Answer: D. The equation requires $2\ \mathrm{mol}$ HCl per $1\ \mathrm{mol}$ H₂, so HCl limits the reaction: $n(\mathrm{H_2})=0.0060/2=0.0030\ \mathrm{mol}$ H⁺ is reduced at the hydrogen electrode, making it the cathode; the cathode is positive in a voltaic cell.
A: gives the correct amount but assigns the hydrogen electrode the negative sign.
B: uses the amount of Mg as though it all reacts, despite insufficient HCl.
C: uses the Mg amount without checking the limiting reagent and also assigns the cathode a negative sign.
Remember: in a voltaic cell, reduction occurs at the positive cathode.