Ch. 20 Oxidation-Reduction Reactions AKA 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 Numbers 1. 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? SO 2 CO 3-2 Na 2 SO 4 (NH 4 ) 2 S S 2 O 3 Na 2 O 2 P 2 O 5
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 O 2 + 2Hg NH 4 NO 2 (s) N 2 (g) + 2H 2 O(g) PbO 2 (aq) + 4HI(aq) I 2 (aq) + PbI 2 (s) + 2H 2 O(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. CaCO 3 (s) + 2HCl(aq) CaCl 2 (aq) + H 2 O(l) + CO 2 (g) CuO(s) + H 2 (g) Cu(s) + H 2 O(l) 2KMnO 4 + 3KCN + H 2 O 2MnO 2 + 2KOH + 3K(OCN)
Practice: Assign Oxidation Numbers & Identify Which Elements are Oxidized and Reduced 4Na(s) + O 2 (g) 2Na 2 O(s) 2Sr + O 2 2SrO 2Li + H 2 2LiH 2Cs + Br 2 2CsBr
Practice: Assign Oxidation Numbers & Identify Which Elements are Oxidized and Reduced 3Mg + N 2 Mg 3 N 2 4Fe + 3O 2 2Fe 2 O 3 Cl 2 + 2NaBr 2NaCl + Br 2 Si + 2F 2 SiF 4 H 2 + Cl 2 2HCl 5Fe 2+ + MnO 4 - + 8H + Mn 2+ + 5 Fe 3+ + 4H 2 O
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) + H 2 (g)
Practice: Balance the Following Equations Sn 2+ 2- + Cr 2 O 7 + H + Sn 4+ + Cr 3+ +H 2 O - Zn + NO 3 + H 2 O + OH - 2- NH 3 + Zn(OH) 4 - ClO 3 + I - +H + Cl - + I 2 + H 2 O 2- - C 2 O 4 + MnO 4 + H + Mn 2+ + CO 2 + H 2 O Br 2 + SO 2 + H 2 O Br - 2- + SO 4 + H + Zn + As 2 O 3 + H 2 O Zn 2+ +AsH 3 + OH - 2- NiO 2 + S 2 O 3 + H 2 O + OH - 2- Ni(OH) 2 + SO 3