Question

The equilibrium constant for the equation: ${\text{E}}^\circ {\text{ = 0}}{\text{.46V}}$ at 298 K is:
${\text{Cu}}\left( {\text{s}} \right){\text{ + 2A}}{{\text{g}}^{\text{ + }}}\left( {{\text{aq}}} \right) \to {\text{C}}{{\text{u}}^{{\text{2 + }}}}\left( {{\text{aq}}} \right){\text{ + 2Ag}}\left( {\text{s}} \right)$
A.$2.0 \times {10^{10}}$
B.$3.9 \times {10^{15}}$
C.$4.0 \times {10^{10}}$
D.$2.4 \times {10^{10}}$

Answer 1

For a reaction in equilibrium, the equilibrium constant ${{\text{K}}_{\text{c}}}$ is related to the standard EMF ${\text{E}}{^\circ _{{\text{cell}}}}$ of the cell by the equation:
${\text{E}}{^\circ _{{\text{cell}}}} = \dfrac{{{\text{RT}}}}{{{\text{nF}}}}\ln {{\text{K}}_{\text{c}}}$
Here, R is the gas constant, T is the temperature in Kelvin, n is the number of electrons involved in the reaction and F is the Faraday constant.
Knowing the standard EMF of the cell, the equilibrium constant can be calculated using the above equation.

Complete step by step answer:
According to the question, it is given that the standard EMF of the cell is equal to ${\text{0}}{\text{.46V}}$ for the reaction:
${\text{Cu}}\left( {\text{s}} \right){\text{ + 2A}}{{\text{g}}^{\text{ + }}}\left( {{\text{aq}}} \right) \to {\text{C}}{{\text{u}}^{{\text{2 + }}}}\left( {{\text{aq}}} \right){\text{ + 2Ag}}\left( {\text{s}} \right)$
Also given that the temperature of the reaction is 298 K.
We need to find out the value of the equilibrium constant for this reaction.
So, by question, we can write ${\text{E}}{^\circ _{{\text{cell}}}}{\text{ = 0}}{\text{.46V}}$ and T equals to 298 K.
We know that the value of the gas constant R is equal to $8.314{\text{J}}{{\text{K}}^{{\text{ - 1}}}}{\text{mo}}{{\text{l}}^{{\text{ - 1}}}}$ , the value of the Faraday constant F is equal to 96500 Coulombs.
Now, let us find out the number of electrons involved in the reaction. The electrode reactions will be:
At anode: ${\text{Cu}}\left( {\text{s}} \right) \to {\text{C}}{{\text{u}}^{{\text{2 + }}}}\left( {{\text{aq}}} \right){\text{ + 2}}{{\text{e}}^ - }$
At cathode: ${\text{2A}}{{\text{g}}^{\text{ + }}}\left( {{\text{aq}}} \right) + 2{{\text{e}}^ - } \to {\text{2Ag}}\left( {\text{s}} \right)$
So, the number of electrons involved in the reaction, n is equal to 2.
Substitute all these values of R, T, n, F and ${\text{E}}{^\circ _{{\text{cell}}}}$ in the equation. We get
$0.46 = \dfrac{{8.314 \times 298}}{{2 \times 96500}}\ln {{\text{K}}_{\text{c}}}$
Therefore,
$\Rightarrow \ln {{\text{K}}_{\text{c}}} = \dfrac{{0.46 \times 2 \times 96500}}{{8.314 \times 298}} \\\ \Rightarrow \ln {{\text{K}}_{\text{c}}} = 35.83 \\\$
From this, we get
${{\text{K}}_{\text{c}}} = 3.7 \times {10^{15}} \\\ \Rightarrow {{\text{K}}_{\text{c}}} \approx 3.9 \times {10^{15}} \\\$

So, the correct option is B.

Note:
The standard free energy change of a reaction is represented by $\Delta {\text{G}}^\circ$ which is related to the standard cell potential by the equation:
$- \Delta {\text{G}}^\circ = {\text{nFE}}{^\circ _{{\text{cell}}}}$
By using the relation between cell potential and equilibrium constant, the relation between standard free energy change and equilibrium constant can be written as:
$\Delta {\text{G}}^\circ = - {\text{RTln}}{{\text{K}}_{\text{c}}}$