Faraday’s Laws of Electrolysis: First and Second Law

Before studying Faraday’s Laws of Electrolysis, let us first understand the concept of electrolysis. Electrolysis is the process of carrying out non-spontaneous reactions under the influence of electric energy. On passing electricity through water (electrolysis), Hydrogen and Oxygen gases are released. The amount of gas liberated depends on the electrode and electrolyte used, the amount of electricity passed through the solution, etc.

Michael Faraday’s electrochemical research in 1833 came up with Faraday’s laws of electrolysis. Which prove the quantitative relationship between the substance deposited at the electrodes and the amount of electric charge or energy transferred. Let us study more about the first and second laws of electrolysis.

Faraday’s Laws of Electrolysis

Michael Faraday \(\left( {1832} \right)\) studied the phenomenon of electrolysis extensively and established a relationship between the amount of product liberated at the electrode and the quantity of electricity passed in the solution to carry out the electrolysis. He formulated important quantitative laws which govern electrolysis. The laws are commonly known as Faraday’s laws of electrolysis.

What is Electrolysis?

The process of chemical deposition of the electrolyte by the passage of electricity through its molten or dissolved state is called electrolysis.

The device in which the process of electrolysis is carried out is called an electrolytic cell. It consists of an electrolytic tank made up of some non-conducting materials like glass, wood or bakelite. The solution to be electrolysed is filled in this tank.

The electrolyte can also be taken in the fused state. The cell consists of \(2\) metallic graphite rods in the solution of electrolyte and connected to a battery. The rods act as electrodes. The rod connected to the positive terminal of the battery acts as an anode, while that connected to the negative terminal of the battery acts as a cathode.

Mechanism of Electrolysis

Electrolysis is the process that involves the conversion of electrical energy into chemical energy. The process of electrolysis can be explained based on the theory of ionisation. When an electrolyte is passed in water, it splits up into charged particles called ions. The positively charged ion is called cations, while the negatively charged ions are called anions.

The ions are to move about in an aqueous solution. When an electric current is passed through the solution, the ions respond to the applied potential difference, and their movement is directed towards the oppositely charged electrodes. The cations move towards the negatively charged electrode (cathode) while anion moves towards the positively charged electrode (anode).

The formation of products at the respective electrode is due to oxidation loss of electron at anode and reduction gain of an electron at the cathode.

Example: Electrolysis of an aqueous solution of sodium chloride \(\left( {{\rm{NaCl}}} \right)\)
\({{\rm{NaCl}}}\) in aqueous solution ionizes as
\({\rm{NaCl}}\left( {{\rm{aq}}} \right) \to {\rm{N}}{{\rm{a}}^{\rm{ + }}}\left( {{\rm{aq}}} \right){\rm{ + C}}{{\rm{l}}^{\rm{ – }}}\left( {{\rm{aq}}} \right)\)
Reaction at anode, oxidation:
\({\rm{C}}{{\rm{l}}^{\rm{ – }}}\left( {{\rm{aq}}} \right) \to \frac{{\rm{1}}}{{\rm{2}}}{\rm{C}}{{\rm{l}}_{\rm{2}}}\left( {\rm{g}} \right){\rm{ + }}{{\rm{e}}^{\rm{ – }}}\)
Reaction at cathode, reduction:
\({{\rm{H}}_2}{\rm{O}}\left( {\rm{l}} \right) + {{\rm{e}}^ – } \to \frac{1}{2}{{\rm{H}}_2}\left( {\rm{g}} \right) + {\rm{O}}{{\rm{H}}^ – }\left( {{\rm{aq}}} \right)\)
Thus, \({\rm{C}}{{\rm{l}}_2}\) gas is liberated at the anode, whereas \({{\rm{H}}_2}\) gas is liberated at the cathode.

