Physics
a law stating that when the magnetic flux (a measure of the number of magnetic field lines that cross a given area) linking a circuit changes, an electromotive force (emf) is induced in the circuit proportional to the rate of change of the flux linkage.
Chemistry
a law stating that the amount of any substance deposited or liberated during electrolysis is proportional to the quantity of electric charge passed and to the equivalent weight of the substance.

## Faraday’s Laws of Electromagnetic Induction.  What is Faraday’s first law of electromagnetic induction?

The first law describes the induction of emf in a conductor. First Law of Faraday’s Electromagnetic Induction state the mass of the substance generated by electrolysis is proportional to the amount of electricity used.

What is Faraday’s second law of electromagnetic induction?

The second law quantifies the emf produced in the conductor. Second Law of Faraday’s Electromagnetic Induction state that the induced emf is equal to the rate of change of flux linkages (flux linkages is the product of turns, n of the coil and the flux associated with it).
What is Faraday’s first law of electrolysis?
The mass of the substance (m) deposited or liberated at any electrode is directly proportional to the quantity of electricity or charge (Q) passed. The amount of substance produced during electrolysis is directly proportional to the amount of electric charge (measured in coulombs, C) that flows through the electrolyte.
What is Faraday’s second law of electrolysis?
Faraday’s second law of electrolysis states that, when the same quantity of electricity is passed through several electrolytes, the mass of the substances deposited are proportional to their respective chemical equivalent or equivalent weight. The amount of different substances produced by the same amount of electric charge is inversely proportional to the charge on the ions.

What is induced emf?

Induced emf (also known as induced electromotive force, electromagnetic induction, and emf induction) can be defined as the generation of a potential difference in a coil due to the changes in the magnetic flux through it. The applications of induced emf include generators, galvanometers, and transformers.
What is statically induced emf?
Statically induced EMF. This type of EMF is generated by keeping the coil and the magnetic field system, stationary at the same time; that means the change in flux linking with the coil takes place without either moving the conductor (coil) or the field system.

What is the formula of EMF?

The emf is equal to the work done on the charge per unit charge (ϵ=dWdq) when there is no current flowing. Since the unit for work is the joule and the unit for charge is the coulomb, the unit for emf is the volt (1V=1J/C).

## Explanation of Faraday’s Laws of Electrolysis

First, we must understand the process of electrolysis of a metal sulfate. When an electrolyte like metal sulfate is diluted in water, its molecules split into positive and negative ions. The positive ions (or metal ions) move to the electrodes connected with the negative terminal of the battery where these positive ions take electrons from it, becoming a pure metal atom and getting deposited on the electrode.
The negative ions (or sulphate ions) move to the electrode connected with the positive terminal of the battery, where these negative ions give up their extra electrons and become SO4 ^2- radical. Since SO4 ^2- cannot exist in an electrically neutral state, it will attack the metallic positive electrode. Forming a metallic sulfate which will again dissolve in the water.
Faraday’s laws of electrolysis are quantitative (mathematical) relationships that describe the above two phenomena.
The flow of current through the external battery circuit fully depends upon how many electrons get transferred from negative electrode or cathode to positive metallic ion or cations. If the cations have valency of two like Cu^2+ then for every cation, there would be two electrons transferred from cathode to cation. Every electron has negative electrical charge – 1.602 × 10^-19 Coulombs and say it is e-. So for disposition of every Cu atom on the cathode, there would be – 2.e charge transfers from cathode to cation.
Now say for t time there would be total n number of copper atoms deposited on the cathode, so total charge transferred, would be – 2.n.e Coulombs. Mass m of the deposited copper is obviously a function of the number of atoms deposited. So, it can be concluded that the mass of the deposited copper is directly proportional to the quantity of electrical charge that passes through the electrolyte. Hence mass of deposited copper m ∝ Q quantity of electrical charge passes through the electrolyte.
Faraday’s First Law of Electrolysis states that the chemical deposition due to the flow of current through an electrolyte is directly proportional to the quantity of electricity (coulombs) passed through it.
i.e. mass of chemical deposition:
m ∝ Quantity of electricity, Q => m = Z.Q
Where, Z is a constant of proportionality and is known as electrochemical equivalent of the substance.
If we put Q = 1 coulombs in the above equation, we will get Z = m which implies that electrochemical equivalent of any substance is the amount of the substance deposited on the passing of 1 coulomb through its solution. This constant of the passing of electrochemical equivalent is generally expressed in terms of milligrams per coulomb or kilogram per coulomb.
The mass of the chemical, deposited due to electrolysis is proportional to the quantity of electricity that passes through the electrolyte. The mass of the chemical, deposited due to electrolysis is not only proportional to the quantity of electricity passes through the electrolyte, but it also depends upon some other factor. Every substance will have its own atomic weight. So for the same number of atoms, different substances will have different masses.
How many atoms deposited on the electrodes also depends upon their number of valency. If valency is more, then for the same amount of electricity, the number of deposited atoms will be less whereas if valency is less, then for the same quantity of electricity, more number of atoms to be deposited.
So, for the same quantity of electricity or charge passes through different electrolytes, the mass of deposited chemical is directly proportional to its atomic weight and inversely proportional to its valency.
Faraday’s second law of electrolysis states that, when the same quantity of electricity is passed through several electrolytes, the mass of the substances deposited are proportional to their respective chemical equivalent or equivalent weight.
Chemical Equivalent or Equivalent Weight
The chemical equivalent or equivalent weight of a substance can be determined by Faraday’s laws of electrolysis, and it is defined as the weight of that subtenancy which will combine with or displace the unit weight of hydrogen.
The chemical equivalent of hydrogen is, thus, unity. Since valency of a substance is equal to the number of hydrogen atoms, which it can replace or with which it can combine, the chemical equivalent of a substance, therefore may be defined as the ratio of its atomic weight to its valency.
Who Invented Faraday’s Laws of Electrolysis?
Faraday’s Laws of Electrolysis were published by Michael Faraday in 1834. Michael Faraday was also responsible. As well as discovering these laws of electrolysis, Michael Faraday is also responsible for popularizing terminologies such as electrodes, ions, anodes, and cathodes.

