Question

Using the set of smallest whole number coefficients to balance the redox equation

MnO₄^- + NO₂^- ---> MnO₂ + NO₃^-

in basic solution, you get

a) 2 OH^- on the right.

b) 3 OH^- on the left.

c) 1 H₂O on the right

d) 3 H₂O on the left

Answer 1

Steps in balancing redox reaction using the half-reaction method:

1. Separate the equation into two half-reactions:

MnO₄^-  MnO₂ NO₂^-   NO₃^-

2. Balance indiviadually the 2 half equations:

a. Balance the atom that is not hydrogen and oxygen by inspection.

b. Balance oxygen by adding H2O on the side lacking O.

c. Balance hydrogen by adding H⁺

   d. Balance charges by adding electron

MnO₄^-    MnO₂

     MnO₄^- + 4H⁺ + 3e^-   MnO₂ + 2H₂O

  

    NO₂^- + H₂O      NO₃^-   + 2H⁺  + 2e^-

3. Equalize the number of electrons lost and gained then add the 2 equations.

(MnO₄^- + 4H⁺ + 3e^-   MnO₂ + 2H₂O) X 2

(NO₂^- + H₂O      NO₃^-   + 2H⁺  + 2e^-) X 3

2MnO₄^- + 8H⁺ + 6e^- 2MnO₂ + 4H₂O

3NO₂^- + 3H₂O    3NO₃^-   + 6H⁺  + 6e^-

^{------------------------------------------------------------------------------------------------------}

   2MnO₄^- +  2H⁺  +   3NO₂^-      2MnO₂ + H₂O + 3NO₃^-

4. Neutralize the H+ by adding OH- on both sides

2MnO₄^- +  2H⁺  +   3NO₂^- + 2OH^-   2MnO₂ + H₂O + 3NO₃^-   + 2OH^-

       (Note: OH- and H+ combine to form H2O)

2MnO₄^- +  2H₂O + 3NO₂^-   2MnO₂ + H₂O + 3NO₃^-   + 2OH^-

Final Equation:

   2MnO₄^- + H₂O + 3NO₂^-      2MnO₂ + 3NO₃^-   + 2OH^-

Hence, the correctanswer is: a) 2 OH- on the right