QUESTION 11

The standard electrode potential for the Cd/Cd²⁺ couple implies the following electrochemical cell

                    Pt|H₂(g) (P=1 atm) | H⁺(aq) (a=1) ||Cd²⁺(aq)(a_Cd2+) |Cd

The corresponding half-cell reactions are

Right-hand electrode: 1/2 Cd²⁺(aq) + e^- ⇌  1/2 Cd(s)  

Left-hand electrode: H⁺ (eq) + e^-  ⇌   1/2 H₂(g)

Accordingly, the formal cell reaction is

                    1/2 Cd²⁺(aq) + 1/2 H₂(g) ⇌  1/2 Cd(s) + H⁺ (eq) (eqn 1)

For which

△G^o = -FE^o

         = -0.400 F

Similarly, for

                      1/2 Cd[NH₃]₄²⁺(aq) + e^- ⇌  1/2 Cd(s) + 2NH₃ (aq)         E^o = -0.610V

The formal cell reaction is

1/2 Cd[NH₃]₄²⁺(aq) + 1/2 H₂(g) ⇌  1/2 Cd(s) + 2NH₃ (aq) + H⁺ (eq) (eqn 2)

△G^o = -FE^o

= -0.610

Subtracting eqn 1 and 2 gives

1/2 Cd²⁺(aq) + 2 NH₃(aq) ⇌  1/2 Cd[NH₃]₄²⁺(aq)

The Gibbs energy of the above ligand exchange reaction is

△G^o = -F(-0.400 – (-0.610)) = -20.2kJmol^-1

△G^o = -RTln(K)

K = exp (20.2x10³ / 8.31x298) = (a _Cd[NH3]4²⁺)^1/2 / (a _Cd²⁺)^1/2 (a _NH3)²

K = 3500