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For a cell involving two electron change...

For a cell involving two electron changes, `E_("cell")^(@) =0.3V` at `25^(@)C`. The cell equilibrium constant of the reaction is

A

`10^(-10)`

B

`3xx10^(-2)`

C

10

D

`10^(10)`

Text Solution

Verified by Experts

The correct Answer is:
D
`log_(10)K=(-DeltaG^@)/(2.303RT)=(nFE_(cell)^@)/(2.303RT)` (at `25^@C`)
`log_10K=(nE_(cell)^@)/(0.0591)=(2xx0.3)/(0.0591)=10impliesK=10^10`
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