Question

Calculate the equilibrium constant for the following reaction:
${\text{Cu}}\left( {\text{s}} \right){\text{ + 2A}}{{\text{g}}^{\text{ + }}} \rightleftarrows {\text{C}}{{\text{u}}^{{\text{ + 2}}}}\left( {{\text{aq}}} \right){\text{ + Ag}}\left( {\text{s}} \right)$
At 25 $^{\text{0}}{\text{C}}$ , the ${\text{E}}_{{\text{cell}}}^{\text{0}}{\text{ = 0}}{\text{.47}}$ volt, R = ${\text{8}}{\text{.314 J }}{{\text{K}}^{{\text{ - 1}}}}{\text{mo}}{{\text{l}}^{{\text{ - 1}}}}$ , F = 96500 coulombs.

Answer 1

As per the electrochemical series of the elements, the metals that are placed below in the series are reduced by accepting electrons in the presence of the elements that are placed higher in the electrochemical series. The equilibrium constant is the ratio of the concentration of the products to the reactants during equilibrium. We can also find the equilibrium constant using the formula given below.
Formula Used: ${{\text{E}}^{\text{0}}}_{{\text{cell}}}{\text{ = }}\dfrac{{{\text{0}}{\text{.059}}}}{{\text{n}}}{\text{log K}}$
where, ${{\text{E}}^{\text{0}}}_{{\text{cell}}}$ is the cell potential, n is the number of electrons transferred and K is the equilibrium constant.

Complete step by step solution:
Given that,
${\text{Cu}}\left( {\text{s}} \right){\text{ + 2A}}{{\text{g}}^{\text{ + }}} \rightleftarrows {\text{C}}{{\text{u}}^{{\text{ + 2}}}}\left( {{\text{aq}}} \right){\text{ + Ag}}\left( {\text{s}} \right)$
as per this reaction, copper is oxidized in the presence of silver to form cupric solution and the silver ions are discharged at the cathode. The equilibrium constant for the above reaction is as follows:
${\text{K = }}\dfrac{{\left[ {{\text{C}}{{\text{u}}^{{\text{ + 2}}}}} \right]\left[ {{\text{Ag}}} \right]}}{{\left[ {{\text{Cu}}} \right]\left[ {{\text{A}}{{\text{g}}^{\text{ + }}}} \right]}}$
Now, this equilibrium constant is related to the cell potential by the Nernst Equation which is as follows:
${{\text{E}}^{\text{0}}}_{{\text{cell}}}{\text{ = }}\dfrac{{{\text{0}}{\text{.059}}}}{{\text{n}}}{\text{log K}}$ ,
where, ${{\text{E}}^{\text{0}}}_{{\text{cell}}}$ is the cell potential whose value at 25 $^{\text{0}}{\text{C}}$ = $0.47$ volt, R is the universal gas constant which is equal to ${\text{8}}{\text{.314 J }}{{\text{K}}^{{\text{ - 1}}}}{\text{mo}}{{\text{l}}^{{\text{ - 1}}}}$ and n is the number of electrons transferred in the process. Since copper gives up two electrons in the process to get oxidized hence n = 2.
Putting the values of the given parameters in the above equation we get,
${\text{K = antilog}}\dfrac{{2 \times 0.47}}{{{\text{0}}{\text{.059}}}}$
Or, ${\text{K = antilog}}\left[ {15.932} \right]$
Or, K = $8.5 \times {10^{15}}$
So the equilibrium constant for the redox reaction of copper with silver is $8.5 \times {10^{15}}$ . This constant is unitless because it is a ratio of two similar types of quantities.

Note:
The Nernst equation relates the reduction potential of an electrochemical reaction (half –cell or full cell) to the standard reduction potential, temperature and the activities (often approximated by the concentrations) of the chemical species that are undergoing reduction and oxidation. This reaction was named after Walther Nernst, a German chemical physicist who laid this equation.