if the `E^(@)` cell for a given reaction has a negative value of the following gives the correct relationship for the of `DeltaG^(@)` and `K_(eq)`?

Options

A. `DeltaG^(@) lt 0 , K_(eq) gt 1`
B. `DeltaG^(@) lt 0, K_(eq) lt 1`
C. `DeltaG^(@) gt 0, K_(eq) lt 1`
D. `DeltaG^(@) gt 0: K_(eq) gt 1`

Answer

C

Explanation

`DeltaG(@) =-nFE^(@)"cell" , E_("cell")^(@) =(-ve)`
So, `DeltaG^(@) = (+ve), DeltaG gt 0`
Also, `DeltaG^(@) =2.303 RT log K_(eq)`
`therefore K_(eq) lt 1`