Classwork Series and Exercises {Chemistry- SS2}: Faraday’s Laws of Electrolysis

Introduction 

In chemistry, quantitative laws used to express magnitudes of electrolytic effects, first described by the English scientist Michael Faraday in 1833. The quantities of substances produced or consumed by the electrolysis process is dependent upon the following:

• electric current measured in amperes or amps
• time measured in seconds
• the number of electrons required to produce or consume 1 mole of the substance

Faraday’s put forward his two laws of electrolysis, these are:

Faraday’s First Law of Electrolysis states that the mass of an element which is deposited on an electrode during electrolysis is directly proportional to the quantity of electricity which passes through the electrolyte.

Explanation:
If W is the amount of substance which liberates or deposited at the electrode on passing the electricity through the electrolyte and the quantity of electricity is Q, then

or W = ZQ

Z is the electrochemical constant for a given substance.

As Q = I x T

We can write the statement of the first law of electrolysis mathematically as under:
W= ZIt

If 1 ampere electric current passes through the electrolyte for 1 second then W=Z It means that on passing the current of 1 ampere for 1 second the weight of the substance deposited is equal to the electrochemical constant. For doing the calculations of electrochemical problems, we must know the units too.

unit of charge (Q) = Coulomb (C)
unit of mass (m) = Kilogram (kg)
unit of current (A) = ampere (A)
unit of electrochemical equivalent (Z) = kg/C

Note:
Faraday’s first law of electrolysis is written as:

W = ZIt

W= is actually mass and not weight, as mass is commonly called weight.

The Faraday

Electricity is a flow of electrons. For calculation purposes, we need to know how to relate the number of moles of electrons which flow to the measured quantity of electricity.

The charge that each electron carries is 1.60 x 10^-19 coulombs.

1 mole of electrons contains the Avogadro constant, L, electrons – that is 6.02 x 10²³ electrons.

That means the 1 mole of electrons must carry

6.02 x 10²³ x 1.60 x 10^-19 coulombs

= 96320 coulombs

This value is known as the Faraday constant.

The numbers we are using here are rounded off. For exam purposes, the value of the Faraday constant is given as 9.65 x 10⁴ C mol^-1.

That is 9.65 x 10⁴ (or 96500) coulombs per mole. So 96500 coulombs is called 1 faraday.

Example 1: How much electric charge is required to oxidise (a) 1 mole of H2O to O2

and (b)1 mole of FeO to Fe2O3?

Solution:

(a) The oxidation reaction is

H₂O –> 1/2 O₂ + 2H⁺ + 2e^–

1 mole ——-> 2 moles

Q = 2 × F

= 2 × 96500=193000 coulomb

(b) The oxidation reaction is

FeO + 1/2 H₂O –> Fe₂O₃ + H⁺+ e^–

Q = F = 96500 coulomb

Example 2: How much charge is required to reduce (a) 1 mole of Al³⁺ to Al and

(b)1 mole of to Mn²⁺?

Solution:

(a) The reduction reaction is

Al³⁺ + 3e^– —–> Al

1 mole ——-> 3 mole

Thus, 3 moles of electrons are needed to reduce 1 mole of Al³⁺.

Q = 3 × F

= 3 × 96500 = 289500 coulomb

(b) The reduction is

MnO₄^– + 8H⁺ + 5e^– –> Mn²⁺ + 4H₂O

1 mole —–> 5 mole

Q = 5 × F

= 5 × 96500 = 48500 coulomb

Example 3: Calculate the quantity of electricity, Q, obtained when a current of 25 amps runs for 1 minute.

Solution:

Q = ?, I = 25A, t = 1 minute = 60 seconds,

Q = I x t

 Q = 25 x 60 = 1,500 Coulombs

Example 4: Calculate the current needed to provide 30,000 coulombs of electricity in 5 minutes.

Solution:

Q = 30,000C, I = ?,  t = 5 minutes = 5 x 60 = 300 seconds

I = Q ÷ t

I = 30,000 ÷ 300 = 100 amps

Example 5: Calculate the time required to produce 12,000 C of electricity using a current of 10 amps.

Solution:

Q = 12,000C, I = 10A, t = ?

t = Q ÷ I

t = 12,000 ÷ 10 = 1,200 seconds = 1,200 ÷ 60 = 20 minutes

Example 6: Calculate the quantity of electricity obtained from 2 moles of electrons

Solution:

 Q = n x F

Q = ?, n = 2 mol, F = 96,500 C mol^-1

Q = 2 x 96,500 = 193,000C

Example 7: Calculate the moles of electrons obtained from 250 C of electricity

n(e) = ?, Q = 250C, F = 96,500 C mol^-1,

n(e) = Q ÷ F

 n(e) = 250 ÷ 96,500 = 2.59 x 10^-3 mol

Faraday’s Second Law of Electrolysis states that when the same quantity of electricity is passed through different electrolytes, the masses of the elements liberated or deposited are in proportion to the chemical equivalents of these elements. Faraday’s laws are very useful for the determination of electrochemical equivalents of different substances.

Chemical Equivalent:

The chemical equivalent of an element is numerically equal to its relative atomic mass in grams divided by its the valency of the ion.
Faraday’s second law of electrolysis can also be stated as under:
“The mass of different substances liberated or deposited by the same quantity of electricity is proportional to the atomic masses divided by the valencies of their ions.”

Explanation:

Take three solutions of electrolytes: AgNO3, CuSO4 and Al(NO3)3 in a series, pass some quantity of electricity through them for the same time. Now Ag, Cu and Al metals collect at the cathode. Their masses are directly proportional to their equivalent masses.

According to Faraday, if 96,500 Coulombs (or 1 Faraday) is passed through these electrolytes,

we get 

which are the equivalent masses of Ag, Cu and Al respectively.

EXERCISES

Lets see how much you’ve learnt, attach the following answers to the comment below

1. One faraday is equal to __________ (a) 9650 coulombs (b) 96500 coulombs (c) one mole of electrons (d) half a mole of electrons
2. What quantity of electricity is consumed when 10 amperes was consumed in 1 hour during electrolysis? (a) 36 kilo coulombs (b) 3600 coulombs (c) 7200 coulombs (d) 72 kilo coulombs
3. Which of the following solutions will conduct the least amount of electricity? (a) 2.00M aqueous solution of NaOH (b) 0.01M aqueous solution of NaOH (c) 0.01M aqueous solution of hexanoic acid (d) 0.01M aqueous solution of sugar
4. If 1 Faraday of electricity is passed through 1M CuSO4 solution for I minute, which of the following statements is correct? (a) the pH solution at the cathode decreases (b) the pH of the solution at the anode decreases (c) 1 mole of Cu will be deposited at the cathode (d) 60 moles of Cu will be liberated at the anode
5. The time required to produce 12,000 C of electricity using a current of 10 amps is? (a) 200 minutes (b) 2 minutes (c) 20 minutes (d) 0.2 minutes

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