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Redox

Redox. Principles and application. Course outline. Oxidation and reduction Redox reactions explain oxidation and reduction as an electron transfer process calculate oxidation numbers identify and name oxidants and reductants in equations

phoebe-sykes
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Redox

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  1. Redox Principles and application

  2. Course outline Oxidation and reduction Redox reactions • explain oxidation and reduction as an electron transfer process • calculate oxidation numbers • identify and name oxidants and reductants in equations • identify oxidation-reduction reactions using oxidation numbers Oxidation and reduction Redox reactions • describe, write equations for and interpret observations for: • metal displacement reactions • halogen displacement reactions • write balanced simple redox equations (metal/metal ion, metal/hydrogen ion and halogen/halide ion)

  3. Course outline Oxidation and reduction Redox Applications – Electrolytic Cells • describe and explain how an electric current is conducted in an electrolytic cell • describe and explain the following during the operation of an electrolytic cell: • anode processes • cathode processes • role of the electrolyte • direction of ion migration • direction of electron flow in external circuit • electrode product prediction for molten metal halides only • predict and name the electrode products for the electrolysis of molten metal halides only.

  4. OXIDATION Used to be viewed as • Gaining oxygen 2 Mg + O2→ 2 MgO Magnesium gains oxygen • Losing hydrogen CH4 + 2 S → C + 2 H2S Carbon loses hydrogen

  5. REDUCTION Used to be viewed as • Losing oxygen FeO + C → Fe + CO Iron loses oxygen (Note: carbon also gains oxygen = oxidation) • Gaining hydrogen CH4 + 2 O2→ CO2 + 2 H2O Oxygen gains hydrogen (Note: carbon also loses hydrogen and gains oxygen = oxidation)

  6. It was observed that very often when one substance was being oxidised another was being reduced at the same time.

  7. Modern view of oxidation and reduction Oxidation is defined as losing of electrons LEO Reduction is defined as gaining of electrons GER

  8. Modern view of oxidation and reduction Oxidation Is Loss of electrons Reduction Is Gain of electrons OIL RIG

  9. A redox reaction is anelectron transfer process

  10. REDOX = electron transfer 4e- 4e- 4e-` 4e- Reducing agentOxidising agent Loses electrons 2 Fe (ON=0) 4e- + 2 Fe2+O2 – 4e- 4e- O2 (ON=0) Gains electrons 4e- 4e- 4e- 4e-

  11. REDOX = electron transfer 2e- 2e- 2e-` 2e- Reducing agentOxidising agent Loses electrons 2 Na (ON=0) 2e- + 2 Na+Cl– 2e- 2e- Cl2 (ON=0) Gains electrons 2e- 2e- 2e- 2e-

  12. An oxidising agent • Oxidises another substance by taking electrons from it • Is also called the oxidant • Gains electrons, and therefore • Becomes reduced

  13. A reducing agent • Reduces another substance by donating electrons to it • Is also called the reductant • Loses electrons, and therefore • Becomes oxidized

  14. A redox equation analysed Electron transfer S gains electrons Is reduced Is the oxidant Mg + S → Mg2+ + S2– 0 0 –2 +2 Mg loses electrons Is oxidized Is the reductant

  15. What is the ON of P in PH3 ONP + 3ONH = 0 (neutral) ONP + 3 (+1) = 0 ONP + 3 = 0 ONP = –3

  16. ON of P in H3PO4 3H + P + 4O = 0 3(+1) + P + 4(–2) = 0 (+3) + P + (–8) = 0 P + (–5) = 0 P = +5

  17. ON of P in H3PO3 3H + P + 3O = 0 3(+1) + P + 3(–2) = 0 (+3) + P + (–6) = 0 P + (–3) = 0 P = +3

  18. ON of Cr in CrO3 Cr + 3O = 0 Cr + 3(– 2) = 0 Cr + (– 6) = 0 Cr = + 6

  19. ON of Cr in Cr2O72- 2Cr + 7O = – 2 2Cr + 7(– 2) = – 2 2Cr = 14 – 2 2Cr = + 12 Cr = + 6

  20. ON of Cr in CrO42- Cr + 4O = – 2 Cr + 4(– 2) = – 2 Cr = 8 – 2 = + 6

  21. ON of Cr in (NH4)2CrO4 2NH4 + Cr + 4O = 0 2(+1) + Cr + 4(–2) = 0 Cr +2 – 8 = 0 Cr = + 6

  22. ON of C in CH4 C + 4H = 0 C + 4(+1) = 0 C = – 4

  23. ON of C in CH2O C + 2H + O = 0 C + 2(+1) +(– 2) = 0 C + 2 + (– 2) = 0 C = 0

  24. H e e e e e e O C e e e e e e H How can combined C have an ON of zero? H is less electronegative than C O is more electronegative than C C has ‘lost’ these 2 electrons C has ‘gained’ these 2 electrons Carbon has ‘lost’ 2 electrons to Oxygen but ‘gained’ 2 from Hydrogen

