1. Write the cell and identify half-cells

The cell is:

$\text{Pt} | H_2(1\,\text{bar}) | H^+(aq) || Cu^{2+}(aq) | Cu(s)$

Left electrode is the hydrogen electrode and right electrode is the copper electrode.

2. Find standard cell potential

Given:

$E^\circ_{Cu^{2+}/Cu} = 0.34\,\text{V}$

For the standard hydrogen electrode,

$E^\circ_{H^+/H_2} = 0\,\text{V}$

So,

$E^\circ_{cell} = E^\circ_{cathode} - E^\circ_{anode} = 0.34 - 0 = 0.34\,\text{V}$

3. Write the overall cell reaction

Anode (oxidation):

$H_2 \rightarrow 2H^+ + 2e^-$

Cathode (reduction):

$Cu^{2+} + 2e^- \rightarrow Cu$

Overall reaction:

$H_2 + Cu^{2+} \rightarrow 2H^+ + Cu$

Thus,

$n = 2$

4. Write reaction quotient

For the reaction

$H_2 + Cu^{2+} \rightarrow 2H^+ + Cu$

$Q = \frac{[H^+]^2}{P_{H_2}[Cu^{2+}]}$

Since $P_{H_2} = 1$ bar,

$Q = \frac{[H^+]^2}{[Cu^{2+}]}$

5. Apply Nernst equation

At $298\,K$,

$E_{cell} = E^\circ_{cell} - \frac{0.06}{n} \log Q$

$0.31 = 0.34 - \frac{0.06}{2} \log \left(\frac{[H^+]^2}{[Cu^{2+}]}\right)$

$0.31 = 0.34 - 0.03 \log \left(\frac{[H^+]^2}{[Cu^{2+}]}\right)$

6. Use pH = 3

$[H^+] = 10^{-3}$

So,

$[H^+]^2 = 10^{-6}$

Substitute:

$0.31 = 0.34 - 0.03 \log \left(\frac{10^{-6}}{[Cu^{2+}] }\right)$

Rearranging,

$0.03 = 0.03 \log \left(\frac{10^{-6}}{[Cu^{2+}] }\right)$

$\log \left(\frac{10^{-6}}{[Cu^{2+}] }\right) = 1$

Hence,

$\frac{10^{-6}}{[Cu^{2+}]} = 10^1$

$[Cu^{2+}] = 10^{-7}\,M$

7. Compare with given form

Given concentration of $Cu^{2+}$ is $10^{-x}\,M$.

So,

$10^{-x} = 10^{-7}$

Therefore,

$x = 7$

8. Comparison with stored answer

Stored correct answer = 7, which matches our result.