# Chapter7 Redox Reactions Terminology Balancing Redox Reactions

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```					        Chapter 7
Redox Reactions, Terminology;
Balancing Redox Reactions
Oxidation Number
 The oxidation number of an element is the
number of electrons that can be potentially gained
or lost by the element during a redox reaction.
 The oxidation number of an element is usually
equal to its charge. Since some elements can have
more than one charge, these elements can also have
more than one oxidation number.
 In compounds the oxidation numbers of elements
with unknown oxidation numbers maybe calculated
from the oxidation numbers of the other elements
in the compound.
Oxidation Number Calculations
The sum of the oxidation numbers of the elements
that compose the compound is 0. For polyatomic
ions the sum is equal to its overall charge.
+1 +? -2                   +?   -2
K2CrO4                   Cr2O7-2
2Cr + 7(-2) = -2
2(+1) + Cr + 4(-2) = 0
2Cr = 14 - 2 = 12
Cr = 8 - 2 = +6
Cr = 12/2 = +6
+1   +1 +? -2              +? -2
Na2HSO3                   C2O4-2
2C + 4(-2) = -2
2(+1) + 1(+1) + S + 3(-2) = 0
2C = 8 - 2 = 6
S = 6 - 1 - 2 = +3
C = 6/2 = +3
Rules for Assigning Oxidation Numbers
 The oxidation number of a neutral substance that
contains atoms of only one element is 0. For
example H2, Fe, I2 , O2, O3, P4, S8 … all have oxidation
numbers equal to 0.
 Oxygen is assigned an oxidation number of -2 except
in peroxides where it’s -1 and OF2 where it’s +2.
 Hydrogen is assigned a value of +1 except in
hydrides where it’s -1.
 Metals are given positive (+) oxidation numbers.
 For covalently bonded compounds, the more
electronegative elements are given (-) oxidation
numbers.
Exercises
Give the oxidation numbers of the underlined
elements of the following :
1. NH3
2. FeCl3
3. NO2
4. Al2O3
5. PO4-3
6. H2O2
-
7. NO3
8. SO3
9. KMnO4
10. SiO2
Reduction and Oxidation Reactions
 Reactions in which elements lose or gain electrons
are known as reduction oxidation or “Redox”
reactions.
 The process in which elements lose electrons is
called oxidation and the process in which they gain
electron is known as reduction.
 The oxidation number of elements that undergo
reduction is reduced while elements that undergo
oxidation increase.
+2             0
Reduction:      Cu+2 + 2e        Cu
0       +3
Oxidation:      Fe      Fe+3 + 3e
Oxidizing and Reducing Agents
 In a REDOX chemical reaction one element is always
oxidized while another is reduced, i.e. one element
loses electrons while the other gains those electrons.
 The element that effects the reduction of the other
element is called the reducing agent while the element
that causes the oxidation of the other element is called
the oxidizing agent.
Reducing Agent
Oxidation
0                        +2
Zn (s) + H Cl(aq)        Zn+2(aq) + H2(g)
+1                          0
Reduction
Oxidizing Agent
Problems
Determine the
a. element that gets oxidized.
b. element that gets reduced.
c. the oxidizing agent.
d. the reducing agent.
Of the following chemical equations:

1. 3Mg + 2FeCl3        3MgCl2 + 2Fe

2. 2Na + 2H2O      2NaOH + H2

3. 8H+ + 2KMnO4 + 2NaCl  2MnO2 + Cl2 + 2K+ + 6Na+
Activity Series
 The activity series is a list of elements usually
arranged in the order of decreasing oxidation
potential or strength as a reducing agent.
 Any element in the activity series can displace all
the other elements below it from its salt in a single
replacement type reaction. The displaced metal is
reduced to its elemental form
A(s) + BX(aq)  AX(aq) + B(s)
 The topmost elements on the activity series reacts
even with water, sometimes violently, displacing the
hydrogen and producing hydrogen gas.
A(s) + HOH(l)  AOH(aq) + H2(g)
Balancing Redox Reactions (Acid Media)
The Half-Reaction Method:
1.    Divide the reaction into 2 half-reactions; the oxidation half-reaction and
the reduction half-reaction.
2.    Balance each part separately using the usual balancing method except
that the H and O need not be balanced initially.
3.    To balance O, add the necessary H2O molecules to the opposite side of
the equation making sure the number of H2O molecules corresponds to
the number of O atoms.
4.    To balance H add the appropriate number of H+ to the opposite side of
the equation.
5.    To make the charges balance, add e-s to the side of the equation which
have excess + charge.
6.    Multiply each half-reaction by a factor that makes the e-s balance out.
7.    Finally add the 2 reactions together cancelling out substances that occur
on both sides of the equation.
Balancing Redox Equations (Basic Media)
1. Balance the reaction as before (acid media).
2. Add OH- ions corresponding to the number of H+ ions on
the same side of the equation.
3. Combine the H+ and OH- on the same side of the equation
to H2O. Cancel out any H2O appearing on both sides of
the equation.

Problems
7.78 Balance the following half-reactions:
(a) Zn(s) + VO+2(aq)  Zn+2(aq) + V+3 (aq) (acid media)
(b) MnO4- (aq) + IO3- (aq)  MnO2(s) + IO4- (aq) (basic media)
Redox Stoichiometry
Solving redox stoichiometry problems is really no
different from solving any solution stoichiometry
problem and involves the same steps.

Problem
7.86 What is the molar concentration of As(III) in a
solution if 22.35 ml of 0.100 M KBrO3 is needed to
complete reaction with 50.00 ml of the As(III)
solution? The balanced chemical reaction is:
3H3AsO3 (aq) + BrO3- (aq)  Br- (aq) + 3H3AsO4 (aq)

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