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A SEMONAR

ON

FARADAY’S LAW BY

TUSHAL KYADA

ROLL NO: 911

GUIDE: Dr. V.V.MATHANE

ELECTROMETALLURGY AND CORROSION

MATERIAL SCIENCE & METALLURGICAL ENGINEERING DEPARTMENT,

FACULTY OF TECHNOLOGY & ENGINEERING,

KALABHAVAN,BARODA

OCTOBER-2011.



FARADAY’S LAW

INTRODUCTION:

Michel Faraday was an English chemist. In 1833 he established certain

relationships between the quantity of electricity passed through an electrolyte and

the amount of any material liberated/deposited at the electrode. His findings were

known as the two “laws of electrolysis”.

For example, the thickness of the silver plate that is deposited electrochemically on

a spoon is greater if the electrical current is allowed to flow longer.



FIRST LAW

According to it “during electrolysis, the amount of any substance deposited or

evolved at any electrode is proportional to the quantity of electricity passed”.

Quantity of electricity passed (Q) is equal to the product of the current and the time

for which it is passed.

Q = current * time.

Q = I * T.

The practical unit of electricity is coulomb, which is the quantity of electricity

passed through a conductor/electrolyte, when a current of 1 ampere flows for 1

second.

Thus, coulomb =ampere * second

If W is the weight of the substance liberated/deposited at an electrode during

electrolysis then from 1st law

W ∞ Q



But, Q = I * T.

W = Z * I * T.

So, W ∞ I * T.

where,

W = wt. of substance formed at the electrode {gms}

Z = electrochemical equivalent.

I = current {amp.}

T = time {sec.}

In order to use Faraday’s law we need to recognize the relationship between current, time

and the amount of electric charge that flows through a circuit.

By definition, 1 coulomb of charge is transferred when 1-amp current flows for 1 second.

1 coulomb = 1 amp-sec.



APPLICATION:

Converse application of this law permits the use of quantity of electrolytically decomposed

matter as a measure of the quantity of electricity used. The apparatus used for this purpose is

calorimeter.

SECOND LAW:

According to it “the amount of different substance evolveld/deposited by the passage of

same quantity are proportional to their chemical equivalent weights”

W ∞ E.

where,

W = wt. of substance liberated or deposited

E = chemical equivalent weight of substance liberated/deposited.

E = A

------Ne.



It was observed during the process that the quantity of electricity which must be passed

in order to deposit / evolve 1 gram equivalent of any substance is constant and this is

called FARADAY which is equivalent to 96,500 coulomb.

1 faraday = Amount of electricity required to deposit 1 gram equivalent of any substance

From 2nd law,

1 gram equivalent of substance will be deposited by 96,500 coulomb.

Thus, A/Ne gram of substance will be deposited by 96,500.

W gram will be deposited by

F *W-------(A/Ne)

Quantity of electricity = I * T

Thus,

F * W/(A/Ne) =I * T

This equation represents combined form of faraday’s 1st & 2nd law.



APPLICATION:

It helps in determining the equivalent weight of an element.

Example,

If the same amount of electricity is passed through two electrolytic cells containing the

solution of two different electrolytes AB and CD and if the weights of A and C

deposited/evolved on their resp. electrodes are “X” and “Y” grams. Then according to 2 nd

law.

weight of A deposited Chemical equivalent of A----------------------------- = -------------------------------weight of C deposited Chemical equivalent of C

X Ea----- = -----Y Ec

Thus knowing the weights of two elements A and C deposited and equivalent weight of any

one, the equivalent weight of the other can b calculated.



PROBLEMS ON FARADAY’S LAW:

EXAMPLE-1.

Calculate the weight of silver deposited when current of 0.5 ampere strength was passed for

40 minutes in an electrolytic cell, containing silver nitrate solution.

(Chemical equivalent of silver = 108).

Solution:

Quantity of electricity passed = Ampere (I) * time in sec. (T).

= 0.5 * (40 * 60)

Now we know that:

96,500 coulomb deposit = 1 gm equivalent of Ag.

Thus1,200 coulomb deposit = 1,200 * 108 of Ag

-----------------------

96,500

= 1.343 gm of Ag.



EXAMPLE-2.

How long would it take to deposit 100 gm of AL ( at. Wt.=27 ) from an electrolyticcell, containing AL2O3 at a current of 125 amp.?

Solution:

Al + 3e = Al (s).3 mole = 1 mole =27 gm.

therefore, 27 gm of AL needs = 3 mole of electrons.

or,100 gm of AL needs = 3 * 100 mole of electrons

-------------------------------

27= 3 * 100 * 96,500 coulomb---------------------------------

27

= 1.0722 * 10^6 coulomb.

Quantity of electricity passed, Q = 125 amp * t sec.

= 125 t coulomb.

Hence,

125 t = 1.0722 * 10^6

1 0722 * 10^6



t = 1.0722 * 10^6 sec---------------------

12

= 8,577.8 sec

t = 2 hr, 22 min, 57.8 sec.

ADVANTAGE:

Estimation of amount of electricity required for electrolytic process.

Weight of substance evolved/deposited during electrolysis process.

Applicable to Fused salt as well as aqueous solution.

tushal kyada ppt

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