FARADAY’S LAWS OF ELECTROLYSIS

 

Faraday’s First Law of Electrolysis: State that the mass (m) of an element discharged during electrolysis is directly proportional to the quantity of electricity (Q) passing through the electrolyte

Mathematically

M α Q

Q = It

M α It         removing the sign of proportionality we have 

M = ZIt   

Where Z is a constant known as the electrochemical equivalent of the substance.

M = Mass of substance in gram

Q = Quantity of electricity in coulombs

I = Current in ampere

t = Time in seconds

Verification of Faraday's First law of electrolysis

Faraday’s Second Law of Electrolysis: State that when the same quantity of electricity is passed through solutions of different electrolytes, the relative number of moles of the elements discharged at each electrode is inversely proportional to the charges on the ions of each of the element

According to Faraday the minimum quantity of electricity required to liberate one mole of a univalent ion during electrolysis is equal to 1 Faraday and 

1 Faraday = 96500 coulombs 

 

Mass              = Quantity of Electricity

Molar mass       Faraday

 

 

Verification Of Faraday’s Second Law 

Method

 

                               

  

1. Fill two beakers up to ⅔ of their volumes with 1M of copper (II) tetraoxosulphate (VI) solution and 1M solution of silver trioxonitrate (V) solution.

2. Weigh and place two clean plates of copper and silver electrodes in their respective solutions 

3. Connect a battery and complete the circuit as shown above, attach a variable resistor adjusted to maintain a steady current of 0.5A. Allow the current to   pass through the solution for 25 minutes.

4. The cathode is removed, washed with water dried and then reweighed to obtain the masses of copper and silver deposited

5. The ratio of the number of moles of copper and silver deposited is then calculated.

6. The process is repeated to obtain at least three more readings for accuracy.

 

Amounts n (Number of moles) = Mass of element deposited

                                                                  Its relative atomic mass.

Observation: On passing the same quantity of electricity through the solutions, the ratio of the number of moles of copper and silver deposited is 1:2.

This ratio is inversely proportional to the ratio of the charges on the ions, Cu²⁺  and  Ag⁺ , or the number of moles of electrons required to liberate 1 mole each of the ions.

Cu²⁺_(aq)  +  2e^- → Cu(s)             and         Ag⁺_(aq)  + e^-          →         Ag(s)

                    2 moles    63.5g                                   1 mole                     108g

Conclusion:  When the same quantity of electricity is passed through a solution of different electrolytes the relative number of moles of the elements deposited are inversely proportional to the charges on the ions of each of the elements respectively.

 

CALCULATION BASED ON THE FIRST LAW OF ELECTROLYSIS

1. In an electrolysis experiment, the ammeter records a steady current of 1A. The mass of copper deposited in 30mins is 0.66g. Calculate the error in the ammeter reading. [electrochemical equivalent of copper =0.00033gC-1]            

Solution

M = ZIt

M = 0.66g, Z = 0.00033gC-1, t = 30mins = 1800 seconds

I = M/Zt

I = 0.66/0.00033 x 1800

I = 0.66/0.594

I = 1.11A

The error in the ammeter reading is 1.11 – 1 = 0.11A

 

2. Calculated the time in minutes, required to plate a substance of total surface area 300cm², a layer of copper 0.6mm thick, if a constant current of 2A is maintained. Assuming the density of copper is 8.8g/cm3 and one coulomb liberates 0.00033g copper.

Solution

Given that area = 300cm² , thickness = 0.6mm = 0.06cm

Mass = 0.00033g, density = 8.8g/cm3

Density = mass/volume

Mass = density x volume

Mass = 8.8 x 300 x 0.06

Mass = 158.4g

From M = Zit

t = m/ZI

t = 158.4/2 x 0.00033

t = 240000secs

t= 4000mins

 

 

CALCULATION BASED ON THE SECOND LAW OF ELECTROLYSIS

 

6. A current of 4.5A is passed through a solution of gold salt for 1 hour 45 minutes. Calculate

(i) The mass of gold deposited  

(ii) The number of moles of gold deposited

(iii) If the same current is used, find the time taken for 5.5g of gold to be deposited (Au = 197, 1 Faraday = 96500c)

Solution

(i) Au+   +     e-→ Au

    197g       1F       197g

Mass              = Quantity of Electricity
Molar mass       Faraday

M        Q
Mm     F

 

Mass = ?

Molar mass = 197g

Quantity of electricity = I x t =

I = 4.5A

t = 1 hour 45minutes = 105 minutes = 105 x 60 = 6300 seconds

Quantity of electricity = I x t = 4.5 x 6300 = 28,350C

Faraday = 96500F

 

Mass         =    28,350
197                  96500

Mass    =   0.29378
197

Mass of Gold deposited = 57.88g

 

(ii) Number of mole = Mass
                                    Molar mass

 

Number of mole = 57.88
                                197

The number of mole of Gold deposited = 0.30mol

 

(iii) Mass              =    Quantity of Electricity
      Molar mass              Faraday

M        Q
Mm     F

5.5    =  Quantity of Electricity

197                  96500 

0.02792 = Quantity of Electricity

                            96500

Quantity of Electricity = 2694.28C. This is the quantity of electricity (Q) required for 5.5g of Au to be deposited.

