Question

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

Answer 1

Answer:

Explanation:

E°cell = 0.0592/n × log K

In this electrochemical cell, zinc is being oxidised while Nickel is reduced.

The half cell reactions:

Ni2+ (aq) + 2e- --> Ni(s)

Zn(s) --> Zn2+(aq) + 2e-

The reduction potential, E° of Ni2+/Ni = -0.23V.

The reduction potential, E° of Zn2+/Zn = -0.76.

E°cell = E°cathode - E°anode

= -0.23 - (-0.76)

= 0.53 V

Since, number of electrons transferred, n = 2

E°cell = 0.53 V

Log K = (0.53 × 2)/0.0592

Log K = 17.905

K = 8.04 × 10^17

Answer 2

Answer:

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$_{cell}$ = $\frac{0.0592}{n} logK$

Where:

E⁰$_{cell}$ = 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 = $\frac{0.0592}{2} logK$

Therefore log K = 17.905

and K =   $10^{17.9054}$ = 2.04×10¹⁷.