Ch. 20 Oxidation-Reduction ReactionsAKA Redox Reactions

20.1 THE MEANING OF OXIDATION AND REDUCTION

Early Chemistry

•Oxidation: a substance gains oxygen

•Reduction: a substance loses oxygen

•Nothing can be oxidized without something being reduced (and vice versa)

•Called oxidation-reduction reactions or redox reactions

Modern Chemistry

•Oxidation: loss of electrons (complete or partial) or gain of oxygen

•Reduction: gain of electrons (complete or partial) or loss of oxygen

Mnemonic Devices

LEO goes GER OIL RIG

L- loss of O- oxidation

E- electrons is I- is

O- oxidation L- loss

G- gain of R- reduction

E- electrons is I- is

R- reduction G- gain

•The substance that loses electrons is the reducing agent

•Allows the other substance to be reduced while it itself is oxidized

•Substance that accepts electrons is the oxidizing agent

•Allows the other substance to be oxidized while it itself is reduced

HW Review

•A) (loses electrons) oxidation

•B) (loses electrons) oxidation

•C) (gains electrons) reduction

•D) (gains electrons) reduction

Redox with Covalent Compounds

•Complete transfer of electrons does not occur

•In a polar covalent bond electrons are not shared equally, which results in a partial gain or loss of electrons when such a bond is formed

Corrosion

•Occurs more rapidly in presence of salts and acids

•Salts and acids produce conductive solutions that make electron transfer easier and thus accelerate corrosion

Resistance to Corrosion•Not all metals corrode easily

•Some resist losing electrons (ex. Gold, platinum)

•Some are protected from extensive corrosion by an oxide layer coating the surface (ex. Aluminum, chromium)

Controlling Corrosion

•To prevent corrosion, metal surfaces may be coated with oil, paint, plastic, or another metal

•Another method sacrifices one metal to prevent the oxidation of another

•Magnesium and zinc are often used to protect iron

Lesson Check

20.2 OXIDATION NUMBERS

Assigning Oxidation Numbers

•Oxidation number: positive or negative number assigned to an atom to indicate its degree of oxidation or reduction

•Generally, a bonded atom’s oxidation number is the charge it would have if the electrons in the bond were assigned to the more electronegative atom

•In an ionic compound, each ion’s oxidation number is the charge of that ion

Rules for Assigning Oxidation Numbers1. For monatomic ions, the oxidation

number is equal to the charge.2. For H, oxidation number is +1, except in

metal hydrides where it is -1.3. For O, the oxidation number is -2, except

in peroxides where it is -1 and compounds with more electronegative fluorine where it is +1.

4. For elemental form of an element, oxidation number is 0.

Rules for Assigning Oxidation Numbers 5. For neutral compounds, the sum of the oxidation numbers of the atoms must equal 0.

6. For a polyatomic ion, the sum of the oxidation numbers of the atoms must equal the charge of the ion.

Practice

•What is the oxidation number of each kind of atom in the following ions and compounds?

•SO2

•CO3

-2

•Na2SO

4

•(NH4)

2S

•S2O

3

•Na2O

2

•P2O

5

•NO3

-

Oxidation-Number Changes in Chemical Reactions•Use to determine whether oxidation or reduction is happening

•Oxidation: oxidation number increases (becomes more positive)

•Reduction: oxidation number decreases (becomes more negative)

Practice

•Assign oxidation numbers to each element

•Identify which element is oxidized and which is reduced

•Identify the oxidizing agent and reducing agent

2HgO → O2 + 2Hg

NH4NO

2(s) → N

2(g) + 2H

2O(g)

PbO2(aq) + 4HI(aq) → I

2(aq) + PbI

2(s) + 2H

2O(l)

20.3 DESCRIBING REDOX EQUATIONS

Identifying Redox Reactions

•Chemical reaction can be classified into two categories• Redox reactions: electrons are transferred

from one reacting species to another• Ex. Many single replacement, combination,

decomposition, and combustion reactions, many reactions in which color changes

•Non-redox reactions: no electron transfer occurs• Ex. Double-replacement and acid-base

reactions

Practice

Identify which of the following are oxidation-reduction reactions. If a reaction is redox, name the element oxidized and the element reduced.

