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Notes de cours

Physical chemistry

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Chemistry physical subject notes for class is very help full for students board exam.

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Publié le
15 mars 2025
Nombre de pages
28
Écrit en
2024/2025
Type
Notes de cours
Professeur(s)
Bheeshma teja
Contient
Class 12 science

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Tetrahedron Group of Institutions, Tangi, Cuttack Subjective Tetrahedron Group of Institutions, Tangi, Cuttack



TETRAHEDRON GROUP OF INSTITUTIONS
Tangi, Cuttack-754022
SELECTION
Physical Chemistry (Subjective)

E
Electrochemistry or W  Z I T  It - - - (2)
F
1. State & Explain Faraday’s First where, E = equivalent mass of the
law of Electrolysis . (IMP) substance,
(2 marks)
F = Faraday or 96500 C.
Ans: The mass of the substance depos-
ited or liberated at the electrodes 2. What is Electro Chemical
during the process of electrolysis, equivalent and What is its unit .
is directly proportional to the quan- (2 marks) (IMP)
tity of electric charge passed through Ans : From Faraday’s First law of
the electrolyte. Electrolysis
Mathematically, W Q  W  ZQ or W  Z I t
Hence W=Z if Q= 1 coulomb or,
W Q W ZQ or W ZI t I = 1amp. & t = 1sec.
Therefore, the electrochemical
where, W = mass of the substance
equivalent or E.C.E. of any substance
deposited or l i be r at ed at t h e
is defined as the mass of the
electrode, Q = amount of electric
substance deposited or liberated at
charge in coulomb, I = amount of
the electrode when 1 coulomb of
electric current in Ampere,
electric charge is passed or 1 Ampere
t = time in second, of electric current is passed for 1
Z = electrochemical equivalent or second.
E.C.E. of a substance. We know that
Unit of Electrochemical equivalent :
E
Z  W = ZQ
F W
 Z  gm/coulomb gm coulomb1 
Q
Therefore, 1 1
 gm amp sec .
E
W  ZQ  Q - - - (1)
F

1

,Tetrahedron Group of Institutions, Tangi, Cuttack Subjective Tetrahedron Group of Institutions, Tangi, Cuttack


3. State & Explain Faraday’s Second MgCl 2 
 Mg 2   2Cl 
law of Electrolysis . (IMP) Atomic weight 24
(2 marks) E   12
Valency 2
Ans : When the same quantity of Q=0.5F
electricity is passed through different
According to Faraday’s 1st Law.
electrolytic solution connected in
series then the mass of different E 12
W  Q   0.5F
substance deposited at different F F
electrodes are in the ratio of their
5 12
equivalent masses.  12    6 gm
10 2
Explanation :
When the same quantity of electricity 5. Calculate the weight of Copper
is passed through two electrolytic diposited at cathode by passing
solution containing aq. AgNO3 and 50ampere of electric current for
aq. CuSO4 then, 16mins & 5secs through aqueous
solution of copper sulphate.
(atomic weight of Cu= 63.5)
Ans : CuSO 4 
 Cu 2   SO 42 
Atomic weight
E
Valency
63 .5
  31 .75
the mass of silver deposited 2

the mass of copper deposited I= 50 ampere
equivalent mass of Ag F = 96500C

equivalent mass of Cu
t = 16×60+5=965 second.
4. Calculate the weight of
According to Faraday’s 1st law
Magnesium diposited at cathode
by passing 0.5 Faraday of E
W   I t
electricity through molten MgCl2. F
(atomic weight of Mg=24) 31.75
=  50  965
Ans : W = ? 96500

2

, Tetrahedron Group of Institutions, Tangi, Cuttack Subjective Tetrahedron Group of Institutions, Tangi, Cuttack


31 .75 Hence, the unit of conductance is
  15 .875 gm
2 “  –1” or “mho” or “Siemen” or
6. Calculate the charge carried by 2 “  ”.
mole of electrons.
10. What is specific resistance &
Ans : The charge of 1 mole of electron
what is its unit. ( 2 marks) (IMP)
= 1 Faraday
l l
The charge of 2 mole of electron = Ans : R   R , where,
a a
2 Faraday
R = resistance of the conductor,
7. Calculate the charge carried by 1
 = specific r e si s t an ce o r
m ole of A l 3+
ion. resistivity of the conductor,
Ans : Al 
 Al 3  3e  l = length of the conductor in cm,

The charge carried by 1 mole of Al3+ a = area of cross-section of the
conductor in cm2.
= 3 Faraday.
8. What is resistance & what is its R ρ
unit. (2 mark) if l = 1 cm and a = 1 cm2 or V = 1
Ans : The resistance of any conductor is
cm3 = 1 cc = 1 ml.
defined as the properties of the
conductor by which it opposes the Therefore, the specific resistance or
flow of electric current through the resistivity of any conductor is equal
conductor. to resistance of the conductor having
The unit of resistance is “Ohm” or length 1 cm and area of cross-section
“  ”. 1 cm2 or volume of 1 cm3 or 1 cc or
9. What is conductance & what is its
1 cubic cm.
unit. (2 marks)
Ans : It is the properties of the conductor Unit of specific resistance :
by which it allows the flow of electric
current through the conductor. l a
R  R
a l
Or It is the reciprocal of the resistance
of the conductor. cm2
 Ohm  Ohmcm  Ohm. m (in S.I.)
cm
1
Mathematically, C
R

3
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