Quantitative Aspect of Electrolytic Cell: Faraday's First Law is considered one the most difficult concept.
Faraday's Second Law is considered one of the most asked concept.
24 Questions around this concept.

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EASY

How many electrons would be required to deposit 6.35 g of copper at the cathode during the electrolysis of an aqueous solution of copper sulphate ? (Atomic mass of copper = 63.5 u, N_A=Avogadro’s constant) :

$\frac{N_{A}}{20}$

$\frac{N_{A}}{10}$

$\frac{N_{A}}{5}$

$\frac{N_{A}}{2}$

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When 0.1 mol $\text{MnO}_{4}^{-2}$ is oxidised the quantity of electricity required to completely oxidise $\text{MnO}_{4}^{-2}$ to $\text{MnO}_{4}^{-}$ is :

96500 C

2 X 96500 C

9650 C

96.50 C

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Concepts Covered - 2

Quantitative Aspect of Electrolytic Cell: Faraday's First Law

According to the Faraday's first law, "The amount of substance or quantity of chemical reaction at electrode is directly proportional to the quantity of electricity passed into the cell".

$\\\mathrm{W\: or\: m \propto\: q}\\\\\mathrm{W \propto It}\\\\\mathrm{W=ZIt}\\\\\mathrm{Z=\frac{M}{nf}}\\\\\mathrm{Z=\frac{M}{n f}}\\\\\mathrm{Z= Electrochemical \: equivalence}\\\\\mathrm{M\: =\: molar mass}\\\\\mathrm{F=96500}\\\\\mathrm{n\: = Number \: of\: electrons\: transfer} \\\\\mathrm{q\: =\: amount \: of \: charge \: utilized}$

Electrochemical equivalent is the amount of the substance deposited or liberated by one-ampere current passing for one second (that is, one coulomb, I x t = Q or one coulomb of charge.
One gram equivalent of any substance is liberated by one faraday.

$\\\mathrm{Eq. \: Wt. =Z \times 96500}\\\\\mathrm{\frac{W}{E}=\frac{q}{96500}}\\\\\mathrm{w=\frac{E . q}{96500}}\\\\\mathrm{W=\frac{Eit}{96500}}$

As w = a x 1 x d that is, area x length x density
Here a = area of the object to be electroplated
d = density of metal to be deposited
l = thickness of layer deposited
Hence from here, we can predict charge, current strength time, thickness of deposited layer etc.

NOTE: One faraday is the quantity of charge carried by one mole of electrons.

$\\\mathrm{E \: \alpha\: Z}\\\mathrm{E=FZ}\\\mathrm{1 F=1.6023 \times 10^{-19} \times 6.023 \times 10^{23}} \\\mathrm{=96500\: Coulombs}$

Faraday's Second Law

According to Faraday's second law, "When the same quantity of electricity is passed through different electrolytes, the amounts of the products obtained at the electrodes are directly proportional to their chemical equivalents or equivalent weights".

$\\\mathrm{As \: \frac{W}{E}=\frac{q}{96500}= No \: of \: equivalents \: constant}\\\\\mathrm{So}\\\\\mathrm{\frac{E_{1}}{E_{2}}=\frac{M_{1}}{M_{2}} or \frac{W_{1}}{W_{2}}=\frac{Z_{1}}{Z_{2}}}\\\\\mathrm{E_{1}= equivalent\: weight}\\\\\mathrm{E_{2}\: =\: equivalent \: weight}\\\\\mathrm{W \: or \: M= mass\: deposited}$

From this law, it is clear that 96500 coulomb of electricity gives one equivalent of any substance.

Application of Faraday's Laws

It is used in the electroplating of metals.
It is used in the extraction of several metals in pure form.
It is used in the separation of metals from non-metals.
It is used in the preparation of compounds

NOTE:
Current Efficiency: It is the ratio of the mass of the products actually liberated at the electrode to the theoretical mass that could be obtained
$\mathrm{C.E.}=\frac{\text { Actual mass of species liberated}}{\text { Theoretical mass of species liberated}} \times 100 \%$