Get a free home demo of LearnNext

Available for CBSE, ICSE and State Board syllabus.
Call our LearnNext Expert on 1800 419 1234 (tollfree)
OR submit details below for a call back


Products of Electrolysis

Have a doubt? Clear it now.
live_help Have a doubt, Ask our Expert Ask Now
format_list_bulleted Take this Lesson Test Start Test

Products of Electrolysis - Lesson Summary

The products of electrolysis depend on the nature of the electrolyte being electrolysed, and the type of electrodes used.

Inert electrodes do not participate in the chemical reaction.

Platinum, gold or graphite electrodes are commonly used as inert electrodes.

Active electrodes do participate in the chemical reaction. Active electrodes are used in the electrolytic refining of metals.

The products of electrolysis may be different for reactive and inert electrodes.

The products of electrolysis reactions depend on the oxidising and reducing species present in the electrolytic cell, and their standard electrode potentials.

Some electrochemical processes are kinetically very slow. At lower voltages, these processes don't seem to occur at all. In order to make these processes occur, extra potential, called over-potential, must be applied.

The electrolytic decomposition of molten sodium chloride produces sodium metal and chlorine gas.

When electricity passed through the molten NaCl, sodium is deposited at cathode & chlorine gas liberated at anode.

The process can be represented as:

NaCl (l) →Na+ + Cl-

Anode: Cl- →1/2Cl2 + e-

Cathode: Na+ + e- → Na

The electrolytic decomposition of aqueous sodium chloride solution produces sodium hydroxide, chlorine gas and hydrogen gas.


At the cathode, either sodium ions or hydrogen ions may be reduced. As the hydrogen ion has a higher standard electrode potential than sodium, it gets reduced. The hydrogen ions are produced by the dissociation of water.

At Cathode
         Na+(aq) + e- → Na(s)
                  E°cell = - 2.71 V
         H+(aq) + e- → ½ H2(g)
                  E°cell = 0.00 V
    H20(I) ⇔ H+ (aq) + OH- (aq)

At the anode there are two substances that can be oxidised, chloride ions and water molecules. The electrode potential of water is lower than that of chloride ions. But the oxidation of water is a kinetically slow process. Hence, over potential is needed for the production of oxygen. This explains why chlorine but not oxygen is produced during the electrolysis of aqueous sodium chloride.

At Anode
        Cl-(aq) → ½Cl2(g)+e-
             E°cell = 1.36 V
 2h20(l) → o2(g) + 4H+(aq) + 4e-
             E°cell = 1.23 V

The net reactions of electrolysis of an aqueous solution of sodium chloride:

Electrolysis of Aqueous NaCl Solution
                    NaCl(s) → Na+(aq) + Cl (aq)
                    H2O(I) → H + (aq) + OH-(aq)
At Cathode
                     H + (aq) + e - → ½H2(g)
At Anode
                     Cl - (aq) → ½Cl2(g) + e -

Net Reaction
         NaCl(s) + H2O(l) → Na+(aq) + OH-(aq) + ½H2(g) + ½Cl2(g)

Halide ions will undergo oxidation to form the diatomic elements in preference to water.

Chloride ions undergo oxidation in preference to water due to overvoltage of oxygen.

The ability of a cation to be reduced or an anion to be oxidised not only depends on their standard reduction potentials, but also on their concentrations. To account for the effects of concentration, the standard electrode potentials are replaced by the electrode potentials given by the Nernst equation.

Nernst Equation

E(Mn+/M) = E(Mn+/M) - RT/nF ln 1/[Mn+]

In dilute solutions of sulphuric acid, water will be oxidised to release oxygen gas. In concentrated solutions of sulphuric acid, the formation of the peroxydisulphate ion is favoured.

Dilute Solution of Sulphuric Acid
             2H2O(I) → O2(g) + 4H+ +4e-

Concentrated Solution of Sulphuric Acid
             2SO42-(aq) → S2O82-(aq) + 2e-


Feel the LearnNext Experience on App

Download app, watch sample animated video lessons and get a free trial.

Desktop Download Now
Try LearnNext at home

Get a free home demo. Book an appointment now!