Ionic equations and half equations

In a nutshell

Different ionic compounds can react to form new products when molten or in aqueous solution. Net ionic equations can be written to show how the reaction works.

State symbols

When ionic compounds are dissolved in solution or are molten they are able to move freely. In solid ionic compounds, the ions are unable to move freely due to the strong electrostatic forces of attraction between oppositely charged ions. Therefore, solid ionic compounds are not involved in reactions.

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Writing ionic equations

To write a net ionic equation from the full equation, it is important to distinguish between molecules and ionic compounds in an equation:

• A molecule is a compound made up of non-metal atoms joined by covalent bonds. Molecules are not ionic.
  
• To find the ionic compounds, look for the compounds made up of non-metals and metals.

PROCEDURE

Example

Write out the ionic equation for the following reaction:

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​$$sodium~hydroxide+ copper ~sulfate\rightarrow sodium~sulfate+copper~hydroxide\\ \ \\ 2NaOH(aq)+CuSO_4(aq)\rightarrow Na_2SO_4(aq)+Cu(OH)_2(s)$$

Everything except for copper hydroxide is aqueous so write out the ions separately for the aqueous ions but keep copper hydroxide the same:

$$2Na^+(aq) + 2 OH^–(aq) + Cu^{2+}(aq) + SO_4^{2–}(aq)\rightarrow 2 Na^+(aq) + SO_4^{2–}(aq) + Cu(OH)_2(s)$$

Cross out the ions that are the same on both sides (spectator ions):

​​$$\cancel {2Na^+ }(aq)+ 2 OH^–(aq) + Cu^{2+}(aq) + \cancel{SO_4^{2–}}(aq)\rightarrow \cancel{2 Na^+}(aq) + \cancel{SO_4^{2–}}(aq) + Cu(OH)_2(s)$$

The ionic equation is:

$$\underline{ 2 OH^– (aq)+ Cu^{2+}(aq) \rightarrow Cu(OH)_2(s)}$$

Half equations 

Half equations can be written for redox reactions. 

Oxidation

Oxidation is the loss of electrons and gain of oxygen.

Example

​$$sodium ~atom\rightarrow sodium ~ion+1~electron\\ \ \\ Na\rightarrow Na^++e^-$$​​

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Reduction 

Reduction is the gain of electrons and loss of oxygen.

Example

​$$chlorine~atom+ 1~electron\rightarrow chloride ~ion\\ \ \\ \frac 12 Cl_2 +e^- \rightarrow Cl^-$$​​

Balancing half equations

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PROCEDURE

Example

Write down the half equation for the formation of a magnesium ion $$(Mg^{2+})$$ from a magnesium atom $$(Mg)$$.

Write out the equation with the magnesium atom (reactant) on the left-hand side and magnesium ion (product) on the right-hand side:

$$magnesium ~atom\rightarrow magnesium ~ion\\ \ \\Mg\rightarrow Mg^{2+}$$

There is one magnesium atom on the left-hand side and one magnesium ion on the right-hand side so the atoms and ions are already balanced. 

Next, electrons need to be added to balance the charge. This is an oxidation reaction so electrons are added to the right-hand side to show the loss of electrons:

$$\underline{magnesium~atom\rightarrow magnesium~ion+2~electrons }\\ \ \\\underline{Mg\rightarrow Mg^{2+} + 2e^-}$$​​

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The magnesium atom is neutral, giving an overall charge of zero on the left-hand side. The charge of the magnesium ion $$(Mg^{2+})$$​ is balanced with the charge of the two electrons $$(2e^-)$$, giving an overall charge of zero on the right-hand side too. 
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