Products of Electrolysis

Products of Electrolysis

During electrolysis, the reactions occurring at the electrodes are oxidation and reduction reactions. The products of electrolysis depend on the nature of material being electrolysed and the types of electrodes being used. If the electrode is inert such as gold or platinum, it does not take part in the chemical reaction and acts only as a source or sink for electrons.

If the electrode is reactive, it takes part in the electrode reaction. Thus, the products of electrolysis may be different for inert and reactive electrodes.

1) The products of electrolysis mainly depend on the different oxidising and reducing species present in the electrolytic cell and their standard electrode potentials.

2) Some electrochemical processes, though feasible are so slow kinetically that at lower voltages, these do not seem to occur. The slowness of electrode reaction creates electrical resistance at the electrode surface. Therefore, for the occurrence of such reactions some extra potential or voltage is required than the theoretical value of their standard electrode potential. This extra voltage required is called over voltage.

Electrolysis of molten sodium chloride

The molten sodium chloride has Na⁺ and Cl¯ ions,

NaCl ⇔ Na⁺ + Cl¯

During electrolysis, it has been observed that

a) Sodium metal is deposited at the cathode while chlorine gas is liberated at the anode.

b) When electric current is passed through molten sodium chloride, the chloride ions are attracted towards anode. These ions give up one electron each to the anode and become chlorine atoms. The chlorine atoms are unstable and combine in pairs to form chlorine molecules.

c) Na⁺ ions move towards the cathode, take up one electron each and become neutral atoms. As a result, sodium metal is obtained at the cathode.

At cathode

Na+ + e¯ ———-> Na (reduction)

At anode

Cl¯ + e¯ ——-> Cl₂ (oxidation)

Cl + Cl ———-> Cl₂

Thus, the overall reaction is :

2 NaCl ——–> 2 Na + Cl₂

 

Electrolysis of aqueous sodium chloride

In water, sodium chloride ionises as

NaCl ⇔ Na⁺ (aq) + Cl¯ (aq)

Water also dissociates into ions, though to very slight degree as

H₂O (l) ⇔ H⁺(aq) + OH‾(aq)

Thus, the aqueous solution of sodium chloride contains Na⁺, H⁺, OH¯ and Cl‾ ions. When electric current is passed through the solution, Cl₂ gas is evolved at the anode and hydrogen is evolved at the cathode. The resulting solution contains Na⁺ and OH¯ ions.

Na⁺ (aq) + e¯ ———-> Na(s)  E^ø = -2.71 V

H⁺ (aq) + e¯ ———> ½ H₂ (g)  E^ø = 0.00 V

The reaction with higher value of E^ø is preferred and therefore, the reaction at the cathode during is :

H⁺ (aq) + e¯ ———> ½ H₂ (g)

But H⁺ (aq) ions are produced by the dissociation of water as:

H₂O (l) + e¯ ——–> H⁺ (aq) + OH¯ (aq)

Therefore, the net reaction at the cathode may be written as:

H₂O (l) + e¯ ——–> ½ H₂ (g)  + OH¯ (aq)

At the anode, the following reactions are possible :

Cl¯ (aq) ——–> ½ Cl₂(g) + e¯   E^ø = + 1.36 V

2 H₂O (l) ——–> O₂ (g) + 4 H⁺ (aq) + 4 e¯ E^ø = +1.23 V

In concentrated solution of NaCl, oxidation of chloride ions is preferred than water at anode and therefore Cl₂ gas is liberated.

Cl‾(aq) ——–> ½ Cl₂(g) + e¯

The net reactions during the electrolysis of aqueous sodium chloride solution is:

NaCl (aq) ⇔ Na⁺ (aq) + Cl¯ (aq)

At cathode

H₂O (l) + e¯ ——–> ½ H₂ (g)  + OH¯ (aq)

At anode

Cl¯ (aq) ——>½ Cl₂(g) + e¯

The net reactions:

NaCl (aq) + H₂O (l) ———>Na⁺ (aq) + OH¯ (aq) + ½ H₂ (g) + ½ Cl₂(g)

Thus, during the electrolysis of aqueous sodium chloride, H₂, gas is liberated at the cathode and Cl₂ , gas is liberated at the anode and the solution contains sodium hydroxide.

