Value of E_H_2O/H_2 (1atm)Pt at 298 K would be

For water at 298K, \([H^+]  = 10^{-7}M\)

The reduction reaction is

\[H^+ + e^- \rightarrow \frac{1}{2}H_2\]

∴ \(E = -\frac{RT}{F}ln\frac{P_H^{1/2}}{[H^+]} = -0.0591 log\frac{P_H^{1/2}}{[H^+]}\)

\(E =-0.0591 log\frac{1}{10^{-7}}\)

\(E = -0.4137 ≈ -0.414 V\)