Redox reactions involve the transfer of electrons between reactants. In an electrochemical cell, a spontaneous redox reaction occurs between two half-cells separated by a salt bridge. In the oxidation half-reaction, electrons are lost at the anode. In the reduction half-reaction, electrons are gained at the cathode. The standard electrode potential (E0) of a half-reaction indicates its tendency to be reduced or oxidized relative to the standard hydrogen electrode. The cell potential (Ecell) is equal to the cathode potential minus the anode potential and determines if the cell reaction is spontaneous.

2. Redox Reaction - A reaction
where one species gains electrons
and one species loses electrons.
Must have both processes occurring.
Remember acid-base reactions where protons
are transferred. One has to give, the other has to
take.

3. 3
Oxidation is…
–the loss of electrons
–an increase in
oxidation state
–the addition of oxygen
–the loss of hydrogen
2 Mg + O2  2 MgO
notice the magnesium is
losing electrons
Reduction is…
–the gain of electrons
–a decrease in oxidation
state
–the loss of oxygen
–the addition of hydrogen
MgO + H2  Mg + H2O
notice the Mg2+ in MgO is gaining
electrons

4. Oxidation number
Number assigned to each atom to estimate its excess
charge or deficiency.
There is ALWAYS a change in oxidation numbers
in a redox reaction.
2Mg (s) + O2 (g) 2MgO (s)
0 0 2+ 2-

5. 4. The oxidation number of hydrogen is +1
The sum of the oxidation numbers of all the atoms in a
molecule or ion is equal to the charge on the molecule or ion.
5. Group IA metals are +1, IIA metals are +2 and Halogens is always
–1.
2. In monatomic ions, the oxidation number is equal to the charge on
the ion.
Li+, Li = +1; Fe3+, Fe = +3; O2-, O = -2
3. The oxidation number of oxygen is usually –2.
Rules for Determination of Oxidation Number
1. Free elements (uncombined state) and purely covalent
compounds all have atoms with an oxidation number of zero.
Na, Be, K, Pb, H2, O2, P4 = 0
H2O H = +1 and +1; O = -2
LiCl, Li = +1 ; Cl = -1

6. H2CO3
O = -2 H = +1
3x(-2) + 1 + 1 + ? = 0
C = +4
What are the oxidation numbers of all the
atoms in H2CO3 ?
From rules we know:
Can calculate C:

7. What are the oxidation numbers of Mn in the
following compounds and ions ?
 Mn can have at least 5 oxidation states.
Mn - element  oxidation # = 0 metal
Mn 2+ - ion  oxidation # = +2 colorless
MnO2 - compound  oxidation # = +4 brown
+4 + (2x(-2)) = 0
MnO4
- - ion  oxidation # = +7 purple
+7 + (4x(-2)) = -1
MnO4
-2 - ion 
+6 + (4x(-2)) = -2
oxidation # = +6 green

8. Oxidation Numbers Worksheet-1
1. Figure out the oxidation numbers of the atoms in: answers
CH4 , CO2 , N2O5 , Li4C , SO3 , Na2O , Cl-1 , PO4
-3
2. Determine these oxidation numbers: answers
N2O3 , NO , P2O4 , NO2 , NH4
+1, NO3
-1 , OF2 , H2O2 , Na2O2
3. Determine these oxidation numbers: answers
N , N2 , N-3 , Be+2 , F2 , Mg
4. Determine the oxidation numbers of these polyatomic molecules: answers
KNO3, NH4NO3, H2CO3, K3PO4, Al(NO3)3, (NH4) 2SO4
5. Determine the oxidation numbers: answers
K2Cr2O7 , KMnO4 , Fe2O3 , FeO , VO , MnO2
6. Determine the oxidation numbers:
B, N-3, K2O, CH3CH2CH3 , CO3
-2, KClO4 MnO4
-1, Al2(SO4)3 , F2

9. Examples
Zn + Cu SO4 t Zn SO4 + Cu
0 +2 -2 +2 -2 0
Gained e-’s – Reduced – Oxidising agent
Lost e-’s – Oxidised – Reducing agent
Spectator
Ions

10. -1
-1 +1 -1
+3 -1 +1 -2 +2 -1 +1 -1 0
Examples
FeCl3 + H2S → FeCl2 + HCl + S
Lost e-’s – Oxidised – Reducing agent
Gained e-’s – Reduced – Oxidising agent
Spectator
Ions

