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

Mn²⁺ --> MnO^4-

It needs oxygen atom from somewhere, which comes from water.

Mn²⁺ + H₂O --> MnO4^- + H⁺
oxygen is not balanced, so
Mn²⁺ + 4H20 --> MnO4^- + 8H⁺

Left side has a total charge of 2+ and the right hand side has a total of 7+. We add 5 electrons to the right side:

Mn2+ + 4H2O --> MnO4- + 8H+ + 5e-

This is the reduction reaction.

Now we do the oxidation reaction.

NaBiO3 --> Bi3+ + Na+ + 3H2O

We need 6H+ to make the water:

NaBiO3 + 6H+ --> Bi3+ + Na+ + 3H2O

balance the charges now,

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.

We need to remove the electrons from each side, so we multiply the first equation by 2 and the second by 5. This gives 10 electrons on each side so they will cancel out and we will get:


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

Finally,

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

I hope this can help you.