SS2 Chemistry Lesson Note on Faraday Law Of Electrolysis

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TOPIC: ELECTROLYSIS

CONTENT:

1. FARADAY’S LAW OF ELECTROLYSIS.
2. CALCULATIONS BASED ON FARADAY’S LAWS OF ELECTROLYSIS
3. USES OF ELECTROLYSIS AND CORROSION OF METALS: CORROSION TREATED AS A REDOX PROCESS
4. RUSTING OF IRON AND ITS ECONOMIC COSTS, PREVENTION OF RUSTING OF IRON

PERIOD 1: FARADAY’S LAWS OF ELECTROLYSIS

1. Faraday’s first law of electrolysis states that the mass (m) of a substance liberated or deposited at an electrode during electrolysis is directly proportional to the quantity (q) of electricity passing through the electrolyte.

Mathematically;

M α Q ————– (1)

The unit of quantity is coulombs (c). It is the product of current in Amperes and time in seconds

Q = It ——————— (2)

M α It

.: M = k It ——————— (3)

 

Where k = proportionality constant

I = current in ampere (A)

T = time in seconds (S)

It can be shown that the reciprocal of k, which is called charge –to-mass ratio, is Cf/M. It is constant for a particular element

i.e.  =

Re-arranging     =

Where m= mass f element deposited in grams

M = molar mass of the element in g/mol

I = current, in ampere

t = time, in seconds

c = charge on the element (number of Faradays)

F = Faraday (96500Cmol^-1)

The quantity of electricity required for the passage of one mole of electrons is 96500coulombs.It is called one Faraday. i.e. Faraday, F=96500C = 1 mole of electrons 6.02x electrons

2. Faraday’s second law of electrolysis states that when the same quantity of electricity is passed through different electrolytes the relative amounts in moles of the element deposited are inversely proportional to the charge on the ion of respective element.

Mathematically, n α

i.e. nc = constant

Where n is the amount, in moles, of the element liberated, and c is the charge on its ion.

For two different elements: =

CALCULATIONS

Example 1

During the electrolysis of silver trioxonitrate (v) solution, 9650 coulombs of electricity was passed. Calculate (i) the amount of silver deposited in moles (ii) the mass of silver deposited in grains (Faraday = 9650 coulombs, Ag = 108g mol^-1)

Solutions

The equation of reaction is given as

Ag⁺ _(aq) + e^–     →     Ag_(s)

(i) From the above equation, 1 mole of electron is required to deposited 1 mol of Ag but, 1 mol of electron = 1 Faraday 96500C.

i.e. 96500C will deposit 1 mol of Ag

..  1/96500 X 9650/1 = 0.1 mol

. :  Amount of silver deposited = 0.100 mol

• Since 1 mol of silver, (Ag) is 108g

.: 0.1mol of Ag will be 108/1 X 0.1 =10.8g

.:  Mass of silver deposited = 10.8g

Example 2

When a solution of copper (II) salt was electrolyzed 3.20g of copper was deposited. Calculate the quantity of electricity required for electrolysis. (Cu=6; 1 F = 96500C).

Solutions

Equation of the reaction is: Cu²⁺ _(aq)   + 2e^– → Cu_(s)

2F → 64(1 mol)

Amount in mole copper deposited is n =

 

n =      = 0.050 mol

Form the equation, 2 mols of electrons will deposit 1 mol of Cu. But 1 mol of  electrons = Faraday = 96500C .: 2 mol of electrons will require 2 X 96500C to deposit  1 mol of Cu.

1 mol of Cu requires 2 X 96500C

.:  0.05mol of Cu will require 2 X 9500 X 0.05C

Quantity of electricity required = 96500 coulombs.

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