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Redox Reactions

Redox Reactions. Oxidation. In the 1700s, chemists determined that when metals corrode and get a “rust” layer, this rust is a combination of the metal with oxygen. Atoms of the metal join chemically with atoms of oxygen in the air. 4 Fe + 3 O 2  2 Fe 2 O 3.

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Redox Reactions

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  1. Redox Reactions

  2. Oxidation In the 1700s, chemists determined that when metals corrode and get a “rust” layer, this rust is a combination of the metal with oxygen. Atoms of the metal join chemically with atoms of oxygen in the air. 4 Fe + 3 O2 2 Fe2O3

  3. Different Colours of Iron Rust Iron rust, Fe2O3.X H2O, forms various hydrated compounds (with different numbers [X] of water molecules arranged around a central Fe2O3, making many different-coloured rust compounds.

  4. Reduction When iron III oxide (rust) is heated to a high temperature with carbon, the products are pure iron metal and carbon dioxide. In the 1800s this reaction was named a reduction because the iron oxide was reduced to pure iron. (coke) 2Fe2O3 + 3C  4Fe + 3CO2 (iron III oxide – “iron ore”)

  5. The First Meaning of Oxidation and Reduction The first understanding of oxidation was that it involved the joining of oxygen to an element or compound. Reduction was seen as the removal of oxygen from a compound. 4 Fe + 3 O2 2 Fe2O3 In the above reaction, iron is oxidized because oxygen is added to it. 2Fe2O3 + 3C  4Fe + 3CO2 In the above reaction, iron is reduced because oxygen is removed from it.

  6. The Second Meaning for Oxidation and Reduction In the late 1800s, chemists studying oxidation and reduction reactions concluded that oxidation and reduction involved the giving or taking of electrons. Oxidation of iron: 0+3 -2 4 Fe + 3 O2 2 Fe2O3 In the above reaction, each iron atom in the compound gives up or loses three electrons. Reduction of iron: +3 -2 0 2Fe2O3 + 3C  4Fe + 3CO2 In the above reaction, each iron atom in the compound iron is gaining three electrons.

  7. Oxidizing and Reducing Agents An oxidizing agent is the substance helping to oxidize another substance.An oxidizing agent is reduced itself as it helps oxidize another substance. A reducing agent is oxidized as it helps to reduce another substance.

  8. Redox Reactions When oxidation and reduction were understood as the loss and gain of electrons, it was realized that when any substance is oxidized, another substance must be reduced. For this reason oxidation and reduction reactions are called Redox reactions.

  9. Half Reactions In a Redox reaction, half reactions are written to show just the oxidation or reduction reactions of the overall reaction. Example : Redox reaction: Cu (s) + 2AgNO3 (aq)  2Ag (s) + Cu(NO3)2 Reduction half reaction: 2Ag+ + 2e -  2Ag (s) Oxidation half reaction: Cu (s)  Cu2+ + 2e -

  10. Copper and Silver Nitrate Reaction Cu + 2Ag + + 2NO3- 2Ag + Cu2+ + 2NO3 - Reduction half reaction: 2Ag+ + 2e -  2Ag (s) Oxidation half reaction: Cu (s)  Cu2+ + 2e -

  11. LEO says GER The mnemonic LEO says GER is used to remember that a Loss of Electrons is an Oxidation while a Gain of Electrons is a Reduction. Since Zn  Zn2+ + 2 e- (Half reaction showing zinc oxidized) Cl2 + 2 e-  2 Cl- (Half reaction showing chlorine reduced)

  12. Electronegativity The total attraction an atom has for an electron is called its electronegativity. Fluorine, F, has the largest electronegativity, followed by oxygen.

  13. Oxidation Numbers An oxidation number is the charge that an atom would have if the species containing the atom were made up of ions. In reality, this is not true for species which have covalent bonds. However the concept is a useful one, so atoms are assigned oxidation numbers based on electronegativity differences. The greater an atom’s elecronegativity, the stronger its attraction for shared electrons. Shared electrons are assigned to the atom with greater electronegativity which then has a negative sign. Ex: In the compound, Cl2O, the two shared electrons are assigned to oxygen, giving it a 2- oxidation number. Cl has a +1 oxidation number. Oxidation numbers are nearly identical to valence numbers. +1 -2 Cl2O Note F has the largest electroneg.

