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Ch. 20 Oxidation-Reduction ReactionsAKA Redox Reactions
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20.1 THE MEANING OF OXIDATION AND REDUCTION
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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
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Modern Chemistry
•Oxidation: loss of electrons (complete or partial) or gain of oxygen
•Reduction: gain of electrons (complete or partial) or loss of oxygen
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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
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•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
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HW Review
•A) (loses electrons) oxidation
•B) (loses electrons) oxidation
•C) (gains electrons) reduction
•D) (gains electrons) reduction
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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
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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
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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)
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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
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Lesson Check
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20.2 OXIDATION NUMBERS
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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
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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.
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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.
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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
-
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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)
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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)
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20.3 DESCRIBING REDOX EQUATIONS
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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
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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)
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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
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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
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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
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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
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Oxidation-Number Change Example
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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
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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)
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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
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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)
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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-