Step 1: Write the hydrogen electrode half reaction.
The reduction at a hydrogen electrode is \[ 2H^+ + 2e^- \rightarrow H_2(g) \] and its standard potential is $E^{\circ} = 0$ V by definition.
Step 2: Apply the Nernst equation.
For this two electron process, \[ E = E^{\circ} - \frac{0.0591}{2}\log\frac{p_{H_2}}{[H^+]^2} \]
Step 3: Use the given hydrogen pressure.
Since $p_{H_2} = 1$ bar, the pressure term is $1$ and the log simplifies to $\log\frac{1}{[H^+]^2}$, so \[ E = -\frac{0.0591}{2}\times(-2)\log[H^+] \] \[ E = 0.0591\log[H^+] \]
Step 4: Bring in pH.
Because $pH = -\log[H^+]$, this becomes \[ E = -0.0591\,pH \]
Step 5: Insert the neutral pH value.
A neutral solution at $25^{\circ}$C has $pH = 7$, so \[ E = -0.0591 \times 7 \] \[ E = -0.4137 \text{ V} \]
Step 6: State the result.
The reduction potential is about $-0.413$ V, matching option 3.
\[ \boxed{-0.413 \text{ V}} \]