Michael Faraday (1834) stated two laws on the basis of his studies on electrolysis:

According to this law, “The amount of substance liberated at an electrode is directly proportional to the quantity of electricity passed”.

or,   $M or W \propto Q$

Where W or M = amount of substance liberated in gram.

Q = quantity of electricity passed in coulomb.

Since Q = I.t

Where I = Current in ampere

and t = time in seconds

Hence $W \propto I.t \hspace{2mm} or \hspace{2mm}W = ZI.t = ZQ$

where Z = proportionality constant, called electrochemical equivalent.

If I = 1 ampere and t = 1 second then Z = W Therefore electrochemical equivalent may be defined as, “The mass of substance (in grams) liberated at the electrode on passing current of 1 ampere for 1 second or on passing 1 coulomb of electricity is called electrochemical equivalent of the substance”.

$Z = \dfrac{Atomic mass}{n \times 96500} [n = No. of e^- exchanged]$

1F = 9500 columb

$E = Z \times 96500$

So, 1 Faraday [96500 coulomb] of electricity will produce 1 gm equivalent of Ag, Cu and Al at cathode.

2.       Faraday’s second law: According to this law, “if same quantity of electricity is passed through different electrolytes, then the amount of substances liberated at the respective of electrodes are in the ratio of their equivalent masses”.

$\dfrac{W_1}{W_2} = \dfrac{E_1}{E_2} or \dfrac{Z_1}{Z_2} = \dfrac{E_1}{E_2}$

Suppose three cells containing HCI, $AgNO_3 \text{and} CuSO_4$ solutions are connected in series. If same quantity of electricity is passed through these cells, then the amount of hydrogen, silver and copper deposited at the respective cathodes is in the ratio of their equivalent mass.

Hence,

$\dfrac{\text{Amount of substance liberated}}{\text{Equivalent mass of the substance}}$

$= \dfrac{\text{Mass of hydrogen}}{\text{Eq. mass of hydrogen}} = \dfrac{\text{ Mass of silver}}{\text{Eq. mass of silver}}$

$= \dfrac{\text{Mass of coper}}{\text{Eq. mass of copper}}$

$\text{or} \dfrac{\text{Mass of hydrogen}}{1.008} = \dfrac{\text{Mass of silver}}{107.88} = \dfrac{\text{Mass of copper}}{31.78}$

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