Faraday's First Law Of Electrolysis_Statement+Illustration+Numerical Problems ~ CHEMISTRY

Faraday's Laws Of Electrolysis:

A)Faradays First Law Of Electrolysis

Statement:

Mass of substance deposited at the electrode is directly proportional to the quantity of charge passed through the electrolytic solution.

Mathematically,

$m \propto q.........................(i)$

Where,

m=mass of a substance deposited

q=quantity of charge

Illustration:

When the quantity of charge "q" is passed into CuSO₄ solution, the mass of copper deposited (m) is directly proportional to the quantity of charge "q". i.e. ${m_{Cu}} \propto q$

Also,

q=It where I = current in Ampere and t= time in the second

Substituting q in equation (i), equation (i) becomes:

$m \propto It$

Where z is proportionality constant called Electro Chemical Equivalent (E.C.E)

When, I=1A and t=1second,

z=m

Thus, Electro-Chemical Equivalent (E.C.E) is defined as the mass of substance deposited when 1-ampere current is passed to electrolytic solution for 1 second, or it is defined as the mass of substance deposited by 1Coloumb charge.

95600C charge deposits 1gram equivalent (E) of any substance.

1C charge deposits E/96500C of any substance.

Therefore, Electrochemical equivalent (E.C.E) is related to Chemical Equivalent as:

$E.C.E(z) = \frac{E}{{96500}}$ ......................................... (iii)

Replacing z in equation (ii)

$m = \frac{{EIt}}{{96500}}$

Also, Atomic mass (A) is related to Equivalent mass (E) as:

$E = \frac{A}{V}$

Replacing E in equation (iii),

$m = \frac{{AIt}}{{96500V}}$ ...........................................................(iv)

Collection Of Formulas for decoding numerical problems on Faraday First Law Of Electrolysis

$\begin{array}{l} m = zq\\ m = zIt\\ m = \frac{{EIt}}{{96500}}\\ Also,\\ m = Desity(\delta ) \times Area \times Thickness \end{array}$

Tips For solving Numerical Problems On Faradays 1st Law Of Electrolysis:

Mass should be in gram.
Time should be in second.
Area should be in $c{m^2}$ .
Thickness should be in cm.\
Avoid using symbol "A" for both Atomic mass and Area
Avoid using symbol "t" for both time and thickness.

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Complete Solution Of Numerical problems on Faraday's First Law of Electrolysis

1. State Faraday’s laws of electrolysis? Silver electrodeposited on a metal plate of surface area 800cm² by passing 0.2 ampere of current for 3hours. Calculate the thickness of Ag deposited.( Given specific gravity of Ag=10.47 and atomic mass=108)

Given,

I = 0.2A Specific gravity A g = 10.47

$t = 3hours = 3 \times 60 \times 60 = 10800s$ $Area = 800c{m^2}$

A = 108 amu Thickness = ?

V = 1 m = ?

m = ?

have,

$\begin{array}{l} m = \frac{A}{{96500V}}It\\ Or,m = \frac{{108}}{{96500 \times 1}} \times 0.2 \times 10800 = 2.4g\\ Now, \end{array}$

Mass = Specific $gravity \times Area \times Thickness$

$\begin{array}{l} Or,2.4 = 10.47 \times 800 \times Thickness\\ Or,Thickness = 2.87 \times {10^{ - 4}}cm \end{array}$

2. State Faraday’s first law of electrolysis and write the mathematical relation between electrochemical equivalent and chemical equivalent. 1.52g of a trivalent metal M was deposited at cathode by passing a current of 2.5ampere through its salt solution (metal sulphate) for 30 minutes. What is the atomic mass of M?

$\begin{array}{l} Given,\\ m = 1.52g\\ I = 2.5A\\ V = 3\\ t = 30\min utes = 30 \times 60 = 1800s\\ A = ? \end{array}$

have,

$\begin{array}{l} m = \frac{A}{{96500V}}It\\ Or,1.52 = \frac{A}{{96500 \times 3}} \times 2.5 \times 1800\\ Or,A = \frac{{1.52 \times 96500 \times 3}}{{2.5 \times 1800}} = 97.79amu \end{array}$

3. Define one Faraday’s electricity. How many grams of silver could be plated out on a tray by passing electricity through a solution a Ag(I) salt for 8hours at a current of 9 amperes? What is the area of the tray if the thickness of the silver plating is 0.002cm? The density of silver is 10g/cm3. (Atomic mass of Ag=107.8)

Given,

I = 9A $Density = 10gc{m^{ - 3}}$

$t = 8hrs = 8 \times 60 \times 60 = 28800s$ Thickness = 0.002cm

A = 107.8amu Area = ?

V = 1 m = ?

m = ?

