The products of electrolysis of aqueous NaCl at the respective electrodes are :
Cathode : H2
Anode : Cl2 and not O2. Explain.

Dear Student,

Please find below the solution of your asked query:

The products of electrolysis of molten NaCl and aqueous solution of NaCl can be explained as follow
  • Electrolysis of molten NaCl:

Molten NaCl contains only NaCl so gives Na+ and Cl- ions only as

NaCl   Na+ + Cl-

Reaction as cathode:  Na+ + e-   Na

Reaction as anode:  Cl-   Cl + e-

 Cl + Cl   Cl2

So, the product of electrolysis are Na at cathode and Cl2 at anode.

  • Electrolysis of aqueous  NaCl: 

Here, NaCl and Water H2O both are present and both dissociate as

NaCl   Na+ + Cl-

H2O   H+ + OH-

Reaction as cathode: Both Na+ and H+ will compete for cathode but reaction with higher E0 is preferred

 Na+ + e-   Na                     E0 = -2.71 V            -------(1)

H++ e-   1/2 H2                  E0 = 0.00 V              -------(2)

Since, (2) has higher value of E0 So, H2 is product deposited at cathode. 

Note-The equation 2 can also be written as  H2O + e-   1/2 H2 + OH-

Reaction as anode: Both Cl- and OH- will compete for anode but reaction with lower E0 is preferred

Cl-   1/2Cl2 + e-                   E0 = +1.36 V                -------(3)

2H2O   O2+ 4H+ + 4 e-      E0 = +1.23 V                -------(4)

The equation 4 is the reaction of OH- at anode. The (4) has lower  E0, it should be preferred. But due to over potential Cl2 is preferred product at anode.

Hence, the product of electrolysis are H2 at cathode and Cl2 at anode.

Hope this information clears your doubts about the topic.

Keep asking!!


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