If the `E_(cell)^(@)` for a given reaction has a positive value, then which of the following gives the correct relationship for the values of `DeltaG^(@)` and `K_(eq)` :-

Options

A. `DeltaG^(@)lt0,K_(eq)lt1`
B. `DeltaG^(@)gt0,K_(eq)lt1`
C. `DeltaG^(@)gt0,K_(eq)gt1`
D. `DeltaG^(@)lt0,K_(eq)gt1`

Answer

B

Explanation

`E_(cell)^(o)lt0`, so it is a non spontaneous process.
`DeltaG^(o)=-nFE^(@)=+ve,` so `DeltaG^(o)gt0`
`DeltaG^(@)=-2.303RT` log K
So, `K lt 1`