We have,

$m = \frac{A}{{96500V}}It = \frac{{107.8}}{{96500 \times 1}} \times 9 \times 28800 = 289.55g$

Again,

${Mass = Density \times Area \times Thickness}$

${Or,289.55 = 10 \times Area \times 0.002}$

${Or,Area = 14477.6c{m^2}}$

4. State and explain Faraday’s laws of electrolysis. How long a current of 3ampere has to be passed through a solution of AgNO3 to coat a metal surface of 80cm2 with 0.005 mm thick layer? (density of Ag is 10.5gcm-3)

Given,

I = 3A Density A g = 10.5

A = 108amu Thickness = 0.005mm = 0.0005cm

V = 1 $Area = 80c{m^2}$

m = ? m = ?

t = ?

have,

$\begin{array}{l} Mass = Density \times Area \times Thickness\\ Or,m = 10.5 \times 80 \times 0.0005\\ Or,m = 0.42g\\ Now, \end{array}$

$\begin{array}{l} m = \frac{A}{{96500V}}It\\ Or,0.42 = \frac{{108}}{{96500 \times 1}} \times 3 \times t\\ Or,t = \frac{{0.42 \times 96500}}{{108 \times 3}} = 125.1s \end{array}$

5. A metallic spoon is coated with silver by passing a current of 5Ampere through AgNO₃ solution for 5hrs. What is the thickness of silver plating if the area of the spoon is 12cm2 (density of Ag is 10.5gcm-3)

Given,

I = 5A $Density = 10.5gc{m^{ - 3}}$

$t = 5hrs = 5 \times 60 \times 60 = 18000s$ $Area = 12c{m^2}$

A = 108amu Thickness = ?

V = 1 m = ?

m = ?

We have,

$m = \frac{A}{{96500V}}It = \frac{{108}}{{96500 \times 1}} \times 5 \times 18000 = 100.73g$

$\begin{array}{l} Again,\\ Mass = Density \times Area \times Thicness\\ Or,100.73 = 10.5 \times 12 \times Thickness\\ Or,Thickness = 0.8cm \end{array}$

6. State Faraday’s law of electrolysis. Establish relation between electro-chemical equivalent and chemical equivalent. 0.197g of copper is deposited by a current of 0.2 A in 50 minutes. Calculate its electrochemical equivalent.

$\begin{array}{l} Given,\\ m = 0.197g\\ I = 0.2A\\ t = 50\min utes = 50 \times 60 = 3000s\\ z = ? \end{array}$

We have,

$\begin{array}{l} m = zIt\\ Or,0.197 = z \times 0.2 \times 3000\\ Or,z = 3.28 \times {10^{ - 4}}g{C^{ - 1}} \end{array}$

7. A current of 2.5 ampere passes through the solution of metal sulphate for 30minutes and deposits 1.52 g of metal at cathode. Find the equivalent weight of the metal.

$\begin{array}{l} Given,\\ m = 1.52g\\ I = 2.5A\\ t = 30\min utes = 30 \times 60 = 1800s\\ E = ? \end{array}$

We have,

$\begin{array}{l} m = \frac{E}{{96500}}It\\ Or,1.52 = \frac{E}{{96500}} \times 2.5 \times 1800\\ Or,E = \frac{{1.52 \times 96500}}{{2.5 \times 1800}} = 32.6 \end{array}$

8. State and explain Faraday 1st law of electrolysis. An electrolytic cell contains a solution of CuSO4and anode of impose copper. How many kg of copper is deposited at cathode by 150 ampere passing for 12 hours?

$\begin{array}{l} Given,\\ I = 150A\\ t = 12hrs = 12 \times 60 \times 60 = 43200s\\ V = 2\\ A = 63.5amu\\ m = ?\\ m(Kg) = ? \end{array}$

We have,

$\begin{array}{l} m = \frac{A}{{96500V}}It\\ Or,m = \frac{{63.5}}{{96500 \times 2}} \times 150 \times 43200\\ Or,m = 2132.02g = 2.13Kg \end{array}$

9.State Faraday’s 1st Law of electrolysis. What current strength is required to deposit whole copper from 1liter of 1M CuSO4 solution by passing electricity through it in 10 minutes?

1L of 1M $CuS{O_4}solution$ =1mole $CuS{O_4}$

1mole $CuS{O_4}$ deposits 1mole Cu i.e.63.5g so mass of Cu deposited (m)=63.5g

$\begin{array}{l} Given,\\ t = 10\min = 600s\\ A = 63.5amu\\ V = 2\\ I = ? \end{array}$

We have,

$\begin{array}{l} m = \frac{A}{{96500V}}It\\ Or,63.5 = \frac{{63.5}}{{96500 \times 2}} \times I \times 600\\ Or,I = \frac{{96500 \times 2}}{{600}} = 321.67A \end{array}$

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