1. Introductory Chemistry, 3rd Edition Nivaldo Tro
Chapter 16
Oxidation and
Reduction
Roy Kennedy
Massachusetts Bay Community College
Wellesley Hills, MA
2009, Prentice Hall

2. Oxidation–Reduction Reactions
Oxidation–reduction reactions are also called redox reactions.
All redox reactions involve the transfer of electrons from one atom to another.
Spontaneous redox reactions are generally exothermic, and we can use their released energy as a source of energy for other applications.
Convert the heat of combustion into mechanical energy to move our cars.
Use electrical energy in a car battery to start our car engine.
Tro's Introductory Chemistry, Chapter 16

3. Combustion Reactions Combustion reactions are always exothermic.
In combustion reactions, O2 combines with all the elements in another reactant to make the products.
4 Fe(s) + 3 O2(g) → 2 Fe2O3(s) + energy
CH4(g) + 2 O2(g) → CO2(g) + 2 H2O(g) + energy
Tro's Introductory Chemistry, Chapter 16

4. Reverse of Combustion Reactions
Since combustion reactions are exothermic, their reverse reactions are endothermic.
The reverse of a combustion reaction involves the production of O2.
energy + 2 Fe2O3(s) → 4 Fe(s) + 3 O2(g)
energy + CO2(g) + 2 H2O(g) → CH4(g) + 2 O2(g)
Reactions in which O2 is gained or lost are redox reactions.
Tro's Introductory Chemistry, Chapter 16

5. Oxidation and Reduction: One Definition
When an element attaches to an oxygen during the course of a reaction it is generally being oxidized.
In CH4(g) + 2 O2(g) → CO2(g) + 2 H2O(g), C is being oxidized in this reaction, but H is not.
When an element loses an attachment to oxygen during the course of a reaction, it is generally being reduced.
In 2 Fe2O3(s) → 4 Fe(s) + 3 O2(g), the Fe is being reduced.
One definition of redox is the gain or loss of O, but it is not the best.
Tro's Introductory Chemistry, Chapter 16

6. Another Oxidation–Reduction
Consider the following reactions:
4 Na(s) + O2(g) → 2 Na2O(s)
2 Na(s) + Cl2(g) → 2 NaCl(s)
The reaction involves a metal reacting with a nonmetal.
In addition, both reactions involve the conversion of free elements into ions.
4 Na(s) + O2(g) → 2 Na+2O–(s)
2 Na(s) + Cl2(g) → 2 Na+Cl–(s)
Tro's Introductory Chemistry, Chapter 16

7. Oxidation and Reduction: Another Definition
In order to convert a free element into an ion, the atoms must gain or lose electrons.
Of course, if one atom loses electrons, another must accept them.
Reactions where electrons are transferred from one atom to another are redox reactions.
Atoms that lose electrons are being oxidized, atoms that gain electrons are being reduced.
Leo
Ger
2 Na(s) + Cl2(g) → 2 Na+Cl–(s)
Na → Na+ + 1 e– (oxidation)
Cl2 + 2 e– → 2 Cl– (reduction)
Tro's Introductory Chemistry, Chapter 16

8. Tro's Introductory Chemistry, Chapter 16
Practice—Identify the Element Being Oxidized and the Element Being Reduced.
2 C(s) + O2(g) → 2 CO(g)
Mg(s) + Cl2(g) → MgCl2(s)
Mg(s) + Fe2+(aq) → Mg2+(aq) + Fe(s)
Tro's Introductory Chemistry, Chapter 16

9. Practice—Identify the Element Being Oxidized and the Element Being Reduced, Continued.
2 C(s) + O2(g) → 2 CO(g)
Mg(s) + Cl2(g) → MgCl2(s)
Mg(s) + Fe2+(aq) → Mg2+(aq) + Fe(s)
C is oxidized because it is gaining an attachment to O.
O is reduced; there has to be reduction and it’s the only other element. 2+ −
Mg is oxidized because it is becoming a cation by losing electrons.
Cl is reduced because it is becoming an anion by gaining electrons.
Mg is oxidized because it is becoming a cation by losing electrons.
Fe2+ is reduced because it is gaining electrons to become neutral.
Tro's Introductory Chemistry, Chapter 16

