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A bit of a Review of chp . 13.2

A bit of a Review of chp . 13.2. Reminder that the Quest for chp . 13 is March 13. QUICK REVIEW Summary: Five Step method for Predicting Redox Reactions.

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A bit of a Review of chp . 13.2

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  1. A bit of a Review of chp. 13.2 Reminder that the Quest for chp. 13 is March 13

  2. QUICK REVIEW Summary: Five Step method for Predicting Redox Reactions • Step 1: List all entities present and classify each as a possible oxidizing agent, reducing agent, or both. Do not label spectator ions. • Step 2: Choose the strongest oxidizing agent as indicated in a redox table, and write the equation for its oxidation. • Step 3: Choose the strongest reducing agent as indicated in the table, and write the equation for its oxidation • Step 4: Balance the number of electrons lost and gained in the half-reaction equations by multiplying one or both equations by a number. Then add the two balanced half-reaction equations to obtain a net ionic equation. • Step 5: Using the spontaneity rule, predict whether the net ionic equation represents a spontaneous or nonspontaneousredox reaction.

  3. Review: Predicting Redox Reactions #2 given only the reactants to find if it’s a spontaneous redox reaction… Could copper pipe be used to transport a hydrochloric acid solution? • List all entities • Identify all possible OA’s and RA’s • Identify the SOA and SRA • Show ½ reactions and balance • Predict spontaneity Since the reaction is nonspontaneous, it should be possible to use a copper pipe to carry hydrochloric acid

  4. Redox Reactions using Half-Reactions When ½ reaction is not in data booklet (13.2c) Example #3 p.581 • So far we have predicted redox reactions when the ½ reaction was provided to us in the Redox table. • But what if the table does not provide the half reaction??? • We can use our own knowledge to create the equation Rules for Writing Half-Reactions • Write an unbalanced ½ reaction showing formulas for reactants and products • Balance all atoms except H and O • Balance O by adding H2O(l) • Balance H by adding H+(aq) • Balance the charge by adding e- and cancel anything that is the same on both sides For basic solutions only: • Add OH-(aq) to both sides to equal the number of H+(aq) present • Combine H+(aq) and OH-(aq) on the same side to form H2O(l). Cancel equal amounts of H2O(l) from both sides.

  5. Chemistry 30 Chapter 13.3a Oxidation States

  6. Does God hear me… ? • Never a prayer is offered, however faltering, never a tear is shed, however secret, never a sincere desire after God is cherished, however feeble, but the Spirit of God goes forth to meet it. Even before the prayer is uttered or the yearning of the heart made known, grace from Christ goes forth to meet the grace that is working upon the human soul. • –Christ’s Object Lessons p. 206.1 • My soul, wait in silence for God only, For my hope is from Him. He only is my rock and my salvation, My stronghold; I shall not be shaken. On God my salvation and my glory rest; The rock of my strength, my refuge is in God. • Trust in Him at all times, O people; Pour out your heart before Him; God is a refuge for us. • Psalm 62:6-8

  7. Oxidation States • More Complex redox reactions are not adequately described or explained with simple redox theory. • So to describe the oxidation and reduction of molecules and polyatomic ions, chemists developed a method of “electron bookkeeping” to keep track of the loss and gain of electrons. • For atoms in molecules and polyatomic ions, chemists count a shared pair of electrons in a covalent bond as if it belongs entirely to the more electronegative atom in the bond.

  8. Oxidation States • An oxidation state is a useful idea for keeping track of electrons, but it does NOT present an actual charge on an atom. • Oxidation states are arbitrary charges that should NOT be confused with actual electric charges.!! • An oxidation state is defined as the apparentnet electric charge an atom would have if the electron pairs in a covalent bond belonged entirelyto the most electronegative atom.

