If electrons are transferred, it's a redox reaction. OIL RIG.
A loss of electrons -Oxidation
A gain of electrons -Reduction
Reduction and Oxidation happen simultaneously.
An Oxidising agent accepts electrons and get reduced.
A reducing agent donates electrons and gets oxidised.
Na + 1/2 Cl -----> Na Cl
All atoms are treated as ions, even if they are covalently bonded.
Uncombined elements have an oxidation state of 0.
Elements just bonded to identical atoms, like O and H also have an oxidation state of 0.
The oxidation state of a simple monatomic ion ( Na ) is the same as its charge.
In compounds or compound ions, the overall oxidation state is just the ion charge. For example in SO - overall oxidation state = -2 (oxidation state of O = -2 so the total is -8, if we sub this into the equation of orignal compound then Sulpher has oxidation number of +6.
The sum of oxidation states for a neutral compound is 0. For example in Fe O the overall oxidation state = 0. Oxidation state of O is -2 so that totals an overall charge of -6 so if sub this in the oxidation state of a single atom of Fe is +3.
Oxidation states 2
Combined oxygen is nearly always -2 except in peroxides where it's -1.
^ In H O, oxidation state of O = -2, but in H O ,oxidation state of H has to be +1 (an H atom can only lose one electron) so oxidation state of O = -1.
Combined hydrogen is +1, except in metal hydrides where it is -1 (and H where it's is zero)
^ In HF, oxidation state of H = +1 but in NaH where the oxidation state of H = -1.
a. first dividing them into half-reactions; that is, the oxidation reaction and the reduction reaction.
Oxidation: Cl1-(aq) ----> Cl2(g) + 2e-
Reduction: Cr3+(aq) + 3e- ----> Cr(s)
Half equations 2
b. balance each half-reaction with respect to atoms first, then with respect to electrons.
Reduction: 2( Cr3+(aq) + 3e- ----> Cr(s) )
Oxidation: 6Cl1-(aq) ----> 3Cl2(g) + 6e-
Reduction: 2Cr3+(aq) + 6e- ----> 2Cr(s)
Half equations 3
c. add the two half-reactions together cancelling the electrons which are now equal on each side of the arrow.