10. Oxidation–Reduction Oxidation and reduction must occur simultaneously.
If an atom loses electrons, another atom must take them.
The reactant that reduces an element in another reactant is called the reducing agent.
The reducing agent contains the element that is oxidized.
The reactant that oxidizes an element in another reactant is called the oxidizing agent.
The oxidizing agent contains the element that is reduced.
2 Na(s) + Cl2(g) → 2 Na+Cl–(s)
Na is oxidized, Cl is reduced.
Na is the reducing agent, Cl2 is the oxidizing agent.
Tro's Introductory Chemistry, Chapter 16

11. Practice—Identify the Oxidizing and Reducing Agents.
2 C(s) + O2(g) → 2 CO(g)
Mg(s) + Cl2(g) → MgCl2(s)
Mg(s) + Fe2+(aq) → Mg2+(aq) + Fe(s)
C is oxidized because it is gaining attachment to O.
O is reduced; there has to be reduction and it’s the only other element. 2+ −
Mg is oxidized because it is becoming a cation by losing electrons.
Cl is reduced because it is becoming an anion by gaining electrons.
Mg is oxidized because it is becoming a cation by losing electrons.
Fe2+ is reduced because it is gaining electrons to become neutral.
Tro's Introductory Chemistry, Chapter 16

12. Practice—Identify the Oxidizing and Reducing Agents, Continued.
2 C(s) + O2(g) → 2 CO(g)
Mg(s) + Cl2(g) → MgCl2(s)
Mg(s) + Fe2+(aq) → Mg2+(aq) + Fe(s)
C is the reducing agent because it contains the element that is oxidized.
O is the oxidizing agent because it contains the element that is reduced. 2+ −
Mg is the reducing agent because it contains the element that is oxidized.
Cl2 is the oxidizing agent because it contains the element that is reduced.
Mg is the reducing agent because it contains the element that is oxidized.
Fe2+ is the oxidizing agent because it contains the element that is reduced.
Tro's Introductory Chemistry, Chapter 16

13. Tro's Introductory Chemistry, Chapter 16
Electron Bookkeeping
For reactions that are not metal + nonmetal, or do not involve O2, we need a method for determining how the electrons are transferred.
Chemists assign a number to each element in a reaction called an oxidation state that allows them to determine the electron flow in the reaction.
Although they look like them, oxidation states are not ion charges!
Oxidation states are imaginary charges assigned based on a set of rules.
Ion charges are real, measurable charges.
Tro's Introductory Chemistry, Chapter 16

14. Rules for Assigning Oxidation States
Rules are in order of priority.
Free elements have an oxidation state = 0.
Na(s) = 0 and Cl2(g) = 0 in 2 Na(s) + Cl2(g)
2 NaCl(s).
Monoatomic ions have an oxidation state equal to their charge.
Na = +1 and Cl = -1 in NaCl(s).
a. The sum of the oxidation states of all the atoms or ions in a compound is 0.
Na = +1 and Cl = -1 in NaCl, and (+1) + (-1) = 0.
Tro's Introductory Chemistry, Chapter 16

15. Rules for Assigning Oxidation States, Continued
b. The sum of the oxidation states of all the atoms in a polyatomic ion equals the charge on the ion.
N = +5 and O = -2 in NO3–, (+5) + 3(-2) = -1.
a. Group I metals have an oxidation state of +1 in all their compounds.
Na = +1 in NaCl.
b. Group II metals have an oxidation state of +2 in all their compounds.
Mg = +2 in MgCl2.
Tro's Introductory Chemistry, Chapter 16

16. Rules for Assigning Oxidation States, Continued
In their compounds, nonmetals have oxidation states according to the table below.
Nonmetals higher on the table take priority.
Nonmetal
Oxidation state
Example F -1
CF4 H +1
CH4 O -2
CO2
Group 7A
CCl4
Group 6A
CS2
Group 5A -3
NH3
Tro's Introductory Chemistry, Chapter 16

17. Practice—Assign an Oxidation State to Each Element in the Following:
Mg2+
KCl
SO2
PO43−
BaO2
Tro's Introductory Chemistry, Chapter 16

