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Oxidation - Reduction Reactions

Oxidation - Reduction Reactions. There are a whole family of chemical reactions which involve transfers of electrons from one substance to another. A common example is the reaction of copper metal with a solution of silver nitrate. The chemical equation for this reaction can be seen below:.

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Oxidation - Reduction Reactions

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  1. Oxidation - Reduction Reactions There are a whole family of chemical reactions which involve transfers of electrons from one substance to another. A common example is the reaction of copper metal with a solution of silver nitrate. The chemical equation for this reaction can be seen below: Cu(s) + 2AgNO3(aq) 2Ag(s) + Cu(NO3)2(aq)

  2. In this chemical reaction electrons are being transferred from the copper to the silver. This is more evident when the ionic equation is written: Cu(s) + 2Ag1+(aq) + 2NO31-(aq) 2Ag(s) + Cu2+(aq) + 2NO31-(aq) When the spectator ions are removed the net ionic equation is: Electrons are being gained Cu(s) + 2Ag1+(aq) 2Ag(s) + Cu2+(aq) Electrons are being lost

  3. Redox Movie Cu Cu Cu Cu Cu Cu Cu Ag Ag1+ Ag1+ Ag Cu2+ Ag1+ NO31-

  4. When chemical reactions involve transfers of electrons the overall chemical reaction can be written in 2 different steps - the reaction involving the loss of electrons and the reaction involving the gain of electrons Loss Of Electrons (Oxidation) Cu(s) ------> Cu2+(aq) + 2e1- Gain of Electrons (Reduction) 2Ag1+(aq) + 2e1- -----> 2Ag(s)

  5. Notice that electron losses are always written on the product side and electron gains are always written on the reactant side. This is done by convention to avoid confusion. The substance losing electrons, in this example copper, is said to be oxidized and the substance gaining electrons, in this case silver ions, are said to be reduced. Oxidation and reduction reactions always occur together so the name oxidation-reduction reaction is shortened to REDOX.

  6. Cu(s) + 2Ag1+(aq) 2Ag(s) + Cu2+(aq) The substance being oxidized is responsible for the other substance being reduced so the oxidized substance is called the reducing agent. Simlarly the substance being reduced is responsible for the other substance being oxidized so the reduced substance is called the oxidizing agent.

  7. Sample Question When copper metal reacts with chlorine gas the product is copper (II) chloride. 1. Write the overall balanced chemical equation. 2. Write the oxidation and reduction reactions, determine which substance is being oxidized and which is reduced as well as the oxidizing and reducing agents. Cu(s) + Cl2(g) CuCl2(s)

  8. The Cl here exists as Cl1- ions The Clhas gained e1- so it is reduced and is called the oxidizing agent The Cu here exists as Cu2+ ions The Cu haslost e1- so it is oxidized and is called the reducing agent Cu(s) + Cl2(g) CuCl2(s)

  9. OxidationReaction is Cu(s) ------> Cu2+ + 2e1- Reduction Reaction is Cl2(g) + 2e1- -----> 2Cl1- LEO the lion says GER Loss of Electrons Oxidation Gain of Electrons Reduction

  10. Assigning Oxidation Numbers To understand redox reactions chemists have assigned atoms in compounds numbers which indicate the charge an atom would have if the shared electron pairs belonged completely to the more electronegative atoms. Using this principle an element existing in its molecular form always has an oxidation number of zero since neither atom gains control of the shared electron pair(s).

  11. When determining the oxidation number of the atoms in ionic compounds the oxidation number is always the same as the ion’s charge. Here are a list of the basic rules for assigning oxidation numbers: 1. Elements in their pure elemental form have an oxidation number of 0. 2. Elements in group 1A when found in compounds have oxidation numbers of 1+ 3. Elements in group llA when found in compounds have oxidation numbers of 2+

  12. 4. Elements in group llIB when found in compounds have oxidation numbers of 3+ 5. The sum of the oxidation numbers of all of the atoms in a formula must equal the charge written for the formula 6. Fluorine always has an oxidation number of 1- when found in other compounds. 7. Oxygen always has an oxidation number of 2- when found in other compounds. 8. Hydrogen always has an oxidation number of 1+ when found in other compounds, except in peroxides its charge is 1-.

