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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
(b) According to Nernst equation,
`E_(cell)=E_(cell)^(@)-(0.0591)/(n)logK_(c )`
At equilibrium point ,`E_(cell)=0`
`E_(cell)^(@)=(0.0591)/(n) log K_(c )`
`E_(cell)^(@)=(-0.44+0.76)=0.32" V"`.
`0.32=(0.0591)/(2)logK_(c )=0.0295 log K_(c )`
`K_(c )=10^((0.32)/(0.0295))`
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