Respuesta :
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[tex]_{cell}[/tex] = [tex]\frac{0.0592}{n} logK[/tex]
Where:
E⁰[tex]_{cell}[/tex] = 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 = [tex]\frac{0.0592}{2} logK[/tex]
Therefore log K = 17.905
and K = [tex]10^{17.9054}[/tex] = 2.04×10¹⁷.
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
