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Summary Notes 3.1.7 - Oxidation, reduction and redox equations

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Summary Notes 3.1.7 - Oxidation, reduction and redox equations

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Taylor Notes Redox Notes
Oxidation, Reduction and Redox
Oxidation is the loss of electrons

Reduction is the gain of electrons

Redox is where oxidation and reduction are happening at the same time

Oxidising agents gain electrons and oxide the other reactant

Reducing agents lose electrons and reduce the other reagent

Redox Reaction
Mg(s) + Cu2+  Mg2+ + Cu

Oxidation Numbers
The Oxidation Number of an atom shows the number of electrons which it has lost or gained because of it forming a compound
1. The Oxidation Number of an uncombined Element is 0
2. Certain Elements have a fixed Oxidation Number
 All Group 1 elements are +1
 All Group 2 elements are +2
 Hydrogen is always +1 (except in hydrides)
 Fluorine is always -1
 Oxygen is always -2 (except in peroxides, superoxides and when combined with fluorine)
 Chlorine is always -1 (except when combined with fluorine and oxygen)
3. The Sum of Oxidation Numbers in a Compound is always 0
4. The Sum of Oxidation Numbers in an ion always adds up to the overall charge on an ion

Oxidation Number of Sulphur in the Following:

SO2 = H2S = S2O32- = S2O82- =

Half Equations
When a redox reaction occurs, one substance gains electrons and the other substance loses electrons

These two processes can be considered separately

The reaction between copper and silver nitrate: Cu(s) + 2AgNO3(aq)  Cu(NO3)2(aq) + 2Ag(s)

The Ionic Equation is Cu + 2Ag+  Cu2+ + 2Ag

The Half Equation for the Oxidation Taking Place: Cu  Cu2+ + 2e-

The Half Equation for the Reduction Taking Place: 2Ag+ + 2e-  2Ag

How to Construct More Complex Half Equations
1. Work out the oxidation number of the atom being oxidised/reduced
2. Add electrons on the appropriate side to account for the change in oxidation number
3. Balance oxygen atoms with water molecules
4. Balance hydrogen atoms with H+ ions

Cr2O72-  Cr3+
Combining Half Equations
1. Ensure you have a reduction and an oxidation (not two of the same)
2. Multiply either process so that both oxidation and reduction include the same number of electrons
3. Cancel the electrons so that they do not appear in the overall equation
4. Cancel across any other species that appear on both sides

I2 + 2e- ⇌ 2I- and SO42- + 4H+ + 2e- ⇌ SO2 + 2H2O goes to I2 + SO2 + 2H2O + 2e-  2I- + SO42- + 4H+ + 2e-

Disproportionation Reactions
A disproportionation reaction is where an element undergoes oxidation and reduction simultaneously in a reaction

Hg2Cl2  Hg + HgCl2

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