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Unit 12, Lesson 1 of 3
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Redox Reactions
OBJECTIVES: After this lesson, I will understand the concepts of oxidation and reduction, and I will be able to identify
redox reactions. I will also be able to write half reactions for redox reactions
Oxidation and reduction reactions or redox reactions are another important chemical reaction that occurs in solution.
These reactions are responsible for battery power! They are also responsible for the corrosion of important metals. The
two reactions, oxidation and reduction, are simultaneous and interdependent.
Oxidation:
Reduction:
Redox Reaction:
Scientist developed a system that makes it easy to keep track of the number of electrons lost or gained by an atom in a
reaction. Positive, negative, or neutral values can be assigned to atoms. These values are called oxidation numbers.
Oxidation Number Rules
1. Each uncombined element has an oxidation number of zero. In the chemical equation: 2Na + Cl2  2NaCl, both
the uncombined Na and Cl2 have oxidation numbers of 0.
2. Monatomic ions have an oxidation number equal to the ionic charge. In the equation: 2Na + Cl2  2NaCl, the
sodium in the product NaCl has a charge of 1+ and the oxidation number is +1. The chlorine in NaCl has a
charge of 1- and the oxidation number is -1.
3. The metals of Group 1 always have an oxidation number of +1 in compounds, and the metals of group 2 always
have an oxidation number of +2.
4. Fluorine is always -1 in compounds. The other halogens are also -1 when they are the MOST electronegative
element in the compound.
5. Hydrogen is +1 in compounds unless it is combined with a metal, in which case it is -1. Hydrogen is +1 in HCl
but -1 in LiH.
6. Oxygen is usually -2 in compounds. When it is combined with fluorine, it is +2. Oxygen is -2 in H2O, and +2 in
OF2. IN the peroxide ion, O22- , oxygen is -1.
7.
The oxidation numbers in a polyatomic ion will add to the total charge of the polyatomic ion.
8. The sum of the oxidation numbers in all compounds must be equal to zero.
Unit 12, Lesson 1 of 3
Predict the oxidation number in each of the following compounds by using the rules.
5. MnCl2
1. CO2
Element
Element
C
O
Subscript
Subscript
Oxidation #
Oxidation #
Total:
Total:
2. H2SO4
Element
Subscript
Mn
Cl
6. NaHCO3
H
S
O
Element
Subscript
Na
H
C
O
Oxidation #
Oxidation #
Total:
Total:
3. K2PtCl6
Element
Subscript
K
Pt
Cl
7. AgNO3
Oxidation #
Element
Subscript
Total:
Oxidation #
Ag
N
O
K
Cr
O
Total:
4. P4O10
Element
Subscript
P
O
8. K2CrO4
Oxidation #
Element
Subscript
Total:
Oxidation #
Total:
Spotting Redox Reactions
1. If there is a change in oxidation number for a particular type of atom, the reaction is redox.
2. If an uncombined element appears on one side of an equation and is in a compound on the other side, the reaction
is redox. (single replacement)
3. If the reaction is a double replacement reaction, it is NOT a redox reaction.
Using the above tips, circle the redox reactions.
HCl + KOH  KCl + H2O
HCl + MnO2  MnCl2 + 2H2O + Cl2
HCl + CaCO3  CaCl2 + H2O + CO2
HCl + FeS  FeCl2 + H2S
Zn +HCl  ZnCl2 + H2
C3H8 + O2  CO2 + H2O
Unit 12, Lesson 1 of 3
Oxidation & Reduction: Elements Involved
Once you have spotted a redox reaction, you must be able to identify which element is being oxidized and which element
is being reduced. Remember LEO the lion says GER.
Element Oxidized: The element that is losing electrons. The oxidation number will increase, lower on the reactant side
and higher on the product side.
Reducing Agent:
Element Reduced: The element that is gaining electrons. The oxidation number will decrease, higher on the reactant
side and lower on the product side.
Oxidizing Agent:
Label the following reactions:
A- Element Oxidized, B- Element Reduced, C-Reducing Agent, D- Oxidizing Agent
4HCl + MnO2  MnCl2 + 2H2O + Cl2
Zn +HCl  ZnCl2 + H2
Half Reactions
Chemical equations show the formulas of reactants and products, but they do not show the exchange of electrons. A half
reaction shows either the oxidation or reduction portion of a redox reaction, including the electrons gained or lost.
