1. 2.6 Redox Part 1.
a. demonstrate an understanding of:
i. oxidation number — the rules for assigning oxidation numbers
ii. oxidation and reduction as electron transfer
iii. oxidation and reduction in terms of oxidation number changes
iv. how oxidation number is a useful concept in terms of the classification of reactions as redox and as disproportionation
b. write ionic half-equations and use them to construct full ionic equations.
Connector:
Define the following in terms of electron transfer:
Oxidation
Reduction
Oxidising agent
Reducing agent
Updated_Crowe_2012

2. Terminology for Redox Reactions
OXIDATION—loss of electron(s) by a species; increase in oxidation number; increase in oxygen.
REDUCTION—gain of electron(s); decrease in oxidation number; decrease in oxygen; increase in hydrogen.
OXIDIZING AGENT—electron acceptor; species is reduced. (an agent facilitates something; ex. Travel agents don’t travel, they facilitate travel)
REDUCING AGENT—electron donor; species is oxidized.

3. A way to remember -
OIL RIG
Oxidation Is
Loss
Reduction
Gain

4. Electron Transfer Reactions
Reactions in which electron transfer takes place can be described as being either:
Oxidation reactions
Reduction reactions

5. You can’t have one… without the other!
Reduction (gaining electrons) can’t happen without an oxidation to provide the electrons.
You can’t have 2 oxidations or 2 reductions in the same equation. Reduction has to occur at the cost of oxidation
So both oxidation and reduction occur together in what are described as REDOX reactions

6. REDOX reactions are important in …
Purifying metals (e.g. Al, Na, Li)
Producing gases (e.g. Cl2, O2, H2)
Electroplating metals
Electrical production (batteries, fuel cells)
Protecting metals from corrosion
Balancing complex chemical equations
Sensors and machines (e.g. pH meter)
C3H8O + CrO3 + H2SO4  Cr2(SO4)3 +C3H6O + H2O

7. Oxidation half-reaction (lose e-)
It’s Important to keep a track of electron transfer taking place in a reaction, as charges on both sides of the chemical equation must be balanced.
Oxidation numbers are a way of doing this.
‘these are equal to the charge that an element would have if it were totally ionically bonded’
2Mg (s) + O2 (g) MgO (s)
Oxidation Numbers 2+ 2-
2Mg Mg2+ + 4e-
Oxidation half-reaction (lose e-)
O2 + 4e O2-
Reduction half-reaction (gain e-)

8. More Oxidation Numbers
The sum of the
oxidation numbers on
each side does not
appear to be the same.
….how can we explain
this?

9. More Oxidation Numbers
What are the oxidation numbers for this equation? ?
As we have already seen, these equation scan be broken down into half equations:
As they are half equations, the oxidation numbers on each side of the equation do not balance:

10. Oxidation Numbers and Covalent Compounds
Oxidation numbers can also be applied to reactions involving covalent compounds:
Oxidation numbers can also show shifts in electron density.
As before, uncombined elements are assigned the value 0.
(In these molecules, electron density is equally shared)
When in a compound, elements are given an oxidation number based on what the charge would be if the electrons in each of the bonds they are involved belong to the more electronegative element. ? ?

11. Rules of Oxidation numbers
The charge the atom would have in a molecule (or an
ionic compound) if electrons were completely transferred.
Free elements (uncombined state) have an oxidation number of zero.
Na, Be, K, Pb, H2, O2, P4 = 0
In monatomic ions, the oxidation number is equal to the charge on the ion.
Li+, Li = +1; Fe3+, Fe = +3; O2-, O = -2
The oxidation number of oxygen is usually –2. In H2O2 (peroxides) and O22- it is –1. Positive when it combines with fluorine.

12. The oxidation number of hydrogen is +1 except when it is bonded to metals in binary compounds. In these cases, its oxidation number is –1.
Group I metals are +1, Group 2 metals are +2, Group 3 metals are +3 and fluorine is always –1.
6. Fluorine always has an oxidation number of -1.
7. Chlorine always has an oxidation number of -1, except with fluorine and oxygen when its positive.
8. The sum of the oxidation numbers of all the atoms in a molecule or ion is zero (or equal to the charge on the molecule or ion).

13. Worked Example… HCO3- O = -2 H = +1 3x(-2) + 1 + ? = -1 C = +4
Oxidation numbers of all the atoms in HCO3- ?
HCO3-
O = -2
H = +1
3x(-2) ? = -1
C = +4

14. Now try these:

15. More Examples…

16. Questions

17. Answers

18. Redox Reactions: Metals with Oxygen
e.g.
Can be written as half equations:
What is oxidised and reduced?....
…and which would you describe as the oxidising agent and the reducing agent?
Half equations show what happens to the electrons.
Write the two half equations for the above reaction.
What happens to sodium’s electrons when it reacts?
What does oxygen do with them?
Sodium is oxidised
Oxygen is reduced
Sodium acts as a reducing agent
Oxygen acts as a oxidising agent

19. Redox Reactions: Metals with Water
This equation can be re-written with sodium hydroxide shown as
the ions it is made up of.
Notice how both
equations are
balanced in terms
of charge and the
number and type
of species. 2
Write them out, and label them as oxidation and reduction
2 e-

20. Redox Reactions: Metals with Acid
Write the two half equations.
Identify the oxidation and reduction steps
Then add the two equations together to produce the FULL IONIC EQUATION

21. Adding half equations:
OXIDATION
REDUCTION
The FULL IONIC EQUATION shows the reaction that took place

22. Redox and Displacement Reactions
What is a displacement reaction?
A reaction where a more reactive element displaces a less reactive element in aqueous solution.
Write the two half equations.
Identify the oxidation and reduction steps
Then add the two equations together to produce the FULL IONIC EQUATION

23. FULL IONIC EQUATION OXIDATION REDUCTION
Notice that the sulphate ions do not appear in the above equation.
This is because they are unchanged in the reaction.
Such species are called SPECTATOR IONS.
Displacement reactions are redox reactions

24. Questions

25. Answers

26. Questions

27. Answers

28. Balancing Redox Reactions (Half-equation method)
The half-equation method separates a redox reaction into its oxidation and reduction half reactions.
Overall scheme for the half reaction method:
Step 1: Split reaction into half-reactions (reduction and oxidation)
Step 2: Balance the charge or oxidation number with electrons
Step 3: Balance O by adding H2O
Step 4: Balance H by adding H+
Step 5: Multiply by some integer to make electrons (lost) = electrons (gained)
Step 6: Add half equations and cancel substances on both sides

29. Example 1.
Reaction between manganese (II) ion and the bismuthate ion in acid solution 2+ 5+ 7+ 3+
What are the oxidation numbers for Mn and Bi ?
Write the half equation for Mn 2+
Balance O by adding H2O, then H by adding H+
Repeat the process for the bismuthate ion

30. Balance O by adding H2O, then H by adding H+
We now have the two half equations:
Step 5: Multiply by some integer to make electrons (lost) = electrons (gained)
Step 6: Add half equations and cancel substances on both sides

31. 8 2 2 16 10 10 30 5 5 15
Step 6: Add half equations and cancel substances on both sides

32. Overview

33. Solve these acid solution reactions: :
MnO4-  Mn 2+
2 I-  I2
H2O2  H2O + O2
Cr2O72-  Cr3+

34. 8H+ + MnO4– + 5Fe3+ ➔ 5Fe2+ + Mn2+ + 4H2O
Cr2O72- + 3H2O2 + 8H+  2Cr3+ + 3O2 +7H2O