  9. Oxidation Number • An oxidation number is a positive or negative number corresponding to the oxidation state of the atom in a compound. (These are NOT charges! +2 vs O2+) • Example: In water, which is the most electronegative atom, H or O? • Oxygen is, so we act as if the oxygen owns both electrons in the electron pair. • the oxidation number of an O atom is -2 and that of each hydrogen atom is +1. Each oxygen atom has 8 p+ and 8 e-. But if the oxygen atom gets to count the two hydrogen electrons (red dots) in the two shared pairs, as its own, then it has 8 p+ but 10 e-, leaving an apparent net charge of -2 Each hydrogen atom has 1 p+, but with no additional electron (since oxygen has already counted it), that leaves hydrogen with an apparent net charge of +1

  10. Common Oxidation Numbers (Table 1p. 583) • The sum of oxidation numbers in a compound or ion must equal the total charge of the compound: • zero for neutral compounds • the ion charge for ions (including polyatomic ions) • Note: Finding oxidation states using this method only works if there is just one unknown after referring to Table 1. (If there are 2 or more unknowns, a Lewis formula and electronegativities are required.)

  11. Note: • A monatomic ion is an ion consisting of a single atom. Oxidation # = charge on ion. • (a polyatomic ion, an ion containing more than 1 atom, even if these atoms are of the same element.) • Metals and monatomic anions (-) tend to lose electrons (become oxidized) • Ends in “ide”. Fluoride oxidation # is -1 • Nonmetals and monatomic cations (+) tend to gain electrons (become reduced) • Sodium oxidation # is +1

  12. Rules… • Hydrogen is +1 in all compounds… • for example, in HCl, Cl has a -1 charge and H has a +1 charge to equal zero net charge • EXCEPT H is -1 in Hydrides • eg. lithium hydride LiH • Li has a +1 and H has a -1 oxidation #. H is acting as a – ion (anion). • In Peroxides: Oxygen oxidation # is -1 only when there is an O-O single bond. • Eg. Lithium peroxide Li2O2 or KO2, • Li and K have a +1 so O is -1 • barium peroxide BaO2 • Ba has a +2 oxidation # and so O has a -1 #. • But remember that Oxygen is -2 in OH-, H2O

  13. Oxidation States • Example: What is the oxidation number of carbon in methane CH4? • After referring to Table 1, we assign an oxidation number of +1 to hydrogen (methane is not a hydride so not -1) • So now we have some simple math… • Since a methane molecule is electrically neutral, then the oxidation number of the one carbon atom and the four hydrogen atoms 4(+1) must equal zero. x + 4(+1) = 0 x + (-4) = 0 x = -4 • So the oxidation number of carbon in methane is = -4 • How do we write this?

  14. Oxidation States • Example: What is the oxidation number of manganese in a permanganate ion, MnO4- ? • After referring to Table 1, we assign an oxidation number of -2 to oxygen (it is not a peroxide so the charge is not -1) • Since a permanganate ion has a charge of 1-, then the oxidation number of the one manganese atom and the four oxygen atoms 4(-2) must equal -1. Mn + 4 O = -1 x + 4(-2) = -1 x + -8 = -1 x = -1 + -8 x = -7 • So the oxidation number of manganese is = -7 • Example: What is the oxidation number of sulfur in sodium sulfate? • We know the oxidation numbers of both Na and O, and solve algebraically 2(+1) + x + 4(-2) = 0 2 + x + -8 = 0 So the oxidation number of sulfur is +6

  15. Another way to do the same thing. Tip: • 4*(-2)= -8 • (+8) + (-2) = +6

  16. Redox in Living Organisms • The ability of carbon to take on different oxidation states is essential to life on Earth. • Photosynthesis involves a series of reduction reactions in which the oxidation number of carbon changes from +4 in carbon dioxide to an average of 0 in sugars such as glucose. • The smell of a skunk is caused by a thiol compound (R-SH). To deodorize a pet sprayed by a skunk, you need to convert the smelly thiol to an odourless compound. • Hydrogen peroxide in a basic solution acts as an oxidizing agent to change the thiol to a disulfide compound (RS-SR), which is odourless.

  17. Summary: Determining Oxidation Numbers • Assign common oxidation numbers (Table 1 on page 583) • The total of the oxidation numbers of atoms in a molecule or ion equals the value of the net electric charge of the molecule or ion. • The sum of the oxidation numbers for a compound is zero. • The sum of the oxidation numbers for a polyatomic ion equals the charge of the ion. • Any unknown oxidation number is determined algebraically from the sum of the known oxidation numbers and the net charge on the entity.

