Applications of Electrolysis

 

Table of Content

Determination of Equivalent Masses of Elements Electrometallurgy Manufacture of Non-Metals Thickness of Coated Layer Current Efficiency Related Resources

Electrolysis is a very important and useful phenomenon. The phenomenon of electrolysis has wide range of applications. It is used in large number of industrial manufacturing processes.

Let us discuss some of the many applications of electrolysis

Determination of Equivalent Masses of Elements

• According to second law of electrolysis when the same quantity of electronic current is passed through solutions of salts of two different cells, the amounts of the metals deposited on the cathodes of the two cells are proportional to their equivalent masses of the respective metals.

• If the amounts of the metals deposited on the cathodes be W_A and W_B respectively, then 
  

• Knowing the equivalent mass of one metal, the equivalent mass of the other metal can be calculated from the above relationship.

• The equivalent masses of those non-metals which are evolved at anodes can also be determined by this method.

Electrometallurgy

Electrometallurgy is the process of extraction of metal from ore by electrolysis.

The metals like sodium, potassium, magnesium, calcium aluminum, etc., are obtained by electrolytes of fused electrolytes.

Fused electrolyte	Metal isolated
NaCl + CaCl2 + KF	Na
CaCl2 + CaF2	Ca
Al2O3 + cryolite	Al
MgCi2 (35%) + NaCl (50%) + CaCl2 (15%)	Mg
NaOH	Na
KCl + CaCl2	K

Manufacture of Non-Metals

Non-metals like hydrogen, fluorine, chlorine are obtained by electrolysis.

Electro-refining of metals:

This is the process of refining the metal. i.e. removing impurity from metal by the use of electrolysis method.

The metals like copper, silver, gold, aluminum, tin, etc., are refined by electrolysis.

Manufacture of compounds:

Compounds like NaOH, KOH, Na₂CO₃ KCIO₃, white lead, KMnO₄, etc., are manufactured by electrolysis.

Electroplating:

The process of coating an inferior metal with a superior metal by electrolysis is known as electroplating.

The aims of electroplating are:

• To prevent the inferior metal from corrosion.

• To make it more attractive in appearance.

The object to be electroplated is made the cathode and block of the metal to be deposited is made the anode in an electrolytic bath containing a solution of a salt of the anodic metal. On passing electric current in the cell, the metal of the anode dissolves out and is deposited on the cathode-article in the form of a thin film. The following are the requirements for fine coating:

• The surface of the article should be free from greasy matter and its oxide layer. The surface is cleaned with chromic acid or detergents.

• The surface of the article should be rough so that the metal deposited sticks permanently.

• The concentration of the electrolyte should be so adjusted as to get smooth coating.

• Current density must be the same throughout.

Refer to the following video for electroplating

For electroplating	Anode	Cathode	Electrolyte
With copper With silver With nickel With gold With zinc With thin	Cu Ag Ni Au Zn Sn	Object Object Object Object Iron objects Iron objects	CuSo4 + dilute H2So4 KAg(CN)2 Nickel ammonium sulphate KAu(CN)2 ZnSO4 SnSO4

Thickness of Coated Layer

 Let the dimensions of metal sheet to be coated be (a cm × b cm).

Thickness of coated layer = c cm

Volume of coated layer = (a × b × c) cm³

Mass of the deposited substance = Volume × density =   (a × b × c) × d g

(a × b × c) × d = (I×t×E)/96500

Using above relation we may calculate the thickness of coated layer.

Note:

Sometimes radius of atom of deposited metal is given instead of density, e.g., Radius of silver atom           = 10-8 cm Atomic mass of Ag             = 108 Mass of single silver atom    = 108/(6.023× 1023 ) g Volume of single atom         = 4/3× πR3  = 4/3×3.14×(10-8)3 cm3 Density of Ag  =  (Mass of single atom )/(Volume of single atom)  = (108/6.023 × 1023)/(4/3×3.14×10-8 )3

Current Efficiency

Sometimes the ammeter shows false current due to mechanical fault. In this case,

% Current efficiency =  (Actual current)/(Ammeter current)×100 

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