Balance Redox Basic-Chemistry?

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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)

answer: 1. i got 3.

How to solve?
 
I'm going to be very step by step with this for clarity. I apologize if it is wordier than you want.

Manganese half reaction:

MnO4- ~> MnO2
Mn is being reduced from 7 to 4

MnO4- + 3e- ~> MnO2
MnO4- + 3e- ~> MnO2 + 2 H2O
MnO4- + 3e- + 4 H+ ~> MnO2 + 2 H2O
MnO4- + 3e- + 4 H+ + 4 OH- ~> MnO2 + 2 H2O + 4 OH-
MnO4- + 3e- + 4 H2O ~> MnO2 + 2 H2O + 4 OH-
MnO4- + 3e- + 2 H2O ~> MnO2 + 4 OH-

Nitrogen half reaction:

NO2- ~> NO3-
N is being oxidized from 3 to 5

NO2- ~> NO3- + 2e-
NO2- + H2O ~> NO3- + 2e-
NO2- + H2O ~> NO3- + 2e- + 2 H+
NO2- + H2O + 2 OH- ~> NO3- + 2e- + 2 H+ + 2 OH-
NO2- + H2O + 2 OH- ~> NO3- + 2e- + 2 H2O
NO2- + 2 OH- ~> NO3- + 2e- + H2O

Use multiplication to balance electrons:
(MnO4- + 3e- + 2 H2O ~> MnO2 + 4 OH-) x 2
(NO2- + 2 OH- ~> NO3- + 2e- + H2O) x 3

2 MnO4- + 6e- + 4 H2O ~> 2 MnO2 + 8 OH-
3 NO2- + 6 OH- ~> 3 NO3- + 6e- + 3 H2O

Cancel out electrons, 6 OH- and 3 H2O

2 MnO4- + H2O + 3NO2- ~> 2 MnO2 + 2 OH- + 3 NO3-

There is only 1 water in the final equation.
 
I'm going to be very step by step with this for clarity. I apologize if it is wordier than you want.

Manganese half reaction:

MnO4- ~> MnO2
Mn is being reduced from 7 to 4

MnO4- + 3e- ~> MnO2
MnO4- + 3e- ~> MnO2 + 2 H2O
MnO4- + 3e- + 4 H+ ~> MnO2 + 2 H2O
MnO4- + 3e- + 4 H+ + 4 OH- ~> MnO2 + 2 H2O + 4 OH-
MnO4- + 3e- + 4 H2O ~> MnO2 + 2 H2O + 4 OH-
MnO4- + 3e- + 2 H2O ~> MnO2 + 4 OH-

Nitrogen half reaction:

NO2- ~> NO3-
N is being oxidized from 3 to 5

NO2- ~> NO3- + 2e-
NO2- + H2O ~> NO3- + 2e-
NO2- + H2O ~> NO3- + 2e- + 2 H+
NO2- + H2O + 2 OH- ~> NO3- + 2e- + 2 H+ + 2 OH-
NO2- + H2O + 2 OH- ~> NO3- + 2e- + 2 H2O
NO2- + 2 OH- ~> NO3- + 2e- + H2O

Use multiplication to balance electrons:
(MnO4- + 3e- + 2 H2O ~> MnO2 + 4 OH-) x 2
(NO2- + 2 OH- ~> NO3- + 2e- + H2O) x 3

2 MnO4- + 6e- + 4 H2O ~> 2 MnO2 + 8 OH-
3 NO2- + 6 OH- ~> 3 NO3- + 6e- + 3 H2O

Cancel out electrons, 6 OH- and 3 H2O

2 MnO4- + H2O + 3NO2- ~> 2 MnO2 + 2 OH- + 3 NO3-

There is only 1 water in the final equation.
 
I'm going to be very step by step with this for clarity. I apologize if it is wordier than you want.

Manganese half reaction:

MnO4- ~> MnO2
Mn is being reduced from 7 to 4

MnO4- + 3e- ~> MnO2
MnO4- + 3e- ~> MnO2 + 2 H2O
MnO4- + 3e- + 4 H+ ~> MnO2 + 2 H2O
MnO4- + 3e- + 4 H+ + 4 OH- ~> MnO2 + 2 H2O + 4 OH-
MnO4- + 3e- + 4 H2O ~> MnO2 + 2 H2O + 4 OH-
MnO4- + 3e- + 2 H2O ~> MnO2 + 4 OH-

Nitrogen half reaction:

NO2- ~> NO3-
N is being oxidized from 3 to 5

NO2- ~> NO3- + 2e-
NO2- + H2O ~> NO3- + 2e-
NO2- + H2O ~> NO3- + 2e- + 2 H+
NO2- + H2O + 2 OH- ~> NO3- + 2e- + 2 H+ + 2 OH-
NO2- + H2O + 2 OH- ~> NO3- + 2e- + 2 H2O
NO2- + 2 OH- ~> NO3- + 2e- + H2O

Use multiplication to balance electrons:
(MnO4- + 3e- + 2 H2O ~> MnO2 + 4 OH-) x 2
(NO2- + 2 OH- ~> NO3- + 2e- + H2O) x 3

2 MnO4- + 6e- + 4 H2O ~> 2 MnO2 + 8 OH-
3 NO2- + 6 OH- ~> 3 NO3- + 6e- + 3 H2O

Cancel out electrons, 6 OH- and 3 H2O

2 MnO4- + H2O + 3NO2- ~> 2 MnO2 + 2 OH- + 3 NO3-

There is only 1 water in the final equation.
 
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