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Question : 28
Total: 30
(a) Define the following terms:
(b) Resistance of a conductivity cell filled with 0.1 molL 1 KCl solution is 100 W . If the resistance of the same cell when filled with 0.02 mol L − 1 KCl solution is 520 Ω , calculate the conductivity and molar conductivity of 0.02 mol L − 1 KCl solution. The conductivity of 0.1 mol L − 1 KCl solution is 1.29 × 10 − 2 Ω − 1 cm − 1 .
OR
(a) State Faraday's first law of electrolysis. How much charge in terms of Faraday is required for the reduction of1 mol of Cu 2 + to Cu .
(b) Calculate emf of the following cell at298 K :
Mg ( s ) | Mg 2 + ( 0.1 M ) | | Cu 2 + ( 0.01 M ) | Cu ( s )
[ Give E cell 0 = ± + 2.71 V , F = 96500 C mol − 1 ]
(b) Resistance of a conductivity cell filled with 0.1 mol
OR
(a) State Faraday's first law of electrolysis. How much charge in terms of Faraday is required for the reduction of
(b) Calculate emf of the following cell at
Solution:
(a) (i) When the concentration of the electrolyte approaches zero, the molar conductivity is termed as limiting molar conductivity. It is represented by Λ m .
(ii) Fuel cell: Fuel cells are the galvanic cells or electrochemical cells that transform the chemical energy into electrical energy from fuel combustion by redox reaction, such as hydrogen, methanol, etc.
Given,
For0.1 mol L − 1 KCl solution
Resistance( R ) = 100 Ω
Conductivity( k ) = 1.29 × 10 − 2 Ω − 1 cm − 1
Cell constant( G * ) = k × R
= 1.29 × 10 − 2 Ω − 1 cm − 1 × 100 Ω
= 1.29 cm − 1
For0.02 mol L − 1 KCl solution
Resistance( R ) = 520 Ω
Conductivity =
K =
K = 2.48 × 10 − 3 Ω − 1 cm − 1
( C ) = 0.02 mol L − 1
= 0.02 × 10 − 3 mol cm − 3
Concentration( C ) = 0.02 mol L − 1
= 0.02 × 10 − 3 mol cm − 3
Molar conductivity( Λ m ) =
=
Λ m = 124 Ω − 1 cm 2 mol − 1
OR
(a) Faraday's first law of electrolysis states that "The mass of a substance deposited at any electrode is directly proportional to the amount of charge passed."
m = Z × Q
where,m = mass of a substance deposited or liberated at an electrode.
Q = amount of charge passed through it and
Z = electrochemical equivalent
The reduction of onemol of Cu 2 + to Cu can be represented as:
Cu 2 + + 2 e − → Cu
Since,2 mol of electrons are involved in the reduction, so the amount of charge required is 2 F .
(b) The given cell reaction can be represented as
Mg ( s ) + Cu 2 + ( aq ) → Mg 2 + ( aq ) + Cu ( s )
E cell = E ∘ −
log
E cell = 2.71 −
log
E cell = 2.71 −
log 10
E cell = 2.68 V
(ii) Fuel cell: Fuel cells are the galvanic cells or electrochemical cells that transform the chemical energy into electrical energy from fuel combustion by redox reaction, such as hydrogen, methanol, etc.
Given,
For
Resistance
Conductivity
Cell constant
For
Resistance
Concentration
Molar conductivity
OR
(a) Faraday's first law of electrolysis states that "The mass of a substance deposited at any electrode is directly proportional to the amount of charge passed."
where,
The reduction of one
Since,
(b) The given cell reaction can be represented as
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