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Redox Reactions

Ox idation. Red uction. Redox Reactions. What is the meaning of REDOX?. R E D O X. Examples of Redox Reactions which occur in daily life. Combustion Corrosion/Rusting Photosynthesis Respiration. Defn 1 (In terms of oxygen). Oxidation = addition/gain of oxygen

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Redox Reactions

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  1. Oxidation Reduction Redox Reactions What is the meaning of REDOX? R E D O X

  2. Examples of Redox Reactions which occur in daily life • Combustion • Corrosion/Rusting • Photosynthesis • Respiration

  3. Defn 1 (In terms of oxygen) Oxidation = addition/gain of oxygen A substance that gains oxygen is said to be oxidised. Reduction = removal/loss of oxygen. A substance that loses oxygen is said to be reduced.

  4. Example based on Defn 1 (In terms of oxygen) • Zn + PbO → ZnO + Pb • Zinc (Zn) is oxidised because it gains oxygen to form zinc oxide. • Lead (II) oxide (PbO) is reduced because it loses oxygen to form lead. When one substance is oxidised, another substance must be reduced. Oxidation and reduction take place at the same time. Such reactions are called REDOX reactions.

  5. Example based on Defn 1 (In terms of oxygen) 2. Fe2O3 + 3CO → 2Fe + 3CO2 Carbon monoxide (CO) is oxidised. Iron (III) oxide (Fe2O3) is reduced.

  6. Defn 2 (In terms of hydrogen) Oxidation = removal/loss of hydrogen A substance that loses hydrogen is said to be oxidised. Reduction = addition/gain of hydrogen. A substance that gains hydrogen is said to be reduced.

  7. Example based on Defn 2 (In terms of hydrogen) • H2S + Cl2→ 2HCl + S • The substance oxidised is • hydrogen sulfide (H2S) because it loses hydrogen to form sulfur. • The substance reduced is • chlorine (Cl2) because it gains hydrogen to form hydrogen chloride. • 2. CH4 + Cl2→ CH3Cl + HCl • Substance oxidised is methane (CH4). • Substance reduced is chlorine (Cl2).

  8. Defn 3 (In terms of electrons) Oxidation = loss of electrons A substance that loses electrons is said to be oxidised. Reduction = gain of electrons. A substance that gains electrons is said to be reduced. Remember LEO (Loss of Electrons is Oxidation) GER (Gain of Electrons is Reduction) OR OIL (Oxidation Is Loss of electrons) RIG (Reduction Is Gain of electrons)

  9. Example (In terms of electrons) 1. Mg (s) + 2HCl (aq) → MgCl2 (aq) + H2 (g) Ionic eqn: Recall the steps in writing ionic eqn First, split all compounds which are in (aq) into their ions, do NOT split all those in (s), (l) or (g) as they do not contain free ions. Mg (s) + 2H+ (aq) + 2Cl-→ Mg2+ (aq) + 2Cl- (aq) + H2 (g) Cancel ions which appear on both sides Mg (s) + 2H+ (aq) + 2Cl-→ Mg2+ (aq) + 2Cl- (aq) + H2 (g) Final eqn: Mg (s) + 2H+ (aq) → Mg2+ (aq) + H2 (g)

  10. Example (In terms of electrons) 1. Mg (s) + 2H+ (aq) → Mg2+ (aq) + H2 (g) What happens to Mg in the reaction? From the ionic eqn, we can see that Mg has become Mg2+, this conversion must have involved the transfer of electrons, and can be represented by means of a half-eqn. What are half equations? • Half equations are used to identify and represent redox reactions. In the writing of half-eqn, • The atoms and charges must be balanced. • Electrons are used to balance the charges on the two • sides of the equation. • They can be combined to give an overall ionic eqn for • the reaction.

  11. 0 +2 0 +2 -2 Example (In terms of electrons) Steps in writing half eqn: First from the half eqn, we know that Mg has become Mg2+ So we write, Mg (s) → Mg2+ (aq) Check the charges on both sides of eqn Mg (s) → Mg2+ (aq) Notice that Mg being a atom has a charge of 0 while Mg2+ has a charge of +2. To balance the eqn, charges on both sides must be balanced, and this is done by adding electrons to the eqn. Recall electrons carry negative charge, so to balance the eqn, 2 electrons must be added to the right hand side of the eqn. Mg (s) → Mg2+ (aq)+ 2e- Final half-eqn

  12. Example (In terms of electrons) The electrons can be moved to the left hand side of eqn and the eqn becomes Mg (s) - 2e-→ Mg2+ (aq) It can be seen that each magnesium atom loses 2 electrons to form a magnesium ion, hence magnesium is oxidised.

