1. Redox Reactions

2. Introduction
There are a number of different definitions of oxidation and reduction; such as the loss or gain of oxygen, or the loss or gain of hydrogen.
However, these definitions are only useful in a limited number of reactions.
The most useful definition is given in terms of electrons:
Oxidation is the loss of electrons
Reduction is the gain of electrons

3. Half Equations
If a substance loses electrons then another substance has to gain electrons, so oxidation and reduction always occur together. This is known as a redox reaction
This can be seen more clearly by splitting the overall reaction into half equations
Redox reaction: Zn(s) + Cu2+(aq) → Zn2+(aq) + Cu(s)
Half equations: Zn(s) → Zn2+(aq) + 2e- Oxidation
Cu e- → Cu(s) Reduction

4. Oxidation Number
Oxidation number is a concept which regards all compounds as ionic and assigns charges to the components accordingly.
It provides a guide to the distribution of electrons and relative charges on atoms in covalent compounds and allows us to understand the redox process more easily.
How to determine the oxidation number:
1. Treat the compound as totally ionic – If the compound is ionic then the oxidation number is equal to the charge.
2. The most electronegative atom in a molecule is assigned a negative charge according to its valency to give a full outer shell.
3. The sum of the oxidation numbers of atoms in a molecule must be equal to the overall charge of the molecule.
4. The oxidation number of atoms in an element is 0.
5. Elements in groups 1-3 nearly always have a positive oxidation number equal to their group number.
6. The maximum possible oxidation number for an elements will be its group number.

6. Some elements always have the same oxidation number:
It is not possible to lose more electrons than there are in the valance shell; therefore the maximum possible oxidation number is equal to the group number.
Oxidation number is written with the sign first, i.e. −2 (charge is written with the number first 2−)
Some elements always have the same oxidation number:
Fluorine is the most electronegative element and is therefore always −1
Oxygen is nearly always −2, except in compounds with F or in peroxides (R−O−O−R) and superoxides (O2− anion)
Chlorine is −1 except in compounds with O and F
Hydrogen is +1, except in metal hydrides, e.g. NaH, where its oxidation number is −1

7. Examples
Oxygen is more electronegative so it is given its normal ionic valency of −2
The overall charge on the molecule is 0, therefore sulphur must have an oxidation number of +4 to cancel out the two oxygen atoms
SO2 SO42- Cr2O72-
Oxygen = −2 overall charge = 2− Sulphur = +6
6 – 8 = −2
Oxygen = −2 overall charge = 2− Chromium = +6
12 – 14 = −2

8. Practice
PCl3
H2O2
KMnO4
NaClO3
OF2
NH3
H2SO4
C2H6

9. Transition Metals
The oxidation number of a transition metal in a complex ion is worked out from the charges on the ligands.
Ligands may be either neutral or negatively charged
Examples : H2O and NH3 are both neutral ligands. Cl− and CN− are both −1
A ligand is an ion or molecule (functional group) that binds to a central metal atom to form a coordination complex.
A Dative-bond, also known as a coordinate bond is a type of covalent bond between two atoms in which the bonding electrons are supplied by one of the two atoms

10. Recall that H2O and NH3 are neutral ligands
Complex ions are written in square brackets with the overall charge written outside of the brackets
[Fe(H2O)6]2+
[CuCl4]2−
[Ni(CN)4]2−
[Ag(NH3)2]+
Determine the oxidation numbers of the transitional metals in the complex ions above
Recall that H2O and NH3 are neutral ligands
Fe = +2
Cu = +2
Ni = +2
Ag = +1

11. Naming compounds using oxidation numbers
SO2 and SO3 are commonly referred to as sulphur dioxide and sulphur trioxide, respectively. They are, however, more correctly named using the oxidation state of the sulphur atom:
Note that roman numerals are used for the oxidation numbers in the names of compounds
SO2 Sulphur = +4 = sulphur(IV) oxide
SO3 Sulphur = +6 = sulphur (VI) oxide

12. Self Test
Determine the oxidation numbers for the nitrogen in the following molecules/ions:
a) NF3 b) N2O c) N2H4 d) NO2− e) NH4+
Determine the oxidation number of the species in bold in each of the following compounds:
a) Na2O b) Na2SO3 c) KBrO3 d) K2CrO4 e) NH4NO3
Name the following compounds using oxidation numbers:
a) NO b) SeO2 c) Cr2O3 d) Cl2O7 e) KIO3

