• Created by: Soma
  • Created on: 28-01-11 12:13


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

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Oxidation states

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.

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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.

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Half equations

a. first dividing them into half-reactions; that is, the oxidation reaction and the reduction reaction.

Cr3+(aq)  +  Cl1-(aq) ---->  Cr(s)  +  Cl2(g)

Oxidation:  Cl1-(aq) ---->  Cl2(g)  +  2e-

Reduction:  Cr3+(aq)  +  3e- ---->  Cr(s)

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Half equations 2

b. balance each half-reaction with respect to atoms first, then with respect to electrons.

Oxidation:  3( 2Cl1-(aq) ---->  Cl2(g) +   2e- )

Reduction:  2( Cr3+(aq)  +  3e- ---->  Cr(s) )

Oxidation:  6Cl1-(aq) ---->  3Cl2(g)  +  6e-

Reduction:  2Cr3+(aq)  +   6e- ---->  2Cr(s)

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Half equations 3

c. add the two half-reactions together cancelling the electrons which are now equal on each side of the arrow.

Final Equation:  2Cr3+(aq)  +  6Cl1-(aq) ----->  2Cr(s)  +  3Cl2(g)
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