Faraday’s First Law of Electrolysis

Faraday’s first law of electrolysis states that the mass of any substance deposited or liberated at an electrode is directly proportional to the quantity of electricity passed through the electrolyte (solution or melt).
Thus, if \({\rm{W}}\) gram of the substance is deposited on passing \({\rm{Q}}\) coulombs of electricity, then
\({\rm{W}} \propto {\rm{Q}}\)
\({\rm{W}} = {\rm{ZQ}} ……\left( 1 \right)\)
Where \({\rm{Z}}\) is a constant of proportionality and is called the electrochemical equivalent of the substance deposited.
If  a current of \({\rm{I}}\) ampere is passed for \({\rm{t}}\) seconds, then,
\({\rm{Charge}} = {\rm{Current}}\; \times {\rm{time}}\)
\({\rm{Q}} = {\rm{I}} \times {\rm{t}} ….\left( 2 \right)\)
On substituting eq \(\left( 2 \right)\) in eq \(\left( 1 \right)\)
\({\rm{W}} = {\rm{Z}} \times {\rm{I}} \times {\rm{t}}\)
Thus, if \({\rm{Q}} = 1\) coulomb or \({\rm{I}} = 1\) ampere and \({\rm{t}} = 1\) second,
\({\rm{W}} = {\rm{Z}} \times 1 \times 1\)
\({\rm{W}} = {\rm{Z}}\)
Hence, the electrochemical equivalent of a substance may be defined as the mass of the substance deposited when a current of one ampere is passed for one second, i.e., a quantity of electricity equal to one coulomb is passed.

Faraday’s Second Law of Electrolysis

Faraday’s Second Law of electrolysis states that when the same quantity of electricity is passed through different electrolytes connected in series, then the masses of the substance liberated at the electrode in the ratio of their chemical equivalent masses or the ratio of their electrochemical equivalence.

The chemical equivalent mass of metal can be obtained by dividing its atomic mass by the number of electrons required to reduce its cation.

For example, if the two electrolytic cells A containing silver nitrate \(\left( {{\rm{AgN}}{{\rm{O}}_3}} \right)\) solution and \({\rm{B}}\) be containing copper sulphate \(\left( {{\rm{CuS}}{{\rm{O}}_4}} \right)\) solutions are connected in series, and the same quantity of electricity is passed through the cells. Then the ratio of the mass of copper deposited at the cathode in electrolytic cell \({\rm{B}}\) is \({\rm{x}}\,{\rm{g}}\) to that of silver deposited in a cell \({\rm{A}}\) is \({\rm{y}}\,{\rm{g}}\) is equal to the ratio of their chemical equivalent masses.
\(\frac{{{\rm{Mass\;of\;Cu}}\;\left( {\rm{x}} \right)}}{{{\rm{Mass\;of\;Ag}}\;\left( {\rm{y}} \right)}} = \frac{{{\rm{Chemical\;Equivalent\;Mass\;of\;Cu}}}}{{{\rm{Chemical\;Equivalent\;Mass\;of\;Ag}}}} = \frac{{{{\rm{Z}}_{{\rm{cu}}}}}}{{{{\rm{Z}}_{{\rm{Ag}}}}}}\)
Each copper \({\rm{C}}{{\rm{u}}^{2 + }}\) ion requires \(2\) electrons to form \({\rm{Cu,}}\) and each \({\rm{A}}{{\rm{g}}^{\rm{ + }}}\) needs \(1\) electron to form \({\rm{Ag}}{\rm{.}}\)
Thus, the chemical equivalent mass of \({\rm{Cu}} = \;\frac{{63.5}}{2}.\)
Chemical equivalent mass of \({\rm{Ag}} = \;\frac{{108}}{1}\)
Thus, the ratio \(\frac{{\rm{x}}}{{\rm{y}}}{\rm{ = }}\frac{{{\rm{63}}{\rm{.5}}}}{{{\rm{2 \times 108}}}}\)

Relation Between Faraday, Avogadro’s Constant, and Charge on an Electron

The charge carried by one mole of an electron can be obtained by multiplying the charge present on one electron with Avogadro’s number.
\({\rm{Charge\;carried\;by\;one\;mole\;of\;electron}} = {\rm{Charge\;carried\;by\;an\;electr}}on\; \times {\rm{Avogadro’s\;number}}\)
\({\rm{Charge\;carried\;by\;one\;mole\;of\;electron}} = 1.6021 \times {10^{ – 19}}{\rm{C}} \times 6.022 \times {10^{23}}\)
\({\rm{Charge\;carried\;by\;one\;mole\;of\;electron}} = 96487.84\;{\rm{C}} \cong 96500{\rm{C}}\)
The quantity of electricity, \(96487.84\;{\rm{C}} \cong 96500{\rm{C\;}}\) is called one Faraday. As it is a constant quantity, it is known as Faraday’s constant and is represented by \({\rm{F}}.\)
Hence, Faraday’s constant, \({\rm{F}} = 96487.84\;{\rm{Cmo}}{{\rm{l}}^{ – 1}} \cong 96500{\rm{Cmo}}{{\rm{l}}^{ – 1}}\)
If n electrons are involved in the electrode reaction, the passage of \(‘{\rm{n}}’\) faradays (i.e., \({\rm{n}} \times 96500\;{\rm{C}}\)) of electricity will liberate one mole of the substance.
In terms of gram equivalents, one faraday ( i.e., \(96500\) coulombs) of electricity deposits one gram equivalent of the substance.
Equivalent weight of any element =
\(\frac{{{\rm{Atomic\;weight\;of\;the\;element}}}}{{{\rm{No}}{\rm{.of\;electrons\;gained\;or\;lost\;by\;one\;atom\;or\;ion\;of\;the\;element}}}}\)

