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Physics Lesson 15.8.4 - Faraday's Law for Electrolysis

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Welcome to our Physics lesson on Faraday's Law for Electrolysis, this is the fourth lesson of our suite of physics lessons covering the topic of Miscellaneous, you can find links to the other lessons within this tutorial and access additional physics learning resources below this lesson.

Faraday's Law for Electrolysis

Let's consider again the wet cell discussed in the previous paragraph. When some table salt is poured in water, the Na + and Cl- ions dissolve in water. The Na + ions start to move towards the copper plate (cathode) taking away an electron from it. The Cl- ions do the opposite thing, i.e. they move towards the zinc plate (anode) and become neutral there. As a result, an electric current flows inside the electrolyte and the copper electrode is covered by a zinc layer. In technology, such a process is known as electroplating and it is used to protect metals from corrosion, oxidation or for decoration purpose.

Michael Faraday introduced an equation which represents the relationship between the amount (mass) m of substance decomposed to the anode (deposited in the cathode) and the current I flowing through the electrolyte. This equation, known as the Faraday's Law for Electrolysis is

m = k ∙ Q
= k ∙ I ∙ t

where k is a constant known as the electrochemical constant of electrolysis. It has the unit of [kg/A ∙ s]. I shows the current and t the time during which the electrolysis takes place.

The acid added in water is not the only material that is decomposed. Water also decomposes in two separate elements: hydrogen and oxygen. The salty water simply increases the conductive ability of water. The chemical reaction that takes place during the water electrolysis is

H2 O ⟶ H2 + 1/2 O2

If we consider the two gases produced during the electrolysis as ideal gases (all ideal gases have the same volume for the same number of particles), we can see from the above reaction formula that the volume of hydrogen collected in a tube placed on the cathode is twice the volume of oxygen collected in the tube on the anode.

The process of electrolysis is used to obtain pure hydrogen or oxygen from water decomposition.

Example 2

How much time is needed to obtain 2 kg of copper during an electrolysis process when a 5A current flows through the electrolyte? Take kcopper = 3.3 × 10-7 kg/A ∙ s.

Solution 2

Clues:

m = 2 kg
I = 5 A
k = 3.3 × 10-7 kg/A ∙ s
t = ?

From Faraday's Law of Electrolysis

m = k ∙ I ∙ t

we obtain from the time t after rearranging the above equation

t = m/k ∙ I
= 2 kg/3.3 × 10-7 kg/A ∙ s ∙ (5A)
= 0.1212 × 107 s
= 1 212 121 s

When converted into hours (1h = 3600s) we obtain

t = 336.7 hours

or about 14 days. This result means the process of electrolysis is relatively slow.

You have reached the end of Physics lesson 15.8.4 Faraday's Law for Electrolysis. There are 5 lessons in this physics tutorial covering Miscellaneous, you can access all the lessons from this tutorial below.

More Miscellaneous Lessons and Learning Resources

Electrodynamics Learning Material
Tutorial IDPhysics Tutorial TitleTutorialVideo
Tutorial
Revision
Notes
Revision
Questions
15.8Miscellaneous
Lesson IDPhysics Lesson TitleLessonVideo
Lesson
15.8.1Measurement of Current and Voltage
15.8.2Superconductivity
15.8.3Electric Current in Liquids
15.8.4Faraday's Law for Electrolysis
15.8.5Electric Current in Gases

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