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JEE Advanced Model Paper 8 with solutions for online practice

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

+ 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

Question : 15

Total: 84

From the following values of electrode potentials,
i) (fumarate)2− + 2H+ + 2e− → (succinide)2− , E1° = 0.03 V and
ii) (pyruvate)− + 2H+ + 2e− → (lactate)− , E2° = -0.18 V . Calculate ΔG° for reaction ,
(fumarate)2− + (lactate)− → (succinide)2− + (pyruvate)−

-28.95 kJ
+40.53 kJ
-75.27 kJ
-40.53 kJ

Validate

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