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

Respuesta :

Answer:

[tex]\frac{E_{1}}{E_{2}} =\frac{Z_{1}}{Z_{2}}[/tex]

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,

[tex]W_{1} = Z_{1} \times Q\ .....(1)\\W_{2} = Z_{2} \times Q\ .....(2)[/tex]

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

[tex]\frac{W_{1}}{W_{2}} =\frac{Z_{1}}{Z_{2}}........(3)[/tex]

and according to Faraday's second law,

[tex]W_{1} = E_{1} \times Q\ .....(4)\\W_{2} = E_{2} \times Q\ .....(5)[/tex]

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

[tex]\frac{W_{1}}{W_{2}} =\frac{E_{1}}{E_{2}}........(6)[/tex]

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

[tex]\frac{E_{1}}{E_{2}} =\frac{Z_{1}}{Z_{2}}[/tex]