The emf of the cell,
`Zn|Zn^(2+) (0.01 M)||Fe^(2+) (0.001 M)|Fe` at 298 K is 0.2905 V then the value of equilibrium constant for the cell reaction is :

The emf of the cell, Zn|Zn^(2+)(0.01M)|| Fe^(2+)(0.001M) | Fe at 298 K is 0.2905 then the value of equilibrium constant for the cell reaction is:

The standard e.m.f. of the cell Zn| Zn^(2+) (0.01 M) || Fe^(2+) (0.001 M) | Fe at 298 K is 0.02905 then the value of equilibrium constant for the cell reaction 10^(x) . Find x.

Zn|Zn^(2+)(a=0.1M)||Fe^(2+)(a=0.01M)Fe. The EMF of the above cell is 0.2905 . The equilibrium constant for the cell reaction is

Zn|Zn^(2+)(a=0.1M)||Fe^(2+)(a=0.01M)Fe. The EMF of the above cell is 0.2905 . The equilibrium constant for the cell reaction is