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

Answer

D

Explanation

`E_(cell)^(@)=positive`
`DeltaG^(@)=nF E_(cell)^(@)`
`DeltaG^(@)=Negative`
`DeltaG^(@)=-2.303RTlogK_(eq)`
`"log"K_(eq)=+"ve value"`
`K_(eq)gt1`