Introduction

1. In today’s article, we are going to study about ‘Faraday law of electrolysis ‘ How beneficial it is and what is the first law of Faraday, what is the second law of Faraday, mathematical expression for first law of Faraday, mathematical expression for second law of Faraday, and examples based on faraday law of electrolysis. So, take your notebooks in you hand and get ready to study physics in an easy and sorted way.

In the 19th century, Michael Faraday discovered two laws regarding to electrolysis. These laws co-relate the current passed through the electrolyte and amount of substance deposit on the respective electrodes.

First law of Faraday : This law state that “the amount of any substance that is deposited or liberated at any electrode is directly proportional to the quantity of electricity passed through the electrolyte”.

Mathematically w∝Q

Where w us the mass (in grams) of the substance deposited and Q is the charge (in coulombs) that flows.

Since W= I× t

Or w= Z.I.t         …(1)

Where Z is a constant known as the Electrochemical equivalent of the substance deposited.

If I= 1 ampere

And t= 1sec

Then w=Z         ….(2)

Thus, electro-chemical equivalent of a substance is defined as the amount of the substance deposited or liberated when one ampere current passing for one second (i.e. one coulomb). The unit of Z is gms per coulomb.

The law states that when the same quantity of electricity passes through different electrolytes, the amounts of different substance deposited at the electrodes are directly proportional to their chemical equivalent.

If w1 and w2 are the amounts liberated and E1 and E2 are their respective chemical equivalent, then,

w1/ E1 = w2 /E2

Or ,w1/ w2 = E1/ E2

w1/ w2 = Z1. I.t/ Z2. I.t = E1 /E2

Z1/ Z2 = E1/ E2

Z∝ E      …(3)

### Therefore, Electrochemical equivalent of a substance is proportional to its chemical equivalent (equivalent weight).

Accordingto fig. If same quantity of electricity (1 coulomb) is passed through three different electrolytes viz, dil H2SO4, CuSO4 and AgNO3 solution, liberated quantities of hydrogen, copper and silver have been found is proportional to their chemical equivalent weights i.e. 1:31:78:107.88 respectively.

Practically it has been observed that 1 coulomb of charge deposits 0.0011180 gram of silver. Now, according to first law of Faraday, the mass of an ion deposited is directly proportional to the quantity or electricity. It can be shown that the quantity of electricity required to deposit one gram equivalent of silver will be

Equivalent weight of Ag/ Electrochemical equivalent = 107.88/ 0.00118

= 96493 coulomb

It is obvious from second law of Faraday that 96493 C of electricity would yield one gram equivalent of any substance. This quantity of charge is known as one faraday and is denoted by F. Therefore, Faraday is defined as

#### “This is the quantity of charge which deposits or liberates one gram equivalent of a substance“.

1 F is taken as equal to 96500 coulombs

Mathematically  W=( I× t× E)/ F

Equation (4) can be used to measure the electric current and apparatus used for this purpose is known as Coulometer or Voltameter.

1 Faraday charge is actually the charge on one mole of electrons.

Since charge on one electrons = 1.66×10^-19 C

Therefore charge on one mole of electrons = 1.66×10^-19 C × 6.023 × 10^23

≈ 96500 C

Example. 1 Calculate the quantity of current in amperes required to get 15 grams of chlorine from a solution of KCl in 30 minutes. (Atomic weight of chlorine = 35.5)

Solution: Z for chlorine = 35.5/ 96500

t= 30×60 = 1800 sec

W= 15 gram

I=?

We know that, W = Z. I. t

I= W/ Z.t

And, I= (15× 96500)/ 35.5 × 1800

I= 22.65 ampere

Example 2:  0.2964 gram of Cu is deposited, when 0.5 ampere current is passed for 40 minutes in CuSO4 solution, calculate atomic weight of Cu.

Solution : Charge Q = I.t

Q= 0.5 × 30 × 60

= 900 C

Let E be equivalent weight of Cu

Z= E/ 96500

W= 0.2964 gram

W= Zlt

0.2964 = (E / 96500) × 900

E= 31.78

Atomic weight = Equivalent weight× valency

= 31.78× 2

Therefore, atomic weight of Cu = 63.56

Example 3: how long a current of 3 ampere has to be passed through a solution of AgNO3 to coat a metal surface of 80 cm^2 with a 0.0005 cm thick layer? Density of silver is 10.5 gram/ cm^3 .

Solution: Volume of metal to be coated = 80× 0.0005

= 0.04 cm^3

Mass of silver to be coated = 0.04× 10.5

= 0.420 gn

We know, W= Z.l.t

0.42 =( 108/ 96500 )×3×1

t= 125.09 sec.

##### Conclusion

So our article is finished and after completely reading this article, one can easily tell what is Faraday law of electrolysis. And how we can explain it.

We have also discussed first law of Faraday, mathematical expression for first law of Faraday, second law of Faraday etc.

‌So one can say that they got a detailed information about Faraday law.

IT was an article based on ‘Faraday law’. Your suggestions are invited for more topics of articles.

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