Answer:
6 H⁺ + BrO₃⁻ + 3 Sb³⁺ ⟶ Br⁻ + 3 H₂O + 3 Sb⁵⁺
Step-by-step explanation:
Step 1: Write the unbalanced reaction
BrO₃⁻ + Sb³⁺ ⟶ Br⁻ + Sb⁵⁺
Step 2: Identify both half-reactions
Reduction: BrO₃⁻ ⟶ Br⁻
Oxidation: Sb³⁺ ⟶ Sb⁵⁺
Step 3: Perform the mass balance, adding H⁺ and H₂O where appropriate
6 H⁺ + BrO₃⁻ ⟶ Br⁻ + 3 H₂O
Sb³⁺ ⟶ Sb⁵⁺
Step 4: Perform the charge balance, adding electrons where appropriate
6 H⁺ + BrO₃⁻ + 6 e⁻ ⟶ Br⁻ + 3 H₂O
Sb³⁺ ⟶ Sb⁵⁺ + 2 e⁻
Step 5: Multiply both half-reactions by numbers that assure that the number of electrons gained and lost is the same
1 × (6 H⁺ + BrO₃⁻ + 6 e⁻ ⟶ Br⁻ + 3 H₂O)
3 × (Sb³⁺ ⟶ Sb⁵⁺ + 2 e⁻)
Step 6: Add both half-reactions and cancel what is repeated in both sides
6 H⁺ + BrO₃⁻ + 3 Sb³⁺ ⟶ Br⁻ + 3 H₂O + 3 Sb⁵⁺