Answer to Question #138833 in General Chemistry for Mike

Question

K2Cr2O7+4H2O2+H2SO4 --> 2CrO(O)2+ K2SO4+5H2O

2 CrVI - 8 e- → 2 CrX (oxidation)
8 O-I + 8 e- → 8 O-II (reduction)

Explain step by step how you chemists generate showed redox reaction from given equation?

Answer 1

4 H₂O₂^- + K₂Cr₂⁺⁶O₇^-2 + H₂SO₄^-2→ 2 Cr⁺¹⁰O^-2(O₂^-2)₂ + K₂SO₄^-2+ 5 H₂O^-2

his is a redox reaction:

8 O^- + 8 e- → 8 O^-2(reduction)

2 Cr^+6- 8 e- → 2 Cr⁺¹⁰ (oxidation)

H₂O₂ is an oxidizer, K₂Cr₂O₇ is a reducing agent.