Oxidation and Reduction Definitions of oxidation and reduction Oxidation numbers Redox equations

Oxidation numbers  Metals are typically considered more 'cation-like' and would possess positive oxidation numbers, while nonmetals are considered more 'anion-like' and would possess negative oxidation numbers.  Oxidation number is the number of electrons gained or lost by the element in making a compound

Determine the oxidation number of the elements in each of the following compounds: a. H2CO3 b. N2 c. Zn(OH)42-

Position of element in periodic table determines oxidation number  G1A is +1  G2A is +2  G3A is +3 (some rare exceptions)  G5A are –3 in compounds with metals, H or with NH 4+. Exceptions are in compounds to the right; in which case use rules 3 and 4.  G6A below O are –2 in binary compounds with metals, H or NH 4+. When they are combined with O or with a lighter halogen, use rules 3 and 4.  G7A elements are –1 in binary compounds with metals, H or NH 4+ or with a heavier halogen. When combined with O or a lighter halogen, use rules 3 and 4.

Predicting oxidation numbers  Oxidation number of atoms in element is zero in all cases  Oxidation number of element in monatomic ion is equal to the charge  sum of the oxidation numbers in a compound is zero  sum of oxidation numbers in polyatomic ion is equal to the charge  F has oxidation number –1  H has oxidn no. +1;  Oxygen is usually –2

Oxidation - reduction  Oxidation is loss of electrons  Reduction is gain of electrons  Oxidation is always accompanied by reduction The total number of electrons is kept constantThe total number of electrons is kept constant  Oxidizing agents oxidize and are themselves reduced  Reducing agents reduce and are themselves oxidized

Follow the electrons

Identifying reagents  Those elements that tend to give up electrons (metals) are typically categorized as reducing agents and those that tend to accept electrons (nonmetals) are referred to as oxidizing agents.

Corrosion  The tarnishing of silver is just one example of a broad class of oxidation-reduction reactions that fall under the general heading of corrosion. Another example is the series of reactions that occur when iron or steel rusts. When heated, iron reacts with oxygen to form a mixture of iron(II) and iron(III) oxides.

More active metals are strongly reducing

Predicting results of displacement reactions  In this reaction the element metal A displaces the ion metal B from its compound  This will only occur if A lies above B in the activity series  Displacement reaction exercises

Nuggets of redox processes  Where there is oxidation there is always reduction Oxidizing agent Reducing agent Is itself reduced Is itself oxidized Gains electrons Loses electrons Causes oxidation Causes reduction

In reaction with metals, nonmetals are always oxidizers  Reactions of elements are always redox  The nonmetal gains electrons, becomes a negative ion  The metal loses electrons, becomes a positive ion  Identification is harder when there are no elements involved: oxidation numbers must be used

The Half-Reaction method  Any redox process can be written as the sum of two half reactions: one for the oxidation and one for the reduction

Six habits of the redox equation balancer

 Working with redox reactions is fundamentally a bookkeeping issue. You need to be able to account for all of the electrons as they transfer from one species to another. There are a number of rules and tricks for balancing redox reactions, but basically they all boil down to dealing with each of the two half-reactions individually.

STEP 1: the unbalanced equation  Consider for example the reaction of aluminum metal to form alumina (Al 2 O 3 ) Al(s) + O 2  Al 2 O 3

STEP 2: identify the oxidized and reduced and write the half reactions  Oxidation half-reaction  Al(s)  Al 3+ (s) + 3e -  Reduction half-reaction  O 2 (g) + 4e -  2O 2- (s)  This reaction shows aluminum metal being oxidized to form an aluminum ion with a +3 charge. The half-reaction below shows oxygen being reduced to form two (2) oxygen ions, each with a charge of -2.

 If we combine those two (2) half-reactions, we must make the number of electrons equal on both sides. The number 12 is a common multiple of three (3) and four (4), so we multiply the aluminum reaction by four (4) and the oxygen reaction by three (3) to get 12 electrons on both sides. Now, simply combine the reactions. Notice that we have 12 electrons on both sides, which cancel out. The final step is to combine the aluminum and oxygen ions on the right side using a cross multiply technique:

Taking care of the number of atoms, you should end up with:

STEP 6: Add half reactions and eliminate common items