Question

# Explain faraday's second law of electrolysis in details.

Open in App
Solution

## Faraday's second law of electrolysisMichael Faraday established the second law of electrolysis.It states, "The masses of different ions liberated at the electrodes, when the same amount of electricity is passed through different electrolytes are directly proportional to their chemical equivalent weight."The amount of electricity needed to release one gram-equivalent of any material at any electrode is called a faraday, and it passes through an electrolyte.Faraday's constant is equal to $96485\mathrm{C}{\mathrm{mol}}^{-1}$.Formula of Faraday's second law It follows that the electrochemical equivalent of an element is directly proportional to its equivalent weight.$\begin{array}{rcl}\frac{{W}_{1}}{{W}_{2}}& =& \frac{{E}_{1}}{{E}_{2}}\\ Z& \propto & E\end{array}$${W}_{1},{W}_{2}$ are the deposited amount of any substance.${E}_{1},{E}_{2}$ are the equivalent weights of that substance.$Z$ is the electrochemical equivalent.$E$ is the equivalent weight.ExampleThe same current is passed through the solutions of sulfuric acid $\left({\mathrm{H}}_{2}{\mathrm{SO}}_{4}\right)$, copper sulfate $\left({\mathrm{CuSO}}_{4}\right)$, and silver nitrate $\left({\mathrm{AgNO}}_{3}\right)$ for the same period of time.Then,$\begin{array}{rcl}\frac{\mathrm{Mass}\mathrm{of}\mathrm{copper}\mathrm{deposited}}{\mathrm{Mass}\mathrm{of}\mathrm{silver}\mathrm{deposited}}& =& \frac{\mathrm{Equivalent}\mathrm{mass}\mathrm{of}\mathrm{copper}}{\mathrm{Equivalent}\mathrm{mass}\mathrm{of}\mathrm{silver}}\\ \frac{\mathrm{Mass}\mathrm{of}\mathrm{copper}\mathrm{deposited}}{\mathrm{Mass}\mathrm{of}\mathrm{hydrogen}\mathrm{gas}\mathrm{liberated}}& =& \frac{\mathrm{Equivalent}\mathrm{mass}\mathrm{of}\mathrm{copper}}{\mathrm{Equivalent}\mathrm{mass}\mathrm{of}\mathrm{hydrogen}\mathrm{gas}}\end{array}$

Suggest Corrections
7
Related Videos
Sodium Hydroxide
CHEMISTRY
Watch in App