Faraday’s Law

| March 30, 2012 | 1 Comment

Faraday’s Law

The quantitative relationship between the amount of electricity passed through a cell and the amount of substances discharged at the electrodes was systematized by Michael Faraday in the form of the following laws:

Faraday’s First law

The amount of substance discharged (deposited or dissolved) at the electrode is proportional to the quantity of the electricity passing through the electrolyte

where w is the weight of the substance discharged at an electrode in gram; q is the charge in coulomb, t is the time of flow of electricity in second, I is the current in ampere and z is a constant known as the Electrochemical Equivalent which is defined as the number of grams of the substance deposited or dissolved by one coulomb of electricity

Second law

When the same quantity of electricity is passed through different solutions, the amounts of different substances deposited or dissolved at the electrodes in different electrolytic cells are proportional to their equivalent weights and in an electrolytic cell, chemically equivalent amounts of substances are discharged at both the electrodes.

Interpretation of Faraday’s Second Law:

Let us now interpret the second law of Faraday in a simple manner. 1 electron reduces and deposits 1 M+ ion at an electrode (i.e., M+ + egiving  M)

Therefore, 1 mole of electrons shall reduce and deposit 1 mole of M+ ions if the ion has a valency of nn mole of electrons shall reduce 1 mole of Mn+ ions.

So, 1 mole of electrons shall reduce 1/n mole of Mn+ ions.

For example: 1 mole of electrons reduces or deposits 1 mole of Ag+ or ½ mole of Cu2+ or 1/3 mole

or Al3+

Now, that (number of mole × valency) represents number of equivalents. So, 1 mole of electrons shall reduce or deposit 1 equivalent of Ag+ or Cu2+ or Al3+. In general, 1 mole of electricity (electrons) liberates 1 equivalent of matter.

Again we know:

Charge of 1 mole of electrons   = charge of an electron × Av. const.

                                                 = 1.6021 × 10–19 × 6.022 × 1023 coulomb

                                                 = 96487 coulomb

                                                 = 96500 coulomb

                                                 = 26.8 ampere-hour per equivalent

                                                 = 1 faraday

Thus the essential content of Faraday’s second law is that 1 faraday, which corresponds to 1 mole of electrons, liberates 1 equivalent of matter.

In redox reactions, the amount of the reactant, corresponding to 1 mole of electrons is thus its equivalent mass.

Electrochemical Equivalent and Equivalent Weight

The weight in gram of a substance liberated by 1 coulomb of electricity is called electrochemical equivalent whereas the weight in gram liberated by 96500 (or 1 faraday or 1 mole electricity) is called Gram Equivalent Weight of the substance

From Faraday’s law, we can deduce the relationship between the electrochemical equivalent and equivalent weight

No. of gram equivalents = No. of Faradays of electricity

As 1 gm eq of any substance = 1 F of electricity

Now there are two approaches to solve a problem

First calculate the number of faradays of electricity by using :

(i) Now by using the definition :

1 gm eq. of any substance =  1 F of electricity passed,

Calculate the number of gm. eq. and by using the definition of gm. eq. (gm. eq. = mass/E), determine the amount of substance deposited.

(ii) Using anodic and cathodic reactions as follows:

Let us consider a typical anode reaction:

So in this approach, first write anodic and cathodic reactions and derive the mole Vs faraday relation.

(iii) Using the combined relation obtained from Ist and IInd Laws :

      w = Z I t

      w = E I t/96500

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Category: Electrochemistry

  • http://ebookdownloadhub.blogspot.com Anonymous

    Thank You so much for your post on Faraday’s laws of electrolysis. It is the only article of it’s nature with comprehensive coverage of the subject which also clearly depicts the relationship between Electrochemical equivalent and equivalent weight.

    Your article is very much appreciated!

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