The cell potential for the given cell at 298 K Pt|H$$_2$$(g, 1 bar)|H$$^+$$(aq)||Cu$$^{2+}$$(aq)|Cu(s) is 0.31 V. The pH of the acidic solution is found to be 3, whereas the concentration of Cu$$^{2+}$$ is $$10^x$$ M. The value of x is ______.
(Given: $$E^\ominus_{Cu^{2+}/Cu} = 0.34$$ V and $$\frac{2.303RT}{F} = 0.06$$ V)

The electrochemical cell is Pt|H$$_2$$(g, 1 bar)|H$$^+$$(aq)||Cu$$^{2+}$$(aq)|Cu(s) with $$E_{cell} = 0.31$$ V, pH = 3, $$E^\ominus_{Cu^{2+}/Cu} = 0.34$$ V, and $$\dfrac{2.303RT}{F} = 0.06$$ V. The cell reaction is $$\text{H}_2(g) + \text{Cu}^{2+}(aq) \rightarrow 2\text{H}^+(aq) + \text{Cu}(s)$$ and the standard cell potential is $$E^\ominus_{cell} = 0.34 - 0 = 0.34 \text{ V}, \quad n = 2$$.

Applying the Nernst equation $$E_{cell} = E^\ominus_{cell} - \frac{0.06}{2}\log\frac{[\text{H}^+]^2}{[\text{Cu}^{2+}]}$$ and noting that pH = 3 gives $$[\text{H}^+] = 10^{-3}$$ M, we write $$0.31 = 0.34 - 0.03 \times \log\frac{(10^{-3})^2}{[\text{Cu}^{2+}]} = 0.34 - 0.03 \times \log\frac{10^{-6}}{[\text{Cu}^{2+}]}$$. Solving for $$[\text{Cu}^{2+}]$$, $$-0.03 = -0.03 \times \log\frac{10^{-6}}{[\text{Cu}^{2+}]}$$ gives $$\log\frac{10^{-6}}{[\text{Cu}^{2+}]} = 1,$$ $$\frac{10^{-6}}{[\text{Cu}^{2+}]} = 10,$$ $$[\text{Cu}^{2+}] = \frac{10^{-6}}{10} = 10^{-7} \text{ M}$$. Since $$[\text{Cu}^{2+}] = 10^x$$ M, the value of $$x$$ is $$-7$$.