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The relation between Faraday constant (F), chemical equivalent (E) and electrochemical equivalent (Z) is
A. F=EZ
B. \[F = \dfrac{Z}{E}\]
C. \[F = \dfrac{E}{Z}\]
D. \[F = \dfrac{E}{{{Z^2}}}\]

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Hint: Faraday’s laws of electrolysis were given by Michael Faraday to obtain pure elements. He proposed two laws for electrolysis in which electricity is passed through a solution called electrolyte to form ionic compounds.

Complete Step by Step Answer:
According to Faraday’s first law of electrolysis, mass of a substance that gets deposited at an electrode is directly proportional to the amount of charge passed through the solution. Mathematically, it can be written as
m=ZQ……(i)
Where Z is the electrochemical equivalent of the substance
‘m’ is the mass of the substance liberated
‘Q’ is the amount of electricity passed
According to Faraday’s second law of electrolysis, the mass of substance deposited at any electrode when current is passed through different electrolytes depends on the equivalent weight.
\[w \propto E\]
Where w is the mass of the substance
E is the equivalent weight which is the ratio of atomic weight and valency.
Also it is known that the charge on one mole of electrons is equal to 96500 C also known as 1 Faraday. So, if 1 Faraday of charge is passed through an electrolytic solution, the amount of substance deposited will be equal to 1 gm of equivalent substance. Therefore, it can be written that
\[w = \dfrac{Q}{{96500}} \times E\]
On combining first law and second law we get
\[Z = \dfrac{E}{{96500}}\] or
\[Z = \dfrac{E}{F}\] as 1F=96500C
From the above equation, it can be written that the relation between Faraday constant (F), chemical equivalent (E) and electrochemical equivalent (Z) is
\[F = \dfrac{E}{Z}\]
Hence, Option C is the right answer.

Note: Faraday’s laws of electrolysis were proposed by Michael Faraday in 1833. In electrolysis, a chemical reaction takes place when electric current is passed through an aqueous solution so that a pure substance can be obtained by deposition of ions at the electrodes.