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The e.m.f of the cell Zn| Zn^(2+) (0.01 ...

The e.m.f of the cell `Zn| Zn^(2+) (0.01 M)|| Fe^(2+) (0.001 M) | 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
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