  18. Homework… March 6 • Textbook: • p. 585 #1-5 • P. 595 #1-6 • Diploma: •  you get a break today… • So go over old diploma questions that you got wrong to understand them!!

  19. Chem 30 chp. 13.3b p.585-588

  20. Oxidation Numbers and Redox • Although the concept of oxidation states is somewhat arbitrary, because it is based on assigned charges, it is self-consistent and allows predictions of electron transfer. • Chemists believe that if the oxidation number of an atom or ion changes during a chemical reaction, then an electron transfer (oxidation-reduction reaction) occurs. • To recognize a redox reaction, assign oxidation numbers to each atom or ion, then look for any changes in the numbers. • Based on oxidation numbers, • If the oxidation numbers do not change = no transfer of e-’s = not a redoxrxn • An increase in the oxidation number = oxidation • A decrease in the oxidation number = reduction

  21. Example p.586 • In the complete combustion of pure carbon, carbon is converted to carbon dioxide. • In this reaction the oxidation number of carbon changes from 0 in C(s) to +4 in CO2(g) • Simultaneously, oxygen is reduced from 0 in O2(g) to -2 in CO2(g)

  22. Oxidation Numbers and Redox (p.587) • Example: Identify the oxidation and reduction in the reaction of zinc metal with hydrochloric acid. • First write the chemical equation (as it is not provided) • Determine all of the oxidation numbers • Now look for the oxidation number of an atom/ion that increases as a result of the reaction and label the change as oxidation. There must also be an atom/ion whose oxidation number decreases. Label this change as reduction. (Cl- has an oxidation number of -1 throughout the reaction, thus we ignore it here). Cl- + + Cl- -1 -1 + Cl- Cl- + Spectator ion -1 -1 Cl- + + Cl-

  23. Oxidation Numbers and Redox (p.587) • Example: When natural gas burns in a furnace, carbon dioxide and water form. Identify oxidation and reduction in this reaction. • First write the chemical equation for this combustion reaction (as it is not provided) • Determine all of the oxidation numbers • Now look for the oxidation number of an atom/ion that increases as a result of the reaction and label the change as oxidation. There must also be an atom/ion whose oxidation number decreases. Label this change as reduction.

  24. Remember this example from 13.2 p. 580: Predicting Redox Reactions by Constructing Half Reactions • Example: A person suspected of being intoxicated blows into this device and the alcohol in the person’s breath reacts with an acidic dichromate ion solution to produce acetic acid (ethanoic acid) and aqueous chromium(III) ions. Predict the balanced redox reaction equation. • Create a skeleton equation from the information provided: • Separate the entities into the start of two half-reaction equations • Now use the steps you learned for writing half reactions • Now, balance the electrons lost and gained, and add the half reactions. Cancel the electrons and anything else that is exactly the same on both sides of the equation. Note: a similar example is on p.581 (in basic solution)

  25. Identify oxidation and reduction in this reaction (p.588) • If given this reaction • Assign oxidation numbers: • According to the concept of oxidation states, chromium atoms in are reduced (+6 to +3). Carbon atoms in are oxidized (-2 to 0). oxidation -2 +1 -2 +1 +6 -2 +1 0 +1 0 -2 -2 +1 +3 +1 -2 reduction

  26. Homework… March 7 • Textbook: • Pg. 588 #6 and 7 • Pg. 595 #6 (not this one website…) • Diploma: (make use of oxidation numbers) • EX. #33, 35, 37 • MC 2013 #8 • MC 2012#11 • NR 1997 #4