  13. Al is in group 111A so its oxidation # is Example 1 - the compound calcium chloride has the formula CaCl2. The oxidation numbers of the calcium atom and the chlorine atom are 2+ and 1- respectively. Sample Problems - Assign oxidation numbers to the following atoms in their respective compounds. Notice the total charge is zero 3+ 1- 1- 1- AlF3 Al F F F means

  14. 2+ 2- 2+ 1- 3+ 2- CaO MgI2 Fe2O3 2+ 2- 3+ 2- 1+ 2- SrSe Al2O3 Cu2O 6+ 2- 2- 2- 2- 2- 6+ Remember all charges must add up to -2 SO42- SOOOO

  15. Redox Reactions are chemical changes which involve the loss and gains of electrons. They can be identified by assigning oxidation numbers to every element in a chemical equation and comparing these numbers on the reactant and product side.

  16. 0 to 2, lost electrons, oxidized +4 0 1+ +5 -2 2 +5 -2 -2 1+ -2 Cu + HNO3 ----> Cu(NO3)2 + NO2 + H2O +1 +5 -6 5 to 4, gained electrons, reduced elements by themselves are 0 oxygen is -2, hydrogen is 1 find the other oxidation states using the total for each formula Identify the 2 elements which change, these are the oxidized and reduced substances

  17. 0 to 2, lost electrons, oxidized +4 0 1+ +5 -2 2 +5 -2 -2 1+ -2 Cu + HNO3 ----> Cu(NO3)2 + NO2 + H2O +1 +5 -6 5 to 4, gained electrons, reduced Cu is oxidized so its called the reducing agent N is reduced so the nitric acid is the oxidizing agent

  18. H in HCl is reduced HCl is the oxidizing agent 0 1 -1 2 -1 0 Fe + HCl ----> FeCl2 + H2 Use oxidation numbers to determine what is oxidized and what is reduced Fe is oxidized it is the reducing agent

  19. S in H2SO4 is reduced sulfuric acid is the oxidizing agent 0 1 6 -2 2 6 -2 1 -2 1 -2 Zn + H2SO4 ----> ZnSO4 + H2S + H2O Use oxidation numbers to determine what is oxidized and what is reduced Zn is oxidized it is the reducing agent

  20. N in nitric acid is reduced it is the oxidizing agent 1 5 -2 1 3 -2 1 5 -2 2 -2 1 -2 HNO3 + H3AsO3 ----> H3AsO4 + NO + H2O Use oxidation numbers to determine what is oxidized and what is reduced As is oxidized arsenic acid is the reducing agent

  21. Cl in HOCl is reduced it is the oxidizing agent 1 -1 1 -2 1 1 5 -2 1 -1 NaI + HOCl ----> NaIO3 + HCl Use oxidation numbers to determine what is oxidized and what is reduced I is oxidized iodide ion is the reducing agent

  22. Balancing Redox Equations Using Oxidation Numbers

  23. 1+ 5+ -2 1+ 3+ -2 1+ 5+ 2- 2+ 2- 1+ 2- 2 HNO3 + H3AsO3 -> H3AsO4 + NO + H2O 3 3 2 1+ 6- 3+ 6- 3+ -8 2e1-/H3AsO3 3e1- /HNO3 Step 1 Assign oxidation numbers to each atom Start with O and H, then use these numbers to find the others Step 2 – find the 2 substances which have changing oxidation numbers Step 3 – make the e1- number the same x HNO3 x 2, x H3AsO3 x 3 Step 4 – balance all other atoms

