Question

Show the relation between Electro-chemical equivalent (Z) and Chemical equivalent weight of the element (E) using the Faraday's Laws of Electrolysis.

Answer 1

`(E_(1))/(E_(2)) =(Z_(1))/(Z_(2))`

Step-by-step explanation:

Let us assume W1 g of a substance and W2 g of another substance are produced at two different electrodes when same quantity of electricity is passed through them.

if two elements have their respective electro-chemical equivalents Z1 and Z2 and their corresponding chemical equivalents are E1 and E2.

Then according to Faraday's first law,

`W_(1) = Z_(1) * Q\ .....(1)\\W_(2) = Z_(2) * Q\ .....(2)`

Divide eq.(1) by eq.(2)

`(W_(1))/(W_(2)) =(Z_(1))/(Z_(2))........(3)`

and according to Faraday's second law,

`W_(1) = E_(1) * Q\ .....(4)\\W_(2) = E_(2) * Q\ .....(5)`

Divide eq.(4) by eq.(5)

`(W_(1))/(W_(2)) =(E_(1))/(E_(2))........(6)`

From equation (3) and (6) we get,

`(E_(1))/(E_(2)) =(Z_(1))/(Z_(2))`