Conclusions

1. As one Faraday (\(96,500\) coulombs) deposits one gram equivalent of the substance, hence electrochemical equivalent can be calculated from the equivalent weight, i.e.,
\({\rm{Z}} = \frac{{{\rm{Equivalent\;\;weight\;\;of\;the\;substance}}}}{{96500}}\)

2. Knowing the weight of the substance deposited (\({\rm{W}}\) gram) on passing a definite quantity of electricity (\({\rm{Q}}\) coulombs), the equivalent weight of the substance can be calculated, i.e.,
\({\rm{Equivalent\;\;weight}} = \;\frac{{\rm{W}}}{{\rm{Q}}} \times 96500\)

Can We Combine Faraday’s First and Second Law?

Faraday’s first law and second law can be combined to give a mathematical relation as follows:
\({\rm{W}} = {\rm{ZQ}}\)
\({\rm{W}} = \frac{{\rm{E}}}{{\rm{F}}} \times {\rm{Q}}\)
\({\rm{W}} = \frac{{\rm{Q}}}{{\rm{F}}} \times {\rm{E}}\)
\({\rm{W}} = \frac{{\rm{Q}}}{{\rm{F}}} \times {\rm{E}}\)
\({\rm{W}} = \frac{{\rm{Q}}}{{\rm{F}}} \times \frac{{\rm{M}}}{{\rm{Z}}}\)
\({\rm{W}} = \frac{{{\rm{C}} \times {\rm{t}}}}{{\rm{F}}} \times \frac{{\rm{M}}}{{\rm{Z}}}\)
Where, \({{\rm{Z}}\,{\rm{ = }}}\) Electrochemical equivalent
\({{\rm{Q}}\,{\rm{ = }}}\) Quantity of electricity passed
\({{\rm{E}}\,{\rm{ = }}}\) Equivalent weight of the metal
\({\rm{F}} = 1\) Faraday
\({{\rm{M}}\,{\rm{ = }}}\) Atomic mass of the metal
\({{\rm{C}}\,{\rm{ = }}}\) Current past
\({{\rm{t}}\,{\rm{ = }}}\) Time for which current is passed
\({{\rm{Z}}\,{\rm{ = }}}\) Valency of the metal

Summary

This is all about the concept of Faraday’s first and second laws of electrolysis in detail with chemical equations and application in calculating the amount of substance deposited or liberated on passing electricity.

FAQs

Q.1. Where Faraday’s law of electrolysis is used?
Ans: Faraday’s laws of electrolysis are used to calculate the amount of substance produced or liberated during electrolysis based on the amount of current passed through the electrolyte.

Q.2. What is the value of Faraday’s constant?
Ans: Faraday’s constant \(\left( {\rm{F}} \right) = \;96487.84\;{\rm{C}} \cong 96500{\rm{C}}\)

Q.3. What is Faraday’s first law of electrolysis?
Ans: Faraday’s first law of electrolysis states that the amount of chemical reaction and hence the mass of any substance deposited or liberated at an electrode is directly proportional to the quantity of electricity passed through the electrolyte (solution is in the molten state).

Q.4. What is Faraday’s first and second law?
Ans: Faraday’s first law of electrolysis states that the amount of chemical reaction and hence the mass of any substance deposited or liberated at an electrode is directly proportional to the quantity of electricity passed through the electrolyte (solution is in the molten state).
Faraday’s Second Law of electrolysis states that when the same quantity of electricity is passed through different electrolytes connected in series then the masses of the substance liberated at the electrode in the ratio of their chemical equivalent masses or the ratio of their electrochemical equivalence.

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