•CaCO3(s) + 2HCl(aq) → CaCl

2(aq) + H

2O(l) +

CO2(g)

•CuO(s) + H2(g) → Cu(s) + H

2O(l)

•2KMnO4 + 3KCN + H

2O → 2MnO

2 + 2KOH +

3K(OCN)

Practice: Assign Oxidation Numbers & Identify Which Elements are Oxidized and Reduced

•4Na(s) + O2(g) → 2Na

2O(s)

•2Sr + O2→ 2SrO

•2Li + H2 → 2LiH

•2Cs + Br2 → 2CsBr

Practice: Assign Oxidation Numbers & Identify Which Elements are Oxidized and Reduced

•3Mg + N2 → Mg

3N

2

•4Fe + 3O2 → 2Fe

2O

3

•Cl2 + 2NaBr → 2NaCl + Br

2

•Si + 2F2 → SiF

4

•H2 + Cl

2 →2HCl

•5Fe2+ + MnO4

- + 8H+ → Mn2+ + 5 Fe3+ + 4H2O

Balancing Redox Equations

•Redox reactions are often too complex to balance by trial and error

•Two methods for balancing, both based on the fact that the number of electrons gained in reduction must equal the number of electrons lost in oxidation

•Oxidation-number-change method

•Half-reaction method

Using Oxidation-Number Changes• Step 1: assign oxidation numbers to all atoms in

the equation• Step 2: identify which atoms are oxidized and

which reduced• Step 3: use a bracketing line to connect the atoms

that undergo oxidation and a second bracketing line to connect those that undergo reduction; write the change in oxidation number• Step 4: make the total increase in oxidation

number equal the total decrease in oxidation number by using appropriate coefficients • Step 5: make sure the whole equation is balanced

for both atoms and charge

Oxidation-Number Change Example

Using Half-Reactions

•Half-reactions show only the oxidation or reduction half of a redox reaction

•The two half reactions can be balanced separately before putting them back together for a complete balanced redox reaction

Using Half-Reactions•Step 1: Write the unbalanced equation in ionic

form, separating all ionic compounds into ions•Step 2: Write separate half-reactions for the

oxidation and reduction processes•Step 3: Balance the atoms in the half-reactions•Step 4: Add enough electrons to one side of

each half-reaction to balance the charges (hint, if one reaction has e- on the reactant side, the other reaction will have e- on the product side)

Using Half-Reactions•Step 5: Multiply each half-reaction by the

appropriate numbers to make the numbers of electrons equal in both halves•Step 6: Add the balanced half-reactions to

show an overall equation•Step 7: Add any spectator ions and balance

the equation

Practice: Balance the following equations based

on number of electrons in the half-reactions

Zn(s) + Cu+2(aq) → Zn+2(aq) + Cu(s)

Zn+2(aq) + Cr(s) → Zn(s) + Cr+3(aq)

Ni(s) + Fe+3(aq) → Ni+2(aq) + Fe+2(aq)

Zn(s) + H+(aq) → Zn+2(aq) + H2(g)

Practice: Balance the Following Equations

•Sn2+ + Cr2O

72- + H+ → Sn4+ + Cr3+ +H

2O

•Zn + NO3

- + H2O + OH- → NH

3 + Zn(OH)

42-

•ClO3

- + I- +H+ → Cl- + I2

+ H2O

•C2O

42- + MnO

4- + H+ → Mn2+ + CO

2 + H

2O

•Br2 + SO

2 + H

2O → Br - + SO

42- + H+

•Zn + As2O

3 + H

2O → Zn2+ +AsH

3 + OH-

•NiO2 + S

2O

32- + H

2O + OH- → Ni(OH)

2 + SO

32-