The cell potential for the following cell $$Pt|H_2(g)|H^+(aq)||Cu^{2+}(0.01 \text{ M})|Cu(s)$$ is $$0.576$$ V at $$298$$ K. The pH of the solution is (Nearest integer) ______
(Given : $$E^{\circ}_{Cu^{2+}/Cu} = 0.34$$ V and $$\frac{2.303RT}{F} = 0.06$$ V)

We are given the cell: $$Pt|H_2(g)|H^+(aq)||Cu^{2+}(0.01 \text{ M})|Cu(s)$$ and the cell potential is $$E_{cell} = 0.576$$ V at 298 K.

The cell reaction is: $$ H_2(g) + Cu^{2+}(aq) \rightarrow 2H^+(aq) + Cu(s) $$

The standard cell potential is: $$E^\circ_{cell} = E^\circ_{cathode} - E^\circ_{anode} = E^\circ_{Cu^{2+}/Cu} - E^\circ_{H^+/H_2} = 0.34 - 0 = 0.34$$ V

Using the Nernst equation (with $$n = 2$$ electrons transferred): $$ E_{cell} = E^\circ_{cell} - \frac{2.303RT}{nF} \log \frac{[H^+]^2}{[Cu^{2+}]} $$

Substituting the known values: $$ 0.576 = 0.34 - \frac{0.06}{2} \log \frac{[H^+]^2}{0.01} $$

$$ 0.576 = 0.34 - 0.03 \log \frac{[H^+]^2}{0.01} $$

$$ 0.576 - 0.34 = -0.03 \log \frac{[H^+]^2}{0.01} $$

$$ 0.236 = -0.03 \log \frac{[H^+]^2}{0.01} $$

$$ \frac{0.236}{-0.03} = \log \frac{[H^+]^2}{0.01} $$

$$ -7.867 = \log \frac{[H^+]^2}{0.01} $$

$$ \log [H^+]^2 - \log(0.01) = -7.867 $$

$$ 2\log [H^+] - (-2) = -7.867 $$

$$ 2\log [H^+] = -7.867 - 2 = -9.867 $$

$$ \log [H^+] = -4.933 $$

pH $$= -\log [H^+] = 4.933 \approx 5$$

Therefore, the pH of the solution is 5 (nearest integer).