Introductory Chemistry:
Introductory Chemistry:
Concepts & Connections Concepts & Connections
4th Edition by Charles H. Corwin
Oxidation Oxidation
and and
Reduction Reduction
Chapter 17
Chapter 17 2
Oxidation-Reduction Reactions
• Oxidation-reduction reactions are reactions involving the transfer of electrons from one substance to another.
• We have seen several “oxidation-reduction”
reactions so far.
• Whenever a metal and a nonmetal react, electrons are transferred.
– 2 Na(s) + Cl2(g) → 2 NaCl(s)
• Combustion reactions also are examples of
“oxidation-reduction” reactions.
Example of Oxidation/Reduction
• The rusting of iron is also an example of an oxidation-reduction reaction.
• Iron metal reacts with oxygen in air to produce the ionic compound iron(III) oxide which is
composed of Fe3+ and O2- ions.
– 4 Fe(s) + 3 O2(g) → 2 Fe2O3(s)
• Iron loses electrons and is oxidized
– Fe → Fe3+ + 3 e-
• Oxygen gains electrons and is reduced
Chapter 17 4
Oxidation Numbers
• The oxidation number describes how many electrons have been lost or gained by an atom.
• Oxidation numbers are assigned according to seven rules:
1. A metal or a nonmetal in the free state has an oxidation number of 0.
2. A monoatomic ion has an oxidation number equal to its ionic charge.
3. A hydrogen atom is usually assigned an oxidation number of +1.
Rules for Oxidation Numbers
4. An oxygen atom is usually assigned an oxidation number of -2.
5. For a molecular compound, the more electronegative element is assigned a negative oxidation number
equal to its charge as an anion.
6. For an ionic compound, the sum of the oxidation numbers for each of the atoms in the compound is equal to 0.
7. For a polyatomic ion, the sum of the oxidation
numbers for each of the atoms in the compound is equal to the ionic charge on the polyatomic ion.
Chapter 17 6
Assigning Oxidation Numbers
• What is the oxidation number for magnesium metal, Mg?
– Mg = 0 according to rule #1
• What is the oxidation number for sulfur in the sulfide ion, S2-?
– S = –2 (rule #2)
• What is the oxidation number for barium and chloride in BaCl2?
– Ba is present at Ba2+, so Ba = +2 (rules #2 and #6) – Cl is present as Cl-, so Cl = –1 (rules #2 and #6)
Oxidation Numbers in Compounds
• What are the oxidation numbers for each element in oxalic acid, H2C2O4?
H = +1 (rule #3)
O = -2 (rule #4)
• To find the oxidation number for carbon, recall
that the sum of the oxidation numbers equals zero.
2(+1) + 2(ox no C) + 4(–2) = 0
2 + 2(ox no C) + (– 8) = 0
2(ox no C) = +6
C = +3
Chapter 17 8
Oxidation Numbers in Compounds
• What are the oxidation numbers for each element in carbon tetrachloride, CCl4?
Cl = –1 (rule #5)
• To find the oxidation number for carbon, recall
that the sum of the oxidation numbers equals zero.
(ox no C) + 4(–1) = 0
(ox no C) – 4 = 0 (ox no C) = +4
C = +4
Oxidation Numbers in Polyatomic Ions
• What are the oxidation numbers for chlorine and oxygen in the perchlorate ion, ClO4-?
O = –2 (rule #4)
• To find the oxidation number for carbon, recall that the sum of the oxidation numbers equals the charge on the ion (rule #7).
(ox no Cl) + 4(–2) = –1
(ox no Cl) – 8 = –1 (ox no Cl) = +7
Cl = +7
Chapter 17 10
Redox Reactions
• Recall, a chemical reaction that involves the transfer of electrons is an oxidation-reduction reaction, or a redox reaction.
• For example, iron metal is heated with sulfur to produce, iron(II) sulfide: Fe(s) + S(s) → FeS(s).
• The sulfur changes from 0 to –2 and the iron changes from 0 to +2.
∆
Oxidation and Reduction
• The iron loses electrons and is oxidized.
– Fe → Fe2+ + 2 e-
• The sulfur gains electrons and is reduced.
– S + 2 e- → S2-
Chapter 17 12
Oxidizing & Reducing Agents
• Oxidation is the loss of electrons and reduction is the gain of electrons.
