# Balancing redox equation

So I had these two reactions that confused me .. I tried the standard way of balancing it (Putting oxygen number everywhere, +1,+2,-2 etc.) and then go on with it but it didn't work

So anyways here are the 2 reactions: $$\ce{3 Cu + 8HNO3-> 3Cu(NO3)2 + 2NO + 4H2O}$$ $$\ce{3Ag + 4HNO3 -> 3AgNO3 + NO + 2H2O}$$

I've no idea how do you get those coefficients (numbers).. Please, is there an easy way to do it (perhaps it has to do with electrons added/given? ) and if so - where do I start?

There are a few basic principles that are useful to keep in mind when you want to balance a redox equation.

1) OILRIG - Oxidation Is Loss, Reduction Is Gain (of electrons)

Let's look at the basics of the first equation: $$\ce{Cu + HNO3-> Cu(NO3)2} + ?$$

On the left hand side we have Cu(0) metal and N(V), on the right hand side we have Cu(II) ion and N(V), so from OILRIG we know that the copper is being oxidised... but we know from OILRIG we have to put the electrons from the Cu(0) somewhere...

2) Hydrogen ions are always +1, and oxygen ions are always (there are a few exceptions but this is a good general rule) -2

... So the only thing that can change is the nitrogen. The nitrogen must be being reduced.

Nitrogen actually has a large number of possible oxidation states, taking any value between -3 and +5, so how do we decide which one it goes to in this case? This depends on the temperature and the concentration of the acid. We can probably assume room temperature and bench concentration of acid (otherwise conditions need to be specified), and that the nitrogen isn't going to gain 5 electrons and become nitrogen gas, which means that we are going to form an oxide. Of the oxides, the most thermally stable is NO (nitrogen is in oxidation state +2 here).

What does that look like now?

$$\ce{Cu + HNO3-> Cu(NO3)2 + NO}$$

To balance the electron transfers, we need 3 Cu(O) going to Cu(II) to donate 6 electrons for 2 N(V) to go to N(II). Balancing for electrons and nitrogens, we now have:

$$\ce{3Cu + 8HNO3-> 3Cu(NO3)2 + 2NO}$$

Ok, now we need to look at balancing oxygens and hydrogens.

3) Use water if you can to balance oxygen and hydrogen. If not, start looking at hydroxide or hydrogen ions.

Here we need to somehow account for 8 hydrogens and 8x3 - 3x3x2 + 2 = 4 oxygens... That's 4 water molecules, so we finally have the answer you provided at the start!

$$\ce{3Cu + 8HNO3-> 3Cu(NO3)2 + 2NO + 4H2O}$$

• Good explanation; I would add why one would use water to balance oxygen and hydrogen over hydroxide or hydronium ion (if that is the case, and in what conditions). – Dissenter Jun 19 '14 at 16:04