Compute the equilibrium constant, K, at 25 ∘C for the reaction between Ni2+(aq) and Zn(s), which form Ni(s) and Zn2+(aq)

The answer to the question is;

The equilibrium constant,K, at 25 °C for the reaction between Ni²⁺(aq) and Zn(s), which form Ni(s) and Zn²⁺ is 2.04×10¹⁷.

Explanation:

The half reactions are as follows

Ni²⁺ (aq) + 2e⁻ -> Ni (s)

Zn (s) -> Zn²⁺ (aq) + 2e⁻

For the Ni²⁺/Ni system we have the potential given as -0.23V (Reduction)

For the Zn²⁺/Zn sytem, the potential is -0.76. Here however, we should note that the zinc is being oxidized and therefore the potential is positive, that is;

Zn/Zn²⁺ = 0.76

Therefore the voltage for the sum of the reactions on both sides of the process is

-0.23 V + 0.76 V = 0.53 V    

We then call upon the Nernst equation to calculate the equilibrium constant as follows

E⁰ =

Where:

E⁰ = Standard cell potential = 0.53 V

n = Number of moles of electrons = 2 moles of e⁻

K = Equilibrium constant

Therefore we have

0.53 V =

Therefore log K = 17.905

and K =   = 2.04×10¹⁷.