• An oxidizing agent is a substance that causes
oxidation by accepting electrons. The oxidizing agent is reduced.
• A reducing agent is a substance that causes reduction by donating electrons.
The reducing agent is oxidized.
Redox Reactions
• In a redox reaction,
one substance must be oxidized and one
substance must be reduced.
• The total number of electrons lost is equal to the total electrons gained.
Chapter 17 14
Redox Reactions
• Identify the reducing agent, the oxidizing agent, and the oxidation and reduction in the following reaction:
– CuS(s) + H2(g) → Cu(s) + H2S(g)
• Cu is reduced from +2 to 0.
• H is oxidized from 0 to +1.
∆
Ionic Equations
• Redox reactions in aqueous solution are most often shown in the ionic form.
• Ionic equations readily show us the change in oxidation number.
5 Fe2+(aq)+MnO4-(aq)+8 H+(aq)→5 Fe3+(aq)+Mn2+(aq)+4 H2O(l)
• We can easily tell that the oxidation number of iron changes from +2 to +3; iron is oxidized.
• Manganese is reduced from +7 in MnO4- to +2 in Mn2+; manganese is reduced.
Chapter 17 16
Ionic Equations Continued
• We can map the reaction to show the oxidation and reduction processes and to determine the oxidizing and reducing agents:
Balancing Redox Reactions
• When we balance redox reactions, the number of electrons lost must equal the number of electrons gained.
• We will balance redox reactions using the
oxidation number method which has 3 steps:
1. Inspect the reaction and the substances
undergoing a change in oxidation number.
a) Write the oxidation number above each element.
b) Diagram the number of electrons lost by the oxidized substance and gained by the reduced substance.
Chapter 17 18
Oxidation Number Method
2. Balance each element in the equation using a coefficient. Remember, that the electrons lost must equal the electrons gained. If they are not the same, balance the electrons as follows:
a) In front of the oxidized substance, place a coefficient equal to the number of electrons gained by the
reduced substance.
b) In front of the reduced substance, place a coefficient equal to the number of electrons lost by the oxidized substance.
Oxidation Number Method
3. After balancing the equation, verify that the coefficients are correct.
a) Place a check mark above the symbol for each element to verify that the number of atoms is the same on both sides.
b) For ionic equations, verify that the total charge on the left side of the equation is the same as the total
charge on the right side of the equation.
Chapter 17 20
Balancing a Redox Reaction
• Balance the following redox reaction using the oxidation number method:
Fe2O3(l) + CO(g) → Fe(l) + CO2(g)
• Since the total electrons gained and lost must be equal, we must find the lowest common multiple.
For this reaction, it is 6.
• Each iron gains 3 electrons, so place a 2 in front of the Fe. There are 2 iron atoms in Fe2O3, so no
coefficient is necessary.
• Each carbon loses 2 electrons, so place a 3 in front of CO and CO2.
Fe2O3(l) + 3 CO(g) → 2 Fe(l) + 3 CO2(g)
• Check to see that the number of each type of atom is the same on both sides:
– There are 2 Fe atoms, 6 O atoms, and 3 C atoms on each side.
Balancing a Redox Reaction
√
√ √ √ √ √ √
Chapter 17 22
Balancing Redox Equations
• An alternative method for balancing redox reactions is the half-reaction method.
• A half-reaction shows the oxidation or reduction process of a redox reaction separately.
• The steps are:
1. Write the half-reaction for both the oxidation and reduction processes.
Half-Reaction Method Continued
2. Balance the atoms in each half-reaction using coefficients.
a) Balance all elements except oxygen and hydrogen.
b) Balance oxygen using H2O.
c) Balance hydrogen using H+.
d) For reactions in basic solution, add one OH- to each side for each H+ and combine H+ & OH- to H2O.
e) Balance the ionic charges using electrons.
Chapter 17 24
Half-Reaction Method Continued
3. Multiply each half-reaction by a whole number so that the total number of electrons in each is the same.
4. Add the two half-reactions together and cancel the identical species, including electrons.
5. After balancing, verify that the coefficients are correct by making sure there are the same
number of each atom on each side of the reaction and that the overall charge is the same on both
sides.