  27. REDOX Reactions Reduction Oxidation • Historically, the formation of a metal from its “ore” (or oxide) • I.e. nickel(II) oxide is reduced by hydrogen gas to nickel metal NiO(s) + H2(g) Ni(s) + H2O(l) Ni +2  Nio • A gainof electrons occurs (so the entity becomes more negative) • Electrons are shown as the reactantin the half-reaction • A species undergoing reduction will be responsible for the oxidation of another entity – and is therefore classified as an oxidizing agent (OA) • Decrease in oxidation number • Historically, reactions with oxygen • I.e. iron reacts with oxygen to produce iron(III) oxide 4 Fe(s) + O2(g) Fe2O3(s) Fe 0  Fe+3 • A lossof electrons occurs (so the entity becomes more positive) • Electrons are shown as the productin the half-reaction • A species undergoing oxidation will be responsible for the reduction of another entity – and is therefore classified as an reducing agent (RA) • Increase in oxidation number

  28. Does God hear me… ? • Never a prayer is offered, however faltering, never a tear is shed, however secret, never a sincere desire after God is cherished, however feeble, but the Spirit of God goes forth to meet it. Even before the prayer is uttered or the yearning of the heart made known, grace from Christ goes forth to meet the grace that is working upon the human soul. • –Christ’s Object Lessons p. 206.1 • My soul, wait in silence for God only, For my hope is from Him. He only is my rock and my salvation, My stronghold; I shall not be shaken. On God my salvation and my glory rest; The rock of my strength, my refuge is in God. • Trust in Him at all times, O people; Pour out your heart before Him; God is a refuge for us. • Psalm 62:6-8

  29. Does God love me… ? • Never a prayer is offered, however faltering, never a tear is shed, however secret, never a sincere desire after God is cherished, however feeble, but the Spirit of God goes forth to meet it. Even before the prayer is uttered or the yearning of the heart made known, grace from Christ goes forth to meet the grace that is working upon the human soul. • –Christ’s Object Lessons p. 206.1 • "Are not two sparrows sold for a cent? And yet not one of them will fall to the ground apart from your Father. But the very hairs of your head are all numbered. So do not fear; you are more valuable than many sparrows." ~Matt. 10:29-31

  30. Chemistry 30 13.3c (day 25) Pg. 589-595 Balancing Redox Equations Using Oxidation Numbers

  31. Take Note… • QUEST MARCH 13!! (Day 28)

  32. Balancing Redox Equations Using Oxidation Numbers • Redox reactions are single replacement reactions. • Oxidation numbers and half-reaction equations can be used to balance any redox reaction no matter how complex it is.

  33. Gain = Loss • Electrons are transferred in a redox reaction, • the total number of electrons lost by one atom/ion must = the total number of electrons gained by another atom/ion • The total increase in oxidation number for a particular atom/ion must = the total decrease in oxidation number of another atom/ion. • “Sour” natural gas (hydrogen sulfide) is toxic and smells like rotting eggs. The “flare” or burning of small amounts of sour natural gas to convert it to sulfur dioxide.

  34. Balancing Redox Equations using Oxidation Numbers #1 (p.590) Example: When hydrogen sulfide is burned in the presence of oxygen, it is converted to sulfur dioxide and water vapour. Use oxidation numbers to balance this equation. H2S(g) + O2(g)  SO2(g) + H2O(g) • Assign oxidation numbers to all atoms/ions and look for the numbers that change. Highlight these. • Notice that a sulfur atom is oxidized from -2 to +4. This is a change of 6 meaning 6 e- have been transferred. • An oxygen atom is reduced from 0 to -2. This is a change of 2 or 2e- transferred. • Because the substances are molecules, not atoms, we need to specify the change in the number of e-’s per molecule • The next step is to determine the simplest whole numbers that will balance the number of electrons transferred for each reactant. The numbers become the coefficients of the reactants. (6e- and 4e- must equal –remember ½ reactions- 6e-*2=12e- and 4e-*3=12e-.) • The coefficients for the products can be obtained by balancing the atoms whose oxidation numbers have changed (here S and O) and then any other atoms.

