The products of electrolysis of molten NaCl are Na metal and Cl 2 gas, while the products of electrolysis of aqueous solution of NaCl are NaOH, Cl 2 and H 2 .
Please explain this in detail with the appropriate reactions.
Dear Student,
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 eqn 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 eqn 4 is the reaction of OH- at anode. The (4) has lower E0, it should be preferred. But due to overpotential Cl2 is preferred product at anode.
Hence, the product of electrolysis are H2 at cathode and Cl2 at anode.