# Faraday's Laws of Electrolysis – Statement, Expression and Examples

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Electrolysis is a process which uses the electric current to stimulate a non-spontaneous chemical reaction. Michael Faraday formulated two laws governing the electrolytic processes in 1833. These laws show the quantitative relationship between the substance deposited at the electrodes and the quantity of electric charge passed.

## Faraday's First Law of Electrolysis

Faraday's first law of electrolysis states that “the mass of substance deposited at any electrode in a given time is directly proportional to the quantity of electric charge passing through the electrolyte.”

Mathematically,

$$\mathrm{\mathit{m}\propto \mathit{Q}\:\:\:\cdot \cdot \cdot \mathrm{\left ( \mathrm{1} \right )}}$$

$$\mathrm{\Rightarrow \mathit{m}\propto \mathit{It}\:\:\:\cdot \cdot \cdot \mathrm{\left ( \mathrm{2} \right )}}$$

$$\mathrm{\therefore \mathit{m}\:=\: \mathit{ZIt}\:\:\:\cdot \cdot \cdot \mathrm{\left ( \mathrm{3} \right )}}$$

Where,

• 'Z' is a constant of proportionality called the electro-chemical equivalent. It is measured in grams per coulomb (g/C).

• 'm' is the mass of substance deposited at an electrode and measured in grams.

• 'Q' is the amount of charge passed through the electrolyte in a given time.

• 'I' is the steady current in amperes

• 't' is the time in seconds for which current 'I' flows through the electrolyte

If

$$\mathrm{\mathit{I}\:=\:\mathrm{1}\:\mathrm{A};\:\:\mathit{t}\:=\:\mathrm{1\:sec}}$$

Then,

$$\mathrm{\mathit{Z}\:=\: \mathit{m}\:\:\:\cdot \cdot \cdot \mathrm{\left ( \mathrm{4} \right )}}$$

Thus, the electro-chemical equivalent (Z) of substance is defined as the mass of the substance deposited on passing a steady electric current of 1 A for 1 second through its electrolytic solution. The SI unit of Z is kg/C.

## Numerical Example (1)

Calculate the ampere-hours required to deposit a coating of silver 0.07 mm thick on a sphere of 5 cm radius. Assume the electrochemical equivalent of the silver equal to 0.001118 and the density of the silver to be 10.5.

Solution

The surface area of the given sphere is,

$$\mathrm{\mathit{A}\:=\: \mathrm{4}\mathit{\pi r^{\mathrm{2}}}\:=\:\mathrm{4}\:\times \:\pi \:\times \mathrm{\left ( \mathrm{5} \right )^{\mathrm{2}}}\:=\:\mathrm{314.159\:cm^{\mathrm{2}}}}$$

$$\mathrm{\mathrm{Thickness\: of \:coating},\mathit{t}\:=\:\mathrm{0.07\:mm}\:=\:\mathrm{0.008\:cm}}$$

$$\mathrm{\because \mathrm{Mass\: of \:silver\: to\: be\: deposited},\mathit{m}\:=\:\mathit{A}\:\times \:\mathit{t}\:\times \:\mathrm{density \:of\: silver}}$$

$$\mathrm{\therefore \mathit{m}\:=\:\mathrm{314.159}\:\times \:\mathrm{0.008}\:\times \:\mathrm{10.5}\:=\:\mathrm{26.389\:gm}}$$

$$\mathrm{\mathrm{ECE\: of\: silver},\mathit{Z}\:=\:\mathrm{0.001118\:gm/c}\:=\:\mathrm{0.001118}\:\times \:\mathrm{3600}\:=\:\mathrm{4.0248\:gm/Ah}}$$

$$\mathrm{\therefore \mathrm{Ameper\: hours\: required}\:=\:\frac{\mathit{m}}{\mathit{Z}}\:=\:\frac{\mathrm{26.389}}{\mathrm{4.0248}}\:=\:\mathrm{6.55\:Ah}}$$

## Faraday's Second Law of Electrolysis

Faraday's second law states that “when the same amount of electric charge is passed through several electrolytes, then masses of the substances deposited are proportional to their respective chemical equivalents or equivalent weights.”

