**Faraday’s
Law of Electrolysis**

The process of carrying out nonspontaneous reaction under the influence of electric energy is termed as electrolysis. The relationship between the quantity of electric charge passed through an electrolyte and the amount of the substance deposited at the electrodes was presented by Faraday in 1834 in the form of laws of electrolysis.

**Faraday’s First Law: **It states that the mass of substance deposited
at the cathode during electrolysis is directly proportional to the quantity of
electricity passed through the electrolyte i.e. \(m\propto q\) (or) \(m=zq=zit\), where the constant of proportionality \(z\) is called electrochemical equivalent of the substance.

Therefore we have, \(m=zit\). If \(q=1C\), then we have \(m=z\times 1\) (or) \(z=m\) . Hence, the electrochemical equivalent of substance may be defined as the mass of its substance deposited at the cathode, when one coulomb of charge passes through the electrolyte.

**Faraday’s Second Law: **If same quantity of electricity is passed
through different electrolytes, masses of the substance deposited at the
respective cathodes are directly proportional to their chemical equivalents
i.e. \(m\propto E\Rightarrow
\frac{{{m}_{1}}}{{{m}_{2}}}=\frac{{{E}_{1}}}{{{E}_{2}}}\)

Let \(m\) be the mass of the ions of a substance liberated, whose chemical equivalent is E. Then, according to Faraday’s’ second law of electrolysis, \(m\propto E\) (or) \(m=Constant\times E\) (or) \(\frac{m}{E}=Constant\).

**Relation between chemical equivalent and electrochemical equivalent: **Suppose that on passing same amount of electricity q through two different electrolytes, masses of the two substance liberated are \({{m}_{1}}\) and \({{m}_{2}}\). If \({{E}_{1}}\) and \({{E}_{2}}\) are their chemical equivalents, then from Faraday’s second law, we have: \(\frac{{{m}_{1}}}{{{m}_{2}}}=\frac{{{E}_{1}}}{{{E}_{2}}}\) ; also, from Faraday’s first law, \(\frac{{{m}_{1}}}{{{m}_{2}}}=\frac{{{z}_{1}}}{{{z}_{2}}}\) So, \(\frac{{{z}_{1}}}{{{z}_{2}}}=\frac{{{E}_{1}}}{{{E}_{2}}}\Rightarrow z\propto E\)

**Faraday constant: **As we know, \(E\propto z\) \(E\propto z\Rightarrow E=Fz\Rightarrow z=\frac{E}{F}=\frac{A}{VF}\) ; \(F\) is proportionality constant called Faraday’s constant. As, \(z=\frac{E}{F}\) and \(z=\frac{m}{Q}\) so, \(\frac{E}{F}=\frac{m}{Q}\) hence if \(Q=1\,Faraday\) then \(E=m\) i.e. If electricity supplied to a voltammeter is 1 Faraday then amount of substance liberated or deposited is equal to the chemical equivalent.