl}}^{-}$ takes place preferentially over water.

Conclusion

When molten $\text{NaCl}$ undergo electrolysis, only ${\text{Na}}^{+}$ and ${\text{Cl}}^{-}$ ions are generated whereas when brine is electrolyzed, apart from ${\text{Na}}^{+}$ and ${\text{Cl}}^{-}$ ions, there is an additional possibility of generation of ${\text{H}}^{+}$ and ${\text{OH}}^{-}$ ions. Hence as per the ease of redox reaction or lower discharge potential of ions, product is formed. Therefore, different product formation is observed on electrolysis of brine solution and molten sodium chloride.