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The emf of the cell, Zn|Zn^(2+)(0.01M)...

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)/(0.0591))`

Text Solution

Verified by Experts

The correct Answer is:
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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