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Do now!

Do now!. Can you answer the questions on the experimental sheet you did last lesson?. Objectives. To know how to carry out electrolysis experiments. To work out what happens to ions at each electrode. To be able to write successful half-equations. To predict the products of electrolysis.

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Do now!

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  1. Do now! Can you answer the questions on the experimental sheet you did last lesson?

  2. Objectives • To know how to carry out electrolysis experiments. • To work out what happens to ions at each electrode. • To be able to write successful half-equations. • To predict the products of electrolysis. • To do electrolysis calculations using the idea of a Faraday

  3. Stick it in!

  4. Electrolysis Lead bromide

  5. Copy please Electrode - an electrical conductor which carries charge to or from a liquid undergoing electrolysis. Electrolysis Electrolysis is the break-down of a substance by electricity Electrolyte- a molten or aqueous solution through which an electrical current can flow. Lead bromide

  6. Let’s try it!

  7. Electrolysis experiments • Electrolysis only happens in: • - molten ionic liquids or • - aqueous solutions containing ions. • There must be a complete circuit. • A lamp or ammeter shows that electricity is flowing around the circuit.

  8. Moving charges Can you stick the sheet in your book?

  9. Electrolysis of zinc chloride

  10. Draw the sentence • Electrolytes contain positive and negative ions. • During electrolysis, positive and negative electrodes are put into the electrolyte. • The positive electrode is called the anode. • The negative electrode is called the cathode. • The negative ions (called anions) are attracted to the anode. • At the anode, the negative ions lose electrons to become atoms/molecules. • The positive ions (called cations) are attracted to the cathode. • At the cathode, the positive ions gain electrons to become atoms/molecules

  11. Draw the sentence • Electrolytes contain positive and negative ions. • During electrolysis, positive and negative electrodes are put into the electrolyte. • The positive electrode is called the anode. • The negative electrode is called the cathode. • The negative ions (called anions) are attracted to the anode. • At the anode, the negative ions lose electrons to become atoms/molecules. • The positive ions (called cations) are attracted to the cathode. • At the cathode, the positive ions gain electrons to become atoms/molecules

  12. Cathode (-) (negative electrode) Positive ions go here (cations). As metal ions are positive, they go to the cathode. Ions gain electrons. They are reducedand become neutral atoms. Anode (+) (positive electrode) Negative ions go here (anions). As non-metal ions are negative, they go to the anode. Ions lose electrons. They are oxidised and become neutral atoms (which react together to form molecules). At the electrodes

  13. Common ions • Li+,Na+,K+, • Mg2+,Ca2+ , Zn2+ • Cu+,Cu2+,Fe2+, Fe3+, Al3+ • NH4+ (ammonium ion) • F-, Cl-, Br-, I- • O2-, S2- • OH- (hydroxide ion), CO32-, NO3- (nitrate ion), SO42- (sulphate ion)

  14. Questions! • Let’s try some easy questions

  15. Half equations • Show what happens at each electrode. • Are balanced equations. • Consider the electrolysis of copper chloride: Cu2+ + 2e-  Cu 2Cl- - 2e-  Cl2

  16. For solutions of highly reactive metals: Hydrogen gas, not the metal, is produced at the cathode. Electrolysis of solutions –Cathode

  17. The product at the anode depends on: The negative anions present in the solution. Electrolysis of solutions – Anode

  18. Electrolysis of aluminium oxide CO2

  19. Electrolysis The Faraday

  20. Objectives • recall that one Faraday represents one mole of electrons • calculate the amounts of the products of the electrolysis of molten salts and aqueous solutions

  21. The Faraday • A Faraday is one mole of electrons, and is equivalent to 96 500C (Coulombs) • A current of 1A = 1C per second flowing • For example, Cu2+ + 2e  Cu 1 mole of Cu2+ ions reacts with 2 Faradays of electrons, to produce 1 mole of Cu

  22. Quantity of electricity in coulombs = current in amps x time in seconds • Q (C) = I (A) x t (s)

  23. Example How much copper is deposited if a current of 0.2 Amps is passed for 2 hours through a copper(II) sulphate solution ?