  14. Rules for Oxidation Numbers The sum of all positive and negative charges must equal the overall charge of the species. F always has a -1 oxidation number. Oxygen always has a 2- Ox.# except with F where it has a 2+ and except in peroxides where it has 1-. 2+ 1- OF2 1+ 2- 1+ 1- Na2O sodium oxide Na2O2 sodium peroxide H always has Ox# = 1 except in metal hydrides (-1). NaH The atoms in an element have an Ox# of zero. Alkali metals have an Ox# = +1, alkaline earths Ox# = +2 Halogens have an Ox# = -1 except when with a more electronegative element (oxygen) where their Ox# = +1 The charge on a single ion is its Ox#

  15. Assigning Oxidation Numbers In a chemical species, the sum of all the oxidation number charges for each individual atom in the species must equal the overall charge of the species. Example: Determine the oxidation number of P in H4P2O7. Solution: H has a 1+ and O a 2-. The overall species charge equals zero. Let x equal Ps Ox#. This means 4(1+) + 2(x) + 7(2-) = 0 Simplified, 4 + 2x -14 = 0 or, 2x -10 = 0 2x = 10 Thus x = +5

  16. What is the Oxidation Number of S in SO42- ? Solution: Let x = sulfur’s Ox# 1(x) + 4(2-) = 2- X - 8 = -2 X = +6

  17. Oxidation Number Changes in Redox Reactions Consider: 0 0 3+ 2- 4 Fe + 3 O2 2 Fe2O3 Note that as iron has been oxidized its Ox# has risen. Note that as oxygen has been reduced its Ox# has been reduced. In redox reactions, elements with rising oxidation numbers are being oxidized while elements with falling oxidation numbers are being reduced.

  18. The Standard Reduction Potentials Chart An excerpt from the Standard Reduction Potentials is included below. Note the following patterns: In general, metals are found on the bottom right of the chart. Halogens and oxyions are at the top right of the chart. Metals with more than one Ox.# have more than one ½ reaction HO is found at the top left and bottom right. In general, top reactions favour the forward reaction. In general, bottom reactions favour the reverse reaction.

  19. Voltages for ½ Reactions in the SRP Chart In the Standard Reduction Potentials (SRP) chart, the voltages given are for forward reduction ½ reactions. Ifthe reactions are written in reverse, the voltage sign must be reversed also. To obtain the theoretical voltage from an oxidation ½ reaction and a reduction ½ reaction, add their voltages. For example, consider Cu2+ + 2e- Cu (+0.34) and Pb  Pb2+ + 2e- (+0.13) The overall theoretical voltage expected from this reaction is: +0.34 + 0.13 = +0.47 . A + voltage like this reaction would be spontaneous, producing energy.

  20. The Spontaneity of Redox Reactions Redox Reactions will be spontaneous if they involve a forward ½ reaction to the right that is above a reverse ½ reaction to the left. Consider these ½ r. Ag+ + e- Ag and Zn  Zn2+ + 2e- 2Ag+ + 2e- 2Ag Zn  Zn2+ + 2e- These two ½ reactions are spontaneous together Net overall reaction: 2Ag+ + Zn  Zn2+ + 2Ag Full chemicals: 2Ag+ + 2NO3- + Zn  Zn2+ + 2Ag + 2NO3-

  21. Nonspontaneous Reactions Consider Cu2+ and Zn2+ as possible reactants in a redox reaction. These reactants are both on the left side of the SRP chart. In this case, there is no possible oxidation reaction for the two possible reduction reactions. No reaction will occur.

  22. Nonspontaneous Reactions : Second Case Consider Ag and Zn as possible reactants in a redox reaction. These reactants are both on the right side of the SRP chart. In this case, there is no possible reduction reaction for the two possible oxidation reactions. No reaction will occur.

  23. Nonspontaneous Reactions : Third Case Consider Cu and Zn2+ as possible reactants in a redox reaction. In this case the forward reduction reaction is below the reverse oxidation reaction. No reaction will occur. Note that the overall voltages are Cu  Cu2+ + 2e- (- 0.34) and Zn2+ + 2e- Zn (- 0.76) Total theoretical V = - 1.1 V which means energy must be added. This is not spontaneous.

  24. The Rule for Spontaneous Reactions A reaction will be spontaneous only if there is a reactant to be reduced (on left side) which is above a reactant to be oxidized (right side). From the SRP table below, find a reactant that will spontaneously react with Ni2+. From the SRP table, find a reactant that will spontaneously react with Pb.

