1. Redox processes
Topic 9 SL Chemistry

2. What is a redox reaction?
Redox reactions are important in almost all chemical processes.
There are 3 main types of reactions that occur in chemistry:
Acid-base reactions
Precipitation reactions
Redox reactions
A redox reaction involves two processes: reduction and oxidation.
They can be thought of in three different ways:
In terms of specific elements – hydrogen and oxygen
In terms of electron transfer
In terms of oxidation number

3. Oxidation: Combining with oxygen
At the simplest level, oxidation can be considered as a reaction in which a substance combines with oxygen. 2Mg(s) + O2(g)  2MgO(s) 2CH3OH(l) + 3O2(g)  2CO2(g) + 4H2O(l) 4Fe(s) + 3O2(g)  2Fe2O3(s)
Reduction: Removal of oxygen or addition of hydrogen
Fe2O3(s), iron (III) oxide, is rust. Rust is an example of corrosion.
The second example of reduction mirrors the first one since oxygen is being removed as well.
Reduction can be viewed as the removal of oxygen:
NiO(s) + C  Ni(s) + CO(g)
Reduction can also be viewed as the addition of hydrogen:
WO3(s) + 3H2(g)  W(s) + 2H2O(g)

4. Oxidation and reduction in terms of electron transfer
Oxidation involves the loss of electrons
Reduction involves the gain of electrons
Remember, since O2 is diatomic, we treat the O atoms as if they were neutral.

5. Oxidation and reduction in terms of electron transfer
Mg(s) + O2(g)  MgO Mg(s) is Group 2 so it loses two electrons to have a complete outer shell. Mg(s)  Mg2+ + 2e- Oxygen is a Group 16 so it gains two electrons to have a complete outer shell. O + 2e-  O2- The overall reaction is: 2Mg(s) + O2(g)  2MgO(s) Two processes: 2Mg  2Mg2+ + 4e- O2 + 4e-  2O2-
OXIDIZED (loses electrons)
REDUCED (gains electrons)
Remember, since O2 is diatomic, we treat the O atoms as if they were neutral.
Oxidizing agent causes the oxidation and is itself reduced.
Reducing agent causes the reduction and is itself oxidized.

6. BEWARE!!
Considering redox processes in terms of electron transfer is a common approach, but you must use caution. This can really only be used with ionic compounds since with covalent compounds electrons are not lost or gained (they are shared). In this reaction: C(s) + O2(g)  CO2(g) there are no ionic bonds, the first method (addition of oxygen) is more appropriate, since carbon is indeed oxidized.

7. Oxidation and reduction in terms of oxidation states
The oxidation state is the apparent charge of an atom in a free element, a molecule, or an ion. There is an oxidation state for all atoms.
Oxidation describe a process in which the oxidation state increases.
Reduction describes a process in which the oxidation state decreases.

8. Rules for assigning oxidation states
Oxidation states of free elements (O2, S8, P4, Na, etc) have a oxidation state of 0.
Group 1 metals always have +1 ox.st. in their ions and compounds. Group 2 metals always have +2 ox.st. in their ions and compounds. Aluminum has a +3 ox.st. in most of its compounds.
Ox.st. of H is +1 when bonded with a nonmetal and -1 when bonded with a metal.
Ox.st. of O is usually -2. Exception: peroxide species have -1 ox.st.
Ox.st. of F is -1 in all it’s compounds. For other halogens, it is usually -1 in binary compounds. Compounds with O in oxoanions and oxoacids the os.st. is positive (ex. In HClO4, O has an ox.st. of +7)

9. Rules for assigning oxidation states
6. In a neutral molecule, the sum of ox.st. is zero. In a polyatomic ion, the sum of ox.st. is equal to the overall charge of the ion. Examples: NH3 is a neutral atom. Ox.st. of N is -3 and ox.st. of H is +1. 1(-3) + 3(+1) = 0 NH4+ is a polyatomic ion. Ox.st. of N is -3 and ox.st. of H is +1 1(-3) + 4(+1) = +1

10. Variable oxidation states
DB Section 14
Variable oxidation states
One characteristic property of transition metals is that they have variable oxidation states. This is due to their incomplete d-subshell.

11. Oxidation states and the nomenclature of transition metal compounds
The Stock nomenclature system is used to identify the oxidation states of transition metals when naming them. This involves using Roman numerals to indicate the oxidation number. Note: the oxidation number is represented as a Roman numeral. The oxidation states in some compounds may not have integer values. The superoxide anion (O2-), for example, has an oxidation state of -1/2.

12. Practice
Deduce the oxidation states of the following atoms (marked x).
K2Crx2O7
MnxO4-
Mg3Nx2
Sx8
[NH4]2[Fex(H2O)6][SO4]2
x=+6
x= +7
x = -3
x = 0
x = +2

13. Practice
Identify the species being oxidized and reduced in each of the following reactions:  1. Cr+ + Sn4+    Cr3+ + Sn2+  Hg2+ + 2 Fe (s)   3 Hg2 + 2 Fe3+  As (s) + 3 Cl2 (g)   2 AsCl3
Oxidized: Cr; Reduced: Sn
Oxidized: Fe; Reduced: Hg
Oxidized: As; Reduced: Cl

14. Cl2(aq) + 2KI(aq)  2KCl(aq) + I2(aq)
Practice
In the following balanced equation:
Cl2(aq) + 2KI(aq)  2KCl(aq) + I2(aq)
Deduce the oxidation states of chlorine and iodine in the reactants and products.
State which element is oxidized and which is reduced.
Identify the oxidizing agent and the reducing agent.
Cl0 in Cl2, I-1 in KI, Cl-1 in KCl and I0 in I2.
Iodine is oxidized (from -1 to 0) and chlorine is reduced (from 0 to -1)
Cl2(aq) is the oxidizing agent and KI(aq), or more precisely, I-(aq), is the reducing agent.

15. Practice
Would you use an oxidizing agent or reducing agent in order for the following reactions to occur?  a. ClO3-   ClO2  b. SO42-   S2-  c. Mn2+    MnO2  d. Zn   ZnCl2 
Reducing agent
Oxidizing agent

16. Expressing redox reactions using half-equations
Half-equations are chemists way of keeping track of electrons in redox reactions. Sometimes these are referred to as net ionic equations. Each half-equation represents either an oxidation or a reduction. There is a typical method used to balance half-equations using oxidation states.

17. Expressing redox reactions using half-equations
Balance for atoms.
Assign oxidation states.
Deduce which is oxidized and which is reduced.
State half-equation for each process.
Balance the electrons gained/lost.
Add half-equations together to get overall.
Check total charge.
Balance charges by adding H+ and H2O where needed.

18. Expressing redox reactions using half-equations
Fe2+(aq)+MnO4-(aq)  Fe3+(aq)+Mn2+(aq)

19. The activity series
The activity series ranks metals according the ease with which they undergo oxidation. Metals higher up on the series can replace lower metals in their salts. As reactivity corresponds with increasing ease of oxidation.