Oxidation Numbers Redox Reactions How to Make Balancing

Oxidation Numbers & Redox Reactions How to Make Balancing Redox Reactions a Relatively Painless Process

What Do We Need to Consider? • Definitions • Rules for Assigning Oxidation Numbers • Balancing Redox Reactions

Redox: LEO says GER • Redox Reactions – Reactions involving the transfer of electrons from one species to another. • Oxidation – Loss of electrons (LEO) • Reduction – Gain of electrons (GER) • A given species is said to be “reduced” when it gains electrons and “oxidized” • when it loses electrons.

Examples

What are Oxidation Numbers? • The oxidation number of an element indicates the number of electrons lost, gained, or shared as a result of chemical bonding. • The change in the oxidation state of a species lets you know if it has undergone oxidation or reduction.

Oxidation can be defined as "an increase in oxidation number". • In other words, if a species starts out at one oxidation state and ends up at a higher oxidation state it has undergone oxidation.

Conversely, • Reduction can be defined as "a decrease in oxidation number". • Any species whose oxidation number is lowered during the course of a reaction has undergone reduction.

Example: Na + Cl 2 2 Na. Cl • The Na starts out with an oxidation number of zero (0) and ends up having an oxidation number of 1+. • It has been oxidized from a sodium atom to a positive sodium ion.

Example: Na + Cl 2 2 Na. Cl • The Cl 2 also starts out with an oxidation number of zero (0), but it ends up with an oxidation number of 1 -. • It, therefore, has been reduced from chlorine atoms to negative chloride ions.

Oxidizing agents • The substance bringing about the oxidation of the sodium atoms is the chlorine, thus the chlorine is called an oxidizing agent. • In other words, the oxidizing agent is being reduced (undergoing reduction).

Reducing agents • The substance bringing about the reduction of the chlorine is the sodium, thus the sodium is called a reducing agent. • Or in other words, the reducing agent is being oxidized (undergoing oxidation).

Redox reactions • Oxidation is ALWAYS accompanied by reduction. • Reactions in which oxidation and reduction are occurring are usually called Redox reactions.

Rules for Assigning Oxidation Numbers • There are several rules for assigning the oxidation number to an element. Learning these rules will simplify the task of determining the oxidation state of an element, and thus, whether it has undergone oxidation or reduction.

Rules 1 and 2 • 1. The oxidation number of an atom in the elemental state is zero. – • Example: Cl 2 and Al both are 0 2. The oxidation number of a monatomic ion is equal to its charge. – Example: In the compound Na. Cl, the sodium has an oxidation number of 1+ and the chlorine is 1 -.

Rule 3 • 3. The algebraic sum of the oxidation numbers in the formula of a compound is zero. • Example: the oxidation numbers in the Na. Cl above add up to 0

Rule 4 • 4. The oxidation number of hydrogen in a compound is 1+, except when hydrogen forms compounds called hydrides with active metals, and then it is 1 -. • Examples: H is 1+ in H 2 O, but 1 - in Na. H (sodium hydride).

Rule 5 • 5. The oxidation number of oxygen in a compound is 2 -, except in peroxides when it is 1 -, and when combined with fluorine. Then it is 2+. • Example: In H 2 O the oxygen is 2 -, in H 2 O 2 it is 1 -.

Rule 6 • 6. The algebraic sum of the oxidation numbers in the formula for a polyatomic ion is equal to the charge on that ion. • Example: in the sulfate ion, SO 42 -, the oxidation numbers of the sulfur and the oxygens add up to 2 -. The oxygens are 2 - each, and the sulfur is 6+.

Application Problems: • – – 1. What is the oxidation number of chromium in a. Na 2 Cr. O 4 ü In Na 2 Cr. O 4 the Cr has an oxidation number of 6+ and it can be said to be in the 6+ oxidation state b. Cr 2 O 72 - ü In Cr 2 O 72 - the Cr also has an oxidation number of 6+ and it can be said to be in the 6+ oxidation state.

Given the unbalanced equation below Cr 2 O 3(s) + Al(s) Cr(s) + Al 2 O 3(s) • a. identify the oxidation state of each element • • • Cr 3+ O 2 - Al 0 Cr 0 Al 3+ O 2 - b. identify the oxidizing agent Cr 3+ in the chromium(III) oxide c. identify the reducing agent Al(s)

Balancing Redox Reactions • Redox reactions often can be most easily balanced by the following method usually called the half-reaction method. • Example: Cr 3+(aq) + Cl 1 -(aq) Cr(s) + Cl 2(g)

first • divide it into half-reactions; that is, the oxidation reaction and the reduction reaction. • Oxidation: Cl 1 -(aq) Cl 2(g) + 2 e • Reduction: Cr 3+(aq) + 3 e- Cr(s)

Next balance • each half-reaction with respect to atoms first, then with respect to electrons. § Oxidation: 2 Cl 1 -(aq) Cl 2(g) + 2 e§ Reduction: Cr 3+(aq) + 3 e- Cr(s) § Oxidation: 3(2 Cl 1 -(aq) Cl 2(g) + 2 e-) § Reduction: 2(Cr 3+(aq) + 3 e- Cr(s))

Which gives § Oxidation: 6 Cl 1 -(aq) 3 Cl 2(g) + 6 e-) § Reduction: 2 Cr 3+(aq) + 6 e- 2 Cr(s)

add • the two half-reactions together cancelling the electrons which are now equal on each side of the arrow • Final Equation: 2 Cr 3+(aq) + 6 Cl 1 -(aq) 2 Cr(s) + 3 Cl 2(g)

Granted, • this example is a simple one, however, the same basic procedure is carried out for most redox reactions, with the addition of a couple of other steps, depending on the reaction conditions.

Example: • copper metal added to concentrated nitric acid • Cu(s) + HNO 3(aq) Cu(NO 3)2(aq) + NO 2(g)

First, • divide the equation into half-reactions • notice that the copper is going from a 0 oxidation state to 2+ which is oxidation • and some of the nitrogen is being reduced from 5+ in the nitrate ion to 4+ in the nitrogen dioxide

Half-reactions • Oxidation: Cu(s) Cu 2+(aq) + 2 e- • Reduction: NO 31 -(aq) + e- NO 2(g)

balance • each half-reaction with respect to atoms first, then with respect to electrons • Oxidation: Cu(s) Cu 2+(aq) + 2 e • Reduction: 2 NO 31 -(aq) + 2 e- 2 NO 2(g) – Add H 2 O on the right to balance out the oxygen atoms. – Then add enough H+ on the left to balance the hydrogens in the H 2 O

balance • Oxidation: Cu(s) Cu 2+(aq) + 2 e • Reduction: 2 NO 31 -(aq) + 4 H+ + 2 e- 2 NO 2(g) + 2 H 2 O • Add the two equations together canceling out the 2 e- on each side

balance • Cu(s) + 2 NO 31 -(aq) + 4 H+ Cu 2+(aq) + 2 NO 2(g) + 2 H 2 O • You now have the net ionic equation that shows that copper reacts with nitric acid to produce copper(II) ions, nitrogen dioxide gas and water. • The other two nitrate ions not shown in this form of the equation give copper(II) nitrate and are spectator ions.