First we need to write a balance equation for this redox reaction
[tex]2Fe _(_s_) + 3Cl_2_(_g_) ==> 2FeCl_3_(_a_q_)[/tex]
In the above reaction [tex]Fe[/tex] on reactant side has oxidation state 0, while on the product side it has an oxidation state of +3. On the other hand,[tex]Cl_2[/tex] has oxidation state of 0 on reactant side and an oxidation state of -1 on product side.
The metal [tex]Fe[/tex] loses electrons to form [tex]Fe^3^+[/tex] ion and is oxidized. So we write an oxidation half reaction.
[tex]2Fe _(_s_) ==> 2Fe^3^+_(_a_q_) + 6e^-[/tex]
and the [tex]Cl_2[/tex] molecule gains electrons to form [tex]Cl^-[/tex] ions and is reduced. The reduction half reaction is
[tex]3Cl_2_(_g_) + 6e^- ==> 6Cl^-_(_a_q_)[/tex].
