Faraday's Laws Of Electrolysis - Practice Questions & MCQ

According to the Faraday's first law, "The amount of substance or quantity of chemical reaction at electrode is directly proportional to the quantity of electricity passed into the cell".

$\begin{aligned} & \mathrm{W} \text { or } \mathrm{m} \propto \mathrm{q} \\ & \mathrm{W} \propto \mathrm{It} \\ & \mathrm{W}=\mathrm{ZIt} \\ & \mathrm{Z}=\frac{\mathrm{M}}{\mathrm{nf}} \\ & \mathrm{Z}=\frac{\mathrm{M}}{\mathrm{nf}} \\ & \mathrm{Z}=\text { Electrochemical equivalence } \\ & \mathrm{M}=\text { molarmass } \\ & \mathrm{F}=96500 \\ & \mathrm{n}=\text { Number of electrons transfer } \\ & \mathrm{q}=\text { amount of charge utilized }\end{aligned}$

Electrochemical equivalent is the amount of the substance deposited or liberated by one-ampere current passing for one second (that is, one coulomb, I x t = Q or one coulomb of charge.
One gram equivalent of any substance is liberated by one faraday.

$\begin{aligned} & \text { Eq. } \mathrm{Wt} .=\mathrm{Z} \times 96500 \\ & \frac{\mathrm{~W}}{\mathrm{E}}=\frac{\mathrm{q}}{96500} \\ & \mathrm{w}=\frac{\mathrm{E} . \mathrm{q}}{96500} \\ & \mathrm{~W}=\frac{\text { Eit }}{96500}\end{aligned}$

As $w=a \times 1 \times d$ that is, area $\times$ length $\times$ density
Here a = area of the object to be electroplated
d = density of metal to be deposited
l = thickness of layer deposited
Hence from here, we can predict charge, current strength time, thickness of deposited layer etc.

NOTE: One faraday is the quantity of charge carried by one mole of electrí            

E $\alpha \mathrm{Z}$

$$
\begin{aligned}
& \mathrm{E}=\mathrm{FZ} \\
& 1 \mathrm{~F}=1.6023 \times 10^{-19} \times 6.023 \times 10^{23} \\
& =96500 \text { Coulombs }
\end{aligned}
$$

According to Faraday's second law, "When the same quantity of electricity is passed through different electrolytes, the amounts of the products obtained at the electrodes are directly proportional to their chemical equivalents or equivalent weights".

As $\frac{W}{E}=\frac{q}{96500}=$ No of equivalents constant
So
$\frac{\mathrm{E}_1}{\mathrm{E}_2}=\frac{\mathrm{M}_1}{\mathrm{M}_2}$ or $\frac{\mathrm{W}_1}{\mathrm{~W}_2}=\frac{\mathrm{Z}_1}{\mathrm{Z}_2 \mathrm{It}}=\frac{\mathrm{Z}_1}{\mathrm{Z}_2}$
$\mathrm{E}_1=$ equivalent weight mass
$\mathrm{E}_2=$ equivalent weight mass
W or $\mathrm{M}=$ mass deposited

From this law, it is clear that 96500 coulomb of electricity gives one equivalent of any substance. 

Application of Faraday's Laws

• It is used in electroplating of metals.
• It is used in the extraction of several metals in pure form.
• It is used in the separation of metals from non-metals.
• It is used in the preparation of compounds

NOTE:
Current Efficiency: It is the ratio of the mass of the products actually liberated at the electrode to the theoretical mass that could be obtained

C.E. $=\frac{\text { desired extent }}{\text { Theoretical extent of reaction }} \times 100 \%$