The Student Room Group
Disregard this
Reply 2
How would I write that in an equation?
That is the equation. Do you mean the ionic half equations?
Reply 4
d/w. I think what I want is just a slightly different way of working it. But i get what you mean now, thanks.
Disregard what I said, I knew something was wrong:


2Fe is an element and therefore oxidation state is 0
3Cl2 would also have oxidation state of 0

2FeCl3. You assign Cl before Fe, so Cls get -1 giving them a total of -6 (because there are 6). The overall charge of the molecule is neutral and there are 2 Fes so the charges of 2Fe must equal -6. Therefore each Fe has an oxidation state of +3.

Fe increased in oxidation number so it was oxidised.
Cl decreased in oxidation number so was reduced



2Fe+3Cl22FeCl32Fe + 3Cl_2 \to 2FeCl_3

Unparseable latex formula:

2Fe \to 2Fe^3^+ + 6e^-



3Cl2+6e2Cl33Cl_2 + 6e^- \to 2Cl_3
2fecl3 cl2=2fecl3