FARADAY’S LAWS OF ELECTROLYSIS

FARADAY’S LAWS OF ELECTROLYSIS

1. What is electrolysis? Explain with an example

Electrolysis:-

The process of decomposition of a chemical compound in a solution when an electric current passes through it is called “Electrolysis”.

Explanation:-

Parts:-
B-Battery, K-plug key, A-Ammeter, Rh-Rheostat

1. Copper sulphate solution is taken in a vessel as an electrolyte.

2. Two copper plates A and C are dipped in the solution and act as anode and cathode respectively.

3. The copper plate A is connected to the positive (+ve) terminal of the battery “B” and copper plate “C” is connected to the negative (-ve) terminal of the battery (B) through an Ammeter “A” and Rheostat (Rh) and a plug-key(K).

4. When K is swithed on the electric current passes through electrolyte.

5. The electrolyte dissociates into respective ions as shown in the following equation.

6. As the copper and sulphate or both divalent, they carry two units of positive and negative charge respectively.

7. Therefore, the cu2+ ions are attracted by the cathode “C” and SO42- ions are attracted by the anode “A”.

8. The SO42- ions react with copper of anode and CuSO4 is formed.

9. This CuSO4 gets dissolved in the solution. Hence the concentration of CuSO4 solution remains same.

10. The positive ions cu2+ ions deposit on the cathode and hence the mass of the cathode increases while the mass of the anode decreases by an equal amount.

2. Voltameter or Electrolytic Cell:-

The vessel which contains an electrolyte and allows electrolysis to take place is called a voltameter or Electrolytic cell.

3. Faraday’s first law of electrolysis:-

The Faraday’s first law of electrolysis states that the mass (m) of ions liberated from an electrolyte is directly proportional to the strength of the current (i) and the time (t) for which the current passes.
mαi and mαt
mαit
or
m=Zit.

4. Verification of Faraday’s first law of electrolysis:-

Parts:-

B-battery, K-plug key, A-ammeter, Rh-Rheostat.

1. Take two copper plates A and C with terminals.

2. Take copper sulphate solution in a glass-trough of suitable dimensions.
3. Dip the copper plates A and C in the solution.
4. Connect the plate A to the positive terminal of a battery B.
5. Determine the weight of the cathode. Let its weight be w1 gm.
6. Connect C to the negative terminal of the battery through a Rheostat (Rh) an Ammeter (A) and a plug-key (K) in series.

7. Close the plug key for a while and adjust the slider-position of in the Rheostat such that the Ammeter shows a Current (i) of about one Ampere.

8. Allow the current to pass through the electrolyte for half-an-hour. Switch off the current and weigh the plate C. Let its weight be w2 gm.

9. Note the mass of the copper deposited m=(w2-w1).

10. Repeat the experiment by increasing the current to 2i and 3i but keeping the time of passage of current (t) as constant.

11. We find that the mass of copper liberated at cathode is 2m and 3m when the current is 2i and 3i respectively. This proves that mαi .

12. Repeat the experiment by increasing the time of passage of current to 2t and 3t but keeping the current constant.

13. Note the mass of the copper liberated at the cathode at each time.

14. We find that the mass of the copper liberated at the cathode at each time is 2m when time is 2t and 3m when time is 3t. This proves that mαt.

Thus, Faraday’s first law of electrolysis is verified.

5. Electro-Chemical equivalent(e.c.e):-

The electro-chemical equivalent (e.c.e) of an element is defined as the mass of its ions liberated at the electrode when one coulumb of electricity is passed through the electrolyte.

6. Describe an experiment to determine (e.c.e) of an element?

1. Take metal plates whose ece is to be determined (say copper)

2. Consider these plates as electrodes anode (A) and cathode (C).

Parts:-
B-battery, K-plug key, A-ammeter, Rh-Rheostat.

3. Clean the copper plate. Dry it and weigh it. Let its mass be w1 gm.
4. Connect the voltameter as shown in the circuit diagram.
5. Switch on the current in the voltameter, record the duration of current flow, time (t) and current (i) by ammeter.