18. Tro's Introductory Chemistry, Chapter 16
Practice—Assign an Oxidation State to Each Element in the Following, Continued:
F2 F = 0 (Rule 1)
Mg2+ Mg = +2 (Rule 2)
KCl K = +1 (Rule 4a) and Cl = -1 (Rule 5)
SO2 O = -2 (Rule 5) and S = +4 (Rule 3a)
PO43− O = -2 (Rule 5) and P = +5 (Rule 3b)
BaO2 Ba = +2 (Rule 4b) and O = -1 (Rule 3a)
Tro's Introductory Chemistry, Chapter 16

19. Oxidation and Reduction: A Better Definition
Oxidation occurs when an atom’s oxidation state increases during a reaction.
Reduction occurs when an atom’s oxidation state decreases during a reaction.
CH O2 → CO2 + 2 H2O –
oxidation
reduction
Tro's Introductory Chemistry, Chapter 16

20. Tro's Introductory Chemistry, Chapter 16
Practice—Assign Oxidation States and Identify the Oxidizing and Reducing Agents in Each of the Following:
3 H2S + 2 NO3– H+ ® 3 S + 2 NO + 4 H2O
MnO2 + 4 HBr ® MnBr2 + Br2 + 2 H2O
Tro's Introductory Chemistry, Chapter 16

21. Tro's Introductory Chemistry, Chapter 16
Practice—Assign Oxidation States and Identify the Oxidizing and Reducing Agents in Each of the Following, Continued:
reducing
agent
oxidizing agent
3 H2S + 2 NO3– H+ ® 3 S + 2 NO + 4 H2O
MnO2 + 4 HBr ® MnBr2 + Br2 + 2 H2O
reduction
oxidation
Oxidizing agent
reducing agent
reduction
oxidation
Tro's Introductory Chemistry, Chapter 16

22. Balancing Redox Reactions
Some redox reactions can be balanced by the method we previously used, but many are hard to balance using that method.
Many are written as net ionic equations.
Many have elements in multiple compounds.
The main principle is that electrons are transferred, so if we can find a method to keep track of the electrons, it will allow us to balance the equation.
Tro's Introductory Chemistry, Chapter 16

23. Balancing Redox Reactions by the Half-Reaction Method
In this method, the reaction is broken down into two half-reactions, one for oxidation and another for reduction.
Each half-reaction includes electrons.
Electrons go on the product side of the oxidation half-reaction—loss of electrons.
Electrons go on the reactant side of the reduction half-reaction—gain of electrons.
Each half-reaction is balanced for its atoms.
Then the two half-reactions are adjusted so that the electrons lost and gained will be equal when added.
Tro's Introductory Chemistry, Chapter 16

24. Balancing Redox Reactions in Acidic Solution
Assign oxidation states and determine element oxidized and element reduced.
Separate into oxidation and reduction half- reactions.
Fe2+ + MnO4– → Fe3+ + Mn2+ +2 +7 -2 +3
oxid
red
Fe2+ → Fe3+
MnO4– → Mn2+
Tro's Introductory Chemistry, Chapter 16

25. Balancing Redox Reactions in Acidic Solution, Continued
3. Balance half-reactions by mass.
First balance atoms other than O and H.
Balance O by adding H2O to side that lacks O.
Balance H by adding H+ to side that lacks H.
Finished if in acidic solution.
If in basic solution, add enough OH− to neutralize the H+, rewrite H+ + OH− as H2O.
Add to both sides.
Then cancel H2O on both sides.
Fe2+ → Fe3+
MnO4– → Mn2+
MnO4– → Mn2+ + 4H2O
MnO4– + 8H+ → Mn2+ + 4H2O
Tro's Introductory Chemistry, Chapter 16

