Răspuns:
3 Hg + 8 H N O3 → 3Hg( NO 3 ) 2 +2 N O + 4H2 O
2 N V + 6 e - → 2 N II (reducere)
3 Hg 0 - 6 e - → 3 Hg II (oxidare)
AgO + H 2 O 2 → Ag + H 2 O + O 2
Ag II + 2 e - → Ag 0 (reducere)
2 O -I - 2 e - → 2 O 0 (oxidare)
6HI + 2HNO 3 → 3 I2 + 2 NO + 4 H 2 O
I-¹ -2e=> I⁰ oxidare ×3
N⁵+ 3e=> N²reducere ×2
3I-¹ -6e=> 3I⁰
2 N⁵ +6e=> 2N²
4N H 3 + 5O2 → 4N O + 6 H2O
4 N -³ - 20 e - → 4 N² (oxidare)
O⁰ +2e=> O -² reducere
4 NH 3 + 3 O2 → 2 N2 + 6 H 2 O
4 N -III - 12 e - → 4 N 0 (oxidare)
6 O 0 + 12 e - → 6 O -II (reducere)
