Which ion will be migrating towards anode during electrolysis of concentrated NaCl solution?
OR
What will be the reaction taking place at anode during the electrolysis of concentrated NaCl solution?
Which ion will be migrating towards anode during electrolysis of concentrated NaCl solution?
OR
What will be the reaction taking place at anode during the electrolysis of concentrated NaCl solution?
Dear Student,
There are two type of Nacl solutions.
1)Molton Nacl
2)Aqu Aque Nacl
The products of electrolysis of molten NaCl and aqueous solution of NaCl can be explained as follow
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.
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 E0is preferred
Cl-------------------> 1/2Cl2 + (e-) E0 = +1.36 V (3)
2H2O --------------> (O2) +
(4H+) + (4e-) 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
E0 is just the electric potential which is less important for you basic study...
At anode oxidation is occuring thus cl- at the anode