26. Balancing Redox Reactions in Acidic Solution, Continued
4. Balance each half-reaction, with respect to charge, by adjusting the numbers of electrons.
Electrons on product side for oxidation.
Electrons on reactant side for reduction.
5. Balance electrons between half-reactions.
6. Add half-reactions, canceling electrons and common species.
7. Check.
Fe2+ → Fe e-
MnO4– + 8H+ → Mn2+ + 4H2O +7 +2
MnO4– + 8H+ + 5 e- → Mn2+ + 4H2O
Fe2+ → Fe e-
MnO4– + 8H+ + 5 e- → Mn2+ + 4H2O
} x 5
5 Fe2+ → 5 Fe e-
MnO4– + 8H+ + 5 e- → Mn2+ + 4H2O
5 Fe2+ + MnO4– + 8H+ → Mn2+ + 4H2O + 5 Fe3+
Tro's Introductory Chemistry, Chapter 16

27. Practice—Balance the Following Equation: Cu+ + I2 → Cu2+ + I–
Tro's Introductory Chemistry, Chapter 16

28. Tro's Introductory Chemistry, Chapter 16
Practice—Balance the Following Equation, Continued: Cu+ + I2 → Cu2+ + I– +1 +2 -1
oxid
red
oxid: Cu+ → Cu2+
red: I2 → I–
oxid: Cu+ → Cu2+
red: I2 → 2 I–
oxid: Cu+ → Cu e-
red: I2 + 2 e- → 2 I–
oxid: Cu+ → Cu e- } x 2
red: I2 + 2 e- → 2 I–
2 Cu+ + I2 → 2 Cu2+ + I2
Tro's Introductory Chemistry, Chapter 16

29. Tro's Introductory Chemistry, Chapter 16
Practice—Balance the Following Equation in Acidic Solution: I– + Cr2O72- → Cr3+ + I2
Tro's Introductory Chemistry, Chapter 16

30. Practice—Balancing Redox Reactions
Assign oxidation states and determine element oxidized and element reduced.
Separate into oxidation and reduction half- reactions.
I− + Cr2O72– → I Cr3+ −1 +6 −2 +3
oxid
red
oxid: I− → I2
red: Cr2O72– → Cr3+
Tro's Introductory Chemistry, Chapter 16 30 30

31. Practice—Balancing Redox Reactions, Continued
3. Balance half-reactions by mass.
First balance atoms other than O and H.
Balance O by adding H2O to side that lacks O.
Balance H by adding H+ to side that lacks H.
Finished if in acidic solution.
If in basic solution, add enough OH− to neutralize the H+, rewrite H+ + OH− as H2O.
Add to both sides.
Then cancel H2O on both sides.
oxid: I− → I2
oxid: 2 I− → I2
red: Cr2O72– → Cr3+
red: Cr2O72– → 2 Cr3+
red: Cr2O72– → 2Cr3+ +7H2O
Cr2O72– +14H+ → 2Cr3+ +7H2O
Tro's Introductory Chemistry, Chapter 16 31 31

32. Practice—Balancing Redox Reactions, Continued
4. Balance each half-reaction, with respect to charge, by adjusting the numbers of electrons.
Electrons on product side for oxidation.
Electrons on reactant side for reduction.
5. Balance electrons between half-reactions.
6. Add half-reactions, canceling electrons and common species.
7. Check.
2 I− → I2 + 2e−
2 I− → I2
Cr2O72– + 14H+ → 2Cr3+ + 7H2O
Cr2O72– +14H+ + 6e−→ 2Cr3+ +7H2O
2 I− → I2 + 2e−
6 I− → 3 I2 + 6e−
} x 3
Cr2O72– +14H+ + 6e−→ 2Cr3+ +7H2O
Cr2O72– +14H+ + 6 I−→ 2Cr3+ +7H2O + 3 I2
Tro's Introductory Chemistry, Chapter 16 32

33. Balancing Redox Reactions in Basic Solution
Assign oxidation states and determine element oxidized and element reduced.
Separate into oxidation and reduction half- reactions.
CN− + MnO4– → CNO− + MnO2 +2 +7 -2
+4-3-2 +4
oxid
red -3
C+2 → C+4
MnO4– → MnO2
Tro's Introductory Chemistry, Chapter 16

