CHEMISTRY PROJECT-Ver ifi cati on of faraday ’s 1 s t law of e lectro lys is

Made by - Sidharth Gund Class : 12th – B

Roll no. : 31

Index• Certificate• Acknowledgement• Objective• Theory• Apparatus• Material required• Procedure• Observation• Calculations • Result• Bibliography

Certificate This is to certify that Sidharth Gund of class 12th – B , roll no.-

31 , of GREENWAY MODERN SR. SEC. SCHOOL has satisfactorily completed the project “verification of Faraday’s 1st law of electrolysis” , under my guidance and supervision ,

issued by CBSE during session 2014 -2015 .

____________Mrs. Nidhi Agnihotri

(PGT , Chemistry)School Stamp

Acknowledgement :-

It is my utmost pleasure to express my sincere gratitude to my chemistry teacher Mrs. Nidhi Agnihotri without whose cooperation I

would not have been able to complete my project. It was my privilege to work under her guidance and I heartily thank her for helping me through the course of my project. I am also grateful to Mrs. Manju

Kasana who helped me in selecting the topic for my project.

I would also like thank my parents for their valuable support

Objective :-

“To verify Faraday’s 1st law of electrolysis” (The mass of an elements which is deposited on an

electrode during electrolysis is directly proportional to the quantity of electricity which passes through the

electrolyte.)

Theory :- Faraday's First Law of Electrolysis: Statement -

The mass of an elements which is deposited on an electrode during electrolysis is directly proportional to

the quantity of electricity which passes through the electrolyte.

Explanation-

If M is the amount of substance which liberates or deposited at the electrode on passing the electricity

through the electrolyte and the quantity of electricity is Q, then

M α Qor M = ZQ

Z is the electrochemical constant for a given substance.

As, Q = i x t

therefore , we can write the statement of the first law of electrolysis mathematically as under :

M α i x tOr M= Zit

If 1 ampere electric current passes through the electrolyte for 1 second then M=Zit means that on passing the current of 1

ampere ampere for 1 second the weight of the substance deposited is equal to the electrochemical constant.

Z is known as the electrochemical equivalent (E.C.E) of the substance. If q = 1C, then m = Z.

unit of charge (Q) = Coulomb (C)unit of mass (m) = Kilogram (kg)unit of current (A) = ampere (A)

unit of electrochemical equivalent (Z) = kg/C

Apparatus :- Copper wires, copper plates, beakers, ammeter, rheostat,

weighing machine, battery and a key .

Materials Required :-Copper sulphate solution, Conc. H2SO4 .

Procedure :- • Prepare 500 ml copper sulphate solution and add a few drops

of conc. H2SO4 .• Now, divide the copper sulphate solution into 3 beakers

equally .• Take 6 copper plates :- 3 as cathodes and 3 for anodes .

Measure the weight of each cathode .• Set up of the circuit – Connect 2 beakers in parallel with the third beaker with the

help of copper electrodes and wires. Now, connect the battery, key, ammeter and rheostat .

• Now, close the circuit, note down the reading in ammeter and leave the set up for about 30 minutes.

• Switch off the circuit and weigh the cathodes again.

• Change the resistance with the help of rheostat and repeat the process one more time.

Observation :-

Two cathodes are in parallel with one cathode therefore, current in the cathodes that are parallel will be less than the one in

series Þ According to Faraday’s law mass deposited on the parallel

electrodes will be less than the mass deposited on the one in series.

Observation Table :- Time

(in minutes)

Mass of cathode*(in grams)

Current(mili

amperes)

(Before)M1

(After) M2

(Mass deposited=M2-M1)

M1st 2nd 3rd 1st 2nd 3rd 1st 2nd 3rd

a. 30 300

b. 30 500

* Here 1st cathode is the one in series and 2nd and 3rd are the ones in parallel

Calculations :-According to Faraday’s law – M=zit z=E/F E= equivalent mass (here of Cu2+ ion) = 63.546/2=31.773 F= Faraday’s constant =96500Þ z=31.773/(2x96500)=0.00016 Þ Mass deposited on 1st cathode= a. M=0.00016x300x(10^(-3))x30x60= 0.0864 g b. M=0.00016x500x(10^(-3))x30x60= 0.144 g

Mass deposited on 2nd cathode=a. M=0.00016x150x(10^(-3))x30x60= 0.0432 gb. M=0.00016x250x(10^(-3))x30x60= 0.072 g

Mass deposited on 3rd cathode=c. M=0.00016x150x(10^(-3))x30x60= 0.0432 gd. M=0.00016x250x(10^(-3))x30x60= 0.072 g

Result According to Faraday’s law-Mass α charge or M=zitMass deposited on cathodes is –

Experimentally :- 1st - a. _________g b. __________ g

2nd - a._________g b.__________g

3rd - a._________g b.__________g

Theoretically :- 1st - a. 0.0864 g b. 0.144 g

2nd - a. 0.0432 g b. 0.072 g

3rd - a. 0.0432 g b. 0.072 g

Experimental and theoretical values are equal and also mass deposited on 1st cathode is more than the mass deposited on 2nd and 3rd cathode since, the current is more in 1st cathode.

ÞFaraday’s first law of electrolysis is verified.

Bibliography The information about the project has been

gathered from following –

• www.chemistrysubject.blogspot.in• www.youtube.com