2.3.4 Ionic Equations

• Oxidation numbers are a useful tool for naming compounds as some elements can exist with more than one oxidation number 
• For compound with two elements it is straight forward to name the compound 
• For example
  • PCl₃ is phosphorus(III) chloride or phosphorus trichloride 
  • PCl₅ is phosphorus(V) chloride or phosphorus pentachloride 
  • OF₂ is oxygen difluoride
  • O₂F₂ is dioxygen difluoride 
• In order to name a more complete compound we use Roman numerals for the element that has a variable oxidation number 
  
  • K₂CrO₄ potassium chromate(VI)

Answer:

Answer 1: copper(I) oxide:

The ox. no. of 1 O atom is -2 and Cu₂O has overall no charge so the ox. no. of Cu is +1

Answer 2: manganese(II) sulfate:

The charge on the sulfate ion is -2, so the charge on Mn and ox. no. is +2

Answer 3: sodium chromate(VI):

The ox. no. of 2 Na atoms is +2 so CrO₄ has an overall -2 charge, so the ox. no. of Cr is +6

Answer 4: potassium manganate(VII):

The ox. no. of a K atom is +1 so MnO₄ has overall -1 charge, so the ox. no. of Mn is +7

Answer 5: sodium dichromate(VI):

The ox. no. of 2 Na atoms is +2 so Cr₂O₇ has an overall -2 charge, so the ox. no. of Cr is +6. To distinguish it from CrO₄ we use the prefix di in front of the anion

Metals 

• Metals, in general, will form positive ions by losing electrons 
• Therefore, they are oxidised and the oxidation number increases
• Example 1:
  • When sodium reacts with water, sodium hydroxide and hydrogen gas is formed

2Na (s) + H₂O (l) → 2NaOH (aq) + H₂ (g) 

• The oxidation number of sodium changes from 0 to +1 
• Example 2:
  • When magnesium reacts with hydrochloric acid, magnesium chloride and hydrogen gas is formed 

Mg (s) + 2HCl (l) → MgCl₂ (aq) + H₂ (g)

• The oxidation number of magnesium changed from 0 to +2

Non-metals 

• Non-metals, in general, will form negative ions by gaining electrons 
• Therefore, they are reduced and the oxidation number decreases
• Example:
  • When sodium reacts with oxygen, sodium oxide is formed 

4Na (s) + O₂ (g) → Na₂O (s) 

• The oxidation number of oxygen changes from 0 to -2

Balancing full ionic equations

• Balancing equations using redox principles is a useful skill and is best illustrated by following an example
• It is important to follow a methodical step-by-step approach so that you don't get lost:

Answer

Step 1: Write the unbalanced equation and identify the atoms which change in oxidation number

Step 2: Deduce the oxidation number changes

Step 3: Balance the oxidation number changes

Step 4: Balance the charges

Step 5: Finally, balance the atoms