JEE Main & Advanced Physics Current Electricity, Charging & Discharging Of Capacitors / वर्तमान बिजली, चार्ज और कैपेसिटर का निर Faraday's Law of Electrolysis

(1) First law : It states that the mass (m) of substance deposited at the cathode during electrolysis is directly proportional to the quantity of electricity (total charge q) passed through the electrolyte i.e. \[m\propto q\] or \[\mathbf{m=zq=zit,}\] where the constant of proportionality z is called electrochemical equivalent (E.C.E.) of the substance.

Therefore we have \[m=zit\]. If \[q=1\] coulomb, 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.

S.I. unit of electrochemical equivalent of a substance is kilogram\[coulom{{b}^{1}}(kg-{{C}^{1}})\].  

E.C.E. for certain substances

(2) 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. \[m\propto E\,\Rightarrow \,\frac{{{m}_{1}}}{{{m}_{2}}}=\frac{{{E}_{\mathbf{1}}}}{{{E}_{\mathbf{2}}}}\]

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 \[\frac{m}{E}=\text{ constant}\]

Chemical equivalent E also known as equivalent weight in gm i.e. \[E=\frac{\text{Atomic mass (}A\text{)}}{\text{Valancy (}V\text{)}}\]

(3) Relation between chemical equivalent and electrochemical equivalent : Suppose that on passing same amount of electricity q through two different electrolytes, masses of the two substances 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 \[\frac{{{z}_{1}}}{{{z}_{2}}}=\frac{{{E}_{1}}}{{{E}_{2}}}\]\[\Rightarrow \]\[z\propto E\]

(4) Faraday constant : As we discussed above \[E\propto z\]

\[\Rightarrow \] \[E=Fz\]\[\Rightarrow \]\[z=\frac{E}{F}=\frac{A}{VF}\]. \['F'\] 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 voltameter is 1 Faraday then amount of substance liberated or deposited is (in gm) equal to the chemical equivalent.