Practice Problem 11

Use the following standard-state cell potentials to calculate the complex formation equilibrium constant for the Zn(NH₃)₄²⁺ complex ion.

Zn(NH3)42+ + 2 e- Zn + 4 NH3			Eored = -1.04 V
Zn2+ + 2 e- Zn			Eored = -0.7628 V

Solution

By reversing the first half-reaction and adding it to the second, we can obtain an overall equation that corresponds to the complex formation equilibrium.

Zn + 4 NH3 Zn(NH3)42+ + 2 e-			Eoox = 1.04 V
+ Zn2+ + 2 e- Zn			Eored = -0.7628 V
¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯			¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯
Zn2+ + 4 NH3 Zn(NH3)42+			Eo = 0.28 V

When this reaction comes to equilibrium, the cell potential is zero and the reaction quotient is equal to the complex formation equilibrium constant for the Zn(NH₃)₄²⁺ complex ion.

We now solve this equation for the natural logarithm of the equilibrium constant and calculate the value of K_f for this reaction.