11. -2
0 +4 -2 +2 -2 0
Examples
Mg + S O2 → Mg O + S
Lost e-’s – Oxidised – Reducing agent
Gained e-’s – Reduced – Oxidising agent
Spectator
Ions

12. +1 -2 +1
+1 -2 +4 -2 +1 -2 0
Examples
H2 S + S O2 → H2O + S
Lost e-’s – Oxidised – Reducing agent
Gained e-’s – Reduced – Oxidising agent
Spectator
Ions
Same product

13. +1 -2 -2 +1 -2
0 +1 +6 -2 -2 +2 -2 +1 -2 +4 -2
Examples
Cu + H2 S O4 →CuSO4 + H2O + SO2
Lost e-’s – Oxidised – Reducing agent
Gained e-’s – Reduced – Oxidising agent
Spectator Ions

14. STANDARD
ELECTRODE
POTENTIALS

15. • The more positive E0 the
greater the tendency for the
substance to be reduced
Strongest oxidizing agent
Strongest reducing agent
Zero reference point

19. Half reactions:
Zn0 t Zn2+ + 2e- - lost 2 electrons(OX)
Cu+2 + 2e- t Cu0 - took up 2 electrons (RED)
Cu+2 + Zn0 t Cu0 + Zn2+ - nett reaction
Balancing Redox Equations
Zn + Cu SO4 t Zn SO4 + Cu
0 +2 -2 +2 -2 0
Gained e-’s – Reduced – Oxidising agent
Lost e-’s – Oxidised – Reducing agent
Zn2+ + 2e- t Zn0

20. Half reactions:
H2S t S +2H++ 2e-
Fe+3 + e- t Fe+2
2 Fe+3 + H2S t 2 Fe+2 + S + 2H+
Balancing Redox Equations
S + 2H+ + 2e- t
H2S
-1
-1 +1 -1
+3 -1 +1 -2 +2 -1 +1 -1 0
FeCl3 + H2S → FeCl2 + HCl + S
Lost e-’s – Oxidised – Reducing agent
Gained e-’s – Reduced – Oxidising agent
x2
2 2 2
2 2 2

21. Half reactions:
SO2 + 2H2O t SO4
2- +4H+ + 2e-
Cr2O7
2- + 14H+ + 6e- → 2Cr 3+ + 7H2O
Cr2O7
2- +2H + + 3 SO2 t 2Cr 3++H2O +3SO4
-2
Balancing Redox Equations
SO4
2- + 4H+ + 2e- → SO2 +2H2O
-2 +4 +6 +3+3
K2Cr2O7+SO2+H2SO4→K2SO4+Cr2(SO4)3+H2O
Lost e-’s – Oxidised – Reducing agent
Gained e-’s – Reduced – Oxidising agent
x3
3 6 3 12 6
2 1

22. Balancing Redox Equations
Do yourself
KMnO4 + SO2 + H2O →K2SO4 + MnSO4 + H2SO4
KMnO4+H2S + H2SO4 → K2SO4 + MnSO4+ S+ H2O
K2Cr2O7+ H2S+ H2SO4 → K2SO4+ Cr2(SO4)+S+H2O
MnO2 + HCl → MnCl2 + H2O +Cl2
Mg + Cl2 → MgCl2
Cu + HNO3 (conc) → Cu(NO3)2 + H2O + NO2
Cu + HNO3 (dil) → Cu(NO3)2 + H2O + NO

23. Fe + CuSO4 → Cu + FeSO4
Do yourself:

24. Cu + ZnSO4 → __________
Copper
Zinc sulphate
Observation?
Do yourself:

25. Zn + CuSO4 → Cu + ZnSO4
Zinc
Copper
sulphate
Observation?
Do yourself:

26. Salt bridge - with KCl -
dissosiates into K+ and
Cl- . K+ moves to
beaker B and Cl- to
beaker A - neutralize
charges that develop
Zinc electrode
releases electrons
becomes oxidised
Copper electrode
does not react
Zn →Zn2+ + 2e-
Zn neutral changes
to Zn ions which
dissolves in the
surrounding
solution
Zn electrode gets
smaller (loses mass)
Cotton wool stoppers
Copper sulphate
solution (CuSO4)
Dissosiates into
Cu2+ and SO4
2-
ZnSO4 solution -
Dissosiates into
Zn2+and SO4
2-
Cu2+ takes up
electrons
coming from
Zn electrode
Cu2+ ions in the
solution change to
Cu that clings to the
Cu electrode – Cu
increases in mass
Cu2+ + 2e- t Cu
Voltaic – or Galvanic cell
Cu2+ colours the
solution BLUE and
changes to
COLOURLESS

