Transcript Oxidation and Reduction

Balancing Redox Equations:
following the electrons
Review: Oxidation and reduction
Oxidation numbers
Review: Oxidation - reduction
 Oxidation is loss of electrons
 Reduction is gain of electrons
 Oxidation is always accompanied by reduction
• The total number of electrons is kept constant
 Oxidizing agents oxidize and are
themselves reduced
 Reducing agents reduce and are
themselves oxidized
Nuggets of redox processes
 Where there is oxidation there is always
reduction
Oxidizing agent
Reducing agent
Is itself reduced
Is itself oxidized
Gains electrons
Loses electrons
Causes oxidation
Causes reduction
Oxidation numbers review
 Oxidation number is the number of electrons
gained or lost by the element in making a
compound
 Metals are more 'cation-like'
 Have positive oxidation
numbers
 Nonmetals are 'anion-like'
 Have negative oxidation
numbers.
Predicting oxidation numbers
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Oxidation number of atoms in element is zero in all cases
Oxidation number of element in monatomic ion is equal
to the charge
sum of the oxidation numbers in a compound is zero
sum of oxidation numbers in polyatomic ion is equal to
the charge
F has oxidation number –1
H has oxidn no. +1; except in metal hydrides where it is –
1
Oxygen is usually –2. Except:
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O is –1 in hydrogen peroxide, and other peroxides
O is –1/2 in superoxides KO2
In OF2 O is +2
Position of element in periodic table
determines oxidation number
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G1A is +1
G2A is +2
G3A is +3 (some rare exceptions)
G5A are –3 in compounds with metals, H or with NH4+.
Exceptions are in compounds to the right; in which case
use rules 3 and 4.
 G6A below O are –2 in binary compounds with metals,
H or NH4+. When they are combined with O or with a
lighter halogen, use rules 3 and 4.
 G7A elements are –1 in binary compounds with metals,
H or NH4+ or with a heavier halogen. When combined
with O or a lighter halogen, use rules 3 and 4.
Redox equations
 Net ionic equations summarize the
essentials of a reaction without including all
the particles present
 Redox equations are a subset which involve
electron transfer
 Without being given all the information,
balancing redox equations involves
balancing electron flow
Balancing redox equations:
systematic methods
 Oxidation number method – tracking
changes in the oxidation numbers
 Half-reaction method – tracking changes in
the flow of electrons
 Same principles, different emphasis
 We will examine the half-reaction method
The Half-Reaction method
 Any redox process can be written as the
sum of two half reactions: one for the
oxidation and one for the reduction
Six habits of the redox equation
balancer
STEP 1: the unbalanced equation
 Dichromate ion reacts with chloride ion to
produce chlorine and chromium (III)
Cr2O72 (aq )  Cl  (aq )  Cr 3 (aq )  Cl2 (aq )
STEP 2: identify the oxidized and
reduced and write the half reactions
 Oxidation half-reaction
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Cl (aq)  Cl2 (aq)
 Reduction half-reaction
Cr2O72 (aq)  Cr 3 (aq)
STEP 3: Balance the half reactions
 Oxidation
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2Cl (aq)  Cl2 (aq)
 Reduction
Cr2O72 (aq)  2Cr 3 (aq)
Material balance with H2O and H+ or OH2MnO4 (aq)  5  2Br  (aq)  2Mn 2 (aq)  5Br2 (aq)
 Strategy: add H2O to the side that lacks for
O and add H+ (the reaction is in acid
solution) to the other side
2MnO4 (aq)  10Br  (aq)  16H  (aq)  2Mn 2 (aq)  5Br2 (aq)  8H 2O(l )
 In basic solution we add OH- and H2O
instead of H2O and H+ respectively
2MnO4 (aq)  10Br  (aq)  8H 2O(aq)  2Mn 2 (aq)  5Br2 (aq)  16OH  (aq)
 Test equation for both atoms and charges
STEP 4: Material balance
 Add H2O to the side lacking O and add H+ to
the other side (for reactions in acid solution)
 Oxidation reaction – unchanged
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2Cl (aq)  Cl2 (aq)
 Reduction reaction
14 H  (aq)  Cr2O72 (aq)  2Cr 3 (aq)  7 H 2O(l )
STEP 5: Balance half-reactions for
charge by addition of electrons
 Balance charges on both sides of each halfreaction
2 x -1 = 2 x -1

2Cl (aq)  Cl2 (aq)  2e

14 x +1 + -2 + 6 x -1 = 2 x +3
14 H  (aq)  Cr2O72 (aq)  6e   2Cr 3 (aq)  7 H 2O(l )
STEP 5 cont: Multiply by factors to
balance total electrons
 Overall change in electrons must be zero
 Multiply the oxidation half reaction by 3
3x2=6
32Cl (aq)  Cl2 (aq)  2e
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14 H  (aq)  Cr2O72 (aq)  6e   2Cr 3 (aq)  7 H 2O(l )
STEP 6: Add half reactions and
eliminate common items
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6Cl (aq)  3Cl2 (aq)  6e
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2
7
14 H (aq)  Cr2O
+
(aq)  6e
=
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
 2Cr 3 (aq)  7 H 2O(l )
14 H  (aq)  Cr2O72 (aq)  6Cl  (aq)  2Cr 3 (aq)  3Cl2 (aq)  7 H 2O(l )
Electrons cancel both sides
Atoms and charges balance
Balanced molecular equation
 Add in the spectators: there will always be
space.
Reagents were K2Cr2O7, NaCl and H2SO4
 Net ionic equation
14 H  (aq)  Cr2O72 (aq)  6Cl  (aq)  2Cr 3 (aq)  3Cl2 (aq)  7 H 2O(l )
 Balanced molecular equation
7 H 2 SO4 (aq)  K 2Cr2O7 (aq)  6 NaCl(aq)  Cr2 ( SO4 )3 (aq)  3Cl2 (aq)  K 2 SO4 (aq)  7 H 2O(l )