The emf of the following cell is 0.2905 V at 298 K. `Zn|Zn^(2+) (a= 0.1 M) | |Fe^(2+) (a= 0.01 M) | Fe` The value

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Q 2535534462.     The emf of the following cell is 0.2905 V at 298 K.

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

The value of equilibrium constant for the cell reaction is
JEE 2004 Prelims
A

`10^(0.32//0.0591)`

B

`10^(0.32//0.0295)`

C

`10^(0.26//0.0295)`

D

`e^(0.32//0.295)`

HINT


Solution
(Provided By a Student and Checked/Corrected by EXXAMM.com Team)

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