34. Balancing Redox Reactions in Basic Solution, Continued
CN− → CNO−
CN− + H2O → CNO−
CN− + H2O → CNO− + 2 H+
3. Balance half-reactions by mass.
First balance atoms other than O and H.
Balance O by adding H2O to side that lacks O.
Balance H by adding H+ to side that lacks H.
Finished if in acidic solution.
If in basic solution, add enough OH− to neutralize the H+, rewrite H+ + OH− as H2O.
Add to both sides.
Then cancel H2O on both sides.
CN− + H2O + 2OH− → CNO− + 2H2O
CN− + H2O + 2OH− →
CNO− + 2H+ + 2OH−
CN− + 2OH− → CNO− + H2O
MnO4– → MnO2
MnO4– → MnO2 + 2H2O
MnO4– + 4H+ → MnO2 + 2H2O
MnO4– + 4H2O → MnO2 + 2H2O + 8OH−
MnO4– + 4H+ + 4OH−
→ MnO2 + 2H2O + 8OH−
MnO4– + 2H2O → MnO2 + 8OH−

35. Balancing Redox Reactions in Basic Solution, Continued
4. Balance each half-reaction with respect to charge by adjusting the numbers of electrons.
Electrons on product side for oxidation.
Electrons on reactant side for reduction.
5. Balance electrons between half-reactions.
6. Add half-reactions, canceling electrons and common species.
7. Check.
CN− + 2OH− → CNO− + H2O + 2e−
CN− + 2OH− → CNO− + H2O
MnO4– + 2H2O + 3e− → MnO2 + 4OH−
MnO4– + 2H2O → MnO2 + 4OH−
} x 3
CN− + 2OH− → CNO− + H2O + 2e−
MnO4– + 2H2O + 3e− → MnO2 + 4OH−
} x 2
3CN− + 6OH− → 3CNO− + 3H2O + 6e−
2MnO4– + 4H2O + 6e− → 2MnO2 + 8OH−
3CN– + 2MnO4– + H2O → 3CNO– + 2MnO2 + 2OH–
Tro's Introductory Chemistry, Chapter 16

36. Tro's Introductory Chemistry, Chapter 16
Practice—Balance the Following Equation in Basic Solution: I– + Cr2O72- → Cr3+ + I2
the dominant species in the chromate  dichromate equilibrium is CrO42−under basic conditions – the above problem is included more for practicing balancing than to imply a particular chemistry
Tro's Introductory Chemistry, Chapter 16 36

37. Balancing Redox Reactions
Assign oxidation states and determine element oxidized and element reduced.
Separate into oxidation and reduction half- reactions.
I− + Cr2O72– → I Cr3+ −1 +6 −2 +3
oxid
red
oxid: I− → I2
red: Cr2O72– → Cr3+
Tro's Introductory Chemistry, Chapter 16 37 37

38. Balancing Redox Reactions,Continued
3. Balance half-reactions by mass.
First balance atoms other than O and H.
Balance O by adding H2O to side that lacks O.
Balance H by adding H+ to side that lacks H.
Finished if in acidic solution.
If in basic solution, add enough OH− to neutralize the H+, rewrite H+ + OH− as H2O.
Add to both sides.
Then cancel H2O on both sides.
oxid: I− → I2
oxid: 2 I− → I2
red: Cr2O72– → Cr3+
red: Cr2O72– → 2 Cr3+
red: Cr2O72– → 2Cr3+ +7H2O
Cr2O72– +14H+ → 2Cr3+ +7H2O
Cr2O72– +7H2O → 2Cr3+ +14OH−
Cr2O72– +14H2O →
2Cr3+ +7H2O +14OH−
Cr2O72– +14H+ +14OH− →
2Cr3+ +7H2O +14OH− 38 38

39. Balancing Redox Reactions,Continued
4. Balance each half-reaction with respect to charge by adjusting the numbers of electrons.
Electrons on product side for oxidation.
Electrons on reactant side for reduction.
5. Balance electrons between half-reactions .
6. Add half-reactions, canceling electrons and common species.
7. Check.
2 I− → I2
2 I− → I2 + 2e−
Cr2O72– +7H2O + 6e−→ 2Cr3+ +14OH−
Cr2O72– +7H2O → 2Cr3+ +14OH−
6 I− → 3 I2 + 6e−
2 I− → I2 + 2e−
} x 3
Cr2O72– +7H2O + 6e−→ 2Cr3+ +14OH−
Cr2O72– +7H2O + 6 I−→ 2Cr3+ +14OH− + 3 I2
Tro's Introductory Chemistry, Chapter 16 39

