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Faraday’s Laws of Electrolysis - Practice Questions & MCQ

Edited By admin | Updated on Sep 18, 2023 18:35 AM | #JEE Main

Quick Facts

• Quantitative Aspect of Electrolytic Cell: Faraday's First Law, Faraday's Second Law is considered one of the most asked concept.

• 132 Questions around this concept.

Solve by difficulty

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, NA=Avogadro’s constant) :

A solution of  $\mathrm{Ni}\left(\mathrm{NO}_3\right)_2$  is electrolyzed between platinum electrode 0.3 faraday electricity. How many mole of $\mathrm{Ni}$  will be deposited at the Cathode.

The anodic half - cell of lead - acid battery is recharged using electricity of 0.13 Faraday. The amount of  $\mathrm{\mathrm{Pb}SO_4}$  electrolyzed in  g during the process is

Four Faraday of electricity is passed through a solution of  $\mathrm{CuSO_4}$ .The mass of copper deposited at the cathode is-

How many electrons would be required to deposit  $\mathrm{12.7 g}$  of copper at the cathode during the electrolysis of an aqueous solution of copper sulphate?

The atomic weight of $\mathrm{Al}$ is 27. When current of 6 Faraday is passed through a solution of $\mathrm{\mathrm{Al}^{3+}}$  ion, the weight of $\mathrm{\mathrm{Al}}$ deposited is

How many coulombs are provided by a current of  $0.020 \mathrm{~mA}$ in the calculator battery that can operate for 1100 hours?

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How many minutes are required to deliver $2.81 \times 10^6$  Coulombs using a current of  $450 \mathrm{~A}$ used in the commercial production of chlorine

The cost of 5 Rs/KWh of operating an electric motor for 9 hours takes 8 amp at 115 v is.

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$5.64 \mathrm{~F}$ of electricity is passed through A solution of Cushy. The mass of copper deposited at cathode is

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".

$\\\mathrm{W\: or\: m \propto\: q}\\\\\mathrm{W \propto it}\\\\\mathrm{W=Zit}\\\\\mathrm{Z=\frac{M}{nf} = \frac{Eq.wt}{96500}}\ \\\\\mathrm{Z= Electrochemical \: equivalent}\\\\\mathrm{M\: =\: Molar \ Mass}\\\\\mathrm{F=96500C}\\\\\mathrm{n\: = Number \: of\: electrons\: transfered} \\\\\mathrm{q\: =\: amount \: of \: charge \: utilized}$

Electrochemical equivalent is the amount of the substance deposited or liberated by one-ampere current passing for one second (that is, one coulomb 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 l 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.

$\\\mathrm{1 F=1.6 \times 10^{-19} \times 6.023 \times 10^{23}} \\\mathrm{\simeq 96500\: Coulombs}$

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.

• 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
$\mathrm{C.E.}=\frac{\text { Actual mass of species liberated}}{\text { Theoretical mass of species liberated}} \times 100 \%$

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Quantitative Aspect of Electrolytic Cell: Faraday's First Law