The branch of chemistry that deals with the interrelationship between electrical energy and chemical transformation is called electrochemistry.
Conductors and non conductors
The substance which allows the passing of electricity through them is called conductors. eg. metals, alloys, electrolytes, etc. The substance that does not flow the passing of electricity through it is called a non-conductor or insulator. eg. plastic, rubber, etc. Conductors are further categorized into two types:
- Metallic conductor:
Conductance is due to the flow of electron. This takes place fast. Metals, alloys are examples. Transportation of matter does not take place. Conducting power decreases on increasing temperature.
- Electrolytic conductor:
Conductance is due to the flow of ions. This takes place slowly. The aqueous solution of acid, base and salts are examples. Transportation of matter takes place. Conducting power increases on increasing temperature.
Electrolyte
The chemical substance which produces ions in solution or in the molten state is called an electrolyte. They are good conductor of electricity as they dissociate into ions. eg HCl, NaCl, H2SO4, CuSO4, etc.
The electrolyte which undergoes complete ionization in an aqueous solution is called a strong electrolyte. eg. HCl, H2SO4, HNO3, NaOH, KOH, NaCl, KCl, CH3COONa, NH4Cl, etc.
An electrolyte that is partially ionized in an aqueous solution and partly remains in undissociated molecules is called a weak electrolyte. eg. HCOOH, CH3COOH, HCN, H2CO3, H3PO4, NH4OH, Ca(OH)2, etc.
Electrolysis
The process of decomposition of an electrolyte by passing electricity through its aqueous or molten state is called electrolysis.
Qualitative aspects of electrolysis
This process is carried out in an apparatus called an electrolytic cell. In the cell, two metallic rods or plates are dipped into the electrolyte. These metallic rods are connected to the direct current(DC) source like a battery. These metallic rods are called electrodes. The electrode which is connected to the positive terminal to the battery is called an anode while another electrode that is connected to the negative terminal of the battery is called the cathode. A simple diagram for the electrolytic process is given below:

On passing electricity, cations migrate towards the cathode where they get reduced by gaining electron from the cathode. Similarly, anion migrates towards anode where they get oxidised by losing electrons to the anode.
The final product formed during electrolysis depends upon the following factors:
- Nature of electrolyte (strong or weak)
- State of electrolyte (aqueous or molten)
- Concentration of electrolytic solution
- Types of the electrode (attackable or non-attackable)
- Amount of electricity passed
Examples:
- Electrolysis of molten NaCl using iron cathode and graphite anode
NaCl ⇌ Na+ + Cl–
At anode (oxidation): 2Cl– → Cl2 + 2e–
At cathode (reduction): Na+ + e– → Na
Cell reaction: 2NaCl → 2Na + Cl2
- Electrolysis of aqueous NaCl in presence of platinum electrode
NaCl ⇌ Na+ + Cl–
H2O ⇌ H+ + OH–
At anode (oxidation): 2Cl– → Cl2 + 2e–
At cathode (reduction): 2H+ + 2e– → H2
Cell reaction: NaCl + H2O → NaOH + H2 + Cl2
- Electrolysis of aqueous CuSO4 using attackable (Cu) electrode
CuSO4 ⇌ Cu++ + SO4– –
H2O ⇌ H+ + OH–
At anode (oxidation): Cu → Cu++ + 2e–
At cathode (reduction): Cu++ + 2e– → Cu
Quantitative aspect of electrolysis
The quantitative relationship between the amount of electricity passed and the mass of ions discharged was first investigated by Michael Faraday in the form of following two laws known as Faraday’s law of electrolysis:
1. Faraday’s first law of electrolysis
It states that “The mass of substance liberated or deposited on electrodes during electrolysis is directly proportional to the quantity of charge passed through the solution”.
m α Q
or, m = ZQ
m = ZIt (I = Q/t or Q = It)
where, m = mass of substance (gm)
I = current (Ampere)
t = Time (sec.)
Q = charge (columb)
Z = proportionality constant known as electrochemical equivalent (ECE)
Electrochemical equivalent (ECE)
From Faraday’s first law of electrolysis,
m = ZIt
If I = 1 amp and t = 1 sec, then m = Z.
Hence, ECE is defined as the mass of substance in gram deposited or liberated in electrode during electrolysis on passing 1-ampere current in unit time.
or
We have, m = ZQ
If Q = 1 Columb, m = Z
Hence, ECE is also defined as the mass of substance in gram deposited or liberated in electrode during electrolysis on passing 1 coulomb of charge.
Faraday’s second law of electrolysis
It states that “When an equal amount of electricity is passed through a different electrolyte, the mass of different substances liberated are deposited on the respective electrode during electrolysis are directly proportional to their chemical equivalent “.
