If `E_(cell)^(ɵ)` for a given reaction is negative, which gives the correct relationships for the values of `DeltaG^(ɵ)` and `K_(eq)`?

Options

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

Answer

A

Explanation

(a) Given, `E_(cell)^(@)=-ve`
The relation between `DeltaG^(@) and E_(cell)^(@)` is given as
`DeltaG^(@)=-nFE_(cell)^(@)`…(i)
If `E_(cell)^(@)` is negative, so `DeltaG^(@)` comes out to be positive. Again, relation between `DeltaG^(@)` and `K_(eq)` is given as
`DeltaG^(@) =- 2.303 nRT log K_(eq)` ...(ii)
From Eq. (i) we get that `DeltaG^(@)` is positive . Now , if `DeltaG^(@)` is positive then `K_(eq)` comes out to be negative from eq (ii).
i.e `DeltaG^(@) gt 1 and K_(eq) lt 1`
Short trick As `E_(cell)^(@)` is negative so reaction is non-spontaneous or you can say reaction is moving in backward direction. For non-spontaneous reaction. `DeltaG^(@)` is always positive and `K_(eq)` is always less than 1.