How many grams of MnCl2 will be oxidized by 10.0g of NaBiO3?

Question

Sodium bismuthate (NaBiO3) oxidizes MnCl2 to MnO4- and is consequently reduced to Bi(OH)3. How many grams of MnCl2 will be oxidized by 10.0g of NaBiO3? (show your work.)

Answer 1

Let's look at the reaction taking place:

MnCl₂ + NaBiO₃ ==> MnO₄^- + Bi(OH)₃

This is a redox reaction which now must be balanced...

Looking at the half reactions (without spectator ions) we have...

Mn²⁺ + 4H₂O + 8OH^- ==> MnO₄^- + 8H₂O + 5e-

BiO₃^- + 3H₂O + 2e- ===> Bi(OH)₃ + 3OH^-

multiply reduction reaction by 5 and oxidation reaction by 2 to equalize electrons, and we have...

2Mn²⁺ + 8H₂O + 16OH^- ==> 2MnO₄^- + 16H₂O + 10e-

5BiO₃^- + 15H₂O + 10e- ===> 5Bi(OH)₃ + 15OH^-

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2Mn²⁺ + 8H₂O + 16OH^- + 5BiO₃^- + 15H₂O + 10e- => 2MnO₄^- + 16H₂O + 10e- + 5Bi(OH)₃ + 15OH^-

2Mn²⁺ + 7H₂O + OH^- + 5BiO₃^- ==> 2MnO₄^- + 5Bi(OH)₃ BALANCED EQUATION

10.0 g NaBiO₃ x 1 mol NaBiO₃/280 g x 2 mol MnCl₂/5 mol NaBiO₃ x 126 g MnCl₂/mol MnCl₂ = 0.788 g MnCl₂