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Understanding Redox Reactions and Oxidation States

Learn about redox reactions, oxidation and reduction, and how to assign oxidation numbers. Discover the role of oxidizing and reducing agents in chemical equations.

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Understanding Redox Reactions and Oxidation States

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  1. Redox Equations • Assign oxidation numbers to atoms in elements, compounds and ions. • Understand that a Roman numeral can be used to indicate oxidation states. • Writing formulae using oxidation numbers. • Describe oxidation and reduction in terms of electron transfer and oxidation number. • Recognise that metals form positive ions by losing electrons – increasing their oxidation number. • Recognise that non-metals form negative ions by gaining electrons – decreasing their oxidation number. • Describe redox reactions of metals with either dilute hydrochloric or sulfuricacid. • Interpret and make predictions from equations regarding oxidation numbers and electron loss/gain.

  2. REDOX REACTIONS OXIDATION AND REDUCTION IN TERMS OF ELECTRONS REDOXWhen reduction and oxidation take place OXIDATIONRemoval (loss) of electrons ‘OIL’ species will get less negative or more positive REDUCTIONGain of electrons ‘RIG’ species will become more negative or less positive

  3. REDOX REACTIONS OXIDATION AND REDUCTION IN TERMS OF ELECTRONS REDOXWhen reduction and oxidation take place OXIDATIONRemoval (loss) of electrons ‘OIL’ species will get less negative or more positive REDUCTIONGain of electrons ‘RIG’ species will become more negative or less positive REDUCTION in O.S. Species has been REDUCED go down the ladder ! INCREASE in O.S.Species has been OXIDISED go up the ladder !

  4. Redox Equations Using oxidation states to identify what's been oxidised and what's been reduced • Oxidation involves an increase in oxidation state • Reduction involves a decrease in oxidation state

  5. Example 1: the reaction between magnesium and hydrochloric acid Have the oxidation states of anything changed?   The magnesium's oxidation state has increased - it has been oxidised. The hydrogen's oxidation state has fallen - it has been reduced. The chlorine is in the same oxidation state on both sides of the equation - it hasn't been oxidised or reduced.

  6. Example 2: The reaction between sodium hydroxide and hydrochloric acid is: Nothing has changed. This isn't a redox reaction.

  7. This is a disproportionation reaction. Example 3: The chlorine is the only thing to have changed oxidation state. Has it been oxidised or reduced? A reaction is one in which a single substance is both oxidised and reduced is called a disproportionation reaction

  8. DEFINITIONS OF OXIDATION AND REDUCTION (REDOX) Oxidation and reduction in terms of oxygen transfer Definitions Oxidation is gain of oxygen. Reduction is loss of oxygen.

  9. 2- 2- 0 4+ 3+ 2- 2+ the extraction of iron from its ore reduction and oxidation are going on side-by-side, this is known as a redox reaction Oxidising agents give oxygen to another substance. Reducing agents remove oxygen from another substance. iron(III) oxide is the oxidising agent. the carbon monoxide is the reducing agent.

  10. Oxidation and reduction in terms of hydrogen transfer • Definitions • Oxidation is loss of hydrogen. • Reduction is gain of hydrogen. • these are exactly the opposite of the oxygen definitions. • ethanol can be oxidised to ethanal:

  11. Oxidation and reduction in terms of electron transfer Definitions Oxidation is loss of electrons. Reduction is gain of electrons. :

  12. Copper Oxide & Magnesium . rewrite this as an ionic equation the oxide ions are spectator ions

  13. An oxidising agent oxidises something else. Oxidation is loss of electrons (OIL RIG). That means that an oxidising agent takes electrons from that other substance. So an oxidising agent must gain electrons. Or you could think it out like An oxidising agent oxidises something else. That means that the oxidising agent must be being reduced. Reduction is gain of electrons (OIL RIG). So an oxidising agent must gain electrons.

  14. Electron-half-equations What is an electron-half-equation? When magnesium reduces hot copper(II) oxide to copper, the ionic equation for the reaction is:

  15. The reaction between chlorine and iron(II) ions Chlorine gas oxidises iron(II) ions to iron(III) ions. In the process, the chlorine is reduced to chloride ions Starting with the chlorine . . . .

  16. Now you repeat this for the iron(II) ions . . . . they are oxidised to iron(III) ions. Write this down:

  17. Combining the half-reactions to make the ionic equation for the reaction What we've got at the moment is this:

  18. the iron reaction will have to happen twice for every chlorine molecule that reacts. . . . Allow for that, and then add the two half-equations together.