In other words, Faraday's second law of electrolysis can be stated as “the mass of a substance deposited at any electrode on passing a certain amount of electric charge is directly proportional to its chemical equivalent weight.”

Mathematically,

$$\mathrm{\mathit{W}\propto \:\mathit{E}\:\:\:\cdot \cdot \cdot \mathrm{\left ( \mathrm{5} \right )}}$$

Where,

• 'W' is the mass of the substance,

• 'E' is the equivalent weight of the substance.

For two different electrolytes, this law can also be expressed as,

$$\mathrm{\frac{\mathit{W_{\mathrm{1}}}}{\mathit{W_{\mathrm{2}}}}\:=\:\frac{\mathit{E_{\mathrm{1}}}}{\mathit{E_{\mathrm{2}}}}\:\:\:\cdot \cdot \cdot \mathrm{\left ( \mathrm{6} \right )}}$$

The equivalent weight or chemical equivalent of a substance can be defined as the ratio of its atomic weight and its valency, i.e.

$$\mathrm{\mathrm{Chemical\: equivalent},\mathit{E}\:=\:\frac{\mathrm{Atomic\: weight}}{\mathrm{Valency}}\:\:\:\cdot \cdot \cdot \mathrm{\left ( \mathrm{7} \right )}}$$

## Numerical Example (2)

In a copper sulphate voltmeter, the copper cathode is increased in weight by 0.06 kg in 2 hours, when the electric current was maintained constant. Calculate the value of this current. Given that −

• Atomic weight of Copper = 63.5;

• Atomic weight of Hydrogen = 1;

• Atomic weight of Silver = 108;

• ECE of Silver = 111.8 × 10-8kg/C

Solution

$$\mathrm{\mathrm{Mass \:of\: copper\: deposited},\mathit{m_{\mathit{cu}}}\:=\:\mathrm{0.06\:kg}\:=\:\mathrm{60\:gm}}$$

$$\mathrm{\mathrm{Time\: of\: current\: flow},\mathit{t}\:=\:\mathrm{2}\:\times \:\mathrm{3600}\:=\:\mathrm{7200\:sec}}$$

Therefore,

$$\mathrm{\mathrm{Chemical \:equivalent\: of\: silver},\mathit{E_{\mathit{Ag}}}\:=\:\frac{\mathrm{Atomic\: weight}}{\mathrm{Valency}}\:=\:\frac{\mathrm{108}}{\mathrm{1}}\:=\:\mathrm{108}}$$

Similarly,

$$\mathrm{\mathrm{Chemical \:equivalent\: of\: Copper},\mathit{E_{\mathit{Cu}}}\:=\:\frac{\mathrm{Atomic\: weight}}{\mathrm{Valency}}\:=\:\frac{\mathrm{63.5}}{\mathrm{2}}\:=\:\mathrm{31.75}}$$

Therefore, the electro-chemical equivalent of copper is,

$$\mathrm{\mathrm{ECE\:of\:Copper},\mathit{Z_{\mathit{cu}}}\:=\:\mathit{Z_{\mathit{Ag}}}\:\times \:\frac{\mathit{E_{\mathit{Cu}}}}{\mathit{E_{\mathit{Ag}}}}}$$

$$\mathrm{\Rightarrow \mathit{Z_{\mathit{Cu}}}\:=\:\mathrm{111.8\:\times \:10^{-8}\:\times }\:\frac{\mathrm{31.75}}{\mathrm{108}}\:=\:\mathrm{3.287\:\times \:10^{-7}\:kg/C}}$$

Now, the value of current is,

$$\mathrm{\mathit{I}\:=\:\frac{\mathit{m_{\mathit{Cu}}}}{\mathit{Z_{\mathit{Cu}}\times\: t}}\:=\:\frac{\mathrm{0.06}}{\mathrm{3.287\:\times \:10^{-7}}\:\times \:7200}\:=\:\mathrm{25.35\:A}}$$

Updated on 05-Apr-2022 06:05:05