  13. Example (In terms of electrons) +2 0 +2 -2 0 Which substance is reduced? Write the half eqn which involves the conversion of H+ to H2 Steps: 2H+ (aq) → H2 (g) Balance the no. of H atoms/ions first. Then check the charges on both sides of eqn 2H+ (aq) → H2 (g) To balance the charges on both sides of eqn, electrons must be added to the left hand side of the eqn. 2H+ (aq) + 2e-→ H2 (g) The half-eqn is now complete. From the half eqn, we see that hydrogen ions have gained electrons to form hydrogen gas, hence hydrochloric acid is being reduced.

  14. Example (In terms of electrons) Note: Addition of the 2 half-eqns give the overall ionic eqn for the reaction Mg (s) → Mg (aq) + 2e- Eqn 1 2H+ (aq) + 2e-→ H2 (g) Eqn 2 Adding Eqn 1 and 2, we get Mg (s) + 2H+ (aq) + 2e-→ Mg (aq) + 2e- + H2 (g) Notice 2 electrons appear on both sides of eqn and should be cancelled to yield the final ionic eqn: Mg (s) + 2H+ (aq) → Mg (aq) + H2 (g)

  15. Practice 1 (Pg 3) • Chlorine reacts with iron (II) ions according to the two half-eqn below: • Cl2 + 2e-→ Cl2Eqn 1 • Fe2+→ Fe3+ + e-Eqn 2 • Write the overall ionic eqn for the reaction between chlorine and iron (III) ions. • Notice eqn 1 involves 2e- while eqn 2 involves 1 e-. Hence there is a need to multiply eqn 2 by 2 throughout so that it has the same no. of electron as eqn 1. • 2Fe2+→ 2Fe3+ + 2e- • Now add the two eqns together, we get • Cl2 + 2e- + 2Fe2+→ Cl2 + 2Fe3+ + 2e- • Cancel the 2e- on both sides, the final eqn is: • Cl2 + 2Fe2+ → Cl2 + 2Fe3+

  16. Practice 1 (Pg 3) 2. For the following reaction, identify the substance that is being oxidised and reduced. Explain your answer in terms of gain/loss of electrons and write half equations to illustrate the reduction and oxidation process. Zn (s) + CuSO4 (aq) → ZnSO4 (aq) + Cu (s) Half eqn for oxidation: Zn (s) → Zn2+ (aq) + 2e- Zinc is oxidised because each atom of zinc loses 2 electrons to form zinc ions. Half eqn for reduction: Cu2+ (aq) + 2e-→ Cu (s) Copper (II) sulfate is reduced because each copper (II) ion gains 2 electrons to form a copper atom.

  17. Redox in terms of Oxidation States/Numbers (Pg 3) The oxidation state/number of an element is the charge an atom of an element would have if it existed as an ion in a compound. The charge on a simple ion is said to be its oxidation state. All atoms in elements, ions, or compounds can be given an oxidation no. This no. is obtained by applying the following set of rules.

  18. Redox in terms of Oxidation States/Numbers (Pg 4) Rule 1: The oxidation state of an element in the natural form uncombined with other elements =0. E.g oxidation state of Na = 0. Oxidation state of chlorine in Cl2 = 0.

  19. Redox in terms of Oxidation States/Numbers (Pg 4) Rule 2: The oxidation state of an element in a simple ion is equal to the charge of the ion. E.g oxidation state of Na in Na+ = +1. Oxidation state of sulfur in S2- in =-2 i) oxidation state of oxygen in O2 = ii) oxidation state of oxygen in O2- = iii) oxidation state of Na in Na = iv) oxidation state of Na in Na+ = 0 -2 0 +1

  20. Redox in terms of Oxidation States/Numbers (Pg 4) Rule 3 Hydrogen usually has an oxidation state of +1 when it is covalently bonded to non-metals. Exception: In Gp I hydrides like NaH, hydrogen has oxid no. of -1 while sodium has oxid no. of +1. Rule 4 Oxygen usually has an oxidation state of -2 in the vast majority of its compounds. Exception: In H2O2 the oxid no. of oxygen is -1

  21. Redox in terms of Oxidation States/Numbers (Pg 4) Rule 5 The algebraic sum of the oxidation states/numbers of the elements in a compound equals zero. E.g. see whiteboard Using rule 3-5, calculate the oxidation state of: I) S in H2SO4 Let oxidation state of S be x . 2 (+1) + x + 4 (-2) = 0 2 + x -8 = 0 x = +6 Oxidation state of S in H2SO4 = +6 Try (ii) ad (iii) now ii) Ans: +5 iii) Ans : +3

  22. Redox in terms of Oxidation States/Numbers (Pg 4) Rule 6 The sum of the oxidation states of the elements in a polyatomic ion equals thecharge of the ion. Example: See white board Calculate the oxidation state of I) S in SO32- Let oxidation state of S be x . x + 3 (-2) = -2 x - 6 = -2 x = -2+6 = +4 Oxidation state of S in SO32-= +4 Do part (ii) and (iii) on your own!

