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Zn|Zn^(2+)(a=0.1M)||Fe^(2+)(a=0.01M)|Fe....

`Zn|Zn^(2+)(a=0.1M)||Fe^(2+)(a=0.01M)|Fe`.
The emf of the above cell is 0.2905V. Equilibrium constant for the cell reaction is

A

`10^(0.32//0.0591)`

B

`10^(0.32//0.0295)`

C

`10^(0.26//0.0295)`

D

`10^(0.32//0.295)`

Text Solution

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The correct Answer is:
B
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