6. Switch off the current. Remove the cathode plate, dry and weight it. Let its mass be w2 gm.

7. w2-w1=m is the mass of the metal deposited on the cathode.
8. Record the observations in the following table

S.No Current (i) Amperes Time of passage of current through the cell   t   seconds Mass of the metal deposited on cathode

 w2-w1=m grams

Z=
1

2

3

4

5

       

The value of Z is same for every case. This Z value is called “e.c.e”.

7. Chemical equivalent or Equivalent weight (E):-

The ratio of atomic weight (A) of an element to its valency (V) is defined as chemical equivalent or equivalent weight.

8. Faraday’s Second law of electrolysis:-

Faraday’s second law of electrolysis states that when the same quantity of electricity passes trough different electrolytes, the masses of ions liberated at the respective electrodes are proportional to their chemical equivalents.
m1:m2:m3=z1:z2:z3

9. Verification of Faraday’s second law of electrolysis:

Parts:-

B-battery, K-plug key, A-ammeter, Rh-Rheostat.

1. Arrange three voltameters one containing copper sulphate solution (CuSO4) and the second containing sliver nitrate solution (AgNO3 )and the third Zinc sulphate solution(ZnSO4)

2. Connect the three voltameters as shown in the figure.
3. In all the three voltameters let the electrodes A and C be made of copper only.
4. Determine the weights of individual cathode in each of the voltameters
5. Adjust the Rheostat such that a current of about 1 ampere is read in the Ammeter.
6. As the voltameters are arranged in series same current passes through all of them.
7. Allow the current to pass for half an hour time.
8. At the end, determine the final weights of the individual cathodes.
9. The differences between the corresponding final and initial weights give the masses m1,m2 and m3 of copper, silver and zinc deposited on the respective cathodes.

10. Using the physical table find E1, E2 and E3 the chemical equivalent of copper, Silver and Zinc.

Thus, Faraday’s second law of electrolysis is verified.

10. Applications of Electrolysis:-

1. Metallurgy:- Refining and Extraction of metals:-

Certain metals like Copper, Zinc, tin, lead, gold, chromium, nickel etc are purified and extracted by the method of electrolysis.
In this method, Anode is made of an impure metal or ore of the metal and cathode is made of same but pure metal. The electrolyte should be a solution or a melt containing salt of the same metal as electrodes. On passing the current through the electrolyte, pure metal gets deposited on the cathode.

2. Electroplating:-

Electroplating is a process of coating a thin film of costlier or less corrodable metal on a base metal by the method of electrolysis.
For example, Silver film can be deposited on copper base.

3. Electrotyping:-

Electrotyping is a method of obtaining exact copy of an engraved block containing letters or figures by the method of electrolysis.

II. Problems:-

1. Find the time required to deposit electrolytically one gram of gold by passing a current of 2 Amperes. The e.c.e of gold is 0.0006812 gram / coulomb.

Given,
Mass=m=1 gram
Current=i=2A
ece=z=0.0006812 gm / coulomb.
We know that,

t=734 seconds.
Time required to deposit=734 seconds.

2. In an electrolysis experiment a current of 1.25 A is passed for 1 hour to obtain a deposit of 1.485 gm of copper. Find the ece of copper.

Given,
Current=i=1.25 A
Time=t=1hour=3600 seconds.
Mass=m=1.485 gm
ece=z=?
We know that,

z=0.00033 gm/coulomb
e.c.e of copper =0.00033 gm/coulomb

3. Find the time required to deposit 5 gm of gold by passing a current of 2A in an electroplating experiment? The e.c.e of gold is 0.0006812 gm/coulomb?

Given,
mass=m=5gm.
Current=i=2A
e.c.e=Z=0.0006812 gm/coulomb.
Time required to deposit=t=?
We know that,

t=3670 second
t=1hr1minute10seconds.
Time required to deposit=1hr1minute10sec.

4. A current of 1A flowing for 25min through a silver voltameter deposits 1.44gm of silver. Calculate the e.c.e of silver?

Given,
Current=i=1A
Time=t=25 minutes=1500seconds.
Mass=m=1.44 gm.
e.c.e=Z=?
We know that,

Z=0.00096 gm/coulomb.
The e.c.e of silver=0.00096 gm/coulomb.

