Balance the redox reaction by inserting the appropriate coefficients.

Question

H2O + Br- + Al^3+ --> Al + BrO3^- + H+

Answer 1

H₂O + Br^- + Al³⁺ --> Al + BrO₃^- + H⁺

This is a redox reaction which we will balance using two half reactions and using acid (H⁺).

Br^- ==> BrO₃^- ... oxidation half reaction

Al3+ ==> Al ... reduction half reaction

Br^-+ 3H₂O ==> BrO₃^- ... balanced for Br and O

Br^-+ 3H₂O ==> BrO₃^- + 6H⁺ ... balanced for Br, O and H

Br^-+ 3H₂O ==> BrO₃^- + 6H⁺ + 6e- ... oxidation half reaction balanced for mass and charge

Al³⁺ + 3e- ==> Al ... reduction half reaction balanced for mass and charge

Multiply reduction half reaction by 2 in order to equalize electrons with the oxidation half reaction and add:

2Al³⁺ + 6e- ==> 2Al

Br^-+ 3H₂O ==> BrO₃^- + 6H⁺ + 6e-

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2Al³⁺ + Br^-+ 3H₂O ==> 2Al + BrO₃^- + 6H⁺ ... balanced redox equation



Answer 2

Hello, Caitlyn,

Start with the most complex molecuke. This reaction doesn't have very complex substances, but I'll start with BrO₃^-. Use a pencil and give it a coefficient of 1. We have to have whole molecules, so we'll assume we have one BrO3^- and go from there.

H₂O + Br^- + Al³⁺ --> Al + 1BrO3^- + H⁺

One BrO₃^- means we need 1 Br and 3 O from the reactant side. Pencil in a 1 for the Br^- and 3 for the H₂O.

3H₂O + 1Br^- + Al³⁺ --> Al + 1BrO3^- + H⁺

Now we need to use 6 H atoms on the product side, so add a 6 to the H⁺.

3H₂O + 1Br^- + Al³⁺ --> Al + 1BrO3^- + 6H⁺

We need 2 Al³⁺ to accept the 6 electrons surrendered by the 6H⁺ and become 2Al with no charge.

3H₂O + Br^- + 2Al³⁺ --> 2Al + BrO3^- + 6H⁺

Now it looks balanced.

Bob