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

Answer

A

Explanation

The relationship between `E_("cell")^(@)` for a galvanic cell and `Delta^(@)`, the standard Gibbs energy change for the chemical reaction of the cell is
`Delta_(r)G^(@) = -nfE_("cell")^(@)`
If `E_("cell")^(@)` is negative, then `DeltaG_("cell")^(@)` will be positive, `G_("cell")^(@)gt0`. From standard Gibbs energy of the reacton, we can calculate the equilibrium constant by the equation:
`Delta_(r)G^(@) = -RTInK`
`Delta_(r)G^(@) = -2.303 RTlogK_(eq.)`
If `Delta_(r)G^(@)` is positive, log `K^(eq.)` should be negative. This implies that `K_(eq.)lt1`.