13. 4. Determine the oxidation number of the transitional metal in each of the following complex ions or compounds:
[Cu(Cl4)]2−
[Co(H2O)6]3+
[Fe(CN)6]4−
[Ag(NH3)2]+
[Mn(Br4)]2+

14. Oxidation and Reduction in terms of oxidation numbers
If an atom gains electrons, its oxidation number will become more negative
Therefore, reduction involves a decrease in oxidation number
If an atom loses electrons, its oxidation number will become more positive
Therefore, oxidation involves an increase in oxidation number
For a reaction to be redox, the reaction must involve a change in oxidation state

15. Examples Br2 + SO2 + 2H2O → H2SO4 + 2HBr
Br2 is reduced, as the oxidation number decreases from 0 to −1. The sulphur is SO2 is oxidised as its oxidation number increases from +4 to +6. This is a redox reaction.
6OH− + 3Cl2 → 5Cl− + ClO3− + 3H2O
Five Cl atoms have been reduced and one has been oxidised. This type of reaction, in which the same species has been oxidised and reduced, is called a disproportionation reaction.
Cr2O72− + H2O → 2CrO42− + 2H+
The oxidation number of Cr on both sides of the equation is +6, and no other atom undergoes a change in oxidation number. Therefore, this is not a redox reaction

16. Disproportionation Reactions
A redox reaction in which one species is simultaneously oxidised and reduced
Cl2 + 2OH− → ClO− + Cl− + H2O +1 −1
Reduction
Oxidation

17. Oxidising Agents A substance that oxidises something else
An oxidising agent takes electrons away from something else
Therefore, in the process they are, themselves, reduced.
Examples include halogens, peroxides and nitric acid. Generally, they contain oxygen in the anionic structure. 
A reducing agent is therefore, by definition, the opposite of the above

18. Examples 2Br− + Cl2 → 2Cl− + Br2
Cl2 is the oxidising agent, as It oxidises the Br− to Br2. In terms of electrons, the Cl2 removes electrons from the bromide ions.
The Br− gives electrons away in this reaction and is therefore the reducing agent.

19. Cr2O72− + 6Fe2+ + 14H+ → 2Cr3+ + 6Fe3+ + 7H2O
The Fe2+ is oxidised to Fe3+ by the Cr2O72−, so Cr2O72− is the oxidising agent.
This means that Fe2+ is the reducing agent; it has accepted/gained electrons.

20. Self Test
State whether the following half equations involve oxidation or reduction:
Cl2 + 2e− → 2Cl −
Mn3+ + e− → Mn2+
I2 + 6H2O → 2IO3− + 12H+ + 10e−
2. Identify the element(s) that have been oxidised/reduced and state whether the reaction is a redox reaction.
Cu2+ + 2OH− → Cu(OH)2
2Na+ + 2H2O → 2NaOH + H2
3HgSO4 → Hg2SO4 + Hg + 2SO2+ 2O2
2I− + H+ + HOCl → I2 + H2O + Cl−

21. 3. In each of the following redox reactions, identify the oxidising agent and the reducing agent:
Zn + CuSO4 → ZnSO4 + Cu
I2O5 + 5CO → 5CO2 + I2
2Na2S2O3 + I2 → Na2S4O6 + 2NaI
2KMnO4 + 5Na2C2O4 + 8H2SO4 → 2MnSO4 + 10CO K2SO4 + 5Na2SO4+ 8H2O
6FeSO4 + K2Cr2O7 + 7H2SO4 → 3Fe2(SO4)3 + K2SO Cr2(SO4)3 + 7H2O

22. Redox Equations
A redox equation may be broken down into two half equations
These half equations show the oxidation and reduction processes separately
The half equations must be balanced in terms of the number of atoms on both sides and the total charge on both sides
Br2 + SO2 + 2H2O → H2SO4 + 2HBr
Br2 + 2H+ + 2e- → 2HBr Reduction
SO2 + 2H2O → H2SO4 + 2H+ + 2e- Oxidation
The 2H+ and 2e- cancel out when the half equations are combined to produce the overall redox reaction

23. Balancing Half Equations
Ni2+ → Ni
The number of nickel atoms is the same on both sides, however, the total charge on the left side is 2+ but the total charge on the right side is 0
Ni2+ + 2e- → Ni
Now the number of atoms and the charges are balanced
Br2 → 2Br-
Br2 + 2e- → 2Br-