  19. Page 32 - 33 Extension . . . Divide the equations in 1 & 2 into half equations

  20. Zn → Zn 2+ + 2e- 2 Ag+ + 2e- → 2 Ag Pb 4+ + 2e- → Pb 2+ 2 Cl-→ Cl2+ 2e- 2K → K + + 2e- H2 + 2e- → 2 H- Mg → Mg 2+ + 2e- 2H + + 2e- → H2

  21. P 3+ → P 5+ + 2e- 3 H -→ 3 H + + 6e- 2 O2 + 8e-→ 4 O 2-

  22. 1. The addition of sodium hydroxide produces a gelatinous green precipitate with iron(II) solution and a brown precipitate with iron(III) solution. On standing, oxidation causes the iron (II) hydroxide to turn a brown-yellow colour due to gradual formation of iron(III) hydroxide. Fe2+(aq) + 2OH–(aq) Fe(OH)2(s) 2. The thiocyanate ion gives a deep red colour with iron(III) but should give virtually no colour with iron(II). However, unless it is very pure and freshly prepared, iron(II) will probably give a faint red colour due to the presence of some iron(III).

  23. 3. Iron(III) oxidises iodide ions to iodine which gives the characteristic blue-black colour with starch. Iron(II) should give no reaction unless it contains some iron(III). 2Fe3+(aq) + 2I–(aq) I2(aq) + 2Fe2+(aq) 4. The deep purple colour of manganate(VII) ions gradually diminishes as it is reduced by iron(II) whereas iron(III) has no effect. MnO4– (aq) + 5Fe2+ (aq) + 8H+ (aq) → Mn2+ (aq) + 5Fe3+ (aq) + 4H2O (I) 5. The reaction of silver nitrate and iron(II) ions produces a glittering of metallic silver which is seen using a magnifying glass. There is no corresponding reaction with iron(III) ions.

  24. Iron chemistry – variable oxidation states • Apparatus (per group) • One student worksheet • One clear plastic sheet (egohp sheet) • Magnifying glass. • Chemicals (per group) • Solutions contained in 100 cm3 beakers with plastic pipettes • Sodium hydroxide 1 mol dm–3 • Potassium manganate(VII) 0.01 moldm–3 • Potassium iodide 0.2 mol dm–3 • Iron(II) sulphate 0.2 mol dm–3 • Iron(III) nitrate 0.2 moldm–3 • Silver nitrate 0.2 moldm–3 • Potassium thiocyanate 0.1 moldm–3 • Starch solution (freshly made).

  25. Iron chemistry – variable oxidation states The purpose of this experiment is to compare the chemistry of the two main oxidation states of iron (a first row transition element) and to consider explanations for any differences observed. Carefully follow the instructions below noting down all your observations and trying to give explanations. Instructions 1. Cover the worksheet with a clear plastic sheet. 2. Put one drop of iron(II) solution in each box in the second row. 3. Put one drop of iron(III) solution in each box in the third row. 4. Add two drops of sodium hydroxide solution to each drop in the boxes in the second column. Observe and note whether there are any changes over the next 10 min. 5. Add one drop of potassium thiocyanate solution to each drop in the third column. 6. Add one drop of potassium iodide solution to each drop in the fourth column. After one minute, add one drop of starch solution to each. 7. Add one drop of potassium manganate(VII) solution to each drop in the fifth column. Observe changes over the next 10 min. 8. Add one drop of silver nitrate solution to each drop in the sixth column. Observe closely using a magnifying glass.

  26. Instructions 1. Cover the worksheet with a clear plastic sheet. 2. Put one drop of iron(II) solution in each box in the second row. 3. Put one drop of iron(III) solution in each box in the third row. 4. Add two drops of sodium hydroxide solution to each drop in the boxes in the second column. Observe and note whether there are any changes over the next 10 min. 5. Add one drop of potassium thiocyanate solution to each drop in the third column. 6. Add one drop of potassium iodide solution to each drop in the fourth column. After one minute, add one drop of starch solution to each. 7. Add one drop of potassium manganate(VII) solution to each drop in the fifth column. Observe changes over the next 10 min. 8. Add one drop of silver nitrate solution to each drop in the sixth column. Observe closely using a magnifying glass.

  27. What explanations can you give for your observations ?

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