40. Will a Reaction Take Place?
Reactions that are energetically favorable are said to be spontaneous.
They can happen, but the activation energy may be so large that the rate is very slow.
The relative reactivity of metals can be used to determine if some redox reactions are spontaneous.
Tro's Introductory Chemistry, Chapter 16

41. Single Displacement Reactions
Also known as single replacement reactions.
A more active free element displaces a less active element in a compound.
Metals displace metals or H.
Cu + 2 AgNO3 ® Cu(NO3)2 + 2 Ag
Mg + 2 HCl ® MgCl2 + H2
Nonmetals displace nonmetals.
2 KI + Br2 ® 2 KBr + I2
Carbon displaces metals from oxides.
3 C + Fe2O3 ® 3 CO + 2 Fe
Always redox.
Tro's Introductory Chemistry, Chapter 16

42. Tendency to Lose Electrons
Some metals have a greater tendency to lose electrons than others.
Metallic-free elements are always oxidized.
The greater the tendency of a metal to lose electrons, the easier it is to oxidize.
The greater the tendency of a metal to lose electrons, the harder it is to reduce its cations.
If Metal A has a greater tendency to lose electrons than Metal B, then:
A(s) + B+(aq)  A+(aq) + B(s),
but: A+(aq) + B(s)  no reaction.
Tro's Introductory Chemistry, Chapter 16 42

43. Activity Series of MetalsK Ba Sr Ca Na Mg Al Mn Zn Cr Fe Cd Co Ni Sn Pb H Sb As Bi Cu Hg Ag Pd Pt Au
displace H2
from
cold
H2O
steam
acids
react with O2 in the air to make oxides
Zn is above H,
so Zn will react with acids K Ba Sr Ca Na Mg Al Mn Zn Cr Fe Cd Co Ni Sn Pb H Sb As Bi Cu Hg Ag Pd Pt Au
displace H2
from
cold
H2O
steam
acids
react with O2 in the air to make oxides
Gold is at the
bottom, so it is
very unreactive. K Ba Sr Ca Na Mg Al Mn Zn Cr Fe Cd Co Ni Sn Pb H Sb As Bi Cu Hg Ag Pd Pt Au
displace H2
from
cold
H2O
steam
acids
react with O2 in the air to make oxides
Fe is below Zn, so Zn metal
will displace Fe2+.
Activity Series of Metals K Ba Sr Ca Na Mg Al Mn Zn Cr Fe Cd Co Ni Sn Pb H Sb As Bi Cu Hg Ag Pd Pt Au
displace H2
from
cold
H2O
steam
acids
react with O2 in the air to make oxides
Fe is above Cu, so Cu metal
will not displace Fe2+
Listing of metals by reactivity.
Free metal higher on the list displaces metal cation lower on the list.
Metals above H will dissolve in acid:
Zn H+ ® H2 + Zn2+
Zn + Fe2+ ® Fe + Zn2+
Cu + Fe2+ ® no reaction

44. Tro's Introductory Chemistry, Chapter 16
Mg is above
Cu on the
activity series.
Mg will react with Cu2+ to form Mg2+
and Cu metal.
Cu will not
react with Mg2+.
Tro's Introductory Chemistry, Chapter 16

45. Table of Oxidation Half-Reactions45

46. Table of Oxidation Half-Reactions, Continued
Any oxidation half-reaction that is higher on the list will give a spontaneous reaction when combined with the reverse of a half-reaction that is lower on the list.
The reverse of an oxidation half-reaction is a reduction half-reaction.
Metals will dissolve in acid if their oxidation half-reaction is above H2  2H++ 2e−.
Tro's Introductory Chemistry, Chapter 16 46

