Balance the redox reaction by inserting the appropriate coefficients.

H2O + Br- + Al3- = Al + BrO3- + H+

The balanced equation is given as,

2Al³⁺ + 3H₂O + Br⁻ → 2 Al + BrO³⁻ + 6H⁺

Explanation:

H₂O + Br⁻ + Al³⁺ → Al + BrO₃⁻ + H⁺

Here the half reactions are:

Al³⁺→ Al     [reduction]

Br⁻ → BrO₃⁻ [oxidation]

Now we have to balance the half reactions as,

Al³⁺ + 3e⁻ → Al

To balance H atoms we have to add water and electrons.

3H₂O + Br⁻→ BrO³⁻ + 6H⁺ + 6e⁻

Now we have to balance the electrons as,

2Al³⁺ + 6e⁻ → 2 Al

3H₂O + Br⁻→ BrO³⁻ + 6H⁺ + 6e⁻

Now we have to add  both the equations as,

2Al³⁺ + 6e⁻ + 3H₂O + Br⁻ → 2 Al + BrO³⁻ + 6H⁺ + 6e⁻

6 electrons on both sides of the equation gets cancelled and so the balanced reaction can be written as,

2Al³⁺ + 3H₂O + Br⁻ → 2 Al + BrO³⁻ + 6H⁺

oxygen, selenium, tellurium, and polonium.

they're all from the same group as sulfur (group 16).