MAGNETIC EFFECTS OF ELECTRIC CURRENT

1. Maxwell’s cork-Screw rule:-

When the head of a cork-screw is rotated such that the tip of the screw advances in the direction of the current then the direction of rotation of head represents the direction of the magnetic field around the conductor.

2. Ampere’s right-Hand rule:-

When a current carrying conductor is held in right hand such that the thumb points along the direction of the current, then the remaining fingers indicate the direction of the magnetic field around the conductor.

Note:-

1. The magnetic induction B at a point due to a straight current carrying conductor is inversely proportional to the distance (r).

2. The magnetic induction B at a point near a straight current carrying conductor is directly proportional to the strength of the current (i).
Bαi

3. When a current carrying conductor is placed in a strong external magnetic field, force acting on the conductor is given by
F=ilB
Where,
B=Magnetic field strength.
i=Current in the conductor.
l=Length of the conductor.

3. Fleming’s Left Hand Rule:-

Strech the thumb, forefinger and central finger of left hand perpendicular to each other, such that if the fore finger is in the direction of magnetic field and central finger is in the direction of current then the thumb represents the direction of force on the conductor.

4. Draw the diagram showing Maxwell’s Cork-Screw rule.

5. Draw the diagram explaining Ampere’s Right Hand Rule?

6. Draw the diagram showing Fleming’s Left Hand Rule?

PRINCIPLE OF WORKING OF AN ELECTRIC MOTOR

1. What is an electric motor? Explain its principle?

Electric motor:-

Electric motor is a device which converts electrical energy into mechanical energy.

Principle of working of an electric motor:-

When a current carrying rectangular coil is placed in a uniform magnetic field, it is acted upon by a torque which rotates the coil continuously.

2. Explain the construction and working of an electric motor with a neat sketch showing essential parts.

Electric motor
Parts:-
ABCD-rectangular coil, NS-permanent horse-shoe magnet, B-battery,

K-plug key, B1B2-carbon brushes, C1C2-half rings of the commutator

Sh-shaft,

1. ABCD is a rectangular coil of insulated copper wire wound on a soft-iron core. The coil and iron-core together constitute the armature.

2. NS is a permanent horse-shoe magnet. The coil ABCD is mounted on a shaft (rod), symmetrically between the cylindrical concave poles of the magnet. The position of the shaft is indicated by dotted line (Sh).

3. C1 and C2 are two metallic half-rings and form the commutator. C1 and C2 in the commutator are insulated form each other and are mounted on the shaft of the motor. The two ends of coil ABCD are soldered to C1 and C2 respectively. The commutator rotates with the shaft.

4. The wires from battery ‘B’ are connected to two conducting carbon brushes B1 and B2 which are always in contact with either of the half-rings C1 and C2.

5. The battery ‘B’ the key ‘K’, the brushes, the commutator and the coil form an electric circuit.

Working of an Electric Motor:-

6. The working of coil is explained by Fleming’s Left Hand rule.

7. When the coil is in horizontal position the side AB of the coil experiences an upward force and CD experiences a downward force when current passes through it.

8. These two equal and opposite forces constitute a couple and make the coil to rotate in anti-clockwise direction till it reaches vertical position.

9. In this vertical position current is cut off because the brushes are not in contact with split rings.

10. But still the coil rotates due to the momentum carried to vertical position..

11. As the coil makes another quarter rotation again it becomes horizontal with the split rings interchanging the contact of B1 and B2.

12. Now the side AB of the coil is acted up by upward force and CD by downward force.

13. The couple constituted by these two forces continue to rotate the coil in anti-clock wise direction.

3. What are the factors that influence the speed of an electric motor?

The speed (RPM) of an electric motor depends on

1. Number of turns (n) in the coil (armature).

2. Area (A) of the coil.
3. The magnitude of current (i) and
4. The strength of the magnetic field (B) in which it is placed.

4. What is the essential difference between an AC and DC motor?

1. In AC motor there is no need of a commutator.
2. In a DC motor, commutator is required to reverse the current direction.