  24. 1+ -2 0 4+ 2- 1+ 2- 2 H2Se + O2 -> SeO2 + H2O 2 3 2 2e1-/O 4e1-/O2 6e1- /H2Se Step 1 Assign oxidation numbers to each atom Start with O and H, then use these numbers to find the others Step 2 – find the 2 substances which have changing oxidation numbers Step 3 – make the e1- number the same x H2Se x 2, x O2 x 3 Step 4 – balance all other atoms

  25. Balancing Redox Equations in Acidic and Basic Solutions

  26. acidic 5e1-/I, 10 e1-/I2 5 -2 0 5 -2 ClO31- + I2 ---> Cl1- + IO31- 5 +6H1+ 3H2O + 3 5 6 6e1- common multiple of 6 and 10 is 30 notice O is not balanced, there are 15 on the reactant side and 18 on the product side Use water to balance O Use H1+ to balance H

  27. basic 1e1- -2 1 -2 1 7 -2 4 -2 6 -2 6 CH3OH + MnO41- -->CO32- + MnO42- 6 8H1+ 2H2O + + 8OH1- + 8OH1- 6e1- 25 O's on left, 27 O's on right, add 2 H2O's 8 H's on left, add 8 H1+ 's on the right In a basic solution get rid of the H1+ by adding OH1- to each side

  28. basic 1e1- -2 1 -2 1 7 -2 4 -2 6 -2 6 CH3OH + MnO41- -->CO32- + MnO42- 6 2H2O + + 8OH1- + 8H2O 6e1- Cancel excess waters

  29. basic 1e1- -2 1 -2 1 7 -2 4 -2 6 -2 6 CH3OH + MnO41- -->CO32- + MnO42- 6 8OH1- + + 6H2O 6e1- Read the summary on page 667 and try questions 1-4 on page 668

  30. Balancing Redox Equations Using the Ion-Electron Method (Half-Cell Method) MnO41- + S2O32- Mn2+ + S4O62- Step 1 - Separate the overall equation into 2 half-cell equations, an oxidation and a reduction. MnO41- Mn2+ S2O32- S4O62- Step 2 - Balance atoms other than oxygen and hydrogen. 2S2O32- S4O62-

  31. Step 3 - Balance O using water molecules. O’s are already balanced 2S2O32- S4O62- MnO41- MnO41- Mn2+ Mn2+ + 4 H2O’s + 4 H2O’s There are 4 O’s on the left (O4) so add 4 H2O’s to the right Step 4 - When 4 H2O’s were added it created an imbalance of H’s so next balance the H’s using H1+ ions 8H1+ + 8 H’s on right so add 8H1+’s to left

  32. Step 5 -Balance the charges in each half reaction by adding electrons (e1-) to one side 8H1+ + MnO41- Mn2+ 8H1+ + MnO41- +5e1- + 4 H2O’s Mn2+ 2S2O32- S4O62- + 4 H2O’s The charge on the left is (8x1+)+ (1-)= 7+ The charge on the right is 2+ By adding 5e1-’s to the left its charge is reduced to 2+. This balances the charges. +2e1- Charge on left is 4-(2x2-) , Charge on right is 2- so add 2e1-’s on the right making each side 4-

  33. 16H1+ + 2 MnO41- + 10e1- + 8 H2O’s 2Mn2+ 8H1+ + MnO41- +5e1- + 4 H2O’s Mn2+ + 10e1- 10S2O32- 2S2O32- 5S4O62- S4O62- 16H1+ + 2 MnO41- + 10S2O32- + 8 H2O’s 2Mn2+ + 5S4O62- +2e1- Step 6 -Balance the number of e1-’s on each side. Multiply the first equation by 2 and the 2nd equation by 5 to get 10 e1-’s on each side: Add both equations and simplify to get

  34. 16H1+ + 2 MnO41- + 10S2O32- + 8 H2O’s 2Mn2+ + 5S4O62- To check the answer count the following: charge on each side left side (16+)+(2-)+(20-) = 6- right side (4+)+(10-) = 6- O’s on each side left side 8 + 30 = 38, right side 8 + 30 = 38 H’s on each side left side 16, right side 16 S’s on each side, 20 = 20 Mn’son each side, 2 = 2, it’s balanced

  35. Redox Reactions in Acidic/Basic Solutions Breathalysers used by law enforcement rely on the oxidation of ethanol by dichromate ions. The orange dichromate is changed into green Cr3+ and the ethanol changes into ethanoic acid. Write the balanced chemical equation for this redox reaction. Use water to balance the oxygens and H1+ to balance hydrogens.

