Equilibrium Constant from Nernst Equation
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Q. The emf of the cell Zn|Zn2+(0.1M) || Fe2+|Fe(0.01M) is 0.2905 V. Equilibrium constant for the cell reaction is
- 10320.591
- 100.320.0295
- 100.260.0295
- 100.320.295
Q. The ΔG∘ for the Daniell Cell has been found to be - 212.3 kJ at 25∘C. Calculate the equilibrium constant for the cell reaction.
Q. E∘ for the cell is Zn |Zn2+ (aq)||Cu2+ (aq)| Cu at 25∘C, the equilibrium constant for the reaction Zn+Cu2+(aq)⇌Cu+Zn2+ (aq) is of the order of
- 10−37
- 10−28
- 10+18
- 10+17
Q. For a cell involving one electron, E0cell = 0.59 V at 298 K, the equilibrium constant for the cell reaction is:
[Given that 2.303RTF = 0.059 V at T = 298 K]
[Given that 2.303RTF = 0.059 V at T = 298 K]
- 1.0×102
- 1.0×105
- 1.0×1010
- 1.0×1030
Q. Electra has mastered the basics of Nernst equation. She is now studying equilibrium constants relation with Nernst equation.
Electra was performing experiments with Nernst with a Daniel Cell. She noticed that when the reaction reaches equilibrium, the cell potential goes to 0.
She asked Nernst why this happened. His reply was
Electra was performing experiments with Nernst with a Daniel Cell. She noticed that when the reaction reaches equilibrium, the cell potential goes to 0.
She asked Nernst why this happened. His reply was
- Because there is no displacement reaction
- Because there is change in temperature
- Because there is no change in concentration of Cu2+ and Zn2+
- Because all the ions have been depleted.
Q. Maxium work that can be done by a chemical cell is equal to the change in gibbs free energy.
- True
- False
Q.
The equilibrium constant for the reactions and are and , respectively, the relation between and is
Q. EMF of the cell, Ag|AgNO3(0.1M)|KBr(1N), AgBr(s)|Ag is −0.6 V at 298 K
Calculate the Ksp of AgBr at 298 K.
Calculate the Ksp of AgBr at 298 K.
- Ksp=4.0×10−12
- Ksp=3.5×10−12
- Ksp=4.8×10−12
- Ksp=3.8×10−12