## Experiment procedures for electrochemistry Faraday’s Law

Electrolysis of copper transfers copper atoms from an impure copper anode to a pure copper cathode, leaving the impurities behind. The anode is the impure Cu. The electrolyte is a solution of copper(II) sulfate. The Fe and Zn impurities are more easily oxidized than Cu.   ### Electrolysis

During electrolysis, ionic substances are decomposed into simpler substances when an electric current is passed through them. Electrolysis is used to extract and purify metals.

### Purifying copper by electrolysis:

1. A beaker with pure and impure copper rods dipped into copper(ll) sulfate solution.
2. The pure copper rod is connected to the negative terminal of a battery, and the impure rod is connected to the positive terminal.
3. The pure copper rod has increased in size, while the impure rod has deteriorated, leaving a pool of anode sludge at the bottom of the beaker.
During electrolysis, the anode loses mass as copper dissolves, and the cathode gains mass as copper is deposited.

### The purification of copper by electrolysis:

1. Four Cu ions are attached to the rod on the right, and four Cu2+ ions are floating in the space between the rods.
2. A battery is connected between the rods and the Cu ions are pulled towards the left rod. A half-equation shows what happens at one of the electrodes during electrolysis. Electrons are shown as e-.
3. There are now four Cu ions attached to the left rod, with four Cu2+ ions floating in the middle
These are the half-equations:
anode: Cu -> Cu2+ + 2e- (oxidation)
cathode: Cu2+ + 2e- -> Cu (reduction)
Oxidation happens at the anode because electrons are lost.
Reduction happens at the cathode because electrons are gained.
One way to remember this is by using the mnemonic OIL RIG:
Oxidation Is Loss of electrons, Reduction Is Gain of electrons.

### Procedures

1. Set-up the apparatus as shown in the diagram.

2. Using sand paper, both copper plates were brushed, and then dipped into propanone or acetone, followed by drying drying both plates using hair drier.

3. Both plates were then weigh separately and were recorded as initial mass of anode and cathode in the table prepared.

4. Both weighed plate were clipped using crocodile clips and the batteries were placed in the battery socket. After that, both plates were immersed in copper (lI) solution and the stop watch was started to allow the apparatus to run for 30 minutes.

5. After 30 minutes, the final reading of ammeter was recorded. Then, both plates were taken out from the solution, and were dipped into propanone or acetone, before blow to dry for both plates using hair drier.

6. The dried plates were then weighed and were recorded as final mass accordingly. The mass lost (at anode) and gained (at cathode) were calculated.

### Observations

At Copper (anode):

Initial mass / g, X initial =
Final mass / g, X final =
Difference / g, X =

At Copper (cathode):

Initial mass / g, Y initial =
Final mass / g, Y final =
Difference / g, Y =

Time taken = 30 minutes Currentused : p A

### Discussions

The current used :

Q = It;

Q = (p)(30 x60) = a C

Mol of e-used = Q / F
or
a / 96500 ;

mol of e- = d mol

At anode, since Cu -> Cu2+ + 2e- ;

1 mol of Cu = 2 mol of e-

Mass of copper dissolved =

d mol / 2 x 63.59 = f g

% of purity of Cu =

mass of Cu dissolve / mass lost at anode

= f / X x 100%

[Note that your mass of Cu gained must be lesser than Cu dissolved]