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Question : 5
Total: 12
(a) Calculate ∆ r G ∘ log K c
Ni ( s ) + 2 Ag + ( a q ) → Ni 2 + ( a q ) + 2 Ag ( s )
Given thatE ∘ cell = 1.05 V , 1 F = 96 , 500 C mol − 1
OR
(b) Calculate the e.m.f. of the following cell at298 K
Fe ( s ) | Fe 2 + ( 0.001 M ) | | H + ( 0.01 M ) | H 2 ( g )
(1 bar) | Pt (s)
Given thatE ∘ cell = 0.44 V
[ log 2 = 0.3010 , log 3 = 0.4771 , log 10 = 1 ]
Given that
OR
(b) Calculate the e.m.f. of the following cell at
(1 bar) | Pt (s)
Given that
Solution:
According to the equation,
Ni + 2 Ag + → Ni 2 + + 2 Ag
∆ G = − n FE ∘
where ∆ G = Gibb ′ ' free energy
∆ G = − 2 × 96500 × 1.05
N = No. of electrons gain or lost = 2
∆ G = − 202.650 kJ
F = Faraday's constant = 96500
E ∘ = Standard emf = 1.05 V
The relation between Gibb's free energy and Equilibrium constant is given by equation
E ∘ cell =
log K c
log K c = −
= 35.53
K c = 3.39 × 10 35
OR
According to the equation,
Fe ( s ) + 2 H + ( aq ) → Fe + 2 ( aq ) + H 2 ( g )
E ° cell = E ° cathode − E ° anode
E ° cell = 0 − ( − 0.44 ) V
E ° cell = + 0.44 V
By applying Nernst Equation,
E cell = E ° cell −
log
E cell = 0.44
log
E cell = 0.44 −
log 10
E cell = 0.44 − 0.0295 × 1
E cell = + 0.410 V
The relation between Gibb's free energy and Equilibrium constant is given by equation
OR
According to the equation,
By applying Nernst Equation,
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