An oxidation half reaction: shows an atom or ion losing electrons, while its oxidation number increases.
Fe(s)  Fe3+ + 3eA reduction half reaction: shows an atom or ion gaining electrons, while its oxidation number decreases.
Fe3+ + 3e-  Fe(s)
Half reactions show that in chemical reactions: mass and charge are conserved. As a result, half reactions must be
balanced so that the net charge will be equal on both sides of the equation, but not necessarily equal to zero.
For example, 2Mg(s) + O2(g)  2MgO(s)
2Mg  2Mg2+ + 4eO2 + 4e-  2O2-
Net charge/side = 0
Net charge/side = -4
To write a half-reaction from an equation
1. Assign oxidation numbers to each element to determine which element is being oxidized and which is being
reduced.
2. Write partial half reactions to show the change in oxidation state for both the oxidation and the reduction.
3. Show the number of electrons lost or gained to explain the change in oxidation number.
4. Check to see if the net charge is the same on both sides. If it’s not, balance can be achieved by multiplying one
side by the needed number.
Write correct half reactions for the following redox reactions.
Cu(s) + AgNO3(aq)  Cu(NO3)2(aq) + Ag(s)
Ca(s) + 2H2O(l)  Ca(OH)2(s) + H2(g)
Zn(s) + CuSO4(aq)  ZnSO4(aq)
KClO3(s)  KCl(s) + 3O2(g)
+ Cu(s)
Unit 12, Lesson 1 of 3
Types of Redox Reactions
Based upon the tips for spotting redox reactions, you will find that many of the reaction types we have learned thus far are
redox reactions.
Synthesis Reactions: S(s) + O2(g)  SO2(g)
0
0
+4 -2
Decomposition Reactions: 2NaH(s)  2Na(s) + H2(g)
+1 -1
0
0
Combustion Reactions: C3H8(g) + 5O2(g)  3CO2(g) + 4H2O(l)
0
-2
Single Replacement Reactions: Mg(s) + HCl(aq)  MgCl2(aq) + H2(g)
0
+1 -1
+2 -1
0
There are different types of single replacement reactions. Reactions can be predicted using the metal activity series. The
metals are arranged according to their ability to displace hydrogen from an acid or water. Lithium is the most reactive
metal, gold is the least reactive.
1. Hydrogen Displacement
All alkali metals and some alkaline earth metals will displace hydrogen from cold water.
2Na(s) + H2O(l)  2NaOH(aq) + H2(g)
Ba(s) + H2O(l) 
Many metals, including those that do not react with water, are capable of displacing hydrogen from acids.
Zn(s) + 2HCl(aq)  ZnCl2(aq) + H2(g)
Fe(s) + HCl(aq) 
2. Metal Displacement
A metal in a compound can be displaced by another metal in the elemental state. We have already seen example of
copper replacing silver ions in our famous demonstration. If we reversed the roles of the metals—added pure silver to
copper nitrate solution, no reaction would happen. How do we know this?!
The activity series is an easy way to predict whether a metal or hydrogen displacement will actually occur. An activity
series is a convenient summary of the results of many possible displacement reactions. According to the series, any metal
above hydrogen will displace it from water or from an acid, but metals below hydrogen will not react with either water or
an acid. In addition, the activity series predicts what metals are capable of replacing other metals. Any metal listed in the
series will react with any metal (in a compound) below it.
3. Halogen Displacement
Another activity series summarizes the halogens’ behavior in halogen displacement reactions:
F2 > Cl2 > Br2 > I2
The power of these oxidizing agents decreases as we move down the halogen group—so molecular fluorine can replace
chloride, bromide, and iodide ions in solution. In fact, molecular fluorine is so reactive that it also attacks water; thus
these reactions cannot be carried out in aqueous solutions. On the other hand, molecular chlorine can displace bromide
and iodide ions in aqueous solution.
