#E_(cell)^(0)=+3.34"V"#
You need to look up the relevant standard electrode potentials:
Always list them -ve to +ve:
#Mg^(2+)+2erightleftharpoonsMg# #E^(0)=-2.38"V"#
#NO_3^(-)+4H^(+)+3erightleftharpoonsNO+2H_2O# #E^(0)=+0.96"V"#
The most +ve half cell will take in electrons so the #NO_3^(-)"/"NO# equilibrium will shift left to right.
The #Mg^(2+)"/"Mg# half cell will give out electrons and shift right to left. So this electrode is the -ve side of the cell.
These electrons flow through the external circuit to the #NO_3^(-)"/"NO# half cell.
The overall cell reaction is;
#3Mg+2NO_3^(-)+8H^(+)rarr3Mg^(2+)+2NO+4H_2O#
#E_(cell)^(0)# is an empirically measured quantity so is always +ve.
To work out #E_(cell)^(0)# always subtract the least positive value from the most positive#rArr#
#E_(cell)^(0)=0.96-(-2.38)=+3.34"V"#