  35. Balance When only Reactants are Given in an Acid #2 (p.591) Example: Chlorate ions and iodine react in an acidic solution to produce chloride ions and iodate ions. Balance the equation for this reactions. ClO3-(aq) + I2(aq)  Cl-(aq) + IO3-(aq) • Assign oxidation numbers to all atoms/ions and look for the numbers that change. Highlight these. Remember to record the change in the number of electrons per atom and per molecule or polyatomic ion. • The total # of electrons transferred by each reactant must be the same. Multiply the # of electrons by the simplest whole # to make the totals equal, in this case, 30 e-. The numbers become the coefficients of the reactants. The coefficients for the products can be obtained by balancing the atoms whose oxidation numbers have changed and then any other atoms. • Although Cl and I atoms are balanced, oxygen is not. Add H2O(l) molecules to balance the O atoms. • Add H+(aq) to balance the hydrogen. The redox equation should now be completely balanced. Check your work by checking the total numbers of each atom/ion on each side and checking the total electric charge, which should also be balanced.

  36. Balancing Reactants in a Basic Solution#3 (p.592) Example: Methanol reacts with permanganate ions in a basic solution. The main reactants and products are shown below. Balance the equation for this reaction. • Assign oxidation numbers to all atoms/ions and look for the numbers that change. Highlight these. • Remember to record the change in the number of electrons per atom and per molecule or polyatomic ion. • Determine the simplest whole numbers that will balance the number of electrons transferred for each reactant. Multiply the # of electrons by the simplest whole # to make the totals equal. The numbers become the coefficients of the reactants. The coefficients for the products can be obtained by balancing the atoms whose oxidation numbers have changed and then any other atoms. • Add H2O(l) to balance the oxygen, add H+(aq) to balance the hydrogen. • Then add OH-(aq) to both sides to balance the H+(aq) 8 OH-(aq) + + 8 OH-(aq)

  37. Balancing RedoxDisproportionation Equations using Oxidation Numbers #4 (p.592) Example: Household bleach contains sodium hypochlorite. Some of the hypochlorite ions disproportionate (react with themselves) to produce chloride ions and chlorate ions. Write the balanced redox equation for the disproportionation. For disproportionation reactions, start with two identical entities on the reactant side and follow the usual procedure for balancing equations. 2e- of Cl *2 = 4 4e- of Cl*1=4

  38. Balancing Redox Equations using Oxidation Numbers #5 • Example: A person suspected of being intoxicated blows into this device and the alcohol in the person’s breath reacts with an acidic dichromate ion solution to produce acetic acid (ethanoic acid) and aqueous chromium(III) ions. Balance the equation for this reaction. • Remember that the change in electrons is used to find the coefficients. • (4e- C compound *3=12 e- and 6e-Cr compound*2=12 e-) • Remember to balance the C and Cr first, then add H2O(l) to balance O, H+(aq) to balance H and then stop because this is an acidic solution

  39. Summary: Balancing Redox Equations using Oxidation Numbers • Assign oxidation numbers and identify the atoms/ions whose oxidation numbers change • Using the change in oxidation numbers, write the number of electrons transferred per atom. • Using the chemical formulas, determine the number of electrons transferred per reactant. (Use formula subscripts to do this) • Calculate the simplest whole number coefficients for the reactants that will balance the total number of electrons transferred. Balance the reactants and products. • Balance the O atoms using H2O(l), and then balance the H atoms using H+(aq) • Additionally For Basic Solutions Only: • 6. Add OH-(aq) to both sides equal in number to the number of H+(aq) present • 7. Combine H+(aq) and OH-(aq) on the same side to form H2O(l), and cancel the same number of H2O(l) on both sides

  40. Homework… • Textbook: • P.593 #12-15 • Pg. 595 #7-9,12 • Diploma • Ex. #42 • MC 2013 #11, 12 • MC 2012 #14 • MC 1997 #11 • MC 1996 #31

  41. Chemistry 30 13.4 (day 26) Redox Stoichiometry p. 596-600

  42. Connection… • There are many industrial and laboratory applications of redox stoichiometry: • Mining engineer must know the concentration of iron in a sample of iron ore to decide whether or not a mine would be profitable. • Chemical technicians must monitor the concentration of substances in products (i.e. how much bleach is in a disinfectant) • Hospital lab technicians must detect tiny traces of chemicals in human samples.

