Complete and balance:
Mg(s) + NO3–(aq) → Mg2+(aq) + NO2(g)
Use oxidation numbers to identify the oxidized and reduced species:
Mg(s) + NO3–(aq) → Mg2+(aq) + NO2(g)
Ox. Nos.:0 N, +5; O, –2 +2 N, +4; O, –2
Magnesium is oxidized.
Nitrate is reduced.
Oxidation half-reaction
Mg(s) → Mg2+(aq) + 2 e–
Mg is mass balanced
O is mass balanced
H is mass balanced
charge is balanced
Reduction half-reaction
NO3–(aq) + 2 H+(aq) + e– → NO2(g) + H2O(l)
N is mass balanced
O is mass balanced
H is mass balanced
charge is balanced
The oxidation is a 2–electron process while the reduction is only 1–electron, so the reduction needs to be multiplied by 2:
2×[NO3–(aq)+2 H+(aq) + e– → NO2(g) + H2O(l)]
This gives
2 NO3–(aq) + 4 H+(aq) + 2 e– → 2 NO2(g) + 2 H2O(l)
Add the two half–reactions together:
Mg(s) + 2 NO3–(aq) + 4 H+(aq) + 2 e– → Mg2+(aq) + 2 e– + 2 NO2(g) + 2 H2O(l)
Eliminate the electrons:
Mg(s) + 2 NO3–(aq) + 4 H+(aq) → Mg2+(aq) + 2 NO2(g) + 2 H2O(l)