# Balance the reactionMnO−4 + NO−2--> MnO2 + NO−3in basic solution. What is the sum of thecoefficients?

Asked on by candyyyyy

jeew-m | College Teacher | (Level 1) Educator Emeritus

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Here we have a redox reaction. `MnO_4^(-)` is reduced to `Mn^(2+)` and `NO_2^(-)` to `NO_3^(-)` .

`MnO_4^(-)+8H^+ +5e rarr Mn^(2+)+4H_2O` ----(1)

`NO_2^(-)+H_2O rarr NO_3^(-)+2H^++2e` -----(2)

(1)*2+2*(5) to cancel out electrons

`2MnO_4^(-)+16H^++5NO_2^(-)+5H_2O rarr 2Mn^(2+)+20H_2O+5NO_3^(-)+10H^+`

`2MnO_4^(-)+6H^++5NO_2^(-) rarr 2Mn^(2+)+15H_2O+5NO_3^(-)`

Since this a basic media we have to add OH^(-)

`2MnO_4^(-)+6H^+6OH^(-)+5NO_2^(-) rarr 2Mn^(2+)+15H_2O+5NO_3^(-)+6OH^(-)`

`2MnO_4^(-)+6H_2O+5NO_2^(-) rarr 2Mn^(2+)+15H_2O+5NO_3^(-)+6OH^(-)`

`2MnO_4^(-)+5NO_2^(-) rarr 2Mn^(2+)+9H_2O+5NO_3^(-)+6OH^(-)`

So the balance chemical equation is;

`2MnO_4^(-)+5NO_2^(-) rarr 2Mn^(2+)+9H_2O+5NO_3^(-)+6OH^(-)`

Sum of coefficients = 2+5+2+9+5+6 = 29

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