Răspuns:
a) 4 KMnO4 → 2 K2O + 4 MnO2 + 3 O2
4 MnVII + 12 e- → 4 MnIV (reducere- agent oxidant)
6 O-II - 12 e- → 6 O0 (oxidare - agent reducator)
b) K2Cr2O7 + 3 SO2 + H2SO4 → Cr2(SO4)3 + K2SO4 + H2O
3 SIV - 6 e- → 3 SVI (oxidare - agent reducator)
2 CrVI + 6 e- → 2 CrIII (reducere - agent oxidant)
c) Br2 + SO2 + 2 H2O → 2 HBr + H2SO4
2 Br0 + 2 e- → 2 Br-I (reducere - agent oxidant)
SIV - 2 e- → SVI (oxidare - agent reducator)
d) Br2 + 5 HClO + H2O → 2 HBrO3 + 5 HCl
2 Br0 - 10 e- → 2 BrV (oxidare - agent reducator)
5 ClI + 10 e- → 5 Cl-I (reducere - agent oxidant)
Explicație: