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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)/(0.295))`

C

`10^((0.26)/(0.0295))`

D

`10^((0.32)/(0.0591))`

Text Solution

Verified by Experts

The correct Answer is:
B
`Zn+Fe^(2+)toZn^(2+)+Fe(n=2)`
`E=E^(@)-(0.0591)/(n)logK_(c)`
`therefore0.2905=E^(@)-(0.0591)/(2)"log"(0.01)/(0.001)`
or `E^(@)=0.2905+0.0295=0.32` volt
`E^(@)=(0.0591)/(n)"log"K_(eq)`
`therefore0.32=(0.0591)/(2)"log "K_(eq)thereforeK_(eq)=10^((0.32)/(0.0295))`
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