The e.m.f. of the cell `Zn|Zn^(2+)(0.01M)||Fe^(2+)(0.001M)|Fe` at 298K is 0.2905 then the value of equilibrium for the cell reaction isA. `e^((0.32)/(0.0295))`B. `10^((0.32)/(0.0295))`C. `10^((0.26)/(0.0295))`D. `10^((0.32)/(0.0591))`

The e.m.f. of the cell `Zn|Zn^(2+)(0.01M)||Fe^(2+)(0.001M)|Fe` at 298K

1.	The e.m.f. of the cell `Zn\|Zn^(2+)(0.01M)\|\|Fe^(2+)(0.001M)\|Fe` at 298K is 0.2905 then the value of equilibrium for the cell reaction isA. `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 For this cell, reaction is:` Zn+Fe^(2+)toZn^(2+)+Fe` `E=E^(o)-(0.0591)/(n)"log"(c_(1))/(c_(2)),E^(o)+(0.0591)/(n)"log"(c_(1))/(c_(2)) ` `=0.2905+(0.0591)/(2)"log"(10^(-2))/(10^(-3))=0.32V`. `E^(o)=(0.0591)/(2)log" "K_(c),log" "K_(c)=(0.32xx2)/(0.0591)=(0.32)/(0.0295)` `thereforeK_(c)=10^((0.32)/(0.295))`.

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