JEE Advanced Model Paper 8 with solutions for online practice

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Paragraph for Questions Nos. 14 to 16

Electrical energy produced by a reversible electrochemical cell is given by the free energy decrease (-ΔG) of the reaction occurring in the cell. According to Gibbs- Helmholtz equation, decrease in freeenergy is given by -ΔG = -ΔH - T [
δ(ΔG)
δT
]
p
where -ΔH is the decrease in enthalpy of the cell reaction at constant pressure. EMF of the cell, E =
ΔH
nF
+ T [
δE
δT
]
p
. By measuring the emf of the cell and its temperature co-efficient,thermodynamic quantities like ΔH, ΔG and ΔS can be determined. Standard emf of the cell is related to equilibrium constant of the cell reaction as E° =
2.303RTlogk
nF
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Question : 14 of 84
 
Marks: +1, -0
EMF of the cell ,
A
(s)
|
A2+
(aq)
1M
|
|
B+
(aq)
0.1M
|
B
(s)
is found to be 1.475 Volt. The equilibrium constant of the cellreaction at 25°C is (approximately)
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