Cl2(g) + 2KBr(aq)  2KCl(aq) + Br2(l)
Cl2(g) + 2NaI(aq)  2NaCl(aq) + I2(s)
Disproportionation Reactions
A special type of redox reaction is a disproportionation reaction, a reaction in which one element in one oxidation state is
simultaneously oxidized and reduced. One reactant in a disproportionation reaction must contain an element that can have
at least three oxidation states. The element itself is an intermediate oxidation state; that is both higher and lower oxidation
states exist for that element in the products.
Consider the decomposition of hydrogen peroxide:
2H2O2(aq)  2H2O(l) + O2(g)
Consider: Cl2(g) + 2OH-(aq)  ClO-(aq) + Cl-(aq) + H2O(l)
Unit 12, Lesson 1 of 3
Redox Titrations
These titrations involve the titration of an oxidizing agent with a reducing agent or vice versa. There must be a
sufficiently large difference between the oxidizing and reducing capabilities of these agents for the reaction to undergo
completion with a sharp end point. The endpoint or equivalence point is reached when the reducing agent is completely
oxidized by the oxidizing agent. Like acid-base titrations, redox titrations normally require an indicator that clearly
changes color. In the presence of large amounts of reducing agent, the color of the indicator is characteristic of its
reduced form. The indicator assumes the color of its oxidized form when it is present in an oxidizing medium. At or near
the equivalence point, a sharp change in the indicator’s color will occur as it changes from one form to the other.
Two common oxidizing agents are KMnO4 and K2Cr2O7. The colors of the anions are distinctly different from those of
the reduced species. Thus, these oxidizing agents can themselves be used as an internal indicator in a redox titration.
The process for solving redox titration calculations is similar to acid-base titration problems in that the amount of the
oxidizing agent is dependent upon the amount of reducing agent. However, you should recall that acid-base reactions
required a 1:1 ratio of H+ ion to OH- ions because neutralization must occur. Redox titrations are not so simple, and as a
result, we cannot assume that the reaction requires equal moles of each agent, because as you can see above, each redox
reaction is different.
Practice Problems
1. A 16.42 mL volume of 0.1327 M KMnO4 solution is needed to oxidize 25.00 mL of a FeSO4 solution in an acidic
medium. What is the concentration of FeSO4 solution in terms of molarity?
The net ionic equation is: 5Fe2+ + MnO4- + 8H+  Mn2+ + 5Fe3+ + 4H2O
2. How many milliliters of a 0.206 M HI solution are needed to reduce 22.5 mL of a 0.374 M KMnO4 solution according
to the following equation: 10HI + 2KMnO4 + 3H2SO4  5I2 + 2MnSO4 + K2SO4 + 8H2O
3. Iron(II) can be oxidized by an acidic K2Cr2O7 solution according to the net ionic equation: Cr2O72- + 6Fe2+ + 14H+
 2Cr3+ + 6Fe3+ + 7H2O
If it takes 26.0 mL of 0.0250 M K2Cr2O7 to titrate 25.0 mL of a solution containing Fe2+, what is the molar concentration
of Fe2+?
Unit 12, Lesson 1 of 3
4. The SO2 present in air is mainly responsible for the acid rain phenomenon. Its concentration can be determined by
titrating against a standard permanganate solution as follows: 5SO2 + 2MnO4- + 2H2O  5SO42- + 2Mn2+ + 4H+
Calculate the number of grams of SO2 in a sample of air if 7.37 mL of 0.00800 M KMnO4 solution are required for the
titration.
5. A sample of iron ore (containing Fe2+ ions, only) weighing 0.2792g was dissolved in dilute acid solution, and all the
Fe(II) was converted to Fe(III) ions. The solution required 23.30 mL of 0.0194 MK2Cr2O7 for titration. Calculate the
percent by mass of iron in the ore.
The oxidation of iron is as follows: Cr2O72- + 6Fe2+ + 14H+  2Cr3+ + 6Fe3+ + 7H2O
6. For each reaction, determine which element is being oxidized and which element is being reduced. Write and balance
the half reactions for oxidation and reduction.
A) 2Na + FeCl2  2NaCl + Fe
B) 2C2H2 + 5O2  4CO2 + 2H2O
C) 2PbS + 3O2  2SO2 + 2PbO
D) 2H2 + O2  2H2O