TERM׃ 3RD TERM
WEEK SEVEN
Class: Senior Secondary School 2
Age: 16 years
Duration: 40 minutes of 5 periods each
Date:
Subject: Chemistry
Topic:- Electrolysis 2
SPECIFIC OBJECTIVES: At the end of the lesson, pupils should be able to
INSTRUCTIONAL TECHNIQUES: Identification, explanation, questions and answers,
demonstration, videos from source
INSTRUCTIONAL MATERIALS: Videos, loud speaker, textbook, pictures
INSTRUCTIONAL PROCEDURES
PERIOD 1-2
PRESENTATION |
TEACHER’S ACTIVITY |
STUDENT’S ACTIVITY |
STEP 1 INTRODUCTION |
The teacher reviews the laws of Electrolysis to the students.
|
Students pay attention |
STEP 2 EXPLANATION |
Teacher carry out calculations involving laws of Electrolysis and also aid students to do same.
|
Students pay attention and participate |
STEP 3 DEMONSTRATIO N |
Teacher discusses the uses of Electrolysis .
|
Students pay attention and participate |
STEP 4 NOTE TAKING |
The teacher writes a summarized note on the board |
The students copy the note in their books |
NOTE
ELECTROLYSIS 2
Faradays laws of electricity
Faraday's First Law of Electromagnetic Induction states that a change in magnetic field within a closed loop induces an electromotive force (EMF) in the loop. This induced EMF creates an electric current if the circuit is closed.
From the definition, we get
m is proportional to Q
Q = I x t
The first law of electrolysis can also be stated as follows; the mass of the element discharged during an electrolysis is directly proportional to,
Hence, m is directly proportional to I x t
m = EIt
Where E is a constant.
Faraday's Second Law quantifies the induced EMF, stating that it is directly proportional to the rate of change of magnetic flux through the loop. Mathematically, it can be expressed as EMF = -dΦ/dt, where EMF is the induced electromotive force, dΦ/dt is the rate of change of magnetic flux, and the negative sign indicates the direction of the induced current according to Lenz's law.
Calculations involving Electrolysis
Example 1; Calculate the mass of silver deposited when a current of 2.6A is passed through a solution of a silver salt for 70minutes. (Ag = 108, 1faraday = 96 500C )
Solution.
Quantity of electricity used = It = (2.6 x 70 x 60) C
Ag+(aq) + e- forward arrow Ag(s)
108g. 1F. 108g
96500 C liberate108g of Ag1
(2.6 x 70 x 60) C will liberate 2.6 x 70 x 60 x 108
96500 = 12.22g
Therefore the mass of silver deposited is 12.22g.
Example 2. Calculate the volume of oxygen evolved at 285 K and 0.9 x 105Nm-2 when a current of 2.5 A is passed through acidified water for 1.5 min. (Molar volume of a gas is 22.4dm³, standard pressure = 1.01 x 10⁵Nm-²,
1faraday = 96500C)
Solution
Quantity of current used = I t = ( 2.5 x 1.5 x 60) C
4OH- (aq) forward arrow 2H2O(l) + O2(g) + 4e-
22.4dm³. 4F
4 x 96500C of electricity liberate 22.4 dm³ oxygen at stp.
Thus 2.5 x 1.5 x 60 x 22.4. dm³
4 x 96500
= 2.5 x 1.5 x 60 x 22400 cm³
4 x 96500
= 13.06cm³ at stp.
Now convert the volume at STP to it's volume at 285K and 0.91 x 10⁵ Nm-²
p1V1 = p2V2
T1. T2
Where p1 = STP = 1.01 x 10⁵Nm-²
V1 = 13.06cm³
T1 = 273K
p2. = 0.91 x 10⁵ Nm-²
T2 = 285K
V2 = p1V1T2
T1
= 1.01 x 10⁵ x13.06 x 285 cm³
0.91 x 10⁵ x 273
= 15.13cm³.
The volume of oxygen evolved at 0.91 x 10⁵Nm-² and 285K is 15.13cm³.
Uses of Electrolysis
Electrolysis, the process of using an electric current to drive a non-spontaneous chemical reaction, has various practical applications:
EVALUATION: 1. State faraday’s first law of electrolysis.
CLASSWORK: As in evaluation
CONCLUSION: The teacher commends the students positively