Topic 9 Oxidation and reduction
9.1.1 Define oxidation and reduction in terms of electron loss and gain.
Oxidation: the loss of electrons
Reduction: the gain of electrons
9.1.2 Deduce the oxidation number of an element in a compound.
Always determine elements that never change their oxidation number then ensure the charge of whole molecule is right.
N.B. Atoms in elemental state have oxidation numbers of 0
9.1.3 State the names of compounds using oxidation numbers.
Example of this in iron, can be iron(II) or iron(III)
9.1.4 Deduce whether an element undergoes oxidation or reduction in reactions using oxidation numbers.
If an elements oxidation number decreases from reactants to products then it is being reduced, and vice versa for oxidation.
9.2.1 Deduce simple oxidation and reduction half-equations given the species involved in a redox reaction.
Example: Zn + Cu2+ → Zn2+ + Cu
Oxidation: Zn → Zn2+ + 2e
Reduction: Cu2+ + 2e → Cu
9.2.2 Deduce redox equations using half-equations.
Example: NO3- + Cu → NO + Cu2+
1. Balance atoms other than H and O
Cu → Cu2+
NO3- → NO
2. Balance O by adding H2O as needed
NO3- → NO +2H2O
3. Balance H by adding H+ as needed
4H+ + NO3- → NO +2H2O
4. Balance charges by adding e
Cu → Cu2+ + 2e
3e + 4H+ + NO3- → NO + 2H2O
5. Equalise electrons in half equations
3Cu → 3Cu2+ + 6e
6e + 8H+ + 2NO3- → 2NO + 4H2O
6. Add half equations together
3Cu + 8H+ + 2NO3- → 3Cu2+ + 2NO + 4H2O
9.2.3 Define the terms oxidizing agent and reducing agent.
Oxidising agent: accepts electrons and becomes reduced
Reducing agent: donates electrons and becomes oxidised
9.2.4 Identify the oxidizing and reducing agents in redox equations.
Same as figuring 9.1.4
9.3.1 Deduce a reactivity series based on the chemical behaviour of group of oxidizing and reducing agents.
A species is more reactive than another if it easily loses it electrons
e.g. Zn + Cu2+ → Zn2+ + Cu, Zn is more reactive than Cu
9.3.2 Deduce the feasibility of a redox reaction from a given reactivity series. For a redox reaction to
Oxidation: the loss of electrons
Reduction: the gain of electrons
9.1.2 Deduce the oxidation number of an element in a compound.
Always determine elements that never change their oxidation number then ensure the charge of whole molecule is right.
N.B. Atoms in elemental state have oxidation numbers of 0
9.1.3 State the names of compounds using oxidation numbers.
Example of this in iron, can be iron(II) or iron(III)
9.1.4 Deduce whether an element undergoes oxidation or reduction in reactions using oxidation numbers.
If an elements oxidation number decreases from reactants to products then it is being reduced, and vice versa for oxidation.
9.2.1 Deduce simple oxidation and reduction half-equations given the species involved in a redox reaction.
Example: Zn + Cu2+ → Zn2+ + Cu
Oxidation: Zn → Zn2+ + 2e
Reduction: Cu2+ + 2e → Cu
9.2.2 Deduce redox equations using half-equations.
Example: NO3- + Cu → NO + Cu2+
1. Balance atoms other than H and O
Cu → Cu2+
NO3- → NO
2. Balance O by adding H2O as needed
NO3- → NO +2H2O
3. Balance H by adding H+ as needed
4H+ + NO3- → NO +2H2O
4. Balance charges by adding e
Cu → Cu2+ + 2e
3e + 4H+ + NO3- → NO + 2H2O
5. Equalise electrons in half equations
3Cu → 3Cu2+ + 6e
6e + 8H+ + 2NO3- → 2NO + 4H2O
6. Add half equations together
3Cu + 8H+ + 2NO3- → 3Cu2+ + 2NO + 4H2O
9.2.3 Define the terms oxidizing agent and reducing agent.
Oxidising agent: accepts electrons and becomes reduced
Reducing agent: donates electrons and becomes oxidised
9.2.4 Identify the oxidizing and reducing agents in redox equations.
Same as figuring 9.1.4
9.3.1 Deduce a reactivity series based on the chemical behaviour of group of oxidizing and reducing agents.
A species is more reactive than another if it easily loses it electrons
e.g. Zn + Cu2+ → Zn2+ + Cu, Zn is more reactive than Cu
9.3.2 Deduce the feasibility of a redox reaction from a given reactivity series. For a redox reaction to