Complete and balance the equation for this reaction in acidic solution. equation: MnO_{4}^{-} + HNO_{2} -> NO_{3}^{-} + Mn^{2 + } MnO−4+HNO2⟶NO−3+Mn2+

Answer 1

Reduction half reaction:

MnO₄^- ==> Mn²⁺ + 4H₂O ... balanced for Mn and O

MnO₄^- + 8H⁺ ==> Mn²⁺ + 4H₂O ... balanced for Mn, O and H by adding acid

MnO₄^- + 8H⁺ + 5e- ==> Mn²⁺ + 4H₂O ... balanced for mass and charge = balanced reduction reaction

Oxidation half reaction:

NO₂^- + H₂O==> NO₃^- ... balanced for N and O

NO₂^- + H₂O==> NO₃^- +2H⁺ ... balanced for N, O and H by adding acid

NO₂^- + H₂O==> NO₃^- +2H⁺ + 2e- ... balanced for mass and charge = balanced oxidation reaction

To equalize electrons transferred, multiply reduction reaction by 2 and oxidation reaction by 5:

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

5NO₂^- + 5H₂O==> 5NO₃^- +10H⁺ + 10e-

Add these two together and combine/cancel like terms to get:

2MnO₄^- + 5NO₂^- + 6H⁺ ==> 2Mn²⁺ 5NO₃^- + 3H₂O ... balanced redox equation