Question

The equilibrium constant (K) for the reaction\[Cu(s)+2A{{g}^{+}}(aq)\to C{{u}^{2+}}(aq)+2Ag(s)\]will be [Given, ${{E}^{0}}_{cell}=0.46V$]
(a) ${{K}_{c}}=$ Antilog 15.6
(b) ${{K}_{c}}=$Antilog 2.5
(c) ${{K}_{c}}=$ Antilog 1.5
(d) ${{K}_{c}}=$ Antilog 12.2

Answer 1

Hint: The answer here is based on the concept of physical chemistry that uses the formula to calculate equilibrium constant which is given by Nernst equation, ${{E}_{cell}}={{E}^{0}}_{cell}-\dfrac{0.0591}{n}\log {{K}_{c}} $

Complete step – by – step answer:
We have studied the problems regarding concepts that deal with the finding of the emf of the cell by various methods in our previous classes.
Let us now deal with the same problem which is given in the above question.
Here, in this question we are supposed to find out the equilibrium constant.
- Equilibrium constant is nothing but the value of its reaction quotient, a state that is approached by the chemical system after sufficient elapsed time where its chemical composition has no measurable tendency for further change.
- Nernst equation is the one that deals with the relationship between reduction potential of an electrochemical reaction to that of standard electrode potential, activities of species which undergo oxidation and reduction.
Here, in the above reaction it is half cell reaction and is given by,
At anode, oxidation takes place as
$Cu\left( s \right)\to C{{u}^{2+}}+2{{e}^{-}}$
and at cathode reduction takes place as
$2A{{g}^{+}}+2{{e}^{-}}\to 2Ag\left( s \right)$
Therefore value of n=2
According to Nernst equation,
${{E}_{cell}}={{E}^{0}}_{cell}-\dfrac{0.0591}{n}\log {{K}_{c}}$
At equilibrium ${{E}_{cell}}$=0
\[0=0.46-\dfrac{0.0591}{2}\log {{K}_{c}}\]
\[\Rightarrow \log {{K}_{c}}=\dfrac{0.46\times 2}{0.0591}\]
$\Rightarrow {{K}_{c}}=Anti\log 15.567\approx Anti\log 15.6$

Therefore, the correct answer is option (a) ${{K}_{c}}=$ Antilog 15.6

Note: Note that the solids do not participate in cell reaction at equilibrium conditions as it is considered to be having the value of unity because the molar concentrations of pure solids and liquids are unity.