27. Zn t Zn2+ + 2e-
Cu2+ + 2e- t Cu
Zn + Cu2+  Zn2+ + Cu
Chemical energy to Electrical energy
Redcat
Cathode -
reduction
An Ox
Anode-
oxidation
Zn / Zn2+ // Cu2+ / Cu
Anode (ox) s.bridge Cathode (red)
EMF = E(cell) = E(cathode) - E(anode)
= 0.34 - (-0.76)
= 1.1V

28. Chapter 14 Name: .........................................................................
1 A standard cell is constructed by connecting a Cu / Cu 2+
half-cell to a Cl2.Pt / Cl-
half-cell, as shown in
the sketch.
X(aq)
1.1 Write down the formula of a compound that X can be. ........................................................................(2)
1.2 Write down, for the reaction that takes place in this cell, the balanced
(a) oxidation half-reaction. (2)
................................................................................................................................................................
(b) reduction half-reaction. (2)
................................................................................................................................................................
(c) complete (overall) reaction. (3)
................................................................................................................................................................
1.3 Calculate the emf of this cell. (4)
................................................................................................................................................................
................................................................................................................................................................
................................................................................................................................................................
................................................................................................................................................................
2 The following reactions take place by oxidation-reduction:
A- Cu + HNO3 (conc.) B- Mg + HNO3 (conc.) C - Cu + HNO3 (dil.)
2.1 Use table B and state which one of the above three reactions will take place .....
2.1.1 the fastest? ......... Reason? ......................................................................................................................
......................................................................................................................
2.2 Write down the half reaction ......
2.2.1 which occurs at the anode in reaction A
................................................................................................................................................................
2.2.2 which demonstrates oxidation in reaction C
................................................................................................................................................................
3 A zinc rod is placed in a acidified solution of KMnO4.
3.1 Is this a spontaneous reaction? Give a reason............................................................................................
................................................................................................................................................................
3.2 Why must the KMnO4 solution be acidified?.............................................................................................
3.3 What happens to the mass of the Zn rod? ................................................................................................
3.4 Give a possible explanation for your answer in 3.3 ...................................................................................
................................................................................................................................................................
3.5 What happens to the colour of the KMnO4 solution? ................................................................................
3.6 Give a possible explanation for your answer in 3.5 ...................................................................................
................................................................................................................................................................

29. Electrochemical Cells
19.2
spontaneous
redox reaction
anode
oxidation
cathode
reduction
Attracts anions Attracts cations

30. Electrochemical Cells
19.2
The difference in electrical
potential between the anode
and cathode is called:
• cell voltage
• electromotive force (emf)
• cell potential
Cell Diagram
Zn (s) + Cu2+ (aq) Cu (s) + Zn2+ (aq)
[Cu2+] = 1 M & [Zn2+] = 1 M
Zn (s) | Zn2+ (1 M) || Cu2+ (1 M) | Cu (s)
anode cathode
Use || to separate half cells
Use | to separate reactants/phases in each half cell

31. Standard Electrode Potentials
19.3
Zn (s) | Zn2+ (1 M) || H+ (1 M) | H2 (1 atm) | Pt (s)
2e- + 2H+ (1 M) H2 (1 atm)
Zn (s) Zn2+ (1 M) + 2e-
Anode (oxidation):
Cathode (reduction):
Zn (s) + 2H+ (1 M) Zn2+ + H2 (1 atm)

32. Standard Electrode Potentials
19.3
Standard reduction potential (E0) is the voltage associated
with a reduction reaction at an electrode when all solutes
are 1 M and all gases are at 1 atm.
Defines E0 = 0 V
Standard hydrogen electrode (SHE)
2e- + 2H+ (1 M) H2 (1 atm)
Reduction Reaction
Use as a reference to measure
all other potentials. 

33. 19.3
E0 = 0.76 V
cell
Standard emf (E0 )
cell
0.76 V = 0 - EZn /Zn
0 2+
EZn /Zn = -0.76 V
0 2+
Zn2+ (1 M) + 2e- Zn E0 = -0.76 V
E0 = EH /H - EZn /Zn
cell
0 0
+ 2+
2
Standard Electrode Potentials
E0 = Ecathode - Eanode
cell
0 0
Zn (s) | Zn2+ (1 M) || H+ (1 M) | H2 (1 atm) | Pt (s)
Convention:
E° > 0 spontaneous reaction
reduction oxidation