47. Tro's Introductory Chemistry, Chapter 16
Example—Complete and Balance the Following Reaction: Al(s) + NiCl2(aq) ®
Check the activity series. Al more reactive.
If more reactive metal uncombined, then you will get reaction.
Reaction.
Determine what ion the uncombined metal will form. Al ® Al+3
Determine formula of new compound. AlCl3
Complete and balance; check solubility of salt.
2 Al(s) + 3 NiCl2(aq) ® 3 Ni(s) + 2 AlCl3(aq)
Tro's Introductory Chemistry, Chapter 16 47

48. Practice—Predict the Products and Balance the Equation.
Mg + H3PO4 ®
Cu + H2SO4 ®
Al + Fe2+ ®
Tro's Introductory Chemistry, Chapter 16

49. Practice—Predict the Products and Balance the Equation, Continued.
3 Mg + 2 H3PO4 ® Mg3(PO4)2 + 3 H2
Cu + H2SO4 ® No reaction.
2 Al + 3 Fe2+ ® 2 Al Fe
Tro's Introductory Chemistry, Chapter 16

50. Tro's Introductory Chemistry, Chapter 16
Electrical Current
When we talk about the current of a liquid in a stream, we are discussing the amount of water that passes by in a given period of time.
When we discuss electric current, we are discussing the amount of electric charge that passes a point in a given period of time.
Whether as electrons flowing through a wire or ions flowing through a solution.
Tro's Introductory Chemistry, Chapter 16 50

51. Redox Reactions and Current
Redox reactions involve the transfer of electrons from one substance to another.
Therefore, redox reactions have the potential to generate an electric current.
In order to use that current, we need to separate the place where oxidation is occurring from the place that reduction is occurring.
Tro's Introductory Chemistry, Chapter 16 51

52. Electric Current Flowing Directly Between Atoms52

53. Electric Current Flowing Indirectly Between Atoms
Tro's Introductory Chemistry, Chapter 16 53

54. Electrochemical Cells
Electrochemistry is the study of redox reactions that produce or require an electric current.
The conversion between chemical energy and electrical energy is carried out in an electrochemical cell.
Spontaneous redox reactions take place in a voltaic cell.
Also known as galvanic cells.
Batteries are voltaic cells.
Nonspontaneous redox reactions can be made to occur in an electrolytic cell by the addition of electrical energy.
Tro's Introductory Chemistry, Chapter 16

55. Electrochemical Cells, Continued
Oxidation and reduction reactions kept separate.
Half-cells.
Electron flow through a wire, along with ion flow through a solution, constitutes an electric circuit.
Requires a conductive solid (metal or graphite) electrode to allow the transfer of electrons.
Through external circuit.
Ion exchange between the two halves of the system.
Electrolyte.
Tro's Introductory Chemistry, Chapter 16

56. Tro's Introductory Chemistry, Chapter 16
Electrodes
Anode
Electrode where oxidation occurs.
Anions attracted to it.
Connected to positive end of battery in electrolytic cell.
Loses weight in electrolytic cell.
Cathode
Electrode where reduction occurs.
Cations attracted to it.
Connected to negative end of battery in electrolytic cell.
Gains weight in electrolytic cell.
Electrode where plating takes place in electroplating.
Tro's Introductory Chemistry, Chapter 16

57. Voltaic Cell

58. Tro's Introductory Chemistry, Chapter 16
Current and Voltage
The number of electrons that flow through the system per second is the current.
Electrode surface area dictates the number of electrons that can flow.
The amount of force pushing the electrons through the wire is the voltage.
The farther the metals are separated on the activity series, the larger the voltage will be.
Tro's Introductory Chemistry, Chapter 16

59. Tro's Introductory Chemistry, Chapter 16
Current
The number of
electrons that pass
a point each second
is called the current
of the electricity.
The amount of
water that passes
a point each second
is called the current
of the river.
Tro's Introductory Chemistry, Chapter 16

60. Tro's Introductory Chemistry, Chapter 16
Voltage
Voltage is the
force pushing
the electrons
down the wire.
Gravity is the
force pulling
the water down
the river.
Tro's Introductory Chemistry, Chapter 16

61. Tro's Introductory Chemistry, Chapter 16
Dead Battery
As the reaction
proceeds, the
reactants get
consumed and
the voltaic cell
“dies.” The
current decreases
until electrons
can no longer
flow through
the wire.
Tro's Introductory Chemistry, Chapter 16

