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

The emf of the cell,
`Zn|Zn^(2+) (0.01 M)||Fe^(2+) (0.001 M)|Fe` at 298 K is 0.2905 V then the value of equilibrium constant for the cell reaction is :

A

`e^((0.32)/(0.0296))`

B

`10^((0.32)/(0.0296))`

C

3.26 V

D

`10^((0.320)/(0.0592))`

Text Solution

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The correct Answer is:
B
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