34. Standard Electrode Potentials
19.3
Pt (s) | H2 (1 atm) | H+ (1 M) || Cu2+ (1 M) | Cu (s)
2e- + Cu2+ (1 M) Cu (s)
H2 (1 atm) 2H+ (1 M) + 2e-
Anode (oxidation):
Cathode (reduction):
H2 (1 atm) + Cu2+ (1 M) Cu (s) + 2H+ (1 M)
E0 = Ecathode - Eanode
cell
0 0
E0 = 0.34 V
cell
Ecell = ECu /Cu – EH /H
2+ +
2
0 0 0
0.34 = ECu /Cu - 0
0 2+
ECu /Cu = 0.34 V
2+
0

35. 19.3
• E0 is for the reaction as
written
• The half-cell reactions are
reversible
• The sign of E0 changes
when the reaction is
reversed (E° red = -E°oxid)
• Changing the stoichiometric
coefficients of a half-cell
reaction does not change
the value of E0

36. • The more positive E0 the
greater the tendency for the
substance to be reduced
Strongest oxidizing agent
Strongest reducing agent
Zero reference point

37. E0 = 0.76 V
cell E0 = 0.34 V
cell
Combine
Zn (s) Zn2+ (1 M) + 2e-
2e- + Cu2+ (1 M) Cu (s)
Zn (s) + Cu2+ (1 M) Cu (s) + Zn2+ (1 M)
E0 = 0.76 V + 0.34 V = 1.10 V
cell

38. What is the standard emf of an electrochemical cell made
of a Cd electrode in a 1.0 M Cd(NO3)2 solution and a Cr
electrode in a 1.0 M Cr(NO3)3 solution?
Cd2+ (aq) + 2e- Cd (s) E0 = -0.40 V
Cr3+ (aq) + 3e- Cr (s) E0 = -0.74 V
Cd is the stronger oxidizer
Cd will oxidize Cr
2e- + Cd2+ (1 M) Cd (s)
Cr (s) Cr3+ (1 M) + 3e-
Anode (oxidation):
Cathode (reduction):
2Cr (s) + 3Cd2+ (1 M) 3Cd (s) + 2Cr3+ (1 M)
x 2
x 3
E0 = Ecathode - Eanode
cell
0 0
E0 = -0.40 – (-0.74)
cell
E0 = 0.34 V
cell
19.3
 spontaneous

39. Batteries
19.6
Leclanché cell
Dry cell
Zn (s) Zn2+ (aq) + 2e-
Anode:
Cathode: 2NH4 (aq) + 2MnO2 (s) + 2e- Mn2O3 (s) + 2NH3 (aq) + H2O (l)
+
Zn (s) + 2NH4 (aq) + 2MnO2 (s) Zn2+ (aq) + 2NH3 (aq) + H2O (l) + Mn2O3 (s)

40. Batteries
Zn(Hg) + 2OH- (aq) ZnO (s) + H2O (l) + 2e-
Anode:
Cathode: HgO (s) + H2O (l) + 2e- Hg (l) + 2OH- (aq)
Zn(Hg) + HgO (s) ZnO (s) + Hg (l)
Mercury Battery
19.6

41. Batteries
19.6
Anode:
Cathode:
Lead storage
battery
PbO2 (s) + 4H+ (aq) + SO2- (aq) + 2e- PbSO4 (s) + 2H2O (l)
4
Pb (s) + SO2- (aq) PbSO4 (s) + 2e-
4
Pb (s) + PbO2 (s) + 4H+ (aq) + 2SO2- (aq) 2PbSO4 (s) + 2H2O (l)
4

42. Batteries
19.6
Solid State Lithium Battery

43. Batteries
19.6
A fuel cell is an
electrochemical cell
that requires a
continuous supply of
reactants to keep
functioning
Anode:
Cathode: O2 (g) + 2H2O (l) + 4e- 4OH- (aq)
2H2 (g) + 4OH- (aq) 4H2O (l) + 4e-
2H2 (g) + O2 (g) 2H2O (l)

44. Corrosion
19.7
Dissolved oxygen in
water causes oxidation
E°red = -0.44 V
Rust Fe2O3
E°red = 1.23 V
Since E°red (Fe3+) < E°red (O2)
Fe can be oxidized by oxygen

45. Cathodic Protection of an Iron Storage Tank
19.7
E°red = -2.37
V
E°red = 1.23 V
Mg reduction potential is too negative to be overcome by oxygen.

46. 19.8
Electrolysis is the process in which electrical energy is used
to cause a nonspontaneous chemical reaction to occur.

47. Electrolysis of Water
19.8