What is the equation for the electrolysis of NaCl?

We will begin the explanation of the Chlor-Alkali process by determining the reactions that occur during the electrolysis of NaCl.   Because NaCl is in an aqueous solution, we also have to consider the electrolysis of water at both the anode and the cathode. Therefore, there are two possible reduction equations and two possible oxidation reactions

Reduction     

Na+ (aq) + 2e- ---> Na(s)                     E0_Na+/Na= -2.71 V                        (1)                

2 H20 (l) + 2e- ---> H2(g) +2OH- (aq)         E⁰_H20/H2= -.83 V                         (2)

Oxidation:

2Cl^-(aq) ---> Cl₂ + 2e^{-                                  -}E⁰_Cl2/Cl-= -(1.36 V)            (3)                

2H₂0(l) ---> O₂ + 4H⁺ + 4e^{-                   -}E⁰_O2/H20= -(1.23 V)                  (4)

As we can see, due to the very much more negative electrode potential,  the reduction of Sodium ions is much less likely to occur than the reduction of Water, so we can assume that that in the electrolysis of NaCl, the reduction that occurs is reaction (2).  Therefore, we should try to determine what the oxidation reaction that occurs is.  Lets say we have reaction (2) as the reduction and reaction (3) as the oxidation.  We would get:

reduction: 2 H20 (l) + 2e- ---> H2(g) +2OH- (aq) E⁰_H20/H2= -.83 V  (2)

oxidation: 2Cl-(aq) ---> Cl2 + 2e-  -E⁰_Cl2/Cl-= -(1.36 V)        (3)

overall: 2 H₂0 (l) + 2Cl^-(aq) ---> H₂(g) +2OH^- (aq) +Cl₂    E⁰_H20/H2-E⁰_Cl2/Cl^- =-.83 + (-1.36)= -2.19       (5)

On the other hand, we could also have reaction (2) with reaction (4)

reduction: 2[2 H20 (l) + 2e- ---> H2(g) +2OH- (aq)]   E_H20/H2⁰=-.83 V  (2)

oxidation: 2H₂0(l) ---> O2 + 4H+ + 4e-  ^-EO_2/H20⁰= -(1.23 V) (4)

overall: 2H20 (l) ---> 2H2(g) + O2 (g) E⁰_H20/H2 - E⁰O2/H20= -.83 V -(1.23 V)= -2.06(6)

At first glance it would appear as though reaction (6) would occur due to the smaller (less negative) electrode potential.  However, O₂ actually has a fairly large overpotential, so instead Cl₂  is more likely to form, making reaction (5) the most probable outcome for the  electrolysis of NaCl.  

Chlor-Alkali Process in a Diaphragm Cell 

Depending on the method used, there can be several different products produced through the Chlor-Alkali process. The value of these products is what makes the Chlor-Alkali process so important.   The name comes from the two main products of the process, chlorine and  the alkali, Sodium Hydroxide (NaOH).  Therefore, one of the main uses of the Chlor-Alkali process is the production of NaOH. As described earlier,the equation for the Chlor-Alkali process, that is, the electrolysis of NaCl,  is as follows: 

reduction: 2 H20 (l) + 2e- ---> H2(g) +2OH- (aq) E_H20/H2⁰= -.83

oxidation: 2Cl-(aq) ---> Cl2 + 2e-  E⁰Cl2/Cl^-= -1.36

Overall: 2Cl- + 2H20(l) ---> 2 OH- + H2(g) +Cl2 (g) = E⁰_H2O/H2- E⁰Cl/l2^-= -83V- (-1.36 V)= -2.19 V

• The Sodium Chloride is put into the anode compartment, in aqueous form.
• the actual physical anode is made of either graphite or titanium 
• in the anode compartment, Cl₂ gas is produced as Cl^- is oxidized

cathode

• on the cathode side, OH^- (aq) and H₂ gas are formed as water is reduced.
• You may wonder, why are there sodium and chloride ions of the cathode side if we put the sodium chloride into the anode compartment? To answer this question, we can consider the difference in solution levels between the anode and cathode.  This causes the gradual flow of NaCl into the cathode, and prevents the backflow of NaOH into the anode.  If Cl₂ and NaoH come into contact, Cl₂ turns into ClO^-,ClO₃-, Cl^- ions.  The water level difference prevents this contact and also encourages the flow of NaCl to the cathode side, so NaOH (aq) can form.  
• Now you may notice, that the solution with NaOH in the cathode will also have aqueous NaCl mixed with it due to the flow of NaCl from the anode to the cathode. Therefore, if we want very pure NaOH for uses such as rayon manufacture, we need to somehow purify the salt out of the NaOH. In general, before purification, the solution is about 14-16% NaCl(aq) and 10-12% NaOH (aq). However, the NaOH(aq)  can be concentrated and the NaCl (s) can be crystallized to form a solution with 50% NaOH(aq) and about 1% NaCl (aq)

Chlor-Alkali Process in Mercury Cell

To even further improve the purity of NaOH, a mercury cell can be used for the location of electrolysis, opposed to a diaphragm cell

Figure 2:

anode side

• the anodes are placed in the aqueous NaCl solution, above the liquid Mercury
•  the reduction of Cl^- occurs to produce chlorine gas, Cl₂ (g)

cathode side

• A layer of Hg(l) at the bottom of the tank serves as the cathode 
• With a mercury cathode, the reaction of H₂O (l) to H₂ has a fairly high over potential, so the reduction of Na⁺ to Na occurs instead. The Na is soluble in Hg(l) and the two combine to form the Na-Hg alloy amalgam. This amalgam can be removed and then mixed with water to cause the following reaction:

  • 2Na(in Hg) + 2H2O ---> 2 Na+ + 2 OH- + H2(g) + Hg(l)

• The Hg(l) that forms is recycled back into the liquid at the bottom of the tank that acts as a cathode
•  H₂ gas is released 
• NaOH is left in a very pure, aqueous form 

**Some of the main problems with the mercury cell are as follows:

• The reaction needs a higher voltage than the diaphragm cell: 4.5 V in the mercury cell compared to 3.5 in a diaphragm cell.
• Requires quite a bit of electrical energy, as it need 3400 kWh/ton Cl₂ opposed to 2500 in a diaphragm cell.
• Potential damage to the environment due to mercury deposits.  Luckily deposits were as large as 200 g Mercury per ton Chlorine gas, but now, they never exceed 0.25 g per ton Chlorine gas.   

Membrane Cell Process

One final way to make even more pure NaOH is to use a membrane cell. It is preferred over the diaphragm cell or mercury cell method because it uses the least amount of electric energy and produces the highest quality NaOH.  For instance, it can produce NaOH with a degree of chlorine ion contamination of only 50 parts per million. An ion-permeable membrane is used to separate the anode and cathode. 

anode: 

• Saturated brine is fed to the compartment . 
• Current passed through the cell splits the sodium chloride into its constituent components, Na⁺ and Cl^-.
• Chloride ions are oxidized to chlorine gas at the anode
• The membrane passes Na⁺ ions to the cathode compartment

 cathode:

• H₂O is reduced to form OH^- and H₂ gas

  • 2H2O --> H2 + 2OH- 

• Na⁺ ions that had flowed over and OH^- produced through the reduction of water react to form aqueous NaOH.  
• H₂ gas also is produced as a byproduct