Balancing a Redox Equation
• Balance the following redox reaction using the half-reaction method:
Fe2+(aq) + MnO4-(aq) → Fe3+(aq) + Mn2+(aq)
• The two unbalanced half-reactions are:
Fe2+ → Fe3+
MnO4- → Mn2+
• We balance the two half-reactions as follows:
Fe2+ → Fe3+ + e-
5 e + 8 H + MnO → Mn + 4 H O
Chapter 17 26
Balancing a Redox Equation
• Since Fe2+ loses 1 electron and MnO4- gains 5 electrons, we have to multiply the iron half- reaction by 5:
5 Fe2+ → 5 Fe3+ + 5 e- 5 e- + 8 H+ + MnO4- → Mn2+ + 4 H2O
• We add the two half-reactions together and cancel out the 5 electrons on each side to get the
balanced equation:
5 Fe2+ + 8 H+ + MnO4- → 5 Fe2+ + Mn2+ + 4 H2O
Spontaneous Redox Reactions
• Chemical reactions that occur without any input of energy are spontaneous.
• The reaction of zinc metal with aqueous copper sulfate is spontaneous:
Zn(s) + Cu2+(aq) → Zn2+(aq) + Cu(s)
• Cu2+ has a greater tendency to gain electrons than Zn2+.
• We can compare metals and arrange them in a series based on their ability to gain electrons.
Chapter 17 28
Reduction Potentials
• The tendency for a
substance to gain electrons is its reduction potential.
• The strongest reducing agent is the most easily oxidized.
• This is a table of reduction potentials for several
metals.
Spontaneous Reactions
• A species can be reduced by any reducing agent
lower in the table.
• Any metal below H2 can react with acid and be oxidized.
• A species can be oxidized by any
oxidizing agent above it on the table.
Chapter 17 30
Predicting Spontaneous Reactions
• A reaction will be spontaneous when the stronger oxidizing and reducing agents are the reactants and the weaker oxidizing and reducing agents are the products.
• Predict whether the following reaction will be spontaneous:
Ni2+(aq) + Sn(s) → Ni(s) + Sn2+(aq)
stronger oxidizing agent
weaker oxidizing agent stronger
reducing agent
weaker reducing agent
• The reaction is spontaneous as written.
Voltaic Cells
• The conversion of chemical energy to electrical energy in a redox reaction is electrochemistry.
• If we can physically separate the oxidation and reduction half-reactions, we can use the electrons from the redox reaction to do work. This is an
electrochemical cell.
• Lets look at the reaction of zinc metal with copper(II) sulfate:
Zn(s) + CuSO4(aq) → Cu(s) + ZnSO4(aq)
Chapter 17 32
Voltaic Cells
• We place a zinc electrode in aqueous ZnSO4 and a copper electrode in aqueous CuSO4. The
electrodes are connected by a wire to allow the flow of electrons.
• A salt bridge is used to complete the circuit.
• Zinc metal is
oxidized and copper ions are reduced in each half cell.
Voltaic Cells
• Oxidation occurs at the anode of an electrochemical cell.
• Reduction occurs at the cathode of an electrochemical cell.
• Electrons flow through the wire from the anode to the cathode in a voltaic cell.
• Negatively charged ions travel through the salt bridge away from the cathode and towards the anode in a voltaic cell.
Chapter 17 34
Electrolytic Cells
• Electrolytic cells are electrochemical cells that do not operate spontaneously. The process is
referred to as electrolysis.
• A source of electricity is required to drive an electrolytic cell.
• An example of an electrolysis reactions is the recharging of the battery in a cell phone.
Conclusions
• A redox reaction is a reaction involving the transfer of electrons from one substance to another.
• The oxidation number describes how many electrons have been lost or gained by an atom.
• Oxidation is the loss of electrons.
• Reduction is the gain of electrons.
Chapter 17 36
Conclusions Continued
• An oxidizing agent is a substance that causes
oxidation by accepting electrons. The oxidizing agent is reduced.
• A reducing agent is a substance that causes
reduction by donating electrons. The reducing agent is oxidized.
• In redox reactions, the number of electrons lost must equal the number of electrons gained.
Conclusions Continued
• There are two methods to balance redox reactions:
– The Oxidation Number Method – The Half-Reaction Method
• The tendency for a substance to gain electrons is its reduction potential.
• The conversion of chemical energy to electrical energy in a redox reaction is electrochemistry.
• We can physically separate oxidation and
reduction half-reactions and use the electrons from the redox reaction to do work in an
