(NH_{4})_{2}Cr_{2}O_{7} --> Heat --> N_{2} + 4H_{2}O + Cr_{2}O_{3}

Is there a good way to describe the formula here? Could you run through both sides and show how it "balances"?

I never understood it at school, perhaps I'll get it here!

1. Find the oxidation numbers (O.N.) of all the atoms and take notice of those atoms which change its O.N. In this case we have: N

^{(-III)} in (NH

_{4})

_{2}Cr

_{2}O

_{7} and N

^{(0)} in N

_{2} so every N atom has lost 3 electrons --> N has been oxydized. Cr

^{(+VI)} in (NH

_{4})

_{2}Cr

_{2}O

_{7} and Cr

^{(+III)} in Cr

_{2}O

_{3} so Cr has taken 3 electrons --> Cr has been reduced.

2. Write the 2 semi-reactions of oxidation and reduction:

2NH

_{4} - 6e

^{-} --> N

_{2}; here it's 6e

^{-} because there are 2 atoms of N.

Cr

_{2}O

_{7} + 6e

^{-} --> Cr

_{2}O

_{3}; same as up.

3. Balance those semi-reactions. In this case thy are already balanced: the number of electrons involved in the first semi-reaction is = number in the other. If it was, e.g. 3e

^{-} in the first and 5e

^{-} in the second, you multiply all the first semi-reaction by 5 and all the second by 3, so you have 15 e

^{-} in both (minimum common multiple of 3 and 5).

4. Write the reaction with the coefficients you have already computed. In this case they are: 2 for (NH

_{4}), 1 for N

_{2}, 1 for Cr

_{2}O

_{7} and 1 for Cr

_{2}O

_{3} (in the other hypotetical case, you should have multiplied the first two coefficients by 5 and the other two by 3).

So you write the still unbalanced reaction but with these coefficients:

(NH

_{4})

_{2}Cr

_{2}O

_{7} --> N

_{2} + Cr

_{2}O

_{3}.

5. Balance charges, if there are. Not in this case, so it's already balanced.

6. Balance atoms. You have: 2 atoms of N at left and 2 at right. Ok. 8 H atoms at left and 0 at right, you have to balance adding H

_{2}O molecules, so you add 4 H

_{2}O molecules at right. 2 Cr atoms at left and 2 at right. Ok. 7 O atoms at left and 4 + 3 = 7 at right. Ok.

Everything balanced, you can write the reaction:

(NH

_{4})

_{2}Cr

_{2}O

_{7} --> N

_{2} + 4H

_{2}O + Cr

_{2}O

_{3}.

This was a very simple case. In other cases could be less simple to find O.N. and the semi-reactions. Ask if you want to know more.