Balance the following equation in basic solution using the lowest possible integers and give the coefficient of water. PbO(s) + NH3(aq) → N2(g) + Pb(s)

Question

I'm having trouble understanding these type of problems, is there a simple way to approach or solve this?

Answer 1

This is similar to the other problem of yours that I answered, except it used base (OH-) instead of acid (H+). The steps are pretty much the same.

PbO ==> Pb ... reduction half reaction b/c Pb goes from 2+ to zero

PbO ==> Pb + H₂O ... balanced for Pb and oxygen

PbO + 2H₂O ==> Pb + H₂O + 2OH^- ... balanced for hydrogen using OH^- and H₂O

PbO + 2H₂O + 2e- ==> Pb + H₂O + 2OH^- ... balanced for charge

NH₃ ==> N₂ ... oxidation half reaction b/c N goes from 3- to zero

2NH₃ ==> N₂ ... balanced for mass of nitrogen

2NH₃ + 6OH^- ==> N₂ + 6H₂O ... balanced for hydrogen using OH^- and H₂O

2NH₃ + 6OH^- ==> N₂ + 6H₂O + 6e- ... balanced for charge

Multiply reduction reaction by 3 to equalize electrons and then add the two equations together.

3PbO + 6H₂O + 6e- ==> 3Pb + 3H₂O + 6OH^- ... reduction reaction

2NH₃ + 6OH^- ==> N₂ + 6H₂O + 6e- ... oxidation reaction

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3PbO + 2NH₃ ==> 3Pb + 3H₂O + N₂ ... balanced redox (balanced for mass and charge)