At $$298 \text{ K}$$, the equilibrium constant is $$2 \times 10^{15}$$ for the reaction:
$$Cu(s) + 2Ag^+(aq) \rightleftharpoons Cu^{2+}(aq) + 2Ag(s)$$
The equilibrium constant for the reaction
$$\dfrac{1}{2}Cu^{2+}(aq) + Ag(s) \rightleftharpoons \dfrac{1}{2}Cu(s) + Ag^+(aq)$$
is $$x \times 10^{-8}$$. The value of $$x$$ is ______.

We are given that the equilibrium constant for the reaction:

$$Cu(s) + 2Ag^+(aq) \rightleftharpoons Cu^{2+}(aq) + 2Ag(s)$$

is $$K_1 = 2 \times 10^{15}$$ at 298 K.

We need to find the equilibrium constant for:

$$\frac{1}{2}Cu^{2+}(aq) + Ag(s) \rightleftharpoons \frac{1}{2}Cu(s) + Ag^+(aq)$$

The second reaction is the reverse of the first reaction, divided by 2.

When a reaction is reversed, the new equilibrium constant is the reciprocal: $$K_{reverse} = \frac{1}{K_1}$$. When a reaction is divided by 2, the new equilibrium constant is the square root: $$K_2 = \sqrt{K_{reverse}}$$.

Therefore:

$$K_2 = \sqrt{\frac{1}{K_1}} = \frac{1}{\sqrt{K_1}} = \frac{1}{\sqrt{2 \times 10^{15}}}$$

Calculating the value gives:

$$K_2 = \frac{1}{\sqrt{2 \times 10^{15}}} = \frac{1}{\sqrt{2} \times 10^{7.5}} = \frac{1}{1.414 \times 3.162 \times 10^7} = \frac{1}{4.472 \times 10^7}$$ $$K_2 = 2.236 \times 10^{-8} \approx 2 \times 10^{-8}$$

Since $$K_2 = x \times 10^{-8}$$, we have $$x = 2$$.

Therefore, the correct answer is 2.