Electrolysis

It is the process of decomposition of an electrolyte when electric current is passed through either its aqueous solution or molten state.

Example

One of the simplest electrolytic cell consists of two copper strips dipping in an aqueous solution of copper sulphate.

If a DC voltage is applied to the two electrodes, then Cu²⁺ ions discharge at the cathode (negatively charged)

Cu²⁺⁽aq) + 2e^– → Cu (s)

At the anode, copper is converted into Cu²⁺ ions,

Cu(s) → Cu²⁺⁽aq) + 2e^–

Important points regarding electrolytic cell

i. In electrolytic cell both oxidation and reduction takes place in the same cell

ii. Anode is positively charged and cathode is negatively charged, in electrolytic cell.

iii. During electrolysis of molten electrolyte, cations are liberated at cathode, while anions at the anode.

iv. When two or more ions compete at the electrodes, the ion with higher reduction potential gets liberated at the cathode while the ion with lower reduction potential gets liberated at the anode.

v. For metals to be deposited on the cathode during electrolysis, the voltage required is almost the same as the standard electrode potential. However for liberation of gases, some extra voltage is required on top of the theoretical value of the standard electrode potential. The extra voltage thus required is called over voltage or bubble voltage.

Electrolytic Cell (s)	Galvanic Cell (s)
Electrical energy is transformed into chemical energy.	Chemical energy is transformed into electrical energy.
Anode is positive electrode & Cathode is negative electrode.	Anode is negative electrode & Cathode is positive electrode.
Ions are discharged on both anode and cathode.	Ions are discharged only on the cathode.
If the electrodes are non-reactive or inert, concentration of the electrolyte decreases when the electric current is circulated due to electrolysis.	Concentration on the side of anodic half-cell increases while that of cathodic half-cell decreases when the two electrodes are joined by a wire.
Both the electrodes can be fitted in the same cell.	The electrodes are fitted in different cell.

How to predict the products of electrolysis?

When an aqueous solution of an electrolyte is electrolysed, and if the cation has higher reduction potential than water (-0.83V), cation is liberated at the cathode (e.g. in the electrolysis of copper and silver salts) otherwise H₂ gas is liberated due to reduction of water (e.g., in the electrolysis of K, Na, Ca salts, etc.)

Similarly if anion has higher oxidation potential than water (-1.23V), anion is liberated (e.g., Br^-), otherwise O₂ gas is liberated due to oxidation of water.

Electrolysis of molten NaCl

If we use molten NaCl as electrolyte, the products of electrolysis are sodium metal and Cl₂ gas. Here we have only one cation (Na⁺) which is reduced at the cathode

Na⁺ + e^– → Na

and one anion (Cl^–) which is oxidised at the anode

Cl^–→ ½ Cl₂ + e^–

Electrolysis of aqueous NaCl

During the electrolysis of aqueous sodium chloride solution, the products are NaOH, Cl₂ and H₂. In this case besides Na⁺ and Cl^– ions we also have H⁺ and OH^– ions along with the solvent molecules, H₂O.

At the cathode, there is competition between the following reduction reactions:

Na⁺ (aq) + e^– → Na (s) $=
– 2.71 V$

H⁺ (aq) + e^– → ½ H₂ (g) $=
0.00 V$

The reaction with higher value of E^o_cell is preferred and, therefore, the reaction at the cathode during electrolysis will be,

H⁺ (aq) + e^– → ½ H₂ (g)

But H⁺ (aq) is produced by the dissociation of H₂O, i.e.,

H₂O (l) → H⁺ (aq) + OH^– (aq)

Therefore, the net reaction at the cathode is

H₂O (l) + e– → ½ H₂(g) + OH^–

At the anode the following oxidation reactions compete,

Cl^– (aq) → ½ Cl₂ (g) + e^–, $=
1.36V$

2H₂O (l) → O₂ (g) + 4H⁺(aq) + 4e^– , $=
1.23V$

The reaction at anode with lower value of E^o_cell is preferred, therefore, water should get oxidised in preference to Cl^–(aq). However, on account of overpotential of oxygen (higher extra potential is required to convert to O₂), first reaction is preferred.

Thus, the reactions will be,

In the solution: NaCl (aq) + H₂O → Na⁺ (aq) + Cl^– (aq)

At cathode: H₂O(l) + e^– → ½ H₂(g) + OH^– (aq)

At anode: Cl^– (aq) → ½ Cl₂(g) + e^–

Net reaction, will therefore be,

NaCl(aq) + H₂O(l) → Na⁺(aq) + OH^–(aq) + ½ H₂(g) + ½ Cl₂(g)

The standard electrode potentials are replaced by electrode potentials given by Nernst equation to take into account the concentration effects.

Electrolysis of H₂SO₄

During the electrolysis of sulphuric acid, the following processes are possible at the anode:

2H₂O(l) → O₂(g) + 4H⁺(aq) + 4e^–, Eocell = 1.23V

2S (aq) → S₂ (aq) + 2e^–, ${E^{o}}_{c e l l}$ $= 1.96V$

For dilute sulphuric acid, the first reaction is preferred but at higher concentrations of H₂SO₄, second reaction is preferred.

Effect of nature of electrode on electrolysis

If the electrode is inert (e.g., platinum or gold), it does not participate in the chemical reaction and acts only as source or sink for electrons.

On the other hand, if the electrode is reactive, it participates in the electrode reaction.

Thus, the products of electrolysis may be different for reactive and inert electrodes.

Discharge potential is defined as the minimum potential that must be applied across the electrodes to bring about the electrolysis and subsequent discharge of the ion on the electrode.