The equilibrium constant of the reaction

Cu(s) +2Ag(aq) ---- Cu2+(aq) +2Ag(s) E(standard)cell = 0.46V at 298K is

1- 2 x 10^10 2- 4 x 10^10

3- 4 x 10^15 4- 2.4 x 10^10

We can calculate the equilibrium constant of the reaction as follows:

E^o_cell = 0.46 V
${{\mathrm{E}}^{\mathrm{o}}}_{\mathrm{cell}}=\frac{\mathrm{RT}}{\mathrm{nF}} \ln \mathrm{K}$where, R = gas constant, T = temperature, F = Faraday's constant, n = no. of electrons involved and  K = equilibrium constant.

Substituting the values of R, T and F we get, 

$$\begin{gathered} {{\mathrm{E}}^{\mathrm{o}}}_{\mathrm{cell}}=\frac{0.0592}{\mathrm{n}} \ln \mathrm{K} \\ = \frac{0.0257}{2} \ln \mathrm{K} \\ = 0.0295 \log \mathrm{K} \end{gathered}$$

$$\begin{gathered} 0.46 = 0.0296 \log \mathrm{K} \\ \log \mathrm{K} = 15.54 \\ \mathrm{K} = {10}^{15.54} = 3.46 \times {10}^{15} \end{gathered}$$


Therefore, the correct answer is (3).