_Cuo +H, → _Cu + _H,0
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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 =
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.