62. LeClanché’s Acidic Dry Cell
Electrolyte in paste form.
ZnCl2 + NH4Cl.
Or MgBr2.
Anode = Zn (or Mg).
Zn(s) ® Zn2+(aq) + 2 e-
Cathode = graphite rod.
MnO2 is reduced.
2 MnO2(s) + 2 NH4+(aq) + 2 H2O(l) + 2 e- ® 2 NH4OH(aq) + 2 Mn(O)OH(s)
Cell voltage = 1.5 v.
Expensive, nonrechargeable, heavy, easily corroded.
Tro's Introductory Chemistry, Chapter 16

63. Alkaline Dry Cell
Same basic cell as acidic dry cell, except electrolyte is alkaline KOH paste.
Anode = Zn (or Mg).
Zn(s) ® Zn2+(aq) + 2 e-
Cathode = brass rod.
MnO2 is reduced.
2 MnO2(s) + 2 NH4+(aq) + 2 H2O(l) + 2 e- ® 2 NH4OH(aq) + 2 Mn(O)OH(s)
Cell voltage = 1.54 v.
Longer shelf life than acidic dry cells and rechargeable; little corrosion of zinc.
Tro's Introductory Chemistry, Chapter 16

64. Lead Storage Battery Six cells in series. Electrolyte = 6 M H2SO4.
Anode = Pb.
Pb(s) + SO42-(aq) ® PbSO4(s) + 2 e-
Cathode = Pb coated with PbO2.
PbO2 is reduced.
PbO2(s) + 4 H+(aq) + SO42-(aq) + 2 e- ® PbSO4(s) + 2 H2O(l)
Cell voltage = 2.09 v.
Rechargeable, heavy.
Tro's Introductory Chemistry, Chapter 16

65. Fuel Cells
Like batteries in which reactants are constantly being added.
So it never runs down!
Anode and cathode both Pt-coated metal.
Electrolyte is OH– solution.
Anode reaction: 2 H2 + 4 OH– → 4 H2O(l) + 4 e-.
Cathode reaction: O2 + 4 H2O + 4 e- → 4 OH–.

66. Nonspontaneous Redox Reaction
The reverse of a spontaneous reaction is nonspontaneous.
To get it to run, an outside energy source must be supplied.
Nonspontaneous redox reactions can be made to work by using a battery to force the electrons to flow in the nonspontaneous direction.
Tro's Introductory Chemistry, Chapter 16

67. Tro's Introductory Chemistry, Chapter 16
Electrolysis
Electrolysis is the process of using electricity to break a compound apart.
Electrolysis is done in an electrolytic cell.
Electrolytic cells can be used to separate elements from their compounds.
Generate H2 from water for fuel cells.
Recover metals from their ores.
Tro's Introductory Chemistry, Chapter 16

68. Tro's Introductory Chemistry, Chapter 16
Electrolytic Cell
The + terminal of the battery = anode.
The - terminal of the battery = cathode.
Cations attracted to the cathode; anions attracted to the anode.
Cations pick up electrons from the cathode and are reduced; anions release electrons to the anode and are oxidized.
In electroplating, the work piece is the cathode.
Cations are reduced at the cathode and plate onto the surface.
The anode is made of the plate metal, the anode oxidizes and replaces the metal cations lost from the solution.
Tro's Introductory Chemistry, Chapter 16

69. Electrolytic Cell—Electroplating
Tro's Introductory Chemistry, Chapter 16

70. Tro's Introductory Chemistry, Chapter 16
Corrosion
Corrosion is the spontaneous oxidation of a metal by chemicals in the environment.
Since many materials we use are active metals, corrosion can be a very big problem.
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71. Tro's Introductory Chemistry, Chapter 16
Preventing Corrosion
One way to reduce or slow corrosion is to coat the metal surface to keep it from contacting corrosive chemicals in the environment.
Paint.
Some metals, like Al, form an oxide that strongly attaches to the metal surface, preventing the rest from corroding.
Another method to protect one metal is to attach it to a more reactive metal that is cheap.
Sacrificial electrode.
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