Question

The equilibrium constant $(K)$ for the below reaction will be:
$Cu(s)+2A{{g}^{+}}(aq)\to C{{u}^{2+}}(aq)+2Ag(s)$
[Given, $E{{{}^\circ }_{cell}}=0.46\,V$ ]
A.$antilog\,15.6$
B.$antilog\,2.5$
C.$antilog\,1.5$
D.$antilog\,12.2$

Answer 1

Gibbs free energy is the energy that is associated with a chemical reaction which is used to do work. The temperature in the Gibbs free energy equation must be in Kelvin. In an electrochemical cell, the cell potential can be predicted from the half reaction and its conditions (temperature, concentration and pressure). An electrochemical cell is defined as a device that has the ability to convert a chemical reaction into electrical energy.
Formula used:
1.$\Delta G{}^\circ =-nFE{{{}^\circ }_{cell}}$
here, $\Delta G{}^\circ$ is the standard Gibbs free energy, $n$ is the number of electrons transferred, $F$ is the Faraday’s constant and $E{{{}^\circ }_{cell}}$ is the standard cell potential.
2.$-2.303RT\log K=-nFE{{{}^\circ }_{cell}}$
where, $R$ is the universal gas constant, $T$ is the temperature and $K$ is the equilibrium constant.

Complete step by step answer:
Here, it is given that standard cell potential $(E{{{}^\circ }_{cell}})$ is $0.46\,V$ .
The given chemical reaction is:
$Cu(s)+2A{{g}^{+}}(aq)\to C{{u}^{2+}}(aq)+2Ag(s)$
$\Delta G{}^\circ =-nFE{{{}^\circ }_{cell}}$
where, $\Delta G{}^\circ$ is the standard Gibbs free energy, $n$ is the number of electrons transferred (here, $n=2$
), $F$ is the Faraday’s constant $(F=96500\,C\,mo{{l}^{-1}})$ and $E{{{}^\circ }_{cell}}$ is the standard cell potential.
Here, $n=2$
$\Delta G{}^\circ =-2.303RT\log K$
where, $R$ is the universal gas constant $(R=8.314\,J{{K}^{-1}}mo{{l}^{-1}})$ , $T$ is the temperature and $K$ is the equilibrium constant.
Here, the reaction is in equilibrium, hence the temperature is $298K$ .
Now, equating both the equations, we get,
$-2.303RT\log K=-nFE{{{}^\circ }_{cell}}$
Now, substituting the values in the above formula, we get,
$-2.303\times 8.314\times 298\times \log K=-2\times 96500\times 0.46$
On further solving, we get,
$\Rightarrow \log K=15.6$
$\Rightarrow K=antilog\,15.6$
Equilibrium constant for the given reaction is $antilog\,15.6$
Therefore, the correct option is (A), that is, $antilog\,15.6$

Additional information:
-$\Delta G{}^\circ$ is defined as the energy change under standard conditions like pressure at $1\,atm$ and temperature at $25{}^\circ C$
-$\Delta G{}^\circ$ is the change in the Gibbs free energy which is used for the measurement of energy constant. If the change in the Gibbs free energy is less than zero, then it gives an exothermic reaction. If the change in Gibbs free energy is more than zero, then it gives an endothermic reaction.
-Standard cell potential is defined as the potential of cell under concentration of $1$ mole per litre and pressure of $1\,atm$ at $25{}^\circ C$
-Faraday’s constant is denoted with a symbol $F$ and is defined as the charge in coulombs carried by one mole of the electron.
$1F=96500\,C\,mo{{l}^{-1}}$
$n$ is the number of electrons transferred in the reaction.

Note: The relationship between $\Delta G{}^\circ ,\Delta E{}^\circ ,n,F$ is:
$\Delta G{}^\circ =-nFE{}^\circ$
Standard cell potential is the potential of a cell under the concentration of $1$ mole per litre and pressure at $1\,atm$ and temperature at $25{}^\circ C$ .
gibbs free energy is defined as the change in enthalpy minus the product of entropy and temperature.