Question

Balance by the half-reaction method, both in acid and in base:

Cr₂O₃ (s) + H₂O₂ (aq) -----> CrO₄^2- (aq) + H₂O (l)

Reductant half-reaction:

Oxidant half-reaction:

Net ionic full reaction in acid:

Net ionic full reaction in base:

Answer 1

In acidic media :

+3 -1 +6 -2

Cr₂O₃ (s) + H₂O₂ (aq) -----> CrO₄^2- (aq) + H₂O (l)

Reductant half-reaction: H₂O₂ (aq) -----> H₂O (l)

Balance O atoms : H₂O₂ (aq) -----> H₂O (l) + H₂O(l)

Balance H atoms   H₂O₂ (aq) +2H⁺(aq) -----> 2H₂O (l)

Balance charge : H₂O₂ (aq) +2H⁺(aq) +2e^- -----> 2H₂O (l) ----(1)

Oxidant half-reaction: Cr₂O₃ (s) -----> CrO₄^2- (aq)

Balance Cr atoms : Cr₂O₃ (s)  -----> 2CrO₄^2- (aq)

Balance O atoms : Cr₂O₃ (s)+5H₂O (l) -----> 2CrO₄^2- (aq)

Balance H atoms : Cr₂O₃ (s)+5H₂O (l)  -----> 2CrO₄^2- (aq) + 10 H⁺(aq)

Balance charge : Cr₂O₃ (s)+5H₂O (l)  -----> 2CrO₄^2- (aq) + 10 H⁺(aq) + 6e- -----(2)

[3x(1)] + (2) gives Cr₂O₃ (s)+5H₂O (l) +3H₂O₂ (aq) +6H⁺(aq) -----> 6H₂O (l) + 2CrO₄^2- (aq) + 10 H⁺(aq)

This was the balanced reaction

In basic media :

+3 -1 +6 -2

Cr₂O₃ (s) + H₂O₂ (aq) -----> CrO₄^2- (aq) + H₂O (l)

Reductant half-reaction: H₂O₂ (aq) -----> H₂O (l)

Balance O atoms : H₂O₂ (aq) -----> H₂O (l) + H₂O(l)

Balance H atoms   H₂O₂ (aq) + 2H₂O (l)   -----> 2H₂O (l) + 2OH^-(aq)

Balance charge : H₂O₂ (aq) +2e- -----> 2OH^-(aq) ----(1)

Oxidant half-reaction: Cr₂O₃ (s)  -----> CrO₄^2- (aq)

Balance Cr atoms : Cr₂O₃ (s)  -----> 2CrO₄^2- (aq)

Balance O atoms : Cr₂O₃ (s)+5H₂O (l)  -----> 2CrO₄^2- (aq)

Balance H atoms : Cr₂O₃ (s)+5H₂O (l) +10OH^- (aq) -----> 2CrO₄^2- (aq) + 10 H₂O (l)

Balance charge : Cr₂O₃ (s)+10OH^- (aq) -----> 2CrO₄^2- (aq) + 5H₂O (l)  + 6e- -----(2)

[3x(1)] + (2) gives Cr₂O₃ (s)+10OH^- (aq) +3H₂O₂ (aq) -----> 6OH^-(aq)+ 2CrO₄^2- (aq) + 5H₂O (l)  

This was the balanced reaction