FARADAYS LAWS OF ELECTROLYSIS
Week: six Date: 10-14
/06/2019
Period: Duration: 1 HR
20 MIN. Average
age of learners: 16YEARS
Subject: CHEMISTRY Class:
SS TWO
Topic: ELECTROLYSIS
Sub topic: Reference materials:
(1) ESSENTIAL CHEMISTRY, TONALD PUBLISHERS, I. O ODESINA
(2) NEW SCHOOL CHEMISTRY, AFRICAN FIRST PUBLISHERS, OSEI YAW ABABIO
(3) INTERNET
Instructional
materials:
Entry behavior:
The students have been taught terminologies of electrolysis
Behavioural objective: At the end of the lesson the students should be able to:
i. State faradays laws of electrolysis
ii. How many coulombs make1 faraday
iii. Solve problems on electrolysis
iv. State uses of electrolysis
CONTENT
Faradays Laws of Electrolysis
The relationship between the quantity of electric charge passed
through an electrolyte and the amount of the substance deposited at the
electrodes was presented by Faraday in 1834, in the form of laws of electrolysis.
The number of moles of an element that is liberated during electrolysis depends
on 3 factors:
1. The magnitude of the steady current passed.
2. The time of passing the steady current.
3. The charge on the ion of the element.
Faraday’s First Law
This law states that “the mass of a substance deposited or liberated at any electrode
is directly proportional to the amount of charge passed” i.e., w a q (where w is
the mass of the substance deposited or liberated and q is the amount of
charge passed). This proportionality can be made into an equality by, w = zq
where z is the proportionality constant called the electrochemical
equivalent. It is the mass of the substance in grams deposited or liberated by
passing one coulomb of charge.
Faraday’s Second Law
This law states that “the mass of
a substance deposited or liberated at any electrode on passing a certain amount
of charge is directly proportional to its chemical equivalent weight”.
That is w a E where w is the mass of the substance in grams while
E is its chemical equivalent weight in gms per equivalent =
This law can be explained as follows.
Consider three reactions, such as:
Na+ + e– →
Na
Cu2+ + 2e–- → Cu
Al3+ + 3e– → Al
Assume that these three reactions are occurring in three separate
electrolytic cells connected in series.
When x moles of electrons are passed
through the three cells, the mass of Na, Cu and Al deposited are 23x gms, 31.75x gms and 9x gms respectively.
We can see that 23, 31.75 and 9 gm/eq are the chemical equivalent
weights of the three elements.
w = moles of electrons
The charge possessed by 1 mole of electrons
= 1.6
This charge is called as 1 Faraday.
If we pass one Faraday of charge, it means that we are passing one
mole of electron and by passing 1 Faraday of charge 1gm equivalent weight of
the substance will be deposited or liberated.
By combining the first and second law, we get
|
Note: It should be made clear that the cathode is the electrode in which reduction reaction(s) occurs while the anode is the electrode where the oxidation reaction(s) occurs. Do not relate the sign (positive or negative) of the electrode with the nature of the electrode. |
Coulomb is the unit of electric charge. It is the amount of charge
that moves past may given point in a circuit when a current of 1 ampere is
supplied for one second.
Example 1
If a steady current of 5A
is allowed to pass through a solution for 2 hours 15 minutes (a) calculate the
quantity of electricity (b) how many faradays are used.
Solution:
I =5A, t= 2hrs 15min. = [2x60x60] + [15x60] = 8100s
Q = It = 5A x
8100s = 40500C
If 96500C ------------ 1 faraday
40500C
---------- 1/96500 x 40500 = 0.42
Faraday
Example 2
0.5mole of oxygen gas was produced at the anode
during the electrolysis of an acidified water using steady current of 5A (a)
how many quantity of electricity was used (b) how many faradays of electricity
was used (c) how long does it take to achieve this result.
Solution:
( a) 4OH- à 2H2O + O2 + 4e-
From the equation above:
1 mol of O2 gives 4 mol of electrons
0.5 mol of O2 gives 4 x 96500 x 0.5 = 193500C
(b ) 1 mol of O2 gives 4 mol of electrons
0.5 mol
of O2 gives 4 x 1F x 0.5 =
2F
( c) Q = It,
t = Q/I = 193500C/5A
= 38600s = 10hrs 45 min 20s
Example 3
How long has a current of 3
ampere to be applied through a solution of silver nitrate to coat a metal
surface of 80 cm2 with 0.005 cm thick layer? Density
of silver is 10.5 g/cm3.
Solution:
Mass of silver to be deposited
= Volume × density = Area ×thickness × density
Given: Area = 80 cm2
thickness = 0.0005 cm and density = 10.5 g/cm3
Mass of silver to be deposited = 80 × 0.0005 × 10.5 = 0.42 g
Applying to silver E = Z × 96500
Z=108/96500 g
Let the current be passed for r seconds.
We know that
W = Z × I × t
So, 0.42 = 108/96500 x 3 x t
or t = (0.42 × 96500)/(108×3)=125.09 second
Example 4.
What current strength in ampere will be required to liberate 10 g
of chlorine from sodium chloride solution in one hour?
Solution:
Applying E = Z × 96500 (E for chlorine = 35.5)
35.5 = Z × 96500
or Z = 35.5/96500 g
Now, applying the formula
W = Z × I × t
Where W = 10 g, Z= 35.5/96500 t = 60×60 =3600 second
I = 10x96500/35.5x96500 = 7.55 ampere
USES OF ELECTROLYSIS
I.
Extraction of element: Many metals (e.g Na, K,
Mg, Ca, Al, Zn) and non-metals (eg H2, F2, Cl2) are obtained either by
electrolysis of their ores or of their fused compound or their aqueous
solution.
II.
Purification of metal (eg Cu, Hg, Ag, Au)
III.
Electroplating one metal by another
IV.
Preparation of certain important compounds,
such as sodium hydroxide and sodium trioxochlorate (v)
PRESENTATION
Step I: The teacher states the faradays law of electrolysis.
Step II: The teacher leads the students to solve problems on laws
of electrolysis.
Step III: the teacher explains the uses of electrolysis
EVALUATION
The teacher evaluates the lessons by asking the following
questions:
i. State faradays laws of electrolysis
ii. How many coulombs make1 faraday
iii. State uses of electrolysis
ASSIGNMENT
A current of 2.68 ampere is passed for one hour through an aqueous solution of copper sulphate using copper electrodes. Calculate the change in mass of cathode and that of the anode. (At. mass of copper = 63.5).
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