OXIDATION REDUCTION OxidationReduction Reactions The term oxidation was

OXIDATION – REDUCTION

Oxidation-Reduction Reactions • The term oxidation was originally used to describe reactions in which an element combines with oxygen. • Example: The reaction between magnesium metal and oxygen to form magnesium oxide involves the oxidation of magnesium. • 2 Mg(s) + O 2(g) 2 Mg. O(s)

• The term reduction comes from the Latin stem meaning "to lead back. " Anything that leads back to magnesium metal therefore involves reduction. • The reaction between magnesium oxide and carbon at 2000 C to form magnesium metal and carbon monoxide is an example of the reduction of magnesium oxide to magnesium metal. • Mg. O(s) + C(s) Mg(s) + CO(g)

Practice Problem 1: Determine which element is oxidized and which is reduced when lithium reacts with nitrogen to form lithium nitride. 6 Li(s) + N 2(g) 2 Li 3 N(s)

The Role of Oxidation Numbers in Oxidation-Reduction Reactions • Chemists eventually extended the idea of oxidation and reduction to reactions that do not formally involve the transfer of electrons. • CO(g) + H 2 O(g) CO 2(g) + H 2(g) • What changes in this reaction is the oxidation state of these atoms. The oxidation state of carbon increases from +2 to +4, while the oxidation state of the hydrogen decreases from +1 to 0.

Practice Problem 2: • Determine which atom is oxidized and which is reduced in the following reaction Sr(s) + 2 H 2 O(l) Sr 2+(aq) + 2 OH-(aq) + H 2(g)

Remember this phrase: LEO the lion says GER. • LEO = Loss of Electrons is Oxidation • GER = Gain of Electrons is Reduction • Another way is to simply remember that reduction is to reduce the oxidation number. • Therefore, oxidation must increase the value. • Oxidizing Agent - that substance which oxidizes somebody else. It is reduced in the process. • Reducing Agent - that substance which reduces somebody else. It is oxidized in the process.

Rules for Assigning Oxidation Numbers 1. All free, uncombined elements have an oxidation number of zero. This includes diatomic elements such as O 2 or others like P 4 and S 8. 2. Hydrogen, in all its compounds except hydrides, has an oxidation number of +1 (positive one) 3. Oxygen, in all its compounds except peroxides, has an oxidation number of -2 (negative two).

Practice Problems • • • What is the oxidation number of. . . 1) N in NO 3¯ 2) C in CO 32¯ 3) Cr in Cr. O 42¯ 4) Cr in Cr 2 O 72¯ 5) Fe in Fe 2 O 3 6) Pb in Pb. OH+ 7) V in VO 2+ 8) V in VO 2+ 9) Mn in Mn. O 4¯ 10) Mn in Mn. O 42¯

What is a Half-Reaction? • A half-reaction is simply one which shows either reduction OR oxidation, but not both. Here is an example redox reaction: • Ag+ + Cu ---> Ag + Cu 2+ • It has BOTH a reduction and an oxidation in it. That is why we call it a redox reaction, from REDuction and OXidation.

• What you must be able to do is look at a redox reaction and separate out the two half-reactions in it. To do that, identify the atoms which get reduced and get oxidized. Here are the two halfreactions from the example: Ag+ ---> Ag Cu ---> Cu 2+ • The silver is being reduced, its oxidation number going from +1 to zero. The copper's oxidation number went from zero to +2, so it was oxidized in the reaction. In order to figure out the halfreactions, you MUST be able to calculate the oxidation number of an atom.

• When you look at the two half-reactions, you will see they are already balanced for atoms with one Ag on each side and one Cu on each side. So, all we need to do is balance the charge. • To do this you add electrons to the more positive side. You add enough to make the total charge on each side become EQUAL. • To the silver half-reaction, we add one electron: Ag+ + e¯ ---> Ag • To the copper half-reaction, we add two electrons: Cu ---> Cu 2+ + 2 e¯

Half-reactions NEVER occur alone. • notice that each half-reaction wound up with a total charge of zero on each side. This is not always the case. You need to strive to get the total charge on each side EQUAL, not zero.

