Question

According to Faraday’s first law;
A.${\rm{W}} = \dfrac{{{\rm{96500}} \times {\rm{E}}}}{{{\rm{I}} \times {\rm{t}}}}$
B.${\rm{W}} = \dfrac{{{\rm{E}} \times {\rm{I}} \times {\rm{t}}}}{{{\rm{96500}}}}$
C.${\rm{E}} = \dfrac{{{\rm{I}} \times {\rm{t}} \times {\rm{96500}}}}{{\rm{W}}}$
D.${\rm{E}} = \dfrac{{{\rm{I}} \times {\rm{W}}}}{{{\rm{t}} \times {\rm{96500}}}}$

Answer 1

Hint: We know that, the equation of the Faraday's law can generally be derived by two famous equations one is Maxwell–Faraday equation which usually describes the transformer EMF and other one is the Lorentz force which describes the motional EMF.

Step by step answer:
 As we know that, whenever there occurs electrolysis, the oppositely charged current flows in the solution, a scientist called Faraday studied the value of current and mass of the cations and anion.
We know that the flow of current through any substance always depends on the number of electrons involved in the reaction .We know that electrons have a charge of $ - 1.602 \times {10^{ - 19}}\,C$. So whenever an electron is deposited then an electric charge of $ - 1.602 \times {10^{ - 19}}\,C$is needed. And the rate electron deposited also depends upon the time taken. So, we can say that any mass that is deposited depends on the quantity of electrical charge which has been passed through electrolyte.
In mathematical form,
\[\dfrac{{\rm{M}}}{{\rm{Z}}} = {\rm{Equivalent}}\,{\rm{weight}}\]
$\begin{array}{l}
{\rm{m}} \propto {\rm{quantity}}\,{\rm{of}}\,{\rm{electricity}}\\
{\rm{Q}} = {\rm{m}} = {\rm{ZQ}}
\end{array}$
Where Z is a constant of proportionality and is known as the electrochemical equivalent of the substance.

Charge of electron in 1 mole of electricity is
\[\begin{array}{l}
Charge = {\rm{electronic charge}}\, \times {{\rm{N}}_{\rm{A}}}\\
 charge = 1.6 \times {10^{ - 19}} \times 6.022 \times {10^{23}}\\
 charge = 96500\,{\rm{C}}
\end{array}\]

This charge is being passed by one faraday of electricity,
Amount of charge passing is \[{\rm{q}} = {\rm{i}} \times {\rm{t}}\]
No of faraday passing \[ = \dfrac{{{\rm{It}}}}{{{\rm{96500}}}}\,{\rm{faraday}}\]
Weight;
\[\begin{array}{c}
{\rm{W}} = \dfrac{{{\rm{It}}}}{{{\rm{96500}}}} \times \left( {\dfrac{{\rm{M}}}{{\rm{Z}}}} \right)\\
 = \dfrac{{{\rm{ItE}}}}{{{\rm{96500}}}}
\end{array}\]
Where, \[{\rm{I}}\] is ampere of current passes, \[{\rm{t}}\] is time taken, \[{\rm{Z}}\] is the charge and \[\dfrac{{\rm{M}}}{{\rm{Z}}} = {\rm{Equivalent}}\,{\rm{weight}}\]

Hence, option B is correct.

Note: Each substance has different atomic weights. So, the rate of chemical decomposition is also different for different elements as they have different atomic mass. So, we can also conclude that in an electrolysis mass deposition is directly proportional to atomic weight. This is called faraday second law.