Write the half-cell reduction reaction

$2H^+ + 2e^- \rightarrow H_2(g)$

For the hydrogen electrode, the standard reduction potential is

$E^\circ = 0$

Apply the Nernst equation

For the reaction above,

$E = E^\circ - \frac{0.0591}{2} \log \frac{p_{H_2}}{[H^+]^2}$

Since $E^\circ = 0$,

$E = - \frac{0.0591}{2} \log \frac{p_{H_2}}{[H^+]^2}$

Condition for reduction potential to be negative

We need

$E < 0$

Since the coefficient outside the log is negative, this requires

$\log \frac{p_{H_2}}{[H^+]^2} > 0$

which means

$\frac{p_{H_2}}{[H^+]^2} > 1$

Check each option

Option A

$\frac{p_{H_2}}{[H^+]^2} = \frac{1}{(1)^2} = 1$ $E = 0$ Not negative.

Option B

$\frac{p_{H_2}}{[H^+]^2} = \frac{1}{(2)^2} = \frac{1}{4} < 1$ So $\log(1/4) < 0$, hence $E > 0$. Not negative.

Option C

$\frac{p_{H_2}}{[H^+]^2} = \frac{2}{(1)^2} = 2 > 1$ So $\log 2 > 0$, hence $E < 0$ Negative.

Option D

$\frac{p_{H_2}}{[H^+]^2} = \frac{2}{(2)^2} = \frac{1}{2} < 1$ So $\log(1/2) < 0$, hence $E > 0$. Not negative.

Conclusion

The reduction potential of the hydrogen half-cell is negative only for Option C.