How do I balance this redox

O2+NO2^- = NO3^- in basic solution?
I dont know what to do with the single O2 element, as it is only on one side.

(1/2)O2+NO2^- = NO3^-

O2+2NO2^- = 2NO3^-

It is now balanced for charge and atoms. No need for basic solution. To do it in basic, add this:

O2+NO2^- = NO3^- + OH^-

half-reactions:

O2 = OH^-
NO2^- = NO3^-

balance as if in acidic:
4e- + 2H+ + O2 = 2OH^-
H2O + NO2^- = NO3^- + 2H+ + 2e-

equalize electrons:
4e- + 2H+ + O2 = 2OH^-
2H2O + 2NO2^- = 2NO3^- + 4H+ + 4e-

Add:
O2+ 2H2O + 2NO2^- = 2NO3^- + 2OH^- + 2H+

Cancel two waters from each side:
O2 + 2NO2^- = 2NO3^-

Nice problem.

O2 + 2NO2^ = 2NO3^

So there will be:

On the left side:
O = 2 + 2(2) = 6
N = 2

On the right side:

O = 2(3) = 6
N = 2