The emf of the cell, `Zn|Zn^(2+)(0.01M)|| Fe^(2+)(0.001M)

1.	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:A. `e^((0.32)/(0.0295))`B. `10^((0.32)/(0.0295))`C. `10^((0.26)/(0.0295))`D. `10^((0.32)/(0.059))`
Answer» Correct Answer - B `Zn\|Zn^(2+)(0.01M)\|\|Fe^(2+)(0.001M)\|Fe," "E=0.2905` cell reaction, `Zn+Fe^(2+)hArrZn^(2+)+Fe`. `0.2905=E^(@)-(0.0591)/(2)log((0.01)/(0.001)` `E^(@)=0.32"volt"` At equilibrium `E_(cell)=0` `0=0.32-(0.0591)/(2)logK_(eq)` `K_(eq)=10^(0.32//0.0295)`

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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:A. `e^((0.32)/(0.0295))`B. `10^((0.32)/(0.0295))`C. `10^((0.26)/(0.0295))`D. `10^((0.32)/(0.059))`

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