Question

Faraday's first laws of electrolysis ​

Answer 1

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• In year 1937, Faraday published some of his numerical observations related to electrotherapy.

• They are called Faraday's laws of electrolysis.

• There are two rules under this.

• These rules are presented in different ways in textwriters and scientific literature, but that Shubha prevalent form is something like this.

Answer 2

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Faradays laws of Electrolysis

The relationship between the quantity of electric charge passed through an electrolyte and the amount of the substance deposited at the electrodes was presented by Faraday in 1834, in the form of laws of electrolysis.

FIRST LAW :

“the mass of a substance deposited or liberated at any electrode is directly proportional to the amount of charge passed”

This proportionality can be made into an equality by, w = zq

where z is the proportionality constant called the electrochemical equivalent. It is the mass of the substance in grams deposited or liberated by passing one coulomb of charge.

Some numericals based on Faradays laws of Electrolysis

→Find the charge in coulomb on 1 g-ion of N3-.

Solution:

Charge on one ion of N3-

= 3 × 1.6 × 10-19 coulomb

Thus, charge on one g-ion of N3-

= 3 × 1.6 10-19 × 6.02 × 1023

= 2.89 × 105 coulomb

→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

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

Q = 3 × F = 3 × 96500 = 289500 coulomb

(b) The reduction is

Mn4-+ 8H+ 5e- → Mn2+ + 4H2O

1 mole 5 mole

Q = 5 × F = 5 × 96500 = 48500 coulomb

→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-

Q = 2 × F = 2 × 96500 =193000 coulomb

(b) The oxidation reaction is

FeO + 1/2 H2O → 1/2 Fe2O3 + H+ + e-

Q = F = 96500 coulomb

→faraday electric charge is passed through three electrolytic cells in series, first containing AgNO3, second CuSO4 and third FeCl3 solution. How many gram of rach metal will be deposited assuming only cathodic reaction in each cell?

Solution:

The cathodic reactions in the cells are respectively.

Ag+ + e- → Ag

Cu2+ + 2e- → >Cu

and Fe3+ + 3e- → Fe

Hence, Ag deposited = 108 × 0.4 = 43.2 g

Cu deposited = 63.5/2×0.4=12.7 g

and Fe deposited = 56/3 ×0.4=7.47 g

→An electric current of 100 ampere is passed through a molten liquid of sodium chloride for 5 hours. Calculate the volume of chlorine gas liberated at the electrode at NTP.

Solution:

The reaction taking place at anode is

2Cl- → Cl2 + 2e-

Q = I × t = 100 × 5 × 600 coulomb

The amount of chlorine liberated by passing 100 × 5 × 60 × 60 coulomb of electric charge. =1/(2×96500)×100×5×60×60=9.3264 mole

Volume of Cl2 liberated at NTP = 9.3264 × 22.4 = 208.91 L