Răspuns:
H2SO4 +C=> 2S O 2 + CO2 + 2H2O
C 0 - 4 e - → C IV (oxidare)
2 S VI + 4 e - → 2 S IV (reducere)
6FeSO4 + 2HNO3 + 3H2SO4 → 3 Fe2[ S O 4 ] 3 +2 NO + 4H2O
2 N V + 6 e - → 2 N II (reducere)
6 Fe II - 6 e - → 6 Fe III (oxidare)
Hg + 4HNO3 → Hg ( NO 3 ) 2 +2 N O 2 + 2 H 2 O
2 N V + 2 e - → 2 N IV (reducere)
Hg 0 - 2 e - → Hg II (oxidare)
Ag 2O + H 2 O 2 → 2 Ag + H 2 O + O 2
2 Ag I + 2 e - → 2 Ag 0 (reducere)
2 O -I - 2 e - → 2 O 0 (oxidare)
6 H I + 2 HNO3 → 3 I2 + 2 NO + 4 H2O
6 I -I - 6 e - → 6 I 0 (oxidare)
2 N V + 6 e - → 2 N II (reducere)
4 N H 3 + 5 O 2→ 4 N O + 6 H 2 O
4 N -III - 20 e - → 4 N II (oxidare)
10 O 0 + 20 e - → 10 O -II (reducere)
4 N H 3 + 3 O 2 → 2 N 2 + 6 H 2 O
4 N -III - 12 e - → 4 N 0 (oxidare)
6 O 0 + 12 e - → 6 O -II (reducere)
2 KMnO4+ 16HCI → 2 MnCl2 + 2KCl+ 5Cl2 + 8H2O
2 Mn VII + 10 e - → 2 Mn II (reducere)
10 Cl- I - 10 e - → 10 Cl 0 (oxidare)
2 H 2 S O4 + 2 MnO2 → O 2+ 2 Mn SO4+ 2H2O
2 Mn IV + 4 e - → 2 Mn II (reducere)
2 O -II - 4 e - → 2 O 0 (oxidare)
2KClO3 → 2KCl + 3O2
2 Cl V + 12 e - → 2 Cl -I (reducere)
6 O -II - 12 e - → 6 O 0 (oxidare)
