Question

Explain, why electrolysis of aqueous solution of NaCl gives H₂ at cathode and Cl₂ at anode. Write overall reaction.

[Given: \[\ce{E^{\circ}_{{Na^{+}/{Na}}}}\] = 2.71 V, \[\ce{E^{\circ}_{Cl_2/2Cl^-}}\] = 1.36 V and \[\ce{\frac{1}{2} O2_{(g)} + 2H+_{ (aq)} + 2e- -> H2O_{(l)}}\]; E° = 1.23 V]

Answer 1

In the aqueous electrolysis of NaCl, the products are:

H₂ at the cathode

Cl₂ at the anode

Let’s understand why, based on standard electrode potentials (E°):

Two competing species:

\[\ce{Na+ + e- -> Na}\]; E° = −2.71 V

\[\ce{2H2O + 2e- -> H2 + 2OH−}\] (derived from E° = −0.83 V)

More positive E° indicates easier reduction.

Thus, water is reduced in preference to Na⁺:

\[\ce{2H2O + 2e- -> H2_{(g)} + 2 OH-}\]

Two competing species:

\[\ce{2Cl- ->Cl2 + 2e-}\]; E° = −1.36 V

\[\ce{2H2O -> O2 + 4H+ + 4e-}\]; (Reverse of given E° = +1.23 V)

Though O₂ has a lower E°, Cl⁻ is oxidised due to:

High concentration of Cl⁻:

Overpotential for oxygen evolution from water.

Thus, chloride ions are oxidised:

\[\ce{2Cl- -> Cl2_{(g)} + 2e-}\]

Overall reaction is:

\[\ce{2NaCl_{(aq)} + 2H2O_{(l)} -> 2NaOH_{(aq)} + H2_{(g)} + Cl2_{(g)}}\]