Question

Balance the following redox reaction in acidic solution by adding the appropriate coefficients. H+(aq)+Mn2+(aq)+NaBiO3(s) => H2O(l) +MnO4(aq)+Br3+(aq)+Na+(aq). Show steps please.

Answer 1

First, break the equation into two parts - the oxidation reaction and the reduction reaction.

Start with the Mn²⁺

Mn²⁺ MnO₄^-

It has to get the O from somewhere - this comes from water so

Mn²⁺ + H₂O MnO₄^- + H⁺

but this is not balanced. We need 4O so will need 4H₂O:

Mn²⁺ + 4H₂O 8H⁺ + MnO₄^-

Now the number of atoms is balanced but the charges are not. The left-hand side has a total charge of 2+ and the right-hand side has a total of 7+. We, therefore, add 5 electrons to the right-hand side:

Mn²⁺ + 4H₂O MnO₄^- + 8H⁺ + 5e^-

This is the reduction reaction.

Now we do the oxidation reaction. All O turns into water so we will get 3 molecules of water:

NaBiO₃ Bi³⁺ + Na⁺ + 3H₂O

This is not balanced - we need 6H⁺ to make the water:

NaBiO₃ + 6H⁺ Bi³⁺ + Na⁺ + 3H₂O

We now need to balance the charges again:

the left-hand side has 6+ and the right-hand side has 4⁺ so we need to add 2 electrons to the left-hand side:


NaBiO₃ + 6H⁺ + 2e^- Bi³⁺ + Na⁺ + 3H₂O

So now we have the two equations for the half reactions.

The last step is to put them together. We need to eliminate the electrons from each side, so we multiply the first equation by 2 and the second by 5. This will give 10 electrons on each side so they will cancel out and we will get:


2Mn²⁺ + 8H₂O + 5NaBiO₃ + 30H⁺ 2MnO₄^- + 16H⁺ + 5Bi³⁺ + 5Na⁺ + 15H₂O

You can see that there is H⁺ and H2O on both sides so we eliminate those:

2Mn²⁺ + 5NaBiO₃ + 14H⁺2MnO₄^- + 5Bi³⁺ + 5Na⁺ + 7H₂O