  24. current of 0.2 Amps is passed for 2 hours At the cathode: Cu2+(aq) + 2e-  Cu(s) Q = I x t = 0.2 x (2 x 60 x 60) =1440 Coulombs 1 mole electrons = 96500 Coulombs So, 1440 / 96500 = 0.0149 moles of electrons (Faradays)

  25. Example moles of electrons passed through circuit = 0.01492 Cu2+(aq) + 2e-  Cu(s) • From equation, it takes two moles of electrons to form one mole of copper • moles copper = 0.01492 / 2 = 0.00746

  26. Example • moles Cu = 0.00746 • mass of Cu = moles x Ar = 0.00746 x 64 = 0.4775g of Cu deposited.

  27. ‘How To” Guide • Write out relevant half equation • Work out coulombs of electrons flowing (Q = It) • Convert C into moles of electrons (Faradays) (Q/96500) • Work out moles of product using ratio from equation • Convert into mass (mass = moles x Ar)

  28. In an electrolysis of sodium chloride solution experiment a current of 2 A was passed for 2 minutes. • (a) Calculate the volume of chlorine gas produced. • (b) What volume of hydrogen would be formed? • (c) In practice the measured volume of chlorine can be less than the theoretical value. Why?

  29. Electrode equations: • (-) cathode 2H+ + 2e-  H2 • (+) anode 2Cl- Cl2 + 2e • (a) Calculate the volume of chlorine gas produced. • Q = I x t, so Q = 2 x 2 x 60 = 240 C • 240 C = 240 / 96500 = 0.002487 mol electrons • this will produce 0.002487 / 2 = 0.001244 mol Cl2 (two electrons/molecule) • vol = mol x molar volume = 0.001244 x 24000 = 29.8 cm3 of Cl2

  30. (b) What volume of hydrogen would be formed? • 29.8 cm3 of H2 because two electrons transferred per molecule, same as chlorine. • (c) In practice the measured volume of chlorine can be less than the theoretical value. Why? • chlorine is moderately soluble in water and also reacts with the sodium hydroxide formed.

  31. In the electrolysis of molten sodium chloride 60 cm3 of chlorine was produced. • Calculate ... • (a) how many moles of were chlorine produced? • (b) what mass of sodium would be formed? • (c) for how long would a current of 3 A in the electrolysis circuit have to flow to produce the 60cm3 of chlorine?

  32. (a) how many moles of chlorine produced? • 60 / 24000 = 0.0025 mol Cl2

  33. (b) what mass of sodium would be formed? • from the electrode equations 2 mol sodium will be made for every mole of chlorine • so 0.0025 x 2 = 0.005 mol sodium will be formed. Ar(Na) = 23 • mass = mol x atomic or formula mass = 0.005 x 23 = 0.115g Na

  34. (c) for how long would a current of 3 A in the electrolysis circuit have to flow to produce the 60cm3 of chlorine? • To produce 0.0025 mol of Cl2 you need 0.005 mol of electrons • 0.005 mol electrons = 0.005 x 96500 coulombs = 482.5 C • Q = I x t, so 482.5 = 2 x t, therefore t = 482.5 / 3 = 161 s (to nearest second)

  35. Brine? Brine is salty water (sodium chloride solution)

  36. Electrolysis of brine

  37. Electrolysis of brine • Hydrogen is produced at the cathode • Chlorine is produced at the anode • The solution remaining is sodium hydroxide

  38. Electrolysis of brine • Hydrogen is produced at the cathode • Chlorine is produced at the anode • The solution remaining is sodium hydroxide Cathy’s Ankles (CatHy’s AnCl)

  39. Electrolysis of brine • Hydrogen is produced at the cathode • Chlorine is produced at the anode • The solution remaining is sodium hydroxide Copy please! Cathy’s Ankles (CatHy’s AnCl)

  40. Electrolysis of brine • Cathode (-) 2H+(aq) + 2e- H2(g) (SODIUM IS NOT FORMED (the sodium ion is more stable than the hydrogen ion in water H2O H+ + OH-)) • Anode (+) 2Cl-(aq) – 2e- Cl2(g)

  41. Electrolysis of brine • Cathode (-) 2H+(aq) + 2e- H2(g) (SODIUM IS NOT FORMED (the sodium ion is more stable than the hydrogen ion in water H2O H+ + OH-)) • Anode (+) 2Cl-(aq) – 2e- Cl2(g) Copy please!

  42. Chemicals from salt

  43. Chemicals from salt Copy please!

  44. Let’s try some questions!

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