1. 3.1.7 Redox

2. A redox reaction is one in which both reduction and oxidation take place at the same time. The original definition of oxidation was the formation of an oxide by gaining oxygen. Then this was expanded to include the loss of hydrogen as oxidation. However we now use the more general definitions of: Reduction is GAIN of ELECTRONS and Oxidation is LOSS of ELECTRONS.

3. In the thermite reaction, the reduction of iron(III) oxide occurs as aluminium is oxidised: Fe 2 O 3 + 2Al → Al 2 O 3 + 2Fe The aluminium is described as the reducing agent as it has caused the reduction of the iron(III) oxide. In a redox reaction, the reducing agent gets oxidised, and the oxidising agent gets reduced.

4. Displacement reactions in aqueous solution are examples of redox reactions. For example: Fe(s) + CuSO 4 (aq) → FeSO 4 (aq) + Cu(s)

5. It is easier to understand the redox if the equation is written as an ionic equation. To construct an ionic equation: ① Split any aqueous ionic compound into its ions (leave all molecules, atoms and solids as they are). Fe (s) + Cu 2+ (aq) + SO 4 2- (aq) → Fe 2+ (aq) + SO 4 2- (aq) + Cu (s)

6. ② Cancel any ions that do not change. These are known as spectator ions. Fe (s) + Cu 2+ (aq) + SO 4 2- (aq) → Fe 2+ (aq) + SO 4 2- (aq) + Cu (s) Becomes: Fe(s) + Cu 2+ (aq) → Fe 2+ (aq) + Cu(s)

7. This makes it more clear that the iron has lost electrons to become more positive (been oxidised) and the copper(II) ions have gained electrons to become less positive (been reduced). Note that ionic equations balance for both atoms and charge. The electron transfer is even clearer if the ionic equation is split into two half equations: Fe (s)→ Fe 2+ (aq) + 2e - oxidation Cu 2+ (aq) + 2e - → Cu (s)reduction

8. Oxidation States The oxidation state of an element in a compound is the charge on its ion or the charge that it would have if it existed as a solitary simple ion without bonds to other species. There are some fixed oxidation states. SpeciesOxidation state uncombined elements0 Group 1 metals in compounds+1 Group 2 metals in compounds+2 oxygen in compounds, except peroxides-2 hydrogen in compounds, except metal hydrides+1 hydrogen in metal hydrides oxygen in peroxides (e.g. H 2 O 2 )

9. These values can be used to calculate the oxidation state of an atom in a compound or complex ion. In a compound the total of the oxidation states will be zero, as the compound will be electrically neutral. In an ion the total of the oxidation states will be the same as the charge on the ion.

10. Example 1: what is the oxidation state of sulfur in sulfur trioxide SO 3 ? Answer: The overall charge is zero and each oxygen has a charge of -2. The oxygen ‘atoms’ have a total charge of -6. The sulfur has a charge (oxidation state) of +6

11. Example 2: what is the oxidation state of chromium in Cr 2 O 7 2- ? Answer: The overall charge is -2 and each oxygen has a charge of -2. The oxygen ‘atoms’ have a total charge of -14 so the two chromium ‘atoms’ have a charge +12. Therefore each chromium has a charge (oxidation state) of +6.

12. Redox reactions can be summarised as shown: oxidation X 3- X 2- X-X- XX+X+ X 2+ X 3+ reduction When oxidation occurs, a species becomes more positive (or less negative). When reduction occurs, a species becomes more negative (or less positive).

13. Half Equations for redox reactions In a half equation: Only one element changes oxidation state. The equation balances for atoms The equation balances for charge The equation is cancelled down to its simplest form

14. There a simple rules to construct a half equation: Identify the element that changes oxidation state and ensure that it is balanced. Add electrons to account for the oxidation state change. Balance oxygen using H 2 O. Balance hydrogen using H +.

15. Example: Deduce the half equation for the reduction of NO 3 - ions to NO in acidic conditions. Nitrogen is changing oxidation state from +5 to +2. (i.e. gaining 3 electrons) NO 3 - → NO(nitrogen balances) 3 electrons added NO 3 - + 3 e - → NO Use H 2 O to balance the oxygen atoms NO 3 - + 3 e - → NO + 2H 2 O(oxygen balances) Use H + to balance the hydrogen atoms NO 3 - + 4H + + 3 e - → NO + 2H 2 O(hydrogen balances) You should finally check that both atoms and charges are balanced.

16. Construction of an ionic equation from two half equations The process is: Ensure that equal numbers of electrons are gained by the oxidising agent and lost by the reducing agent. Add together the two half equations. Where the same species appears on both sides of the equation cancel out the correct number.

17. Example: Cu  Cu 2+ + 2e - HNO 3 + H + + e -  NO 2 + H 2 O The Cu atom loses 2 electrons, so the half equation for the reduction of HNO 3 must be doubled. Add together the two half equations. The electrons are the only species to cancel out here. Cu  Cu 2+ + 2e - 2HNO 3 + 2H + + 2e -  2NO 2 + 2H 2 O Cu + 2HNO 3 + 2H + + 2e -  Cu 2+ + 2NO 2 + 2H 2 O + 2e - Cu + 2HNO 3 + 2H +  Cu 2+ + 2NO 2 + 2H 2 O

18. Splitting an ionic equation into half equations Work out the oxidation states to identify the oxidised and reduced species. Construct a half equation for each species.

19. Example: Deduce the two half equations for the redox reaction between copper and cold dilute nitric acid, as given by the equation below: 3Cu + 8H + + 2NO 3 -  3Cu 2+ + 2NO + 4H 2 O Answer: 3Cu + 8H + + 2NO 3 -  3Cu 2+ + 2NO + 4H 2 O 0 +1 +5 -2 +2 +2 -2 +1 -2 The copper atoms have been oxidised and the nitrate ions have been reduced. 3Cu  3Cu 2+ + 6e - Simplifies to Cu  Cu 2+ + 2e - 8H + + 2NO 3 - + 6e -  2NO + 4H 2 O Simplifies to 4H + + NO 3 - + 3e -  NO + 2H 2 O