Complete and balance:
BrO3–(aq) + [Cr(OH)4]–(aq) → CrO42–(aq) + Br–(aq)
Oxidation half–reaction:
[Cr(OH)4]–(aq) → CrO42–(aq) + 4 H+(aq) + 3 e–
Cr is mass balanced
O balance
H balance
charge balance
Reduction half–reaction:
BrO3–(aq) + 6 H+(aq) + 6 e– → Br–(aq) + 3 H2O(l)
Br is mass balanced
O balance
H balance
charge balance
The oxidation requires multiplication by a factor of 2 to equalize the number of electrons:
2 [Cr(OH)4]–(aq) → 2 CrO42–(aq) + 8 H+(aq) + 6 e–
Add the two half–reactions together and eliminate species common to both sides:
BrO3–(aq) + 2 [Cr(OH)4]–(aq) → 2 CrO42–(aq) + Br–(aq) + 2 H+(aq) + 3 H2O(l)
Remove the hydrogen ion by an acid-base reaction:
2 H+(aq) + 2 OH–(aq) → 2 H2O(l)
Combine and eliminate common species:
BrO3–(aq) + 2 [Cr(OH)4]–(aq) + 2 OH–(aq) → 2 CrO42–(aq) + Br–(aq) + 5 H2O(l)