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Electrolysis : Faraday's Laws Of Electrolysis
The quantitative laws of electrolysis, relating the masses of the ions deposited to the charge transported through the electrolyte, were discovered by Faraday (Experimental Researches in Electricity; reprinted in Everyman's Library.). In his first memoir (1832) he made out the relation between the mass of any particular ion deposited and the amount of electricity required for its deposition. The two were proportional, or in Faraday's words: "the chemical action of a current of electricity is in direct proportion to the absolute quantity of electricity which passes." This is the first law of electrolysis.
In his second memoir (1833) he made out the relation between the masses of various ions deposited by the same quantity of electricity. These were in the ratios of the chemical equivalents.
Since galvanometers and units of current were then not related, Faraday used the amount of gas produced by the electrolysis of acidulated water as a measure of the quantity of electricity which passes, and he determined the quantities of other elements liberated in electrolysis by the same current as liberates 1 gram of hydrogen.
These quantities were the ordinary chemical equivalents, and hence he discovered the second law of electrolysis: the weights of the ions deposited by the passage of the same quantity of electricity are in the proportion of their chemical equivalents.
To illustrate Faraday's laws we may connect in series a number of electrolytic cells, containing different electrolytes, with a battery as shown in
Suppose that the first cell contains water acidulated with sulphuric acid, the second a solution of copper sulphate, and the third fused stannous chloride. Fused salts are electrolytes, as well as their solutions.
After the current has passed for a certain time, the volumes of hydrogen and oxygen liberated from the. acidulated water, and the weights of copper and tin deposited from the solution of copper sulphate and the, fused stannous chloride, respectively, are ascertained. If the weights of the other ions which are deposited in the cells whilst 1 gm. of hydrogen is liberated in the first are determined, they are found to be equivalent weights: 7.94 gm. of oxygen, 35.2 gm. of chlorine, 31.5 gm. of copper and 59 gm. of tin.
The quantity of electricity which has passed through the solution is measured by the current strength multiplied by the time. The current strength is measured in amperes, and one ampere passing for one second corresponds with unit quantity of electricity, or one coulomb. A current of C amperes flowing for t seconds conveys Ct coulombs. Hence the weight of an ion deposited in a given time is proportional to the strength of the current. This is Faraday's First Law of Electrolysis.
The international ampere is defined as that current which, flowing uniformly for 1 second, deposits under specified conditions 0.001118 grams of silver from a solution of silver nitrate. This is called the electrochemical equivalent of silver; the weight z gm. of any ion deposited by 1 coulomb is its electrochemical equivalent; hence the weight deposited by a uniform current of C amperes flowing for t seconds is:
W = Czt.
Since the chemical equivalent of silver (O = 16.000) is 107.880, the quantity of electricity required to deposit this amount will be (by Faraday's first law) 107.880/0.001118 = 96,500 coulombs per gm. equiv. very nearly. This fundamental charge is called a faraday, denoted by F (Not to be confused with the farad, the unit of electric capacity.). Faraday's second law shows that 1F will deposit 1 chemical equivalent of any ion. Since Faraday's law will enable us to find the equivalent of an element (e.g., oxygen, copper, silver), it: will enable us to determine the valency of the element in the state of the ion investigated: valency = at. wt. / equivalent.
The laws of electrolysis are conveniently summarised in the statement that 96,500 coulombs liberate one gram-equivalent of any ion in electrolysis.
Thus, one F liberates 1 gm. atom of a univalent element, and nF liberate 1 gm. atom of an n-valent element.
Example. - Find the weight of copper deposited from a solution of copper sulphate by a uniform current of 0.25 amp. flowing for one hour.
Quantity of electricity passed through electrolyte = 0.25 x 60 x 60 =900 cmb.
Copper is bivalent, hence equivalent weight =at. wt. / 2 = 63.5/2 = 31.75.
96,500 cmb. liberate 31.7 gm. of Cu, hence wt. of copper liberated by 900 cmb. = 31.75 x 900/96,500 = 2.95 gm.
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