Question

The equilibrium constant of the reaction;
$Cu(s) + 2A{g^ + }(aq) \to C{u^{2 + }}(aq) + 2Ag(s)$
${E^\circ } = 0.46V\,at\,298K$
A. $2.4 \times {10^{10}}$
B. $2.0 \times {10^{10}}$
C. $4.0 \times {10^{10}}$
D. $4.0 \times {10^{15}}$

Answer 1

To solve for the equilibrium constant of the given reaction, we will go through the equation of Gibbs Free energy. By the help of Gibbs Free Energy we can find the equilibrium constant for the given reaction.

Complete step by step answer:
The given reaction-
$Cu(s) + 2A{g^ + }(aq) \to C{u^{2 + }}(aq) + 2Ag(s)$
${E^\circ } = 0.46V\,at\,298K$
Now, according to the Gibbs Free Energy-
$\log {K_C} = \dfrac{{nF{E^\circ }}}{{RT}}$
here, ${K_C}$ is the equilibrium constant of the given reaction;
${E^\circ }$ is the standard cell potential, which is already given;
$R$ is the gas constant that is needed to find the equilibrium constant;
$T$ is the absolute temperature.
$\Rightarrow {K_C} = \dfrac{{2 \times 0.46}}{{0.059}} \\\ \Rightarrow {K_C} = 4 \times {10^{15}} \\\$
Therefore, the equilibrium constant for the given reaction is $4.0 \times {10^{15}}$ .

So, the correct answer is Option D.

Note: The equilibrium constant of the reactions does not change. Due to the higher concentration of the common product, the product concentrations will be reduced.
Simultaneous equilibrium reactions, having a common product.