Half-Reactions Practice Problems • Balance each half-reaction for atoms and charge: • 1) Cl 2 ---> Cl¯ • 2) Sn ---> Sn 2+ • 3) Fe 2+ ---> Fe 3+ • 4) I 3¯ ---> I¯ • 5) ICl 2¯ ---> I¯ • 6) Sn + NO 3¯ ---> Sn. O 2 + NO 2 • 7) HCl. O + Co ---> Cl 2 + Co 2+ • 8) NO 2 ---> NO 3¯ + NO

Answers • • 1) Cl 2 + 2 e¯ ---> 2 Cl¯ 2) Sn ---> Sn 2+ + 2 e¯ 3) Fe 2+ ---> Fe 3+ + e¯ 4) I 3¯ + 2 e¯ ---> 3 I¯ 5) ICl 2¯ + 2 e¯ ---> I¯ + 2 Cl¯ 6) Sn ---> Sn. O 2 and NO 3¯ ---> NO 2 7) HCl. O ---> Cl 2 and Co ---> Co 2+ 8) NO 2 ---> NO 3¯ and NO 2 ---> NO

Balancing Half-Reactions in Acid Solution • Mn. O 4¯ ---> Mn 2+ in an acid solution • Before looking at the balancing technique, the fact that it is in acid solution can be signaled to you in several different ways: 1) It is explicitly said in the problem. 2) An acid (usually a strong acid) is included as one of the reactants. 3) An H+ is written just above the reaction arrow.

There are three other chemical species available in an acidic solution: H 2 O H+ e¯ water is present because the reaction is taking place in solution • the hydrogen ion is available because it is in acid solution • electrons are available because that's what is transferred in redox reactions • All three will be used in getting the final answer. • •

1. Balance the atom being reduced/oxidized. In our example, there is already one Mn on each side of the arrow, so this step is already done. Mn. O 4¯ ---> Mn 2+ • Balance the oxygens. Do this by adding water molecules (as many as are needed) to the side needing oxygen. In our case, the left side has 4 oxygens, while the right side has none, so: Mn. O 4¯ ---> Mn 2+ + 4 H 2 O

• Balance the hydrogens. Do this by adding hydrogen ions (as many as are needed) to the side needing hydrogen. In our example, we need 8 (notice the water molecule's formula, then consider 4 x 2 = 8). 8 H+ + Mn. O 4¯ ---> Mn 2+ + 4 H 2 O • Balance the total charge. This will be done using electrons. It is ALWAYS the last step. 5 e¯ + 8 H+ + Mn. O 4¯ ---> Mn 2+ + 4 H 2 O

Here is a second half-reaction, also in acid solution: Cr 2 O 72¯ ---> Cr 3+ 1. Balance the atom being reduced/oxidized. Cr 2 O 72¯ ---> 2 Cr 3+ 2. Balance the oxygens. Cr 2 O 72¯ ---> 2 Cr 3+ + 7 H 2 O 3. Balance the hydrogens. 14 H+ + Cr 2 O 72¯ ---> 2 Cr 3+ + 7 H 2 O 4. Balance the total charge. 6 e¯ + 14 H+ + Cr 2 O 72¯ ---> 2 Cr 3+ + 7 H 2 O

Another example (in acid solution) SO 2 ---> SO 42¯ 1. SO 2 ---> SO 42¯ (the sulfur is already balanced) 2. 2 H 2 O + SO 2 ---> SO 42¯ (now there are 4 oxygens on each side) 3. 2 H 2 O + SO 2 ---> SO 42¯ + 4 H+ (2 x 2 from the water makes 4 hydrogens) 4. 2 H 2 O + SO 2 ---> SO 42¯ + 4 H+ + 2 e¯ (zero charge on the left; +4 from the hydrogens and -2 from the sulfate, so 2 electrons gives the -2 charge required to make zero on the right)

