Question

In the electrolysis of aqueous NaCl solution, side reactions taking place are:
$$2\text{O}{\text{H}}^{-}+\text{C}{\text{l}}_2\to \text{2OC}{\text{l}}^{-}+{\text{H}}_{\text{2}}\text{O}$$
$$2\text{Na}+2{\text{H}}_{\text{2}}\text{O}\to \text{2NaOH}+{\text{H}}_{\text{2}}$$
$$4\text{O}{\text{H}}^{-}\to {\text{O}}_2+2{\text{H}}_{\text{2}}+\text{4}{\text{e}}^{-}$$
Select the correct alternate:
A. (I) and (III)
B. (II) and (III)
C. (I) and (II)
D. (I), (II), and (III)

Answer 1

Hint: In the electrolysis of aqueous NaCl solution, one possibility is that two reduction reactions take place at cathode and another possibility is that two oxidation reactions occur at anode. The reduction reaction having higher value of $$E_{\text{cell}}^{\text{o}}$$ will occur at cathode and oxidation reaction having lesser value of $$E_{\text{cell}}^{\text{o}}$$ will occur at anode.

Complete step by step solution:
The process in which an electric current causes a chemical change (conversion of electric energy to chemical energy) is called electrolysis.

In aqueous solution of NaCl, $$\text{N}{\text{a}}^{+}$$, $$\text{C}{\text{l}}^{-}$$ , $${\text{H}}^{+}$$ (from water) and $$\text{O}{\text{H}}^{-}$$(from water) ions are free to move for the conduction of electric current. When electric current is passed, the positive ions get attracted towards the negative electrode that is the cathode and negative ions get attracted towards the positive electrode that is anode.

Thus $$\text{N}{\text{a}}^{+}$$ and $${\text{H}}^{+}$$ ions are attracted to cathode while $$\text{C}{\text{l}}^{-}$$ and $$\text{O}{\text{H}}^{-}$$ ions attracted to anode.

The solution now contains four ions $$\text{N}{\text{a}}^{+}$$, $$\text{C}{\text{l}}^{-}$$ (from NaCl), $${\text{H}}^{+}$$, $$\text{O}{\text{H}}^{-}$$ (from water) and there is a race amongst them for their discharge at their respective electrodes.

At cathode, reduction reactions take place. The reaction which involves reduction of any particular species (gain of electrons or decrease in oxidation number) is known as reduction. At anode, oxidation reactions take place. The reaction which involves oxidation of any particular species (loss of electrons or increase in oxidation number) is known as oxidation.

At cathode, there is race between these two following reaction
$$\begin{array}{c}\text{N}{\text{a}}^{+}\left(\text{aq}\right)+{\text{e}}^{-}\to \text{Na}\left(\text{s}\right);E_{\text{cell}}^{\text{o}}=-2.71\text{V}\\ 2{\text{H}}_{\text{2}}\text{O}\left(\text{l}\right)+2{\text{e}}^{-}\to {\text{H}}_2\left(\text{g}\right)+2\text{O}{\text{H}}^{-};E_{\text{cell}}^{\text{o}}=-0.83\text{V}\end{array}$$

The values of $$E_{\text{cell}}^{\text{o}}$$ for a particular oxidation or reduction reactions remain fixed. We can get these values from standard potential tables provided in the book or from the internet.

The reaction at cathode which is having more value of $$E_{\text{cell}}^{\text{o}}$$ is preferred thus the following reduction reaction occurs at cathode:

$$2{\text{H}}_{\text{2}}\text{O}\left(\text{l}\right)+2{\text{e}}^{-}\to {\text{H}}_2\left(\text{g}\right)+2\text{O}{\text{H}}^{-}\cdot \cdot \cdot \cdot \cdot \cdot \left(\text{I}\right)$$

Similarly, at the cathode there is race between the following oxidation reactions:
$$\begin{array}{c}\text{C}{\text{l}}^{-}\left(\text{aq}\right)\to {\displaystyle \frac{1}{2}}\text{C}{\text{l}}_2\left(\text{g}\right)+2{\text{e}}^{-};E_{\text{cell}}^{\text{o}}=1.36\text{V}\\ 2{\text{H}}_{\text{2}}\text{O}\left(\text{l}\right)\to {\text{O}}_2\left(\text{g}\right)+4{\text{H}}^{+}+4{\text{e}}^{-};E_{\text{cell}}^{\text{o}}=1.23\text{V}\end{array}$$

The values of $$E_{\text{cell}}^{\text{o}}$$ for a particular oxidation or reduction reactions remain fixed. We can get these values from standard potential tables provided in the book or from the internet.

The reaction at anode which having lower value of $$E_{\text{cell}}^{\text{o}}$$ is favoured thus $${\text{H}}_{\text{2}}\text{O}$$ should get oxidised at anode. But formation of $${\text{O}}_2$$ from $${\text{H}}_{\text{2}}\text{O}$$is kinetically very slow. Thus due to overpotential of $${\text{O}}_2$$ (extra potential required to initiate a reaction at a required rate), $$\text{O}{\text{H}}^{-}$$ ions remain in the solution and the following reaction is preferred at anode.
$$\text{C}{\text{l}}^{-}\left(\text{aq}\right)\to {\displaystyle \frac{1}{2}}\text{C}{\text{l}}_2\left(\text{g}\right)+2{\text{e}}^{-}\cdot \cdot \cdot \cdot \cdot \cdot \left(\text{II}\right)$$

From equations (I) and (II), it can be concluded that the side reactions which are taking place are as follows:
$$2\text{O}{\text{H}}^{-}+\text{C}{\text{l}}_2\to \text{2OC}{\text{l}}^{-}+{\text{H}}_{\text{2}}\text{O}$$
$$4\text{O}{\text{H}}^{-}\to {\text{O}}_2+2{\text{H}}_{\text{2}}+\text{4}{\text{e}}^{-}$$

Thus the correct option is A.

Note: These are some cases where electrochemical processes are slow which implies that at lower voltages these processes do not occur. Thus in order for their occurrence, extra potential has to be implemented which is referred to as overpotential.