24. Combining half equations
An oxidation half equation may be combined with a reduction half equation to produce an overall redox reaction
When they are combined, the number of electrons lost in the oxidation reaction must be the same as the number gained in the reduction reaction
MnO4- + 8H+ + 5e- → Mn2+ + 4H2O Reduction
Fe2+ → Fe3+ + e Oxidation
5 electrons are gained in the reduction half equation, but only 1 is lost in the oxidation half equation. Therefore, the oxidation half equation must be multiplied by 5
MnO4- + 8H+ + 5e- → Mn2+ + 4H2O
5Fe2+ → 5Fe3+ + 5e-
MnO4- + 8H+ + 5Fe2+ → Mn2+ + 4H2O + 5Fe3+
All common terms are cancelled out

25. Balancing half equations in acidic solution
The following procedure should be followed:
Separate the half-reactions
Balance all atoms except H and O
Add H2O to the side deficient in O to balance O
Add H+ to the side deficient in H to balance H
Add e- to the side deficient in negative charge
Scale the reactions so that they have an equal amount of electrons
Combine the reactions and cancel out the common terms
Fe2+ + Cr2O72- → Fe3+ + Cr3+
6Fe H+ + Cr2O72- → 6Fe3+ + 2Cr3+ + 7H2O
Oxidation
Reduction
Fe2+ → Fe3+
Cr2O72- → Cr3+
+ e- x6
14H+ + 2
+ 7H2O
+ 6e-

26. Cr2O72- + HNO2 → Cr3+ +NO3− Cr2O72- → Cr3+ HNO2 → NO3- Cr2O72- → 2Cr3+
Cr2O72- → 2Cr3+ + 7H2O
HNO2 + H2O → NO3-
Cr2O H+ + 6e- → 2Cr3+ + 7H2O
HNO2 + H2O → NO3- + 3H+ + 2e-
3HNO2 + 3H2O → 3NO3- + 9H+ + 6e-
Cr2O72- → 2Cr3+
HNO2 → NO3-
Cr2O H+ → 2Cr3+ + 7H2O
HNO2 + H2O → NO3- + 3H+
Multiply by 3

27. Cr2O H+ + 6e- → 2Cr3+ + 7H2O 3HNO2 + 3H2O → 3NO3- + 9H+ + 6e- Cr2O H+ + 6e- + 3HNO2 + 3H2O → 2Cr3+ + 7H2O + 3NO3- + 9H+ + 6e- Cr2O H+ + 3HNO2 → 2Cr3+ + 4H2O + 3NO3-

28. Fe2+ + MnO4- → Fe3+ + Mn2+

29. Balancing half equations in basic solution
Separate the half-reactions
Balance elements other than O and H
Add H2O to balance oxygen
Balance hydrogen with protons
Balance the charge with e-
Scale the reactions so that they have an equal amount of electrons
Ag + Zn2+→Ag2O + Zn
Ag → Ag2O
Zn2+ → Zn 2
+ H2O
+ 2H+
+ 2e-
+ 2e-
Combine the half equations and cancel out the electrons
2Ag + H2O + Zn2+ → Ag2O + 2H+ + Zn

30. 2Ag + H2O + Zn2+ → Ag2O + 2H+ + Zn
Add OH- to balance H+ (add to both sides of the equation)
2Ag + H2O + Zn2+ + 2OH- → Ag2O + 2H+ + Zn + 2OH-
Combine OH- ions and H+ ions that are present on the same side to form water.
2Ag + H2O + Zn2+ + 2OH- → Ag2O + Zn + 2H2O
Cancel common terms
2Ag + Zn2+ + 2OH- → Ag2O + Zn + H2O

31. Fe(CN)63- + Re → Fe(CN)64- + ReO4-

32. Self test Balance the following redox reactions
As2O3(s) + NO3–(aq) → H3AsO4(aq) + NO(g)
NO2–(aq) + Al(s) → NH3(g) + AlO2–(aq)
MnO4- + S2O32- → S4O Mn2+
Cu(NH3) S2O42- → SO32- + Cu + NH3

33. Voltaic Cells
Voltaic cells (electrochemical cells), also known as Galvanic cells, provide us with a way of harnessing redox reactions to generate electricity
This is the basis of batteries