  25. Hydrogen Ion Reductions A number of reactions require H+ to occur. Without the H+, these reactions do NOT occur. There is just ONE reaction in which H+ is the only substance being reduced: The reduction of H has been arbitrarily set to zero as the standard against which all other reductions are measured. (More on this in a later slide)

  26. Balancing Half Reactions The steps in balancing half reactions are as follows: Balance all the MAJOR atoms by inspection (Not O and H) Balance the Oxygen atoms by adding H2O molecules. Balance the Hydrogen atoms by adding H+. Balance the overall charge by adding electrons. The mneumonic, MajorOH- is used to remember the steps. OH-

  27. Sample Balancing Problem 1 Balance the half-reaction, RuO2 Ru The Major Ru is balanced on both sides (no O and H balancing yet) Balance O by adding H2O : RuO2 Ru + 2H2O Balance H by adding H+ : RuO2 + 4H+ Ru + 2H2O Balance the charge by adding electrons to one side (-) . The left side has 4+ while right side has 0. So adding 4- to left side brings it to zero also, balancing the charge. The rule for charge balancing is that the electrons are always added to the side which has the greater charge. RuO2 + 4H+ + 4e- Ru + 2H2O The above reaction occurs in an acid solution What kind of reaction is this half-reaction?

  28. Sample Balancing Problem 2 Balance the half-reaction Cr2O72- Cr3+ Balance the Major Cr: Cr2O72- 2Cr3+ Balance O by adding H2O : Cr2O72- 2Cr3+ + 7H2O Balance H by adding H+ : Cr2O72- + 14H+ 2Cr3+ + 7H2O Balance the charge by adding electrons to one side (-) . The left side has 12+ total while right side has 6+ total. So adding 6- to left side brings it to 6+ like the right side. The rule for charge balancing is that the electrons are always added to the side which has the greater charge. Cr2O72- + 14H+ + 6e- 2Cr3+ + 7H2O The above reaction occurs in an acid solution What kind of reaction is this half-reaction?

  29. Sample Balancing Problem 3 Balance the half-reaction, Pb HPbO2- , in a basic solution Balance the Major Pb: Pb HPbO2- Pb is balanced Balance O by adding H2O: Pb + 2H2O  HPbO2- Balance H by adding H+: Pb + 2H2O  HPbO2- + 3H+ Balance the charge by adding electrons to one side (-) . The left side has 0 total while right side has 2+ total. So adding 2- to the right side brings it to 0 like the left side. Electrons are added to the side with the greater charge. Pb + 2H2O  HPbO2- + 3H+ + 2e- Add the same number of OH- to both sides so that the H+ are neutralized to form water. Pb + 2H2O + 3OH- HPbO2- + 3H++ 3OH- + 2e- Pb + 2H2O + 3OH- HPbO2- + 3H2O + 2e- 7. Cross-cancel H2O : Pb + 3OH- HPbO2- + H2O + 2e- The above reaction is an oxidation happening in an alkaline solution.

  30. Sample Balancing Problem 4 Balance the half-reaction, H2 , in an acidic solution Balance the Major: No major to balance Balance O by adding H2O: No O to balance Balance H by adding H+: H2 2H+ Balance the charge by adding electrons to one side (-) . The left side has 0 total while right side has 2+ total. So adding 2- to the right side brings it to 0 like the left side. Electrons are added to the side with the greater charge. H2 2H+ + 2e- The above reaction is an oxidation happening in an acidic solution.

  31. Balancing Redox Equations : The ½ Reaction Method There are two methods for balancing redox equations. You need only use one, but it is advantageous to know both methods. Balance Os + IO3- OsO4 + I2 (acidic solution) 1. Split the reaction into its half reactions: Os  OsO4 IO3- I2 2. Balance the half reactions using the Major OH- method. Os + 4H2O  OsO4 + 8H+ + 8e- 2IO3- + 12H+ + 10e-  I2 + 6H2O 3. Multiply half reactions to ensure # e lost = # e gained 5(Os + 4H2O  OsO4 + 8H+ + 8e-) 4(2IO3- + 12H+ + 10e-  I2 + 6H2O)

  32. Balancing Redox Eq. : The ½ Reaction Method (Cont.) From step 3: 5(Os + 4H2O  OsO4 + 8H+ + 8e-) 4(2IO3- + 12H+ + 10e-  I2 + 6H2O) 4. Add the two half reactions and simplify: 5Os + 20H2O  5OsO4 + 40H+ + 40e- 8IO3- + 48H+ + 40e-  4I2 + 24H2O Overall Balanced Equation: 5Os + 8IO3- + 8H+  5OsO4 + 4I2 +4H2O