  36. 2- 1+ 2- 1+ 6+ 2- 3+ 1+ 0 1+ 2- 3 C2H5OH + Cr2O72--> Cr3+ + HC2H3O2 2 4 3 +11H2O 4- 5+ 2- 1+ 12+ 14- 1+ 3+ 4- + 16 H1+ 3e1-/Cr 6e1-/Cr2O72- 2e1-/C 4e1-/ C2H5OH LCM for 4 and 6 is 12 Balance other atoms 17 O on the left, 6 O on the right, add 11H2O 18 H’s on the left, 34 H’s on the right, add 16H1+

  37. Basic Solutions – One additional step is needed. Remove H1+ by adding an equal number of OH1- to each side, simplify by cancelling Try this one Nitrite ion is oxidized by iodine. The products are nitrate and iodide ions

  38. + 2H2O 3+ 2- 0 5+ 2- 1- + 2OH1- + 2OH1- NO21- + I2 -> NO31- + I1- 2 + 2H1+ H20 + 4- 6- 2e1-/NO2 1 e1-/I 2 e1-/I2 Try questions 1-4 pg. 668

  39. Galvanic Cells NO31- K1+ K1+ K1+ Zn2+ Cu 2+ 2NO31- Zn2+ 2NO31- Galvanic Cell Movie Zn/Zn(NO3)2(1.0M)//Cu(NO3)2(1.0M)/Cu oxidized substance//reduced substance Salt bridge K1+ NO31- Zn Cu KNO3(s) NO31- NO31- Cu(NO3)2(s) Zn(NO3)2(s)

  40. Salt bridge K1+ NO31- KNO3(s)

  41. Oxidation - Anode Zn(s) ---> Zn2+ + 2e1- The mass of Zn decreases as the [Zn2+] increases. The region around the Zn electrode becomes positive so NO31- ions migrate in from the salt bridge to maintain electrical neutrality. Zn Cu NO31- Zn2+ Cu 2+ NO31-

  42. Reduction - Cathode Cu2+ + 2e1- ---> Cu(s) The mass of Cu increases as the [Cu2+] decreases. The region around the Cu electrode becomes negative so K1+ ions migrate in from the salt bridge to maintain electrical neutrality Zn Cu K1+ Zn2+ Cu 2+ K1+

  43. Click here to view a movie on alkaline batteries Click here to view a movie on common dry cells Click here to view a movie on lead-acid batteries Click here to view a movie on mercury batteries

  44. When 2 half-cells are connected they produce a voltage. Since a half-cell alone cannot produce a voltage it became necessary for the adoption of a standard half-cell. This standard half cell has hydrogen gas, at a pressure of 101.3 kPa, bubbled over a platinum electrode immersed in a 1.0 mol/L H1+ solution at 25oC.

  45. platinum electrode Here is a standard hydrogen half cell connected to a Cu, Cu(NO3)2 half cell. hydrogen gas Cu

  46. platinum electrode Here is a standard hydrogen half cell connected to a Cu, Cu(NO3)2 half cell. hydrogen gas Cu

  47. platinum electrode Here is a standard hydrogen half cell connected to a Cu, Cu(NO3)2 half cell. hydrogen gas Cu

  48. platinum electrode Here is a standard hydrogen half cell connected to a Cu, Cu(NO3)2 half cell. hydrogen gas Cu

  49. platinum electrode Here is a standard hydrogen half cell connected to a Cu, Cu(NO3)2 half cell. hydrogen gas Cu

  50. platinum electrode Here is a standard hydrogen half cell connected to a Cu, Cu(NO3)2 half cell. hydrogen gas Cu

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