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 positiverArr
E_(cell)^(0)=0.96-(-2.38)=+3.34"V"