Mass\ of\ substance\ (m)\ \alpha\ chemical\ equivalent\ (E)\\ or,\ m\ \alpha\ E\\ or,\ m/E = K \\ \frac{mass\ of\ substance}{chemical\ equivalent} = constant \\ \frac{m_{1}}{E_{1}} = \frac{m_{2}}{E_{2}} = \frac{m_{3}}{E_{3}}
Experimental verification of Faraday’s second law of electrolysis:

Let us suppose three electrolytic cells containing acidulated water, an aqueous solution of CuSO4 and an aqueous solution of AgNO3. These three electrolytic cells are connected in series with an electric source as shown in the figure. When the same amount of electricity is passed in three electrolytic cells, after some time we found that:
\frac{mass\ of\ H_{2}\ liberated}{Eq.\ wt.\ of\ H_{2}} = \frac{Mass\ of\ Cu\ liberated}{Eq.\ wt.\ of\ Cu}\\ \frac{m_{H_{2}}}{E_{H_{2}}} = \frac{m_{Cu}}{E_{Cu}}\ ---(i)
Again,
\frac{mass\ of\ H_{2}\ liberated}{Eq.\ wt.\ of\ H_{2}} = \frac{Mass\ of\ Ag\ liberated}{Eq.\ wt.\ of\ Ag}\\ \frac{m_{H_{2}}}{E_{H_{2}}} = \frac{m_{Ag}}{E_{Ag}}\ ---(ii)\\ From\ (i)\ and\ (ii),\ we\ get \\ \frac{m_{H_{2}}}{E_{H_{2}}} = \frac{m_{Cu}}{E_{Cu}} = \frac{m_{Ag}}{E_{Ag}}\\ so,\ we\ can\ write, \\ \frac{Mass\ of\ substance}{Chemical\ equivalent} = constant
Thus, Faraday’s second law of electrolysis is verified.
Faraday
Faraday is defined as the quantity of charge carried by one mole of electron.
We have,
1 mole of electron = 6.023 x 1023 no. of electrons
Again, Charge of 1 electron = 1.602 x 10-19 Coulumb
Charge carried by one mole of electron = 1 mole of electron charge of 1 electron
= 6.023 x 1023 x 1.602 x 10-19
So, 1 Faraday = 96500 Columb
Alternatively, one faraday is defined as the quantity of charge required to deposit or liberate one gram equivalent of any substance on the respective electrode during electrolysis.
Relationship between Electrochemical equivalent and Chemical equivalent
Let mA and mB be the masses of two elements A and B which are discharged at corresponding electrodes on passing the same amount of electricity during electrolysis. From Faraday’s first law of electrolysis:
mA = ZAIt ……(i)
mB = ZBIt ……(ii)
where ZA and ZB are the ECE of A and B resp.
Dividing (i) by (ii),
\frac{m_{A}}{m_{B}} = \frac{Z_{A}}{Z_{B}}
From Faraday’s second law of electrolysis,
\frac{m_{A}}{m_{B}} = \frac{E_{A}}{E_{B}}
where EA and EB are Chemical equivalents of A and B resp.
From (iii) and (iv),
\frac{Z_{A}}{Z_{B}} = \frac{E_{A}}{E_{B}}
This shows that electrochemical equivalent (ECE) is directly proportional to chemical equivalent (E).
Relationship between Z, E and F
As we know that 1 Faraday charge (96500C) discharge 1 gm equivalent of substance at respective electron.
Therefore,
96500 C of charge discharge 1 gm. eq. of any substance
Let 1 gm eq. of any substance = E gm
96500 C charge discharge E gm of any substance
1 C charge discharge E/96500 gm of any substance
Since the mass of substance discharged by passing 1 C charge is called ECE(Z).
So, Z = E/96500
Z = E/F
Solved examples
1. How many coulombs of electric charge are required to deposit?
i. 4.6 gm of sodium ii. 3 mol of aluminium
solution:
i. Na+ + e– → Na
1 mol of Na = 1 mol e–
23 gm of Na = 1 mol e–
4.6 gm of Na = 0.2 mol e–
1 mol of e– = 96500 C
0.2 mol of e– = 19300 C
ii. Al+++ + 3e– → Al
1 mol of Al = 3 mol of e–
3 mol of Al = 9 mol of e–
1 mol of e– = 96500 C
9 mol of e– = 868500 C
2. How many coulombs of electricity are required for the oxidation of one mole of H2O to O2?
solution:
2H2O → O2 ( balancing oxygen)
2H2O → O2 + 4H+ (balancing hydrogen)
2H2O(2 mole) → O2 + 4H+ + 4e– (4 mole) (balancing charge)
Here, 2 moles of H2O = 4 moles of electrons
1 mole of H2O = 2 moles of electrons
Now, Charge of 1 mole of electron = 96500 C
Charge of 2 moles of electron = 2 x 96500 C
3. A current of 2.5 amperes passes through the solution of zinc sulphate for 30 minutes and deposits 1.52 gram of zinc metal at cathode. What is the equivalent weight of zinc metal?
solution:
Time(t) = 30 x 60 sec
Current (I) = 2.5 A
mass(m) = 1.52 gm
From Faraday’s first law,
m = ZIt
or, m = (E/F) It
or, 1.52 = (E/96500) x 2.5 x 1800
E = 32.59
Some Important Questions
- State Faraday’s first and second law of electrolysis.
- Define ECE, Faraday and Chemical equivalent.
- Calculate the number of coulomb of electricity required
a. to discharge 0.3 mol of Zn++ (57900C)
b. to deposit 4 gm of Mg from MgSO4. (32166.6C) - What current is required to deposit whole copper from 1L of 1M CuSO4 solution by passing electricity through it for 10 minutes? (321.66 A)
- A current of 2.5 A is passed through the solution of a metal sulphate for 30 minutes and deposits 1.52 gm of metal at the cathode. Find the equivalent weight of metal. (32.59)