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OILRIG

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. REDOX.

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OILRIG

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  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 • CuSO4(aq) + Zn —> ZnSO4 (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. • MnO4-(aq) —> Mn 2+ (aq) • 2. Balance – both equal in this case. • 3. Balance O by adding water to the other side. • MnO4 -(aq) —> Mn 2+ (aq) + 4H2O (l) • 4. Balance H by adding H+ ions to other side • MnO4- (Aq) + 8 H+ (aq) —> Mn 2+ (aq) + 4 H2O (l)

  5. 5. Balance charge by adding electrons • Mn04- (aq) + 8 H+ (aq)+ 5e —> Mn 2+ (aq) + 4H2O (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) • MnO4- (aq)+ 8H+ (aq) + 5e —> Mn 2+ (aq) + 4 H2O(l) • (purple)( colourless) Fe 2+ (aq) —> Fe 3+ (aq) + e

  9. 2. Balanced redox equation • MnO4- + 8H+ + 5 Fe 2+ —> 5 Fe 3+ + Mn 2+ + 4 H2O • 3.Mole ratio • MnO4:Fe • 1:5 • 4.Work out number of moles of MNO4 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. • MnO4:Fe 2+ • 1:5 • So if MnO4 = 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 Cl2 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 —> H2 (g) We need 2 F ( 2 x 96,500C = 193, 000C) to produce 1 mole of H2 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 CuCl2 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.

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