Answer:
BrO₃⁻(aq) + 3Sb³⁺(aq) + 6H⁺(aq) → Br⁻(aq) + 3Sb⁵⁺(aq) + 3H₂O(l)
Explanation:
At a redox equation, one substance is being oxidized (losing electrons), and the other is being reduced (gaining electrons). In the given reaction:
BrO₃⁻(aq) + Sb³⁺(aq) → Br⁻(aq) + Sb⁵⁺(aq)
When it's at an acidic solution, it must be ions H⁺ on the reactant, which will form water with the oxygen, so the complete reaction is:
BrO₃⁻(aq) + Sb³⁺(aq) + H⁺(aq) → Br⁻(aq) + Sb⁵⁺(aq) + H₂O(l)
As we can see, the antimony is being oxidized (go from +3 to +5), and the Bromo is being reduced. The oxidation number of brome in the reactant, knowing that the oxidation number of O is -2, is:
x + 3*(-2) = -1
x = +5
So, it's going from +5 to -1, and the half-reactions are:
BrO₃⁻(aq) + 6e⁻ → Br⁻(aq)
Sb³⁺(aq) → Sb⁵⁺(aq) + 2e⁻
The number of electrons must be the same, so the second equation must be multiplied by 3:
3Sb³⁺(aq) → 3Sb⁵⁺(aq) + 6e⁻
Thus, the equation will be:
BrO₃⁻(aq) + 3Sb³⁺(aq) + H⁺(aq) → Br⁻(aq) + 3Sb⁵⁺(aq) + H₂O(l)
Now, we verify the amount of the elements, which must be equal on both sides. So, we multiply H₂O by 3, and H⁺ by 6, and the balanced reaction will be:
BrO₃⁻(aq) + 3Sb³⁺(aq) + 6H⁺(aq) → Br⁻(aq) + 3Sb⁵⁺(aq) + 3H₂O(l)
