Help with redox?

Question

In the presence of melted NaOH and NaNo3, Mn (IV) can be oxidized to Mn (VII), with formation of NO2 negative. Can anyone help with the equation for this reaction

Answer 1

2NaOH + 2MnO2 + 3NaNO3 ===> 2NaMnO4 + 3NaNO2 + H2O

I solved it by half-reactions. First I wrote and balanced,

H2O + NaNO3 + 2e- ===> NaNO2 + NaOH

Next,

NaOH + MnO2 ===> NaMnO4 + 3e- + H2O

After balancing the half reactions, I multiplied one by 2 and the other by 3 to get 6 electrons lost and gained in each. Then I added them together. The electrons cancelled.