FARADAY'S LAWS OF ELECTROLYSIS [ELECTROCHEMISTRY] CLASS-12

ELECTROCHEMICAL EQUIVALENT FARADAY'S CONSTANT NUMERICALS 

 

FARADAY'S LAWS OF ELECTROLYSIS

FIRST LAW:-

This law states that
Mass of the substance deposited on the electrode is directly proportional to the quantity of electricity passed .

W ∝ Q ⟹ W = ZQ ⟹ W = ZIt
Where Z is called electrochemical equivalent of the substance deposited and Q = It, I is electric current in ampere and t is time in second.

Electrochemical equivalent(Z)

 As W = ZIt , If I = 1 A and t = 1s then we have , W = Z . Thus, Electrochemical equivalent of a substance is its mass deposited when a current of 1A is passed for 1s through the electrolyte.

SECOND LAW :-

This law states that
On passing same quantity of electricity through solutions of different electrolytes, the masses of the substances deposited at the electrodes are directly proportional to their equivalent weights.

For example, for  CuSO₄ solution and AgNO₃ solution connected in series we have,

  W1/W2 =  E1/E2,
    Where  W1 = Mass of Cu deposited
               W2 =  Mass of Ag deposited
               E1 = Equivalent weight of Cu
               E2 = Equivalent weight of Ag

Equivalent Weight of any element = Atomic weight/ Valency

Faraday's Constant(F):-

The charge carried by one mole of electrons is called Faraday's constant denoted by F.

  1F = Charge of 1 mole of electrons
     = 1 mole of electrons x charge of 1 electron
     = Avogadro's number  x  [ 1.6021 x 10-19 C ]
     = [ 6.022 x 1023 electron mol-1] x [ 1.6021 x 10-19 C per electron ]
     = 96487 C mol-1
     = 96500  C mol-1(approx.)

Quantitative aspects of electrolysis:-

(i)For the calculation of amount liberated or deposited by a certain quantity of the electricity, write balanced electrode reaction.
Number of electrons gives the number of Faradays required for the no. of moles of that substance in the balanced equation.
1 F = 96500 C charge
     2 F = 2 × 96500 C
     3F = 3 × 96500 C
  ∴ n F = n × 96500 C
(ii)If n electrons are involved in the electrode reaction (redox reaction ), the passage of n faradays (i.e. n × 96500 C) of electricity will liberate one mole of the substance.
(a) For the reduction half reaction (at cathode),
  Mⁿ⁺ + ne^- → M
  n faraday deposit 1 mole of M.
  For example:
 Zn²⁺ + 2e^- → Zn,(n=2), 2 faraday deposit 1 mole of Zn i.e. 65.38g     of  Zn.
 Cu²⁺ + 2e^- → Cu,(n=2), 2 faraday deposit 1 mole of Cu i.e. 63.55g of   Cu.
 Al³⁺ + 3e^- → Al,(n=3), 3 faraday deposit 1 mole of Al i.e. 26.98g of Al.
(b) For the oxidation half reaction (at anode),
X^n- → X + ne^-
n faraday liberate 1 mole of X.
(iii) Q = I×t
(iv) 1 F (i.e.,96500 C ) always liberates 1 g equivalent (eq. weight in gram ) of the substance.
Therefore, Electrochemical equivalent,
Z = Weight deposited by 1 C = Eq. wt. of the substance / 96500
Equivalent weight = (W/Q) × 96500
[ If Q coulomb electricity deposits W gram of the substance ]
(v) For the flow of same quantity of electricity through different solutions connected in series, the weights deposited are in the ratio of their equivalent weights. i.e.,
W₁:W₂: W₃ = E₁:E₂:E₃
(vi) Combination of Faraday's first and second laws gives the mathematical relation as-
W = ZQ = (E/F) × Q =(Q/F) × E = (Q/F) × (M/Z) = (It/F) ×M/Z
Where,
Z = Electrochemical equivalent
Q = Quantity of electricity passed
E = Equivalent weight of he metal
F = 1 Faraday
M = Atomic mass of the metal
z = Valency of the metal
I = Current passed
t = Time for which current is passed

QUESTIONS:-

1. How much charge is required for the following reductions ?
(i) 1 mol of Al³⁺ to Al
(ii) 1 mol of cu²⁺ to Cu.
(iii) 1 mol of MnO₄^- to Mn²⁺
2. A solution of CuSO₄ is electrolysed for 10 minutes with a current of 1.5 amperes. What is the mass of copper deposited at the cathode ? ( Molar mass of Cu = 63.5 g mol^-1 )
3. A current of 10 A is passed through molten AlCl₃ for 96.5 seconds. Calculate the mass of Al deposited.
4. How many Faradays are required to produce
(i) 20.0 g of Calcium from molten CaCl₂
(ii) 40.0 g of aluminium from molten Al₂ O₃ ?
5. Two electrolytic cells containing silver nitrate solution and copper sulphate solution are connected in series. A steady current of 2.5 A was passed through them till 1.078 g of Ag were deposited. How long did the current flow? What weight of copper be deposited?
6. When a current of 0.75A is passed through CuSO₄ solution for 25 min, 0.369 g of copper is deposited at the cathod. Calculate the atomic mass of copper.
7. A current of 1.50 A was passed through an electrolytic cell containing AgNO₃ solution with inert electrodes. The mass of the silver deposited at cathode was 1.50 g. How long did the current flow ? ( Atomic mass of Ag = 108 u)