Q = It

t = Q/I

t = 2694
        4.5

t = 598.7

The time taken is 9.98 minutes

 

OBJECTIVE QUESTION

Choose the correct option from A – D.

 1. Faraday’s first law of electrolysis states that the mass of a substance deposited is proportional to the

A. time taken
B. voltage applied
C. quantity of electricity passed
D. temperature

2. The quantity of electricity passed during electrolysis is equal to

A. current × resistance
B. voltage × time
C. current × time
D. resistance × time

3. The SI unit of quantity of electricity is

A. ampere
B. volt
C. coulomb
D. ohm

4. One coulomb is equal to

A. 1 A × 1 s
B. 1 V × 1 s
C. 1 Ω × 1 s
D. 1 J × 1 s

5. According to Faraday’s first law, if the current is doubled, the mass deposited will

A. halve
B. remain constant
C. double
D. become zero

6. Faraday’s second law relates mass deposited to

A. current used
B. time of electrolysis
C. quantity of electricity
D. equivalent weight of the substance

7. The equivalent weight of an element is its

A. atomic mass
B. molecular mass
C. atomic number
D. atomic mass ÷ valency

8. According to Faraday’s second law, equal quantities of electricity will deposit masses proportional to their

A. atomic numbers
B. densities
C. melting points
D. equivalent weights

9. Which of the following will deposit the highest mass for the same quantity of electricity?

A. Na
B. Mg
C. Al
D. Ag

10. A current of 2 A flows for 10 seconds. The quantity of electricity passed is

A. 5 C
B. 10 C
C. 20 C
D. 40 C

11. The unit of current is

A. coulomb
B. ampere
C. volt
D. ohm

12. If the time of electrolysis is tripled, the mass deposited will

A. remain the same
B. halve
C. double
D. triple

13. The formula connecting mass, charge and electrochemical equivalent is

A. m = It
B. m = ZIt
C. Q = It
D. Z = mIt

14. In electrolysis, the electrochemical equivalent (Z) is the

A. mass of substance per second
B. mass per ampere
C. mass deposited per coulomb
D. atomic mass

15. Which of the following obeys Faraday’s laws?

A. Diffusion
B. Neutralization
C. Combustion
D. Electrolysis

16. When a metal ion gains electrons during electrolysis, the process is called

A. oxidation
B. ionization
C. reduction
D. dissociation

17. Which electrode gains mass during electrolysis?

A. anode
B. cathode
C. electrolyte
D. voltmeter

18. The amount of substance deposited depends on all except

A. current
B. time
C. equivalent weight
D. temperature

19. The greater the valency of a metal, the

A. greater the mass deposited
B. smaller the equivalent weight
C. higher the electrochemical equivalent
D. lower the mass deposited

20. Faraday’s laws are used to determine

A. atomic structure
B. chemical bonding
C. amount of substances produced during electrolysis
D. rate of reaction

THEORY QUESTION

1. At what time must a current of 5Amp pass through a solution of zinc sulphate to deposited 1g of zinc. Electrochemical equivalent (e.c.e) = 0.0003387

2. In an electrolysis experiment, a cathode of mass 5g is found to weigh 5.01g, after a current of 5A flows for 50 seconds. What is the electrochemical equivalent for the deposited substance?

 

3. The electrochemical equivalent of silver is 0.0012g/c. if 0.36g of silver is to be deposited by electrolysis on a surface by passing a steady current for 5.0 minutes. Calculate the value of the current.

4. Calculate the current that must be passed into a solution of aluminium salt for 1hr.30minutes in order to deposited 1.5g of Aluminium (Al = 27)

5. 0.222g of a divalent metal is deposited when a current of 0.45A is passed through a solution of its salt for 25 minutes. Calculate the relative atomic mass of the metal. (1 Faraday = 96500 coulombs)

 

6. A given quantity of electricity was passed through three electrolytic cells connected in series containing solutions of Silver trioxonitrate (V), Copper (II) tetraoxosulphate (VI) and Sodium Chloride respectively. If 10.5g of Copper are deposited in the second electrolytic cell. Calculate

(a) The mass of Silver deposited in the first cell.

(b) The Volume of Chloride liberated in the third cell at 180C and 760mmHg pressure. (Ag=108, Cu=63.5, 1Faraday=96500C, molar volume of gases at s.t.p =22.4dm³.)              

                

7. Calculated the time in minutes, required to plate a substance of total surface area 300cm2, a layer of copper 0.6mm thick, if a constant current of 2A is maintained. Assuming the density of copper is 8.8g/cm³ and one coulomb liberates 0.00033g copper.