  33. A Second Example of Redox Equation Balancing Balance MnO4- + C2O42- MnO2 + CO2 (in basic solution) Make ½ reactions MnO4-  MnO2 & C2O42-  CO2 Balance ½ reac. 3e- + 4H+ + MnO4-  MnO2 + 2H2O C2O42-  2CO2 + 2e- 3. Mult for = #e-2(3e- + 4H+ + MnO4-  MnO2 + 2H2O) 3(C2O42-  2CO2 + 2e-) 4. Add ½ reac. 6e- + 8H+ + 2MnO4-  2MnO2 + 4H2O 3C2O42-  6CO2 + 6e- Simp. 8H+ + 2MnO4- + 3C2O42-  2MnO2 + 4H2O + 6CO2 5. Add OH- to both sides for a basic solution 8OH- + 8H++ 2MnO4- + 3C2O42-  2MnO2 + 4H2O + 6CO2 + 8OH- 8H2O+ 2MnO4- + 3C2O42-  2MnO2 + 4H2O + 6CO2 + 8OH- 4H2O+ 2MnO4- + 3C2O42-  2MnO2 + 6CO2 + 8OH-

  34. A Third Example of Redox Balancing: Disproportionation Balance ClO2- ClO3- + Cl- (in basic solution) Make ½ reactions ClO2-  ClO3- & ClO2-  Cl- Balance ½ reac. H2O + ClO2-  ClO3- + 2H+ + 2e- 4e- + 4H+ + ClO2-  Cl- + 2H2O 3. Mult for = #e- 2(H2O + ClO2-  ClO3- + 2H+ + 2e-) 1(4e- + 4H+ + ClO2-  Cl- + 2H2O) 4. Add ½ reac. 2H2O + 2ClO2-  2ClO3- + 4H+ + 4e- 4e- + 4H++ ClO2-  Cl- + 2H2O Simp. 3ClO2-  2ClO3- + Cl- 5. No OH- to add because no H+ present. (Reaction happens in basic solution but equation doesn’t show this) 3ClO2- 2ClO3- + Cl- This reaction is called a disproportionation reaction because one species is both reduced and oxidized.

  35. A Second, Faster Method For Balancing Redox Eq. Balance U4+ + MnO4- Mn2+ + UO22+ Balance majors (done already) Assign ox.#s +4+7+2+6 U4+ + MnO4- Mn2+ + UO22+ 3. Note total changes in ox#s 1U goes up 2 (oxidation) while 1Mn goes down 5 (reduction) ΔU = +2 and ΔMn = -5 4. Find cross-mult. #s so that the product Δ up = product Δdown (Because #e lost = #e gained) 2(-5Δ) = -10 = (Mn Δ) 5(+2Δ) = +10 = (U Δ) 5. Apply cross-mult #s to reactants and then products 5U4+ + 2MnO4- 2Mn2+ + 5UO22+

  36. A Second Method For Balancing Redox (Cont) 6. Balance the O atoms by adding H2O and the H atoms by adding H+ 5U4+ + 2MnO4- 2Mn2+ + 5UO22+ 2 H2O + 5U4+ + 2MnO4- 2Mn2+ + 5UO22+ 2 H2O+ 5U4+ + 2MnO4- 2Mn2+ + 5UO22+ +4H+ Check that the charge is balanced (It should be because the oxidation # rise was set equal to the oxidation # drop). A final charge balance check helps to detect any balancing errors made.

  37. Ex 2 of Redox Equation Balancing by Ox. #s Balance Zn + As2O3  AsH3 + Zn2+ (in basic solution) Balance the majors Zn + As2O3  2AsH3 + Zn2+ 2. Assign ox.#s 0 +3 -3 +2 Zn + As2O3  2AsH3 + Zn2+ Note total changes in ox#s 1 Zn goes up 2 while 2As go down 6 total Δ up (1)(2) = +2Δ total Δ down (2)(-6) = -12Δ 4. Find cross-mult. #s so that the total pr. Δ up = totalΔdown 6(2+Δ) = +12 and 1(-12) = -12 5. Apply cross-mult #s to reactants and then products 6Zn + 1As2O3  1.2AsH3 + 6Zn2+ 6Zn + 1As2O3  2AsH3 + 6Zn2+

  38. Ex 2 of Redox Equation Balancing by Ox. #s (Cont.) 6. At this point O atoms could be balanced by adding H2O and H atoms could be balanced by adding H+ and finally OH- could be added to produce a basic solution. 6. Alternate shortcut: a) Use OH- to charge balance and then b) add H2O to balance O and H atoms at the same time. a) 6Zn + As2O3  2AsH3 + 6Zn2++ 12OH- b) 6Zn + As2O3 + 9H2O 2AsH3 + 6Zn2+ + 12OH- 6Zn + As2O3 + 9H2O  2AsH3 + 6Zn2+ + 12OH-

