SUBJECT: CHEMISTRY
CLASS: SS 2
DATE:
TERM: 1st TERM
REFERENCE MATERIALS
WEEK EIGHT
TOPIC: FARADAY'S LAWS OF ELECTROLYSIS
CONTENT
FARADAY’S FIRST LAW OF ELECTROLYSIS
Faraday’s first law states that the mass (m) of an element discharged during an electrolysis of an electrolyte is directly proportional to the quantity of electricity (Q) passing through it.
Faraday’s first law can be expresses in the following mathematical way
M α Q
But Q = It
M α It
M =EIt or M = EQ
Where E is the constant of proportionality known as Electrochemical equivalent of a substance.
M is measured in gram (g)
I is measured in Amperes (A)
t is measured in seconds (s)
Q is measured in Coulomb(C)
Therefore E is measured in gA-1S-1 or gC-1, for a particular electrolytic process at different time, the law can also be expressed as M1/Q1 = M2/ Q2
OR :- M1 = M2
I1 t1 I2 t2
VERIFICATION OF FARADAY’S FIRST LAW OF ELECTROLYSIS.
The experimental set up is The electrolytic cell above can be used to verify Faraday’s first law.
The copper cathode is weighed at the beginning of the experiment. A steady current of 1A is passed for 10mins and the current supply switched off. The copper cathode is then weighed again to find the gain in mass. This procedure is repeated using different time intervals. A graph of gain in mass against time gives a straight line passing through the origin.
When the experiment is repeated using different currents for a fixed period of time, a graph of gain in mass against the magnitude of the current also gives a straight line passing through the origin.
m(g) m(g)
t(s) I(A )
The two graphs show that the mass of copper deposited is directly proportional the magnitude of the current, i.e M α I
EVALUATION
FARADAY’S SECOND LAW OF ELECTROLYSIS
Faraday’s second law of electrolysis states that when the same quantity of electricity Q is passed through different electrolytes, the relative number of mole of the elements discharged is inversely proportional to the charges on the ions of the elements.
m = Q OR m1/Z1 = m2/Z2
Z/F M M1 M2
The minimum quantity of electricity required to liberate one mole of a univalent ion is 96500C. This quantity of electricity is called the Faraday (F).One Faraday (1F) is one mole of electron which contains Avogadro’s number (6.62 x 1023) of electrons. It follows that two Faraday’s (2F) will be required to discharged 1 mole of ions of a divalent elements or radical such as Pb2+ or S042-
i.e M2+ + 2e- M or A2- A + 2e-
VERIFICATION OF FARADAY’S SECOND LAW OF ELECTROLYSIS
To verify the second law, set up two electrolytic cells.
ii The other with silver electrodes immersed in AgNO3 solution.
The copper and silver cathodes are weighed before the experiment begins
A steady current of 1A is passed into the solution say for 15mins and the current supply switched off. The mass of each cathode is then weighed again to find the gain in mass. The number of moles of silver deposited at the silver cathode will be doubled the number of copper deposited at the copper cathode although the same quantity of electricity was passed through the two electrolytes i.e if Y moles of copper are deposited by 2Y moles of silver are deposited. The charge on the Cu2+ is +2 and that on the Ag+ is +1 hence, the number of moles of copper and silver deposited are inversely proportional to charge on their ions
EXAMPLES
SOLUTION:
:- M = 2g, I = 5A, t = 1hr
M = Z It
2 = Z x 5x 1x 60x 60 = 2/1800
Z = 2.78 x 10-4 g I-1 s-1
2 During electrolysis of a copper solution 5g of copper was deposited in 10min when a current of 1A was passed. Calculate the time required to deposit 20g
SOLUTION:-
M1, = 5g, t1 = 10min = 600sec.
I1 = 1A
M1/ t1 = M2/t2
5/ 600 = 20/t2
t2 = 20 x 600 /5 = 2400sec.
ii I2 = 2A , M2 = 20g, t= ?
M1/I1t1 = M2/ I2 t2
5 = 20
1 x 600 2 x t
t2 = 20 x 1 x 600
2 x 5
t2 = 1200 sec.
EVALUATION
WORKED EXAMPLES
SOLUTION:
Ag++ e- Ag
1 Mole Ag+ is discharged by 1 mole e-
2.5 moles Ag+ are discharged by 2.5 mole e-
2.5 moles of electrons are required
SOLUTION
Cu2+ + 2e- Cu
1 mole of Cu is deposited by 2 moles of e-
3 moles Cu is deposited by 6 moles of e-
For silver:
Ag+ + e- Ag
1 mole of e- deposits 1 moles of Ag
6 moles of e-will deposit 6 moles of Ag
EVALUATION
APPLICATIONS OF ELECTORLYSIS
Electrolysis is applied in the industries in the following areas e.g
EXTRACTION AND PURIFICATION
Very electropositive metalse.gCa, Mg, and Al and electronegative non-metal e.g F and Cl are extracted from their compounds by electrolysis.
The less electropositive metalse.g copper, mercury, silver and gold occur in the uncombined ores mixed with impurities. The pure metals are obtained from the ore by electrolytic purification.
During electrolytic purification of copper ores, the impure copper is made the anode while a pure copper is made the cathode. The electrolyte is usually copper (II)tetraoxosulphate(VI) solution.
At the anode :- The metallic copper atoms in the ore give up two electrons each forming copper II ions and move into the solution.
Cu(s) Cu2+(aq) + 2e-
Electrolytic Purification of Copper
At the Cathode. The copper ions in the solution are discharged at the cathode by receiving electrons to become neutral solid copper.
Cu2+(aq) + 2e- Cu(s)
The net result is that pure copper atoms are transferred from impure ore (anode) to the cathode. The impurities are collected at the base of the cell used.
EVALUATION
ELECTROPLATING
Electroplating is a method of coating the surface of one metal with another metal, usually copper, silver, chromium, nickel, or gold by means of electrolysis.
The main objectives of electroplating are
Article which are usually plated include cutlery, jewellery and iron or steel objects such as the metallic parts of motor cars
Silver – plating of a spoon
In Silver plating of an iron spoon
At the anode: Ag(s) Ag+ + e-
At the cathode: Ag+(aq) + e- Ag(s)
GENERAL EVALUATION/REVISION
READING ASSIGNMENT
New School Chemistry for Senior Secondary Schools by O. Y. Ababio (6th edition) pages 210-212.
WEEKEND ASSIGNMENT
SECTION A: Write the correct option ONLY
1 In silver plating a spoon, the spoon must be made the
2 A current of 20A is passed for 30 minutes during the electrolysis of gold (III) tetraoxosulphate (VI). The mass of gold deposited at the cathode is
3 What quantity of chlorine in moles is liberated when a current of 15A is passed through molten potassium chloride in 1930 seconds?
4 What quantity of silver is deposited when 96500C of electricity is passed through a solution containing silver ions [Ag = 108, F = 96500C]
5 Calculate the volume of oxygen liberated at s.t.p when 96500C of electricity is passed through aqueous solution of H2S04. [IF = 96500C, Volume at s. t. p = 22.4 dm3]
SECTION B
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