  25. ON of C in CH3OH C + 4H + O = 0 C + 4(+1) + (– 2) = 0 C + 4 + (– 2) = 0 C + 2 = 0 C = – 2

  26. ON of C in C6H6 6C + 6H = 0 6C + 6(+1) = 0 6C + 6 = 0 6C = – 6 C = – 1

  27. ON of Cl in HClO4 H + Cl + 4O = 0 (+1) + Cl + 4(– 2) = 0 (+1) + Cl + (– 8) = 0 Cl + (– 7) = 0 Cl = +7

  28. ON of Se in H2SeO3 2H + Se + 3O = 0 2(+1) + Se + 3(– 2) = 0 (+2) + Se + (– 6) = 0 Se + (– 4) = 0 Se = +4

  29. ON of Se in SeO3 Se + 3O = 0 Se + 3(– 2) = 0 Se + (– 6) = 0 Se = +6

  30. ON of Mn in MnO4 - Mn + 4O = – 1 Mn + 4(– 2) = – 1 Mn + (– 8) = – 1 Mn = +7

  31. ON of Mn in MnO4 2 – Mn + 4O = – 2 Mn + 4(– 2) = – 2 Mn + (– 8) = – 2 Mn = +6

  32. ON of Mn in MnO2 Mn + 2O = 0 Mn + 2(– 2) = 0 Mn + (– 4) = 0 Mn = +4

  33. ON of Mn in Mn2O3 2Mn + 3O = 0 2Mn + 3(– 2) = 0 2Mn + (– 6) = 0 2Mn = +6 Mn = +3

  34. In HClO4 the • ON of Cl is +7 • ON of O is – 2 • ON of H is +1 H + Cl + 4O = 0 +1 + Cl – 8 = 0 Cl = 8 – 1 = +7 With the aid of an electron-dot diagram, show this is consistent with the electronegativities of the atoms in each bond. EN: O = 3.4 Cl = 2.8 H = 2.2

  35. e e e e O e e e e e e e e e e e e H O O Cl e e e e e e e e e e e e O e e e e Cl has ‘lost’ these2 electrons And also these2 electrons H has ‘lost’ 1 electron ON = +1 As wellthese 2 electrons And this 1 electron Each O has ‘gained’ 2 electrons ON = – 2 Cl has ‘lost’ 7 electrons ON = +7

  36. In H2SO4 the • ON of S is +6 • ON of O is – 2 • ON of H is +1 2H + S + 4O = 0 +2 + S – 8 = 0 S = 8 – 2 = +6 With the aid of an electron-dot diagram, show this is consistent with the electronegativities of the atoms in each bond. EN: O = 3.4 S = 2.6 H = 2.2

  37. e e e e O e e e e e e e e e e e e O O H S e e e e e e e e e e e e O e e e e H S has ‘lost’ these2 electrons And also these2 electrons Each H has ‘lost’ 1 electron ON = +1 As well as this 1 electron And this 1 electron Each O has ‘gained’ 2 electrons ON = – 2 S has ‘lost’ 6 electrons ON = +6

  38. Write the half equation for the conversion of IO3– to I2 1. Write the ‘skeleton’ equation including the reactant and product species. Determine the oxidation state of each species, to decide which element is being oxidized/reduced. IO3– → I2 I + 3(–2) = –1 (+5)(–2) (0) so I = (+5) I is reduced

  39. Write the half equation for the conversion of IO3– to I2 2. Balance the numbers of atoms of the element that is being oxidized/reduced IO3– → I2 2IO3– → I2

  40. Write the half equation for the conversion of IO3– to I2 3. Balance the number of oxygen atoms by adding H2O to the side that needs more O IO3– → I2 2 IO3– → I2 2 IO3– → I2 + 6 H2O

  41. Write the half equation for the conversion of IO3– to I2 4. Balance the number of hydrogen atoms by adding H+ to the side that needs more H IO3– → I2 2 IO3– → I2 2 IO3– → I2 + 6 H2O 2 IO3– + 12 H+ → I2 + 6 H2O

  42. Write the half equation for the conversion of IO3– to I2 5. Balance the charge by adding electrons to the side that has more positive charge IO3– → I2 2 IO3– → I2 2 IO3– → I2 + 6 H2O 2 IO3– + 12 H+ → I2 + 6 H2O 2 IO3– + 12 H+ + 10e – → I2 + 6 H2O

  43. Write the half equation for the conversion of IO3– to I2 Now amend the equation for this reaction in BASIC solution Add 12 OH– to both sides of the half equation. 2 IO3– + 12 H++ 10e – → I2 + 6 H2O 12 OH– 12 OH–

  44. Write the half equation for the conversion of IO3– to I2 Now amend the equation for this reaction in BASIC solution The OH– and H+ neutralise 2IO3–+ 12H2O+ 10e –→ I2+6H2O +12 OH–

  45. Write the half equation for the conversion of IO3– to I2 Now amend the equation for this reaction in BASIC solution Cancel out excess H2O 6 2IO3–+ 12H2O+ 10e –→ I2+6H2O +12 OH– 2IO3–+ 6H2O+ 10e –→ I2 + 12 OH–

  46. Write the half equation for the conversion of Mn2+ to MnO4– 1. Write the ‘skeleton’ equation including the reactant and product species. Determine the oxidation state of each species, to decide which element is being oxidized/reduced. Mn2+ → MnO4– Mn + 4(–2) = –1 (+2) (+7)(–2) So Mn = +7 Mn is oxidized

  47. Write the half equation for the conversion of Mn2+ to MnO4– 2. Balance the numbers of atoms of the element that is being oxidized/reduced Mn2+ → MnO4– Mn is already balanced

  48. Write the half equation for the conversion of Mn2+ to MnO4– 3. Balance the number of oxygen atoms by adding H2O to the side that needs more O Mn2+ → MnO4– Mn2++4 H2O→ MnO4–

  49. Write the half equation for the conversion of Mn2+ to MnO4– 4. Balance the number of hydrogen atoms by adding H+ to the side that needs more H Mn2+ → MnO4– Mn2++4 H2O→ MnO4– Mn2++4 H2O→ MnO4–+ 8 H+

  50. Write the half equation for the conversion of Mn2+ to MnO4– 5. Balance the charge by adding electrons to the side that has more positive charge Mn2+ → MnO4– Mn2++4 H2O→ MnO4– Mn2++4 H2O→ MnO4– + 8 H+ Mn2+ +4 H2O→ MnO4–+ 8 H++ 5e –

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