Oxidation Numbers and States

This is part of Year 11 HSC Chemistry course under the topic of Predicting Reactions of Metals

HSC Chemistry Syllabus

  • Apply the definitions of oxidation and reduction in terms of electron transfer and oxidation numbers to a range of reduction and oxidation (redox) reactions
  • Construct relevant half-equations and balanced overall equations to represent a range of redox reactions
  • Predict the reaction of metals in solutions using the table of standard reduction potentials
  • Predict the spontaneity of redox reactions using the value of cell potentials

      Oxidation Numbers and States

      This video explains what oxidation numbers are states are. How can you calculate oxidation numbers and states? How do oxidation numbers and states explain to us whether oxidation or reduction has occurred? 

      What Are Oxidation States? 

      An oxidation state, also known as an oxidation number, is a theoretical charge that an atom would have if all its bonds were completely ionic. Understanding oxidation states is crucial in redox chemistry as it aids in identifying the substances getting oxidised or reduced – especially when the application of the OILRIG mnemonic (Oxidation is Loss, Reduction is Gain) is not straightforward. 

      Oxidation Changes:

      • An increase in the oxidation state of a substance signifies that it has been oxidised
      • A decrease in the oxidation state indicates the substance has been reduced

       

      Rules for Determining Oxidation Numbers

      Memorising a set of rules can greatly simplify the process of calculating oxidation numbers. Here is a quick overview:

      1. Elemental Form Rule: Atoms in their elemental form have an oxidation number of zero. Examples include Na (sodium), Mg (magnesium), and Cu (copper) in their neutral states, as well as diatomic gases like `O_2` (oxygen), `N_2` (nitrogen), and `Cl_2` (chlorine). 

      2. Monoatomic Ions: The oxidation state of a monoatomic ion is equal to its charge. Sodium ions (`Na^+`) and magnesium ions (`Mg^{2+}`), for example, have oxidation numbers of +1 and +2 respectively. 

      3. Polyatomic ions: Similar to monoatomic ions, polyatomic ions such as hydrogen (`OH^–`), carbonate (`CO_3^{2–}`), and phosphate (`PO_4^{3–}`) have oxidation numbers equal to their charges of -1, -2, and -3 respectively. 
      4. Electronegativity rule: When analysing the oxidation state of two covalently bonded atoms, we consider the bond as being purely ionic, with the more electronegative atom taken to have all the electrons in the bond(s). For example, in carbon dioxide `O=C=O`, the four electrons in each of the `C=O` double bonds are assigned to the oxygen atom, which is more electronegative than carbon. So, carbon would be considered to have 'lost' the four valence electrons it contributed, giving it an oxidation state of +4. When the covalently bonded atoms are identical and thus have the same electronegativity, the electrons are considered to be evenly split amongst these atoms. This is consistent with the elemental form rule for diatomic gases.

      Oxidation Numbers in Compounds 

      To understand how to determine the oxidation number in compounds, let's examine some examples: 

       

      Sulfate Ion (`SO_4^{2-}`)

      Following the rules, the sulfate ion as a whole has an oxidation state of –2. Oxygen typically has an oxidation state of –2, except in peroxides. With four oxygen atoms, we have a total of –8. Consequently, sulfur must exhibit an oxidation state of +6 to achieve the –2 state for the entire ion. 

       

      Dichromate Ion (`Cr_2O_7^{2–}`)

      The dichromate ion, with a total charge of –2 and seven oxygens, leads to oxygen having a total of –14. To balance this, each chromium must have an oxidation state of +6. 

       

      Nitrate Ion (`NO_3^–`)

      For the nitrate ion, with a charge of –1 and three oxygens contributing –6, nitrogen has an oxidation state of +5. 

       

      Permanganate Ion (`MnO_4^–`)

      The permanganate ion has a charge of –1. With four oxygens totalling –8, manganese's oxidation state is +7

       

      Applying Oxidation Numbers to Reactions

      In compounds, oxidation numbers must sum up to the compound's overall charge. For example, in calcium carbonate (`CaCO_3`), calcium has an oxidation number of +2, while the carbonate ion has –2, balancing the compound to a neutral charge. 

       

      Redox Example: Magnesium and Copper Sulfate Reaction 

      Consider the reaction between magnesium and copper sulfate forming magnesium sulfate and elemental copper:

      $$Mg(s) + CuSO_4(aq) \rightarrow MgSO_4(aq) + Cu(s)$$


      Here's how we apply oxidation states: 

       

      • Magnesium (Mg) starts with an oxidation state of 0.
      • Copper (Cu) in copper sulfate has an oxidation state of +2.
      • Sulfate (`SO_4^{2–}`) maintains an oxidation state of –2 throughout the reaction. 

       

      After the reaction:

       

      • Magnesium appears in magnesium sulfate (`MgSO_4`) with an oxidation state of +2
      • Copper is reduced to its elemental form with an oxidation state of 0.

       

      Conclusion: The increase from 0 to +2 for the oxidation state of magnesium means it has been oxidised, while the decrease from +2 to 0 for copper means it has been reduced. 

       

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