1. OILRIG Oxidation is loss of electrons. Reduction is gaining electrons. REDOX – oxidation and reduction happening together. Example Mg(s) —> Mg 2+ (aq) + 2e Oxidation Cu 2+ (aq) + 2 e —> Cu (s) Reduction Cl 2 (g) + 2e —> 2Cl - (aq) Reduction

2. REDOX CuSO 4 (aq) + Zn —> ZnSO 4 (aq) + Cu (s) The Cu ions gain electrons to form Cu atoms Cu 2+ (aq) + 2e —> Cu (s) Reduction The Zn atoms loose electrons forming Zn ions Zn (s) —> Zn 2+ (aq) + 2e REDOX ( we combine the oxidation and Reduction equations) Cu 2+ (aq) + 2e + Zn (s) —> Cu (s) + Zn 2+ (aq) + 2e We rewrite eq without electrons since they cancel each other out.

3. An oxidising agent is a substance that accepts electrons. In the previous example the Cu 2+ ions are the oxidising agent. A reducing agent is one that donates electrons. The Zn atoms act as a reducing agent.

4. More complex ion electron equations! Example ( data book p 11) Permanganate ions reducing to form Manganese ions. 1. Write formula of each ion. MnO 4 - (aq) —>Mn 2+ (aq) 2. Balance – both equal in this case. 3. Balance O by adding water to the other side. MnO 4 - (aq) —> Mn 2+ (aq) + 4H 2 O (l) 4. Balance H by adding H + ions to other side MnO 4 - (Aq) + 8 H + (aq) —> Mn 2+ (aq) + 4 H 2 O (l)

5. 5. Balance charge by adding electrons Mn0 4 - (aq) + 8 H + (aq)+ 5e —> Mn 2+ (aq) + 4H 2 O (l) + Total charge on LHS Total charge on RHS 1 - + 8 + = 7+ 2+ So if we add 5e to LHS both side now have a charge of 2+

6. Volumetric Titrations This is when we work out the volume or concentration of an acid required to neutralise a fixed conc. / volume of an alkali. C1xV1 = C2xV2 Example What is the conc. of HCl if 25 cm3 neutralises 20 cm3 of 2 mol/l NaOH? C1xV1= C2xV2 C1 = C2xV2/V1 = 2 x 0.02/ 0.025 = 1.6 mol/l

7. Redox Titrations We use redox titration to find out the volume or concentration of a reducing agent. We need to know: Accurate volumes of reactants. Concentration of oxidising agent. Balanced redox equation. Recognisable end pioint – e.g.colour change!

8. Example of Redox Titration Calculate the concentration of Iron(ii) sulphate solution if 20 cm3 of it react with 24 cm3 of Potassium permanganate – concentration 0.02 mol/l. 1. Get redox equation ( p11) MnO 4 - (aq)+ 8H + (aq) + 5e —> Mn 2+ (aq) + 4 H 2 O(l) (purple)( colourless) Fe 2+ (aq) —> Fe 3+ (aq) + e

9. 2. Balanced redox equation MnO 4 - + 8H + + 5 Fe 2+ —> 5 Fe 3+ + Mn 2+ + 4 H 2 O 3.Mole ratio MnO 4 :Fe 1:5 4.Work out number of moles of MNO 4 used M = C xV = 0.02 x 0.024 = 0.00048 ( 4.8 x 10 –4 )

10. 5. Use mole ratio to work out number of moles of Iron used. MnO 4 :Fe 2+ 1:5 So if MnO 4 = 0.00048 moles, Fe 2+ will be; 0.00048 x 5 = 0.0024 moles We can now find the concentration of the Fe 2+ solution; C = M/V = 0.0024/ 0.02 = 0.12 mol/l

11. End Point We can identify the end point using the colour change – when all the iron ions have been used up there will be excess permanganate ions and so we will see a purple colour. We call this a self indicating reaction – using one of the reactants to indicate end point.

12. Quantative Electrolysis Electrolysis – breaking down a compound using electricity. Oxidation occurs at the at the positive electrode. Reduction occurs at the negative electrode. In the redox reaction – the total number of electrons lost must be the same as the total number of electrons gained. 1 mole of electrons requires 96,500 C of charge. ( This is also known as 1 Faraday = 1F)

13. We can calculate the quantity of electricity required to produce 1 mole of a product – in electrolysis. Q = I xT Q = quantity of electricity ( coulombs) I = current( amps) T = time ( s)

14. Worked Example Calculate the mass of Sodium produced in the electrolysis of NaCl – using a current of 10 A for 32 mins. Q = IT = 10 x ( 32 x 60) 10 x 1920 = 19200 C Na + (aq) + e —> Na(s) To produce 1 mole of Sodium( 23g) we need 1 mole of electrons – 1 F = 96,500C.

15. Therefore if we have 19200C ( from Q=IT) We will produce 19200/96500 x 23 = 4.57g of Na Example 2 Calculate the mass of Cu produced in the electrolysis of Cu Cl 2 if a current of 20A is passed for 25 minutes. Q = IT = 20 x ( 25 x 60 ) = 20 x 1500. = 30000C

16. Cu 2+ (aq) + 2e —> Cu(s) 1 mole of Cu ( 63g) requires 2 moles of electrons = 2F = 2 x 96,500C = 193,000C. Therefore – 30000C ( from Q=IT) will produce: 30000/193,000 x 63 = 9.79g of Cu Example 3 Calculate the volume of Hydrogen gas produced( molar volume 24l) when a current of 15A is passed through a solution of sulphuric acid for 25 minutes.

17. Q = IT = 15 x ( 25 x 60 ) = 15 x 1500 = 22500C 2H + (aq) + 2e —> H 2 (g) We need 2 F ( 2 x 96,500C = 193, 000C) to produce 1 mole of H 2 gas – 24 l

18. Therefore - 22500C will produce : 22500/193,000 x 24 = 2.79litres. Example 4 How long will it take to produce 6.3 g of Cu at the positive electrode in the electrolysis of CuCl 2 using a current of 10A? Rearrange Q = IT to find T T = Q/I

19. I mole of Cu requires 2 mole of electrons = 2F = 2 x 96,500C. ( 193,00C) Cu 2+ (aq) + 2e —> Cu (s) To find 1 mole = 63g of Cu : T = Q/I = 193,000/10 = 19300s. For 6.3 g = 6.3/63 = 0.1 moles. 0.1 x 19300 = 1930 seconds.