Explain faraday's second law of electrolysis in details

Faraday’s second law states that, the amount of substances deposited by the passage of same amount of electric current will be proportional to their equivalent weights.

According to Faraday’s first law, m = \(frac{EQ}{96485}=frac{E,I,t}{96485}\)

When, the electric current (It or Q) remains the same, m α E.

This law is useful, in calculating the amount of reaction or mass involved for different substances for the passage of same quantity of current.

Let the equivalent weight of two electrolytes, electrolyzed with the same amount of electric current as E1 and E2 respectively. Then the mass M1 and M2 of the substances deposited at electrodes will be-


For example, when to aqueous electrolytes, say silver nitrate and copper sulphate are connected in series with an electric supply, Faraday’s law can calculate amount of silver and copper, deposited in their respective electrolytic cells.

According to Faraday’s second law, m α E

Equivalent weight of silver = Molecular weight of silver = 108

Equivalent weight of copper = Molecular weight of copper/2 = \(frac{63.54}{2}\) = 31.77

For one faraday charge, mass of silver and copper deposited will be 108g and 31.77g respectively.

Mcu α Ecuand MAg α EAg

\(frac{text{Mag}}{text{Mcu}}~=~~frac{Eag}{Ecu};frac{text{Mag}}{text{Mcu}}~=~~frac{108~times ~2}{63.54}\) = 3.4

Understanding Faraday’s second Law of electrolysis

  1. Electric current of 0.6F was, passed through three separate electrolytic cells containing aqueous

solutions of silver nitrate, copper sulphate and Aluminum sulphate simultaneously. Calculate the percentage ratio of silver , copper and iron metal deposited at the corresponding cathodes.

Reactions of the metallic ions are-

Ag++ e → Ag

Cu2+ + 2eCu

Al3+ + 3eAl


Atomic weight

Number of faraday needed for reducing one by atomic weight

Weight reduced by 0.6F













Ratio of weight of silver: copper: aluminum deposited = 64.8:31.77:9 = 7.2: 3.5:1 = 7:3.5:1

  1. A constant DC current of 1.5amperes as passed through, three electrolytic serially connected cells

Containing, solutions of ZnSO4, AgNO3 and CuSO4. If, 1.45 g of silver was deposited from the silver

nitrate solution, calculate the time taken for the deposition and the mass of copper and zinc that will be deposited from their respective cells.

  1. i) Time of current flow;

Mass of nickel deposited = \(frac{E,I,t}{96500}\)

Equivalent eight of nickel = Atomic weight/2 =58.7/2

Mass = \(frac{58.7times 1.5times t}{2times 96500}\) = 1.45gm

Time = \(frac{1.45times 2times 96500}{58.7times 1.5}\) = 3178sec = 53min

  1. ii) Mass of copper and zinc deposited;
\(frac{text{Mag}}{text{Eag}}~=~~frac{Mcu}{Ecu};frac{1.45}{108}~=~~frac{2xMcu}{63.54}\) Mass of copper deposited = \(frac{63.54text{x}1.45}{2text{ }!!~!!text{ x }!!~!!text{ }108}\) = 0.427g

\(frac{text{Mag}}{text{Eag}}~=~~frac{Mzn}{Ezn};frac{1.45}{108}~=~~frac{2times Mzn}{65.38}\) Mass of zinc deposited = \(frac{65.38text{x}1.45}{2text{ }!!~!!text{ x }!!~!!text{ }108}\) = 0.44g

  1. The weight of silver (at.wt=10) displaced by a quantity of electricity which displaces 5600mL of O2 at STP
  2. a) 5.4g b) 10.8g c) 54g d) 108g

For silver, Atomic weight = Equivalent weight =108g

For oxygen, one mole (32g) of oxygen occupy a volume of 22400mL

βΈͺ Equivalent weight of oxygen = 8g = \(frac{22400times 8}{32}\) = 5600mL

Electric current that displaces one equivalent volume of oxygen will also displace one equivalent weight

of silver=108

Answer = d“

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