The potential for the given half cell at 298K is $$- x \times 10^{-2}$$ V.
$$2H^+_{(aq)} + 2e^- \rightarrow H_2(g)$$, $$[H^+] = 1$$ M, $$P_{H_2} = 2$$ atm.
(Given $$2.303\ RT/F = 0.06$$ V, $$\log 2 = 0.3$$). The value of $$x$$ is:

For the half-cell reaction $$2H^+(aq) + 2e^- \rightarrow H_2(g)$$ with $$[H^+] = 1$$ M and $$P_{H_2} = 2$$ atm, we wish to find $$x$$ such that the half-cell potential is $$-x \times 10^{-2}$$ V.

According to the Nernst equation, the cell potential $$E$$ is related to the standard potential $$E°$$ and the reaction quotient $$Q$$ by $$ E = E° - \frac{2.303RT}{nF}\log Q $$. For the hydrogen electrode, $$E° = 0$$ V by definition, and $$n = 2$$ electrons are transferred.

The reaction quotient for $$2H^+ + 2e^- \rightarrow H_2$$ is given by $$ Q = \frac{P_{H_2}}{[H^+]^2} = \frac{2}{1^2} = 2 $$.

Substituting the given value $$\frac{2.303RT}{F} = 0.06\text{ V}$$ along with $$n = 2$$ and $$\log 2 = 0.3$$ into the Nernst equation yields $$ E = 0 - \frac{0.06}{2} \times \log 2 = -0.03 \times 0.3 = -0.009 \text{ V} $$.

This value may be written as $$ E = -0.009 \text{ V} = -9 \times 10^{-3} \text{ V} \approx -1 \times 10^{-2} \text{ V} $$. Since $$-0.009 = -0.9 \times 10^{-2}$$, rounding to the nearest integer gives $$x = 1$$.

Therefore, the answer is $$x = 1$$.