  23. Redox in terms of Oxidation States/Numbers (Pg 4) Rule 6 ii) +5 iii) +6 Rule 7 The oxidation state of Group I metals in ionic compounds = +1 The oxidation state of Group II metals in ionic compounds = +2 The oxidation state of Group III metals in ionic compounds = +3

  24. Redox in terms of Oxidation States/Numbers (Pg 5) Rule 9 In the writing of oxidation states,signmust be written. Rule 10 Fluorine always has an oxidation state of -1 in its compounds. Try practice 2 on page 5

  25. 0 • 0 • +2 • -3 • +6 • +4 • +5 • +4 • +1 • +5 Answers to practice 2 (Pg 5)

  26. Redox in terms of change in oxidation states/No (Pg 6) Oxidation is defined as the increase in oxidation state/number. Reduction is defined as the decrease in oxidation state/number. • Examples • State the substance oxidised and reduced in each of the following reactions. • Zn + H2SO4→ ZnSO4 + H2 • Zinc is oxidised because the oxidation no. of zinc increases from 0 (in Zn) to +2 (in ZnSO4). • Sulphuric acid is reduced because the oxidation no. of hydrogen decreases from +1 (in H2SO4) to 0 (in H2)

  27. Redox in terms of change in oxidation states/No (Pg 6) Oxidation is defined as the increase in oxidation state/number. Reduction is defined as the decrease in oxidation state/number. Examples State the substance oxidised and reduced in each of the following reactions. 2. 2Fe2+ + Cl2→ 2Fe3+ + 2Cl- Iron (II) ion is oxidised because the oxidation no. of iron increases from +2 (in Fe2+) to +3 (in Fe3+). Chlorine is reduced because the oxidation no. of chlorine decreases from 0 (in Cl2) to -1 (in Cl-).

  28. Practice 3 (Pg 6) 1. Cu2+ + Zn   Cu + Zn2+ +2 0 0 +2 Zinc, Zn is oxidised because the oxidation no. of zinc increases from 0 (in Zn) to +2 (in Zn2+). Copper (II) ion, Cu2+ , is reduced because the oxidation no. of copper decreases from +2 (in Cu2+ ) to 0 (in Cu).

  29. Practice 3 (Pg 6) 2. S + O2 SO2 0 0 +4 -2 Sulfur is oxidised because the oxidation no. of sulfur increases from 0 (in S) to +4 (in SO2). Oxygen is reduced because the oxidation no. of oxygen decreases from 0 (in O2) to -2 (in SO2).

  30. Names of Compounds, Pg 6 Oxidation states are included in some chemical names if the element has more than one oxidation state. Examples 1 copper (II) oxide Oxidation state of Cu = +2 2. Potassium manganate (VII) Oxidation state of Mn = +7 3. Potassium dichromate (VI) Oxidation state of chromium = +6 4. Sulfate (VI) ion Oxidation state of S = +6

  31. Try question on page 7 1. a) lead (IV) oxide [Pb4+ O2-] PbO2 b) chromium (III) oxide [Cr3+ O2-] Cr2O3 c) vanadium (V) oxide [V5+ O2-] V2O5

  32. Oxidising and Reducing Agents (Pg 7) • Oxidising Agent • causes another substance to undergo oxidation • itself undergoes reduction • Reducing Agent • causes another substance to undergo reduction • itself undergoes oxidation

  33. +2 0 0 +1 Oxidising and Reducing Agents (Pg 7) Eg, identify the oxidising and reducing agent in the following reaction CuO + H2 Cu + H2O Hydrogen is oxidised, because the oxidation no. of hydrogen increases from 0 (in H2) to +1 (in H2O) Copper (II) oxide is reducedbecause the oxidation no. of copper decreases from +2 (in CuO) to 0 (in Cu) Hence the oxidising agent is copper (II) oxide. The reducing agent is hydrogen.

  34. Common Oxidising Agents (Pg 7) • Acidified potassium dichromate (VI), K2Cr2O7; Acid used here is dilute sulphuric acid. • In the presence of a reducing agent, the acidified potassium dichromate (VI) changes from orange to green. • Half eqn: • Cr2O72- (aq)+ 14H +(aq)+ 6e -  2Cr3+ (aq)+7H2O (l) • Potassium dichromate (VI) is reduced because • the oxidation no. of chromium decreases from +6 (in Cr2O72- ) to +3 (in Cr3+).

