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# Balance Redox reactions

1. NaNO3 = NaNO2 + O2

Let us write half reaction of the oxidation and reduction. Initially nitrogen has charge +5 And at the end of reaction it has charge +3. How do we calculate that? In NaNO3 oxygen has charge -2. Sodium has charge +1. The molecule of NaNO3 is neutral. It means that negative charges inside NaNO3 molecule must be equal to positive charges.

Na (+1) + O3 (-2 x 3) = 1 - 6 = - 5. Then nitrogen has to be + 5 to make the molecule neutral.

In NaNO2, nitrogen has charge +3. Nitrogen must receive 2 negative electrons to change its charge from +5 to +3. 5 + ( - 2) = 3 So we can write

N5+ + 2e = N3+ | 2

Oxygen initially has charge -2. At the end of reaction it becomes neutral and has charge 0.

So, we can write 2O2- - 4e = O2 | 4

Combine two half reactions and get:

N5+ + 2e = N3+ | 2
2O2- - 4e = O2 | 4

Since 2 and 4 can be divided by 2 and reduced to 1 and 2

N5+ + 2e = N3+ | 2 1
2O2- - 4e = O2 | 4 2

Now switch positions of 1 and 2

N5+ + 2e = N3+ | 2 1 2
2O2- - 4e = O2 | 4 2 1

From the above N5+ and N3+ should have coefficient 2 and oxygen should have coefficient 1.

2NaNO3 = 2NaNO2 + O2

Check the equation balance: from both sides of the equation we have 2 Na, 2 N, and 6 O. We are done!