Oxidation and reduction: redox reactions

In a nutshell

Redox reactions refers to the transfer of electrons. Oxidation and reduction both occur in a redox reaction. Displacement reactions are one type of redox reaction. To show where redox reactions occur ionic equations are used. 

Redox reactions

The term redox is derived from reduction and oxidation reactions. Oxidation is the loss of electrons whilst reduction is the gain of electrons. A good way to remember this is the mnemonic OIL RIG. Oxidation Is ​Loss and Reduction Is Gain. 

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Examples 

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​​$$magnesium\ +\ hydrogen \ ions \ \rightarrow\ magnesium\ ions\ +\ hydrogen\\ \ \\Mg(s)\ +\ 2H^+(aq)\ \rightarrow\ Mg^{2+}(aq)\ + H_{2}(g): Redox\ reaction\\ \$$​

​$$magnesium\ \rightarrow\ magnesium\ ions\ +\ 2\ electrons\\ \ \\ \\Mg(s)\ \rightarrow\ Mg^{2+}(aq)\ +\ 2e^-\ :Oxidation \ reaction\\ \$$

$$hydrogen\ ions+\ 2\ electrons\ \rightarrow hydrogen\\\ \ \\2H^+(aq)\ +\ 2e^-\ \rightarrow\ H_2(g)\ : Reduction \ reaction$$

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Displacement reactions

Displacement reactions are reactions where a more reactive metal displaces a less reactive metal in solution. This causes the less reactive metal to be "kicked out" of solution and the more reactive metal to take its place.

Example

​$$magnesium\ +\ copper\ sulfate\ \rightarrow\ magnesium\ sulfate\ +\ copper\\ \ \\Mg(s)\ +\ CuSO_4(aq)\ \rightarrow\ MgSO_4(aq)\ +\ Cu(s)$$​​

In this case $$Mg$$​ will displace $$Cu$$​, this is because it is more reactive and therefore $$Cu$$​ will be displaced.  

Ionic equations

Ionic equations remove ions which do not change in the reaction, these are known as spectator ions. This leaves the important parts of the reaction, which is where there is actual chemical change. As you can see in the example below, the $$SO_{4}^{2-}$$ ion is not included, this is because this does not undergo a reaction.

Example 

​$$magnesium + copper\ ions → magnesium\ ions + copper\\ \ \\Mg(s) + Cu^{2+} (aq) → Mg ^{2+} (aq) +Cu(s)$$ 

Ionic equations can be further broken down into ionic half equations. A good tip to ensure that this step is done correctly is to check the 'charge balance'. This is simply the amount of electrons present in both half equations at different sides of the arrow.

Examples 

​$$magnesium\ \rightarrow\ magnesium\ ions +\ 2\ electrons\\ \ \\Mg(s)\ \rightarrow\ Mg^{2+}(aq)\ + 2e^-$$​​

​​​$$copper\ ions\ +\ 2\ electrons\ \rightarrow\ copper\\ \ \\Cu^{2+}(aq)\ + 2e^-\ \rightarrow\ Cu(s)$$​​