  43. General Stoichiometry Procedures: • 1. Write a balanced chemical equation with measurements and conversion factors. • 2. Convert given measurements into chemical amounts. • 3. Calculate the amount of the required substance using the mole ration. • 4. Convert this calculated amount to the final requested quantity.

  44. Redox Stoichiometry • How is this different from Chemistry 20 stoichiometry? • We will need to predict the redox equation that will occur, and then we will use the quantities provided to answer the question. • The math is the same as Chem 20, we will just be using our knowledge of redox to start the question. • (Stoichiometry textbook chp. 7 p. 274-303 and chp. 8 p. 320-337)

  45. RedoxStoichiometry • In a titration one reagent (the titrant) is slowly added to another (the sample) until an abrupt change in a solution property (the endpoint) occurs. • In acid-base titrations, the titrant is generally a strong acid or base… • In redox titrations, the titrant is always a strong oxidizing or reducing reagent.

  46. RedoxStoichiometry • When a titration is used to analyze the concentration of a sample, the concentration of the titrant used must be accurately known. • If the titrant is not a standard solution, the titrant is standardized by calculating its concentration using evidence from analysis with a primary standard… • Standardizing involves performing an initial titration with a solution prepared from a solid (so the exact concentration is known) to determine the exact concentration of the titrant. • In a redox titration, it is often necessary to standardize the titrant. Due to the reactive nature of the oxidizing agents used as the titrant, they often react with themselves in their storage container. • A primary standard is a chemical that can be used directly to prepare a standard solution – a solution of precisely known concentration.

  47. Redox Titration • In redox titration, no indicator is required because the titrant is a strong oxidizing agent that has a very significant colour change when it undergoes reduction. • Common OA’s used in redox titration are MnO4-(aq)and Cr2O72-(aq) , both in acidified solutions (in data booklet).

  48. Redox Stoichiometry • Example #1 • A strong acid is painted onto a copper sheet to etch a design. If 500 mL of a 0.250 mol/L solution is used, what mass of copper will react? • List entities present, identify SOA and SRA: H+(aq)Cu(s) H2O(l) • Write oxidation and reduction half reactions. Balance the number of electrons gained and lost and add the reactions 2H+(aq) + 2e- H2(g) Cu(s) Cu2+(aq) + 2e- 2H+(aq) + Cu(s) H2(g) + Cu2+(aq) V= 500mL m= ???g 0.250mol/L M=63.55g/mol 0.500 L x0.25 molH+(aq)x1 mol Cu(s)x63.55g = 3.97 gCu(s) L 2 mol H+(aq) mol Cu(s) SOA SRA

  49. Redox Stoichiometry • Example #2 • Nickel metal is oxidized to Ni2+(aq) ions by an acidified potassium dichromate solution. If 2.50g of metal is oxidizes by 50.0 mL of solution, what is the concentration of the K2Cr2O7(aq) solution? • List entities present, identify SOA and SRA: Ni(s)H+(aq)K+(aq) Cr2O72-(aq) H2O(l) • Write oxidation and reduction half reactions. Balance the number of electrons gained and lost and add the reactions 3 [Ni(s) Ni2+(aq) + 2e-] Cr2O72-(aq) + 14 H+(aq)+ 6 e- 2Cr3+(aq) + 7H2O(l) 3Ni(s)+ Cr2O72-(aq) + 14 H+(aq) 3Ni2+(aq + 2Cr3+(aq) + 7H2O(l) 2.50g50.0mL M=58.62g?mol/L 2.50 gx mol Ni(s)x1 mol Cr2O72-(aq)x __1__= 0.284 mol/L Cr2O72-(aq) 58.69 g 3 mol Ni(s) 0.0500L SOA SRA

  50. Example Lab Exercise 13.C (pg. 598) • What is the amount concentration of tin(II) ions in a solution prepared for research on toothpaste? • The titration evidence collected is below. • Average volume added 12.4mL + 12.3mL +12.5mL =12.4mL KMnO4(aq) 3 Titration of 10.00mL of acidic Sn2+(aq) with 0.0832 mol/L KMnO4(aq)

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