Question

balance the redox reaction

NO₂^- + MnO₄^- + H⁺ -------> NO₃^- + Mn²⁺ + H₂O

show work please

Answer 1

We have to first break the equation into two halves: Oxidation half reaction and reduction half reaction.

As nitrite (NO₂^-) is converted to nitrate (NO₃^-), it is oxidized and permanganate (MnO₄^-) is reduced to Mn²⁺. We will balance these equations one by one.

NO₂^- --------------------> NO₃^-

balance O by H₂O

NO₂^- + H₂O ------------> NO₃^-

balance H by H⁺

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

balance charge by e-

NO₂^- + H₂O ------------> NO₃^- + 2 H⁺ + 2 e- (1)

Now come to reduction half reaction.

MnO₄^- -------> Mn²⁺

balance O by H₂O

MnO₄^- -------> Mn²⁺ + 4 H₂O

balance H by H⁺

MnO₄^- + 8 H⁺ -------> Mn²⁺ + 4 H₂O

balance charge by e-

MnO₄^- + 8 H⁺ + 5e- -------> Mn²⁺ + 4 H₂O (2)

loss of electrons must be equal to gain of electrons so we will multiply equation (1) by 5 and equation (2) by 2. and then add both equations.

(NO₂^- + H₂O ------------> NO₃^- + 2 H⁺ + 2 e-)x5

5 NO₂^- + 5 H₂O ------------> 5 NO₃^- + 10 H⁺ + 10 e-

and

(MnO₄^- + 8 H⁺ + 5e- -------> Mn²⁺ + 4 H₂O)x2

2 MnO_4^- + 16 H⁺ + 10e- -------> 2 Mn²⁺ + 8 H₂O

Now adding the two resultant equations:

5 NO₂^- + 5 H₂O ------------> 5 NO₃^- + 10 H⁺ + 10 e-

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

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5 NO₂^- + 2 MnO₄^- + 6 H⁺ ---------------> 5 NO₃^- + 2 Mn²⁺ + 3 H₂O