• • • Practice Problems 1) Re ---> Re. O 2 2) Cl 2 ---> HCl. O 3) NO 3¯ ---> HNO 2 4) H 2 Ge. O 3 ---> Ge 5) H 2 Se. O 3 ---> Se. O 42¯ 6) Au ---> Au(OH)3 (this one is a bit odd!) 7) H 3 As. O 4 ---> As. H 3 8) H 2 Mo. O 4 ---> Mo 9) NO ---> NO 3¯ 10) H 2 O 2 ---> H 2 O

Answers: 1. 2 H 2 O + Re ---> Re. O 2 + 4 H+ + 4 e¯ 2. 2 H 2 O + Cl 2 ---> 2 HCl. O + 2 H+ + 2 e¯ 3. 2 e¯ + 3 H+ + NO 3¯ ---> HNO 2 + H 2 O 4. 4 e¯ + 4 H+ + H 2 Ge. O 3 ---> Ge + 3 H 2 O 5. H 2 O + H 2 Se. O 3 ---> Se. O 42¯ + 4 H+ + 2 e¯ 6. 3 H 2 O + Au ---> Au(OH)3 + 3 H+ + 3 e¯ 7. 8 e¯ + 8 H+ + H 3 As. O 4 ---> As. H 3 + 4 H 2 O 8. 6 e¯ + 6 H+ + H 2 Mo. O 4 ---> Mo + 4 H 2 O 9. 2 H 2 O + NO ---> NO 3¯ + 4 H+ + 3 e¯ 10. 2 e¯ + 2 H+ + H 2 O 2 ---> 2 H 2 O

Balancing Half-Reactions in Basic Solution • Before looking at the balancing technique, the fact that it is in basic solution can be signaled to you in several different ways: • It is explicitly said in the problem. • A base (usually a strong base) is included as one of the reactants. • An OH¯ is written just above the reaction arrow.

There are three other chemical species available in a basic solution besides the ones shown above. They are: H 2 O OH¯ e¯ water is present because the reaction is taking place in solution • the hydroxide ion is available because it is in basic solution • electrons are available because that's what is transferred in redox reactions. • All three will be used in getting the final answer. • •

Pb. O 2 ---> Pb. O It is to be balanced in basic solution. • Step One to Four: Balance the half-reaction AS IF it were in acid solution. 1. Balance the atom being reduced/oxidized. 2. Balance the oxygens (using H 2 O). 3. Balance the hydrogens (using H+). 4. Balance the charge. When you do that to the above half-reaction, you get: 2 e¯ + 2 H+ + Pb. O 2 ---> Pb. O + H 2 O

• Step Five: Convert all H+ to H 2 O. • Do this by adding OH¯ ions to both sides. • The side with the H+ will determine how many hydroxide to add. In our case, the left side has 2 hydrogen ions, while the right side has none, so: 2 e¯ + 2 H 2 O + Pb. O 2 ---> Pb. O + H 2 O + 2 OH¯

• Notice that, when the two hydroxide ions on the left were added, they immediately reacted with the hydrogen ion present. The reaction is: H+ + OH¯ ---> H 2 O

• Step Six: Remove any duplicate molecules or ions. • In our example, there are two water molecules on the left and one on the right. • This means one water molecule may be removed from each side, giving: 2 e¯ + H 2 O + Pb. O 2 ---> Pb. O + 2 OH¯ • The half-reaction is now correctly balanced.

Here is a second half-reaction, also in basic solution: Mn. O 4¯ ---> Mn. O 2 • Step One: Balance the half-reaction AS IF it were in acid solution. • 3 e¯ + 4 H+ + Mn. O 4¯ ---> Mn. O 2 + 2 H 2 O • Step Two: Convert all H+ to H 2 O. • 3 e¯ + 4 H 2 O + Mn. O 4¯ ---> Mn. O 2 + 2 H 2 O + 4 OH¯ • Step Three: Remove any duplicate molecules or ions. • 3 e¯ + 2 H 2 O + Mn. O 4¯ ---> Mn. O 2 + 4 OH¯