Chapter 21, Problem 50A

Interpretation Introduction

Interpretation: Chemical changes occurring during the electrolysis of brine needs to be explained.

Concept Introduction: In the electrolysis of brine (NaCl solution), three important chemicals can be obtained that is $\text{NaOH}$ , ${\text{Cl}}_{\text{2}}$ and ${\text{H}}_{\text{2}}$ . Here, reduction occurs at the cathode and oxidation at the anode.

Explanation of Solution

At the anode, oxidation takes place. In the electrolysis of brine (NaCl solution), the two possible oxidation reactions are as follows:

1. $2{\text{Cl}}^{-}\left(aq\right)\to {\text{Cl}}_{\text{2}}\left(g\right)+2e\text{ 1}\text{.36 V}$
2. $2{\text{H}}_{\text{2}}\text{O}\left(l\right)\to {\text{O}}_{\text{2}}\left(g\right)+4{\text{H}}^{+}\left(aq\right)+4e\text{ 2}\text{.07 V}$

Here, the reduction electrode potential for reaction 1 is lower than for reaction 2. Thus, from the two possible reactions, chlorine ions can easily get oxidized to chlorine gas. Therefore, the following reaction occurs at the anode:

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

Similarly, at the cathode, reduction takes place. In the electrolysis of brine, the two possible reduction reactions are as follows:

1. ${\text{Na}}^{+}\left(aq\right)+e\to \text{Na}\left(s\right)\text{ -2}\text{.71 V}$
2. $2{\text{H}}_{\text{2}}\text{O}\left(l\right)+2e\to {\text{H}}_{\text{2}}\left(g\right)+2{\text{OH}}^{-}\left(aq\right)\text{ -0}\text{.83 V}$

Here, the reduction electrode potential for reaction 2 is higher than reaction 1; thus, from the two possible reactions, water molecules can easily get reduced to hydrogen gas. Therefore, the following reaction occurs at the cathode:

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