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Second Law of Electrolysis

 Second law of electrolysis

Statement:

" When same quantity of current is passed through two or more that two electrolytic solution connected in series, then mass of substance deposited at each electrode is directly proportional to equivalent mass." Mathematically,



Illustration:
Let us consider AgNO3, CuSO4 and ZnSOsolution. These solutions are connected in series and is connected with source of electricity, the mass of Ag, Cu and Zn deposited at its electrode are directly proportional to their equivalent masses:



Numerical Problems:



1.State Faraday’s 2nd Law of electrolysis. Equal amount current was passed through an aqueous solution of tri-valent salt and dil. H2 SO4. The volume of H2 liberated was 96.5ml at 27℃ and 765 mmHg pressure. The weight of the metal deposited was 0.74g. Calculate the atomic weight of the metal.

2. Electrolysis is carried out by passing same quantity of electricity through an acidulated water and through of solution of
MSO_4. 0.573 g of the metal M and 203 cc of H2 at NTP are produced at the respective electrodes of the cell. Calculate the atomic mass of M.

3. Three electrolytic cells A, B and C containing electrolytes
ZnSO_4, AgNO_3and CuSO_4 respectively were connected in series. A steady current of 1.50A was passed through them until 1.45g of Ag were deposited at the cathode of the cell B. How long did the current flow? What masses of Cu and Zn were deposited?

4. A metal forma an oxide MO. The same quantity of electricity deposited 0.862 g of silver and 0.350g of the metal (M). Calculate the atomic mass of the metal.

5. An electric current is passed through three cells in series containing respectively solutions of copper
sulphate, silver nitrate and potassium iodide. What masses of silver and iodide will be liberated while 1.25 g of copper is deposited? (At. Masses are: Cu=63.5, Ag=108, I=127

6. The same current was passed through 
H2SO4 and a solution of divalent metallic salt. It liberated 40cc of H2gas at 15℃, and 720 torr pressure and deposited 0.101 g of the metal. Calculate atomic mass.
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Solution Of Numerical Problems:
1.State Faraday’s 2nd Law of electrolysis. Equal amount current was passed through an aqueous solution of tri-valent salt and dil. H2 SO4. The volume of H2 liberated was 96.5ml at 27℃ and 765 mmHg pressure. The weight of the metal deposited was 0.74g. Calculate the atomic weight of the metal.



2. Electrolysis is carried out by passing same quantity of electricity through an acidulated water and through of solution of 
MSO4. 0.573 g of the metal M and 203 cc of H2 at NTP are produced at the respective electrodes of the cell. Calculate the atomic mass of M.



3. Three electrolytic cells A, B and C containing electrolytes 
ZnSO4, AgNO3and CuSO4 respectively were connected in series. A steady current of 1.50A was passed through them until 1.45g of Ag were deposited at the cathode of the cell B. How long did the current flow? What masses of Cu and Zn were deposited?



4. A metal formula of an oxide MO. The same quantity of electricity deposited 0.862 g of silver and 0.350g of the metal (M). Calculate the atomic mass of the metal.




5. An electric current is passed through three cells in series containing respectively solutions of copper 
sulphate, silver nitrate and potassium iodide. What masses of silver and iodide will be liberated while 1.25 g of copper is deposited? (At. Masses are: Cu=63.5, Ag=108, I=127

 
6. The same current was passed through H2SO4 and a solution of divalent metallic salt. It liberated 40cc of H2gas at 15℃, and 720 torr pressure and deposited 0.101 g of the metal. Calculate atomic mass.

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