[Borek]

1. You have two atoms of Cr on both sides of the oxidation equation - but you have lost 2 on the RHS of the full equation.

2. No idea why do you think it should be 6 INSTEAD of 3. 3 is wrong - your reduction reaction after being multiplied by 3 will give you 6 atoms of Br (more precisely - six Br^- anions) on the RHS.

Seems like you did the same mistake twice.

[Evaldas]

First, what is RHS?

[Borek]

Right Hand Side, Left Hand Side (of the equation).

[Evaldas]

There are two P atoms in a product molecule, so it will be easier to write phosphorus part as

2P⁰ -> 2P⁺⁵ + 10e^-

(you did the same for oxygen, just the other way around).

First of all, my teacher says you're suppsed to write P⁰ -> P⁺⁵ + 5 e^-, in the P + O₂ -> P₂O₅ reaction.

[Evaldas]

I still haven't cracked this topic and I still need help, does anyone know a good web page that explains in a detailed way how to balance the more difficult reactions?

[AWK]

I still haven't cracked this topic and I still need help, does anyone know a good web page that explains in a detailed way how to balance the more difficult reactions?

Try these examples
Pb(N₃)₂ + Cr(MnO₄)₂ = Cr₂O₃ + MnO₂ + Pb₃O₄ + NO
P₂I₄ + P₄ + H₂O = PH₄I + H₃PO₄
K₄[Fe(CN)₆] + Ce(NO₃)₄ + KOH = K₂CO₃ + Fe(OH)₃ + Ce(OH)₃ + KNO₃ + H₂O