Electrolysis of molten lead bromide using platinum electrodes

In the molten state, PbBr₂ exists as 

PbBr₂ ⇔ Pb²⁺ + 2 Br‾

The electrolysis may be carried out by taking solid lead bromide in a silica crucible. Two electrodes of graphite are used. When a D.C. voltage source is applied, no current is observed. However, when the crucible is heated so that lead bromide metals the current is found to pass. During electrolysis, metallic lead is produced at the cathode while bromine gas is liberated at the anode. When electric current is passed through the molten lead bromide Pb²⁺ ions move towards the cathode while Br‾ ions move towards the anode.

At anode

2 Br¯ ——-> 2 Br + 2e¯

2Br —–> Br₂(g)

At cathode

Pb²⁺ +2 e¯ ———-> Pb

The overall reaction is

PbBr₂  ———> Pb + Br₂(g)

Electrolysis of water

Water containing a few drops of acid or alkali becomes good conductor. The electrolysis of water may be carried out by taking some water in an electrolytic cell. Introduce two graphite electrodes into it.

Connect these electrodes through an ammeter to a battery. It is observed that ammeter does not show any deflection. This shows that water is a bad conductor of electricity.

However, when a few drops of dil. H₂SO₄ are added to water, the ammeter shows deflection and electrolysis starts.

Water is only weakly ionized but in the presence of an acid, its degree of ionization increases. Upon passing the electric current, it dissociates as:

H₂O (l) ⇔ H⁺(aq) + OH‾(aq)

At cathode

2H⁺(aq) + 2e¯ ————->2H or H₂

At anode:

OH‾(aq) ———> OH + e¯

4OH ——> 2H₂O + O₂

 

In case we add a few drops of dilute H₂SO₄ to water before carrying electrolysis, the SO₄^2_¯ ions will also be formed at the anode along with OH¯ ions.

In this case, both OH¯ and SO₄^2_¯ ions move towards anode. However, SO₄^2_¯ ions will not be released since their discharge potential is more as compared to OH¯ ions.

The gases oxygen and hydrogen are collected at the anode and cathode respectively. On measuring the volumes, the volume of hydrogen collected is found to be twice the volume of oxygen. This shows that the two elements H and O are present in water in the ratio of 2: 1.

Electrolysis of aqueous copper sulphate solution

Copper sulphate and water ionise as:

CuSO₄ (aq) ⇔ Cu²⁺(aq) + SO₄^2_¯

H₂O (l) ⇔ H⁺(aq) + OH¯ (aq)

During electrolysis, copper is deposited at the cathode while oxygen is liberated at the anode. 

At anode (oxidation)

2 H₂O (l) ——-> 4 H⁺ + O₂ +4 e¯

At cathode (reduction)

Cu²⁺ (aq) +2e‾ ———> Cu(s) x 2

The overall reaction is :

2Cu²⁺ (aq) + 2 H₂O (l) ———> Cu(s) + 4 H⁺ + O₂ (g)

2Cu²⁺ (aq) + 2SO₄^2_¯ + 2 H₂O (l) ——-> 2Cu(s) + 4 H⁺ + O₂ (g) + 2SO₄^2_¯

Electrolysis Of Sulphuric acid

During the electrolysis of sulphuric acid, hydrogen is liberated at cathode:

At cathode

H⁺ + e¯ ——-> ½ H₂

The following two reactions occur at anode:

2 H₂O (l) ———–> O₂ (g) + 4 e¯   E^ø = 1.23 V

2 SO₄^2_¯  (aq) ——> S₂O8₂¯ (aq) + 2 e¯  E^ø = 1.96 V