Balance the following equation in basic solution using the lowest possible integers and give the coefficient of water.

Question

Balance the following equation in basic solution using the lowest possible integers and give the coefficient of water.

MnO₄^-(aq) + NO₂^-(aq) → MnO₂(s) + NO₃^-(aq)

Where does water come in?

I got the equation below but I don't understand where water comes in

MnO₄^- + 2NO₂^- --> MnO₂^- + 2NO₃^-

the answer is: 1

please show steps so I can follow along, thank you!

Answer 1

Using the half reaction method:

MnO₄^- ==> MnO₂

NO₂^- ===> NO₃^-

MnO₄^- ==> MnO₂ + 2H₂O ... to balance the oxygens

NO₂^- + H₂O ===> NO₃^-

MnO₄^- + 4H⁺ ==> MnO₂ + 2H₂O ... to balance the hydrogens

NO₂^- + H₂O ===> NO₃^- + 2H⁺

MnO₄^- + 4H⁺ + 4OH^- ==> MnO₂ + 2H₂O + 4OH^- ... converting to basic solution

NO₂^- + H₂O + 2OH^- ===> NO₃^- + 2H⁺ + 2OH^-

MnO₄^- + 4H⁺ + 4OH^- + 3e- ==> MnO₂ + 2H₂O + 4OH^- ... to balance charge

NO₂^- + H₂O + 2OH^- ===> NO₃^- + 2H⁺ + 2OH^- + 2e-

2MnO₄^- + 8H⁺ + 8OH^- + 6e- ==> 2MnO₂ + 4H₂O + 8OH^- ... to equalize electrons

3NO₂^- + 3H₂O + 6OH^- ===> 3NO₃^- + 6H⁺ + 6OH^- + 6e-

Adding the two half reactions and forming H₂O from H⁺ and OH^- ...

2MnO₄^- + 8H₂O + 6e- + 3NO₂^- + 3H₂O + 6OH^- => 2MnO₂ + 4H₂O + 6H₂O + 3NO₃^- + 6H⁺ + 2OH^- + 6e-

Combining like terms and canceling common terms on opposite sides of equation we get...

2MnO₄^- + 3NO₂^- + 1H₂O ==>2MnO₂ + 3NO₃^- + 2OH^- ... balanced for mass and charge

I hope this all made sense. If not, add a comment and I'll try to explain better.