The emf of the cell Zn|Zn2+(0.1M) || Fe2+|Fe(0.01M) is 0.2905 V. Equilibrium constant for the cell reaction is

The emf of the cell Zn|Zn2+(0.1M) || Fe2+|Fe(0.01M) is 0.2905 V. Equil

1.	The emf of the cell Zn\|Zn2+(0.1M) \|\| Fe2+\|Fe(0.01M) is 0.2905 V. Equilibrium constant for the cell reaction is
Answer» The emf of the cell $Z n \| Z n^{2 +} (0.1 M) \| \| F e^{2 +} \| F e (0.01 M)$ is $0.2905 V$ . Equilibrium constant for the cell reaction is