  39. Ex 3 of Redox Equation Balancing by Ox. #s Balance this disproportionation P4 H2PO2- + PH3 (acid) Write equation so same reactant can undergo ox. and red. P4 + P4 H2PO2- + PH3 Balance majors P4 + P4 4H2PO2- + 4PH3 Assign ox.#s 0 0 +1 -3 P4 + P4 4H2PO2- + 4PH3 4. Note total changes in ox#s 4 P go up 1 while 4 P go down 3 total Δup = (4)(+1 Δ)= +4 Δ, total Δ down = (4)(-3 Δ) = -12 Δ 5. Find cross-mult. #s so that the total Δ up = totalΔdown 3(4+Δ) = +12 and 1(-12) = -12 Apply cross-mult #s to reactants and then products 3P4 + 1P4 12H2PO2- + 4PH3

  40. Ex 3 of Redox Equation Balancing by Ox. #s (Cont.) Apply cross-mult #s to reactants and then products 3P4 + 1P4 12H2PO2- + 4PH3 Simplify and reduce to lowest terms 4P4  12H2PO2- + 4PH3 P4  3H2PO2- + PH3 8. Balance O atoms by adding H2O P4 + 6H2O 3H2PO2- + PH3 Balance H atoms by adding H+ P4 + 6H2O  3H2PO2- + PH3+3H+ Check charge balance to detect mistakes P4 + 6H2O  3H2PO2- + PH3 +3H+

  41. Redox Titrations Redox reactions can be used to accurately determine unknown concentrations of substances. The end point of the redox reaction is determined by a colour change in the agent as it is oxidized or reduced.

  42. Oxidizing Agents for Redox Titrations In an acidic solution of KMnO4, the ion, MnO4- is a strong oxidizing agent (has a strong tendency to reduce) as shown by this reaction: MnO4- + 8H+ + 5e- Mn2+ + 4H2O ;Eo = 1.49V MnO4- is purple in solution while Mn2+ is colourless. As the MnO4- is added to a solution, it is oxidized and becomes clear Mn2+ until the end point where all the reducing agent is used up and a single added drop of MnO4- remains purple.

  43. Example of Redox Titration Using MnO4- When 25.00 mL of a solution of Fe2+ is titrated to an endpoint with acidic KMnO4 , the titration requires 17.52 mL of acidified 0.1000 M KMnO. What is the Fe2+ in the solution? MnO4- + 8H+ + 5Fe2+ Mn2+ + 4H2O + 5Fe3+ The mols MnO4- = (0.1000 mmol/mL)(17.52 mL) = 1.752 mmol MnO4- The mols Fe2+ = (1.752 mmol MnO4-)(5mmol Fe2+/1 mmolMnO4-) = 8.760 mmol Fe2+ [Fe2+] = 8.760 mmol Fe2+ / 25.00 ml = 0.3504 M (mol/L)

  44. Reducing Agents for Redox Titrations NaI or KI are commonly used as reducing agents in redox titrations. The I- ion (reducing agent) is oxidized as follows: 2I- I2 + 2e- Two steps are generally used: Titrations using I- to reduce a given species produces I2, and (2) this I2 is then generally reduced back to I- by a second reducing agent like S2O32- (thiosulfate ion). For example, to determine the concentration of the hypochlorite ion (OCl-) in bleach (NaOCl), an excess of KI solution is added to acidified bleach to consume all the moles of OCl-. The I2 turns the solution brownish. 2 H+ + OCl- + 2I- Cl- + H2O + I2 + 

  45. Reducing Agents for Redox Titrations (Cont.) The brownish I2 solution is then reacted with a known concentration of Na2S2O3 until a pale yellow colour remains which indicates just a small amount of I2 left. Only the I2 reacts with the S2O32-. The excess I- does not react with the S2O32-. 2S2O32-+ I2 S4O62- + 2I- Starch is added to the yellow solution (solution becomes dark blue) and then more S2O32- is added until a single drop turns the mixture from blue to clear (the endpoint).

  46. Sample Redox Titration Problem With I- and S2O32- A 25.00 mL sample of bleach is reacted with excess KI. The resulting brown solution takes 46.84 mL of 0.7500 M Na2S2O3 to reach the endpoint. What was the [OCl-] in the bleach? 2 H+ + OCl- + 2I- Cl- + H2O + I2 2S2O32-+ I2 S4O62- + 2I-

  47. End of Presentation

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