1. Write the cell reaction

The given cell is

$\text{Pt }| H_2(g)|H^+(aq) || Cu^{2+}(0.01\,M)|Cu(s)$

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

• Anode (oxidation): $H_2 \rightarrow 2H^+ + 2e^-$

• Cathode (reduction): $Cu^{2+} + 2e^- \rightarrow Cu$

Overall reaction: $H_2 + Cu^{2+} \rightarrow 2H^+ + Cu$

2. Find standard cell potential

Given: $E^\circ_{Cu^{2+}/Cu} = 0.34\,V$ $E^\circ_{H^+/H_2} = 0\,V$

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

3. Apply Nernst equation

For the overall reaction, $Q = \frac{[H^+]^2}{P_{H_2}[Cu^{2+}]}$

Assuming $H_2$ gas is at $1\,atm$, $Q = \frac{[H^+]^2}{[Cu^{2+}]}$

Given $[Cu^{2+}] = 0.01 = 10^{-2}$, $Q = \frac{[H^+]^2}{10^{-2}}$

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

Here $n=2$, so $0.576 = 0.34 - \frac{0.06}{2} \log\left(\frac{[H^+]^2}{10^{-2}}\right)$

$0.576 = 0.34 - 0.03\log\left(\frac{[H^+]^2}{10^{-2}}\right)$

4. Solve for $[H^+]$

Rearranging, $0.576 - 0.34 = -0.03\log\left(\frac{[H^+]^2}{10^{-2}}\right)$

$0.236 = -0.03\log\left(\frac{[H^+]^2}{10^{-2}}\right)$

$\log\left(\frac{[H^+]^2}{10^{-2}}\right) = -\frac{0.236}{0.03} \approx -7.867$

So, $\frac{[H^+]^2}{10^{-2}} = 10^{-7.867}$

$[H^+]^2 = 10^{-2}\cdot 10^{-7.867} = 10^{-9.867}$

$[H^+] = 10^{-4.9335}$

Therefore, $\text{pH} = -\log[H^+] = 4.9335$

Nearest integer: $\boxed{5}$

5. Comparison with stored answer

Derived answer is $5$, which matches the stored correct answer.