  35. Common Oxidising Agents (Pg 8) 2. Acidified potassium manganate (VII), KMnO4; Acid used here is dilute sulphuric acid. In the presence of a reducing agent, the acidified potassium manganate (VII) solution changes from purple to colourless.The acidified potassium manganate (VII) is said to be decolourised. Half eqn: MnO4- (aq)+ 8H +(aq)+ 5e -  Mn2+ (aq)+ 4H2O (l) Potassium manganate (VII) is reduced because the oxidation no. of manganese decreases from +7 (in MnO4-) to +2 (in Mn2+).

  36. Common Oxidising Agents (Pg 8) 3. Oxygen gas, O2 4. Chlorine and other halogens. Oxidising property decreases down the group; fluorine is the most powerful oxidising agent in group VII. Half eqn for Cl2 : Cl2 + 2e- 2Cl- 5. Acidified iron (III) solution, Fe3+ The iron (III) solution changes from yellow to green. Half eqn: Fe3+ + e- Fe2+

  37. Common Oxidising Agents (Pg 8) • Hydrogen peroxide • Manganese (IV) oxide,MnO2 (a black solid) • 8. lead (IV) oxide, PbO2 • 9. Concentrated sulfuric acid • 10. Concentrated nitric acid

  38. Common Reducing Agents (Pg 8) 1. Aqueous potassium iodide, KI In the presence of oxidising agent, the colour of aqueous potassium iodide changes from colourless to brown. Half eqn: 2I- I2 + 2e-

  39. Common Reducing Agents (Pg 8) 2. Hydrogen 3. Carbon 4. Carbon monoxide, CO

  40. Common Reducing Agents (Pg 8-9) • 5. Most metals like magnesium and sodium • Acidified iron (II) solution • Half eqn: Fe2+  Fe3+ + e- • 7. Hydrogen peroxide • 8. sodium sulfite, Na2SO3 • sodium thiosulfate, Na2S2O3 • Sulfur dioxide, SO2

  41. Question on page 9 Which substance can act as a reducing agent and oxidising agent? Hydrogen peroxide, H2O2

  42. Tests for oxidising & Reducing agents (Pg 10) Reducing agents are used to test for the presence of oxidising agent. Oxidising agent is usually identified by a colour change. Oxidisng agents are used to test for the presence of reducing agent. Reducing agent is usually identified by a colour change.

  43. The most common way to test for oxidising/reducing agent is : To test whether an aqueous solution is a reducing agent, acidified potassium manganate (VII) or acidified potassium dichromate (VI) can be used. Colour change: For KMnO4: The acidified aqueous potassium manganate (VII) turns from purple to colourless. For K2Cr2O7: The acidified aqueouspotassium dichromate (VI) turns from orange to green.

  44. To test whether an aqueous solution is an oxidising agent, (acidified) aqueous potassium iodide can be used. Colour change: The aqueous potassium iodide turns from colourless to brown. (Due to liberation of iodine) 2I-  I2+ 2e – Try practice 4, Pg 11

  45. Practice 4 (Pg 11) • The following reactions involve both oxidation and reduction. • A Fe2O3 + 3H2→ 2Fe + 3H2O • D Cl2 + 2I-→ I2 + 2Cl- The following types of reactions are NOT redox: 1. Precipitation 2. Neutralisation 3. Acid + carbonate

  46. Practice 4 (Pg 11) • The reaction between iron (II) ions and dichromate (VI) ions can be represented by the following eqn: • 6Fe2+ (aq) + Cr2O72- (aq) + 14H+ (aq) → 6Fe3+ (aq) + 2Cr3+(aq) + 7H2O (l) • Is the above reaction redox? Explain your ans in terms of oxidation no. • The reaction is redox. Iron (II) ions are oxidised as the oxidation no. of iron increases from +2 (in Fe2+) to +3 (in Fe3+). • Dichromate (VI) ions are reduced as the oxidation no. of chromium decreases from +6 (in Cr2O72-) to +3 (in Cr3+) • Since both oxidation and reduction occur, the reaction must be redox.

  47. Practice 4 (Pg 11) 3. When hydrogen peroxide decomposes, water and oxygen are formed. a) Write a balanced eqn for the reaction. 2H2O2→ 2H2O + O2 b) Is the above reaction redox? Explain your ans in terms of oxidation no. The reaction is redox. Hydrogen peroxide is oxidised as the oxidation no. of oxygen increases from -1 (in H2O2) to 0 (in O2). Hydrogen peroxide is also reduced as the oxidation no. of oxygen decreases from -1 (in H2O2) to -2 (in H2O) Since both oxidation and reduction occur, the reaction must be redox. Notice in this example, hydrogen peroxide is both oxidised and reduced. A reaction whereby a substance is both oxidised and reduced is known as disproportionation.

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