1.

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 :A. `e^(0.32//0.0295)`B. `10^(0.32//0.0295)`C. `10^(0.26//0.0295)`D. `10^(0.32//0.0591)`

Answer» Correct Answer - B


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