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 :

Options

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

Answer

B

Explanation

For this cell , reaction is `Zn + Fe^(2+) to Zn ^(2+) + Fe`
`E = E^(@) - (0.0591)/(n) "log" (c_(1))/(c_(2)) , E^(@) = E + (0.0591)/(n) "log" (c_(1))/(c_(2))`
`E^(@) = 0.2905 + (0.0591)/(2) "log" (10^(-2))/(10^(-3)) = 0.32 V `
`E^(@) = (0.0591)/(2) "log" K_(c)`
`log K_(c) = (0.32 xx 2)/(0.0591) = (0.32)/(0.0295)`
`therefore K_(c) = 10^((0.32)/(0.0295))`