I am having trouble figuring out a chemistry question on how to balance a redox reaction using the half-reaction method.

2H2O + NO --> NO3- + 4H+

2H2O + NO --> NO3- + 4H+ +3e-

Step 6. Multiply both of the half-reactions by a whole number so that the number of electrons gained and lost is equal.

14H++Cr2O72- + 6e- --> 2Cr3+ + 7H2O

2H2O + NO --> NO3- + 4H+ +3e- X 2 [because 2X3= 6 electrons]

4H2O + 2NO --> 2NO3- +8H+ +6e-

Step 7. Add the half-reactions. Subtract any chemicals that are common to both sides.

Cr2O72- + 2NO+6H+ --> 2Cr3+ + 2NO3-+ 3H2O