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Na2Cr2O7 + H2O2 + H2SO4 → Na2SO4 + Cr2(SO4)3 + H2O +O2
Cr2O7^2- + 14H^+ +6e^- = 2Cr^3+ + 7H2O | · 1
H2O2 = O2 + 2H^+ + 2e^- | · 3
Na2Cr2O7 + 3H2O2 + 4H2SO4 → Na2SO4 + Cr2(SO4)3 + 7H2O + 3O2
Drugi redox:
Na2Cr2O7 + Na2S + HCl
Cr(VI) redukuje sie do Cr(III), a S(-II) utlenia się do wolnej siarki.
Na2Cr2O7 + Na2S + HCl → NaCl + CrCl3 + S + H2O
Cr2O7^2- + 14H^+ + 6e^- = 2Cr^3+ + 7H2O | ·1
S^2- = S + 2e^- | ·3
Na2Cr2O7 + 3Na2S + 14HCl → 2CrCl3 + 8NaCl + 3S + 7H2O
proszę :)