Răspuns:
2 KMnO4 + 10KI+ 8 H2SO4 → 2 MnSO4 + 6 K2SO4 + 5 I2 + 8 H2O
2 MnVII + 10 e- → 2 MnII (reducere)
10 I-I - 10 e- → 10 I0 (oxidare)
K2Cr2O7+ 6 NaBr + 7 H2SO4 → 3 Na2SO4 + 3 Br2 + K2SO4 + 7 H2O + Cr2(SO4)3
6 Br-I - 6 e- → 6 Br0 (oxidare)
2 CrVI + 6 e- → 2 CrIII(reducere)
