SCH 4C - Electrochemistry - Oxidation Numbers

Competition for Electrons

Many chemical reactions are electrical in nature and involve a competition for electrons.

2Mg(s)  +  O₂(g)  →  2MgO(s)

There is a competition for electrons between the magnesium and the oxygen atoms.  Since oxygen has a higher electronegativity, it gains two electrons from the magnesium. 

• magnesium – loses 2 electrons (and undergoes oxidation)

• oxygen – gains 2 electrons (and undergoes reduction)

This is the basis of reduction-oxidation reactions (or redox reactions, for short).

Oxidation Numbers

How can you tell a redox reaction has occurred?

Look for a change in the oxidation number of an element during the course of a reaction. 

Oxidation numbers are used by chemists to keep track of the positive or negative character of atoms or ions in redox reactions.  Then, it is possible to determine which species has been oxidized (lost electrons) and which has been reduced (gained electrons). 

An oxidation number is the real or apparent charge an atom or ion has when all bonds are assumed to be ionic.  You are already familiar with oxidation numbers – up to this point, we have called them valences or charges.

Rules for Determining Oxidation Numbers

• In free elements, each atom has an oxidation number of zero.

ex.  Cu⁰, Zn⁰, Fe⁰, H₂⁰

• For ions consisting of a single atom, the oxidation number is equal to the charge on the ion.

ex.  Cl^- (-1), Na⁺ (+1), Pb²⁺ (+2), Fe³⁺ (+3)

• The oxidation number of H is +1 and of O is -2 in most compounds (except in hydrides where the oxidation number of H is -1 and peroxides where the oxidation number of O is -1).

• The algebraic sum of the oxidation numbers in a neutral compound must be zero; in a polyatomic ion, the sum must be equal to the ion charge.

ex. HSO₃^-       ON(H) + ON(S) + 3ON(O) = -1

                            +1   +      x     +   3(-2)    = -1

                                                                 x = +4

TryIts!. Find the oxidation number of the underlined element in each case: HNO₃, MnO₄^-, S₂O₃^2-

Answers are found at the end of the lesson.

TryIts! Answers:

HNO₃   +1 + x +3(-2) = 0         MnO₄^-     x + 4(-2) = -1       S₂O₃^2-     2x + 3(-2) = -2

             x = +5                                         x = +7                                  x = +2

Homework

Answers: