(N/A) In a galvanic cell,the cell potential is calculated as $E^{\circ}_{cell} = E^{\circ}_{cathode} - E^{\circ}_{anode}$.
Given that the hydrogen half-cell is the cathode,$E^{\circ}_{cathode} = E^{\circ}_{H^+/H_2} = 0.00 \ V$.
Therefore,$E^{\circ}_{cell} = 0.00 \ V - E^{\circ}_{anode} = -E^{\circ}_{anode}$.
If the anode is a metal with a negative standard reduction potential (e.g.,$Zn/Zn^{2+}$ where $E^{\circ}_{Zn^{2+}/Zn} = -0.76 \ V$),then $E^{\circ}_{cell} = 0.00 \ V - (-0.76 \ V) = +0.76 \ V$,which is positive.
However,if the anode is a metal with a positive standard reduction potential (e.g.,$Cu/Cu^{2+}$ where $E^{\circ}_{Cu^{2+}/Cu} = +0.34 \ V$),then $E^{\circ}_{cell} = 0.00 \ V - (+0.34 \ V) = -0.34 \ V$,which is negative.
Thus,the sign of the standard potential depends on the standard reduction potential of the metal electrode used as the anode.