# 4.3 Balancing redox reactions

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## Balancing redox reactions

Half reactions can be used to balance redox reactions. We are going to use some worked examples to help explain the method.

Magnesium reduces copper (II) oxide to copper. In the process, magnesium is oxidised to magnesium ions. Write a balanced equation for this reaction.

1. $Mg\to M{g}^{2+}$

2. You are allowed to add hydrogen ions (H ${}^{+}$ ) and water molecules if the reaction takes place in an acid medium. If the reaction takes place in a basic medium, you can add either hydroxide ions (OH ${}^{-}$ ) or water molecules. In this case, there is one magnesium atom on the left and one on the right, so no additional atoms need to be added.

3. Charges can be balanced by adding electrons to either side. The charge on the left of the equation is 0, but the charge on the right is +2. Therefore, two electrons must be added to the right hand side so that the charges balance. The half reaction is now:

$Mg\to M{g}^{2+}+2{e}^{-}$

4. The reduction half reaction is:

$C{u}^{2+}\to Cu$

The atoms balance but the charges don't. Two electrons must be added to the right hand side.

$C{u}^{2+}+2{e}^{-}\to Cu$

5. No multiplication is needed because there are two electrons on either side.

6. $Mg+C{u}^{2+}+2{e}^{-}\to M{g}^{2+}+Cu+2{e}^{-}$ (The electrons on either side cancel and you get...)

$Mg+C{u}^{2+}\to M{g}^{2+}+Cu$

7. In this case, it is.

Chlorine gas oxidises Fe(II) ions to Fe(III) ions. In the process, chlorine is reduced to chloride ions. Write a balanced equation for this reaction.

1. $F{e}^{2+}\to F{e}^{3+}$

2. There is one iron atom on the left and one on the right, so no additional atoms need to be added.

3. The charge on the left of the equation is +2, but the charge on the right is +3. Therefore, one electron must be added to the right hand side so that the charges balance. The half reaction is now:

$F{e}^{2+}\to F{e}^{3+}+{e}^{-}$

4. The reduction half reaction is:

$C{l}_{2}\to C{l}^{-}$

The atoms don't balance, so we need to multiply the right hand side by two to fix this. Two electrons must be added to the left hand side to balance the charges.

$C{l}_{2}+2{e}^{-}\to 2C{l}^{-}$

5. We need to multiply the oxidation half reaction by two so that the number of electrons on either side are balanced. This gives:

$2F{e}^{2+}\to 2F{e}^{3+}+2{e}^{-}$

6. $2F{e}^{2+}+C{l}_{2}\to 2F{e}^{3+}+2C{l}^{-}$

7. The equation is balanced.

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