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Oxidation-Reduction
Reactions
Oxidation and Reduction
 Oxidation-reduction reactions always occur
simultaneoulsy.
 Redox Reactions
 Oxidation
 Loss of electrons
 Gain oxygen
 2Fe2O3
 Reduced
+
3C2
 Reduction
 Gain Electrons
 Loss of Oxygen
 4Fe + 3CO2
Oxizided
Redox Reactions that Form Ions
 Between metal and nonmetals, electrons are
transferred from the metal to the nonmetal.
 Increases stability
 Mg + S  Mg2+ + S2 Oxidation: Mg  Mg2+ + 2e- (loss of electrons)
 Reduction: S + 2e-  S2-
(gain of electrons)
Oxidizing and Reducing Agents
 Reducing Agent: substance that loses electrons.
 Oxidizing Agent: substance that accepts the electrons
is the oxidizing agent.
 Mg + S  Mg2+ + S2 Mg: reducing agent, oxidized
 S: oxidizing agent, reduced
 LEO the lion goes GER
 LEO: Losing Electrons is Oxidation
 GER: Gaining Electrons is Reduction
Redox with Covalent Compounds
 In covalent compounds, polar molecules involve
unequal sharing of electrons
 The shift in electrons is redox for it is the partial gain
and loss of electrons
 H 2O
 Oxygen: electrons shift toward
 Reduced, oxidizing agent
 Hydrogen: electrons shift away
 Oxidized, reducing agent
Processes Leading to Redox





Oxidation
Complete loss of electrons
(ionic reactions)
Shift of electrons away from
an atom in covalent bond
Gain of Oxygen
Loss of Hydrogen by a
covalent compound
Increase in oxidation number





Reduction
Complete gain of electrons
(ionic reactions)
Shift of electrons toward from
an atom in covalent bond
Loss of Oxygen
Gain of Hydrogen by a
covalent compound
Decrease in oxidation
number
Corrosion
 Iron, corrodes by being oxidized to ions of iorn by
oxgyen
 2Fe +O2 + H2O  2Fe(OH)2
 To protect iron, a piece of magnesium is placed in
electrical contact.
 When oxygen or water attack the iron object, iron lose
electrons.
 Because Mg is more easily oxidized, the Mg
immediately transfers electrons to the iron, preventing
their oxidation to iron ions.
Oxidation Numbers
Oxidation Numbers
 A positive or negative number assigned to an atom to
indicate its degree of oxidation or reduction.
 Rule of Thumb: when bonded, the oxidation number
is the same as its ionic charge.
 In a chemical reaction:
 Increase in oxidation number  oxidation
 Decrease in oxidation number  reduction
Rules for Assigning Oxidation Numbers
 Monatomic ions equal ionic charge; Br1-: -1
 H: compounds is +1; metal hydrides is -1
 H2O: +1, NaH: -1
 O: compounds is -2; peroxides is -1, or in compounds with F it’s +
 H2O: -2, H2O2: -1
 Atoms in elemental form or diatomic is 0.
 S: 0, H2: 0
 For compound, the sum of oxidation numbers must equal 0.
 H2O 
H(+1), O(-2) 
2(+1) + 1(-2) =0
 For a polyatomic ion, the sum of the oxidation numbers must
equal the ionic charge of the ion.
 NO32- : N(4), O(-2)  1(+4) + 3(-2) = 2-
Oxidation Number Practice
 NaCl
 (+1,-1)
 H 2O
 (+1,-2)
 SO2
 (+4, -2)
 CO32 (+4,-2)
 Na2SO4
 (+1,+6,-2)
Oxidation-Number Changes in
Chemical Reactions
 +1 +5 -2
0
+2 +5 -2
0
 2AgNO3 +Cu Cu(NO3)2 + 2Ag
 Ag: reduced
 Cu: oxidized
Let’s Try These:
 Cl2 + 2HBr  2HCl +Br
 2H2 + O2  2H2O
 2KNO3  2KNO2 + O2
Answers
Let’s Try These:
0
+1 -1
+1 -1 0
 Cl2 + 2HBr  2HCl +Br
 0
0
+1 -2
 2H2 + O2  2H2O
 +1 +5 -2
+1 +3 -2 0
 2KNO3  2KNO2 + O2
Balancing Redox Reactions
How to tell if it’s a redox rxn
 If the oxidation number of an element in a reacting
species changes
 0
0
+2 -2
 N2 + O2  2NO
Balancing by Oxidation No.
1) Assign oxidation numbers to all the atoms
+3 -2
+2 -2
0
+4 -2
Fe2O3 + CO  Fe + CO2
2) Identify which atoms are oxidized and reduced.
3) Use brackets to connect that atoms undergoing
oxidation, and other set for those reduced.
+2 Oxidation
+3 -2
+2 -2
0
+4 -2
Fe2O3 + CO  Fe + CO2
-3 reduction
Balancing by Oxidation No.
 Make the total increase in oxidation number equal the
total decrease using coefficients
(+2)x3=6
+3 -2
+2 -2
0
+4 -2
Fe2O3 + 3CO  2Fe + 3CO2
(-3)x2=6
 Make sure the equation is balanced for both atoms and
charge
Let’s Practice
 KClO3  KCl + O2
 HNO2 + HI  NO + I2 + H2O
 Bi2S3 + HNO3  Bi(NO3) 3 + NO + S + H2O
 SbCl5 + KI  SbCl3 +KCl + I2
Redox Reactions
Half-Reactions
 Equation showing just the oxidation or reduction
portion of the redox reaction.
 S + HNO3  SO2 + NO +H2O

0
+4 -2
 Oxidation Half: S  SO2

+5 -2
+2 -2
 Reduction Half: NO3-  NO
Balancing Half-Reactions
 To balance:
 Write the unbalanced ionic equation
 Write separate half reactions for oxidation & reduction
 Balance atoms in each half-reaction
 Add enough electrons to one side of each half-reaction
to balance the charges
 Multiply each half-reaction by an appropriate number to
make the numbers of electrons equal
 Add the half reaction to show the overall equation
 Add the spectator ions and balance the equation
Half-Reactions
 S + HNO3  SO2 + NO +H2O
 Ionic Form: S + H+ + NO3-  SO2 + NO +H2O

0
+4 -2
 Oxidation Half: S  SO2

+5 -2
+2 -2
 Reduction Half: NO3-  NO
 Balancing Atoms in Half –Reactions
 2H2O + S  SO2 + 4H+
 4H+ + NO3-  NO + 2H2O
Half-Reactions
 Add e- to each side of half reactions to balance charges
 Oxidation: 2H2O + S  SO2 + 4H+ + 4e Reduction: 4H+ + NO3- + 3e-  NO + 2H2O
 Multiply each half reaction by an appropriate number
to make the numbers of electrons equal
 Oxidation: 6H2O + 3S  3SO2 + 12H+ + 12e Reduction: 16H+ + 4NO3- + 12e-  4NO + 8H2O
 Subtract the terms that appear on both sides and add
in the spectator ions
 6H2O + 3S + 16H+ + 4NO3- + 12e-  3SO2 + 12H+ + 12e-+ 4NO + 8H2O
 3S + 4HNO3  3SO2 + 4NO + 2H2O
Half-Reactions
 Let’s Practice:
 KMnO4 + HCl  MnCl2 + Cl2 + H2O + KCl