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Oxidation-Reduction Reactions

Oxidation-Reduction Reactions. Oxidation Numbers Oxidation Numbers and Nomenclature Identifying Oxidation-Reduction Reactions Writing Equations for Oxidation-Reduction Reactions Oxidation-Reduction Titrations Oxidation by Oxygen. Oxidation Numbers. Definition:

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Oxidation-Reduction Reactions

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  1. Oxidation-Reduction Reactions • Oxidation Numbers • Oxidation Numbers and Nomenclature • Identifying Oxidation-Reduction Reactions • Writing Equations for Oxidation-Reduction Reactions • Oxidation-Reduction Titrations • Oxidation by Oxygen

  2. Oxidation Numbers • Definition: • An oxidation number is a number that reflects the electrons gained, lost, or shared when an element reacts. • Remember that when you lose electrons you become more positive and when you gain electrons you become more negative.

  3. Rules for Assigning Oxidation Numbers • The oxidation # of any uncombined element is zero. • The oxidation # of a monoatomic ion equals its’ charge. • The more electrongative element in a binary compound is assigned the number equal to the charge it would have if it were an ion. • The oxidation number of fluorine is always -1. • Oxygen has an oxidation number of -2. • Hydrogen has an oxidation number of +1. • In compounds, Group 1 and 2 elements and aluminum have oxidation numbers of +1, +2, and +3 respectively. • The sum of the oxidation numbers of atoms in a compound is zero. • The sum of the oxidation # in a polyattomic ion is equal to the charge of the ion.

  4. Oxidation number or state • When dealing with simple ions, this is easy to determine. It is simply the charge on the ion. • Examples • Group IA (1) +1 • Group IIA (2) +2 • Group VIIA (17) -1 • Oxygen -2 usually • Hydrogen +1 if bonded to nonmetal • Hydrogen -1 if bonded to metal

  5. Oxidation number • For elements in their elemental state, the oxidation number is also pretty straightforward. • Since all of the atoms are the same, the electrons are shared equally so the oxidation number is zero. • Examples • The atoms in N2, Na, P4, H2 and O2 all have oxidation numbers of zero.

  6. Oxidation numbers • With polar covalent bonds electrons are shared but not equally. • For electrons that are shared in these compounds, we assign the shared electrons to the most electronegative element. • We are just acting as though the electronegativity difference was large enough for the transfer of electrons to occur.

  7. Example • Assign the oxidation states for all elements in water. • The electronegativities are: • H = 2.2, O = 3.5 • The electrons from both hydrogen are assigned to the oxygen. • Oxidation numbers: O = -2 • H = +1

  8. Oxidation numbers • Many elements have more than one possible oxidation number. • Often, it is possible to determine the oxidation number of those elements in a compound simply by looking at what you do know. • Follow the previous rules and then assign an oxidation number that insures that the overall compound has no net charge.

  9. Example • Find the oxidation state for all elements in: HNO3 Hydrogen we know - it must be +1 Oxygen should be -2 in this case. What about nitrogen?

  10. Example • See what you do know and find the difference. • HNO3 We know that H is assigned a value of +1 Oxygen is -2 and we have 3 of them = -6 OK, so what's left over?? (+1) + (-6) + charge on nitrogen = 0 Nitrogen must have a value of +5

  11. Determining Oxidation Numbers • SO2 NO3- • PCl5 SO42- • H2O NH4+ • H2SO3 MnO4-

  12. Li +1 Be +2 Na +1 Mg +2 K +1 Ca +2 Rb +1 Sr +2 Cs +1 Ba +2 Fr +1 Ra +2 B +3 C +4 -2 -4 N +5 +4 +3 +2 +1 -3 O -1 -2 F -1 Ne H +1 He Al +3 Si +4 -4 P +5 +3 -3 S +6 +4 +2 -2 Cl +7 +5 +3 +1 -1 Ar Sc 3+ Ti +4 +3 +2 V +5 +4 +3 +2 Cr +6 +3 +2 Mn +7 +6 +4 +3 +2 Fe +3 +2 Co +3 +2 Ni +2 Cu +2 +1 Zn +2 Ga +3 Ge +4 -4 As 5+ 3+ 3- Se 6+ 4+ 2- Br +5 +1 -1 Kr +4 +2 Y +3 Zr +4 Nb +5 +4 Mo +6 +4 +3 Tc +7 +6 +4 Ru +8 +6 +4 +3 Rh +4 +3 +2 Pd +4 +2 Ag +1 Cd +2 In +3 Sn +4 +2 Sb +5 +3 -3 Te +6 +4 -2 I +7 +5 +1 -1 Xe +6 +4 +2 Lu +3 Hf +4 Ta +5 W +6 +4 Re +7 +6 +4 Os +8 +6 Ir +4 +3 Pt +4 +2 Au +3 +1 Hg +2 +1 Tl +3 +1 Pb +4 +2 Bi +5 +3 Po +2 At -1 Rn Lr +3 Common oxidation numbers

  13. Oxidation numbers and the periodic table • Some observed trends in compounds. • Metals have positive oxidation numbers. • Transition metals typically have more than one oxidation number. • Nonmetals and semimetals have both positive and negative oxidation numbers. • No element exists in a compound with an oxidation number greater than +8. • The most negative oxidation numbers equals 8 - the group number

  14. Oxidation number and nomenclature • Stock system • For metals with several possible oxidation numbers, use Roman numeral in the name. • FeSO4 iron(II) sulfate • Fe2(SO4)3 iron (III) sulfate • Cu2O copper(I) oxide • CuO copper(II) oxide • PbCl2 lead(II) chloride • PbCl4 lead(IV) chloride

  15. Identifying oxidation-reduction reactions. • Oxidation-Reduction - REDOX • A chemical reaction where there is a net change in the oxidation number of one or more species. • Both an oxidation and a reduction must occur during the reaction. Mg (s) + Cl2 (g) MgCl2 (s) Here the oxidation number of Mg has changed from zero to +2. Cl has changed from zero to -1.

  16. REDOX reactions • Oxidation • An increase in oxidation number. • Reduction • A decrease in oxidation number. • If the oxidation number of any element changes in the course of a reaction, the reaction is oxidation-reduction. • Example. • 2 Fe(NO3)3 (aq) + Zn(s) 2 Fe(NO3)2 (aq) + Zn(NO3)2 (aq)

  17. Redox Reaction • Fe(s) + O2(g)  Fe2O3(s) • Iron is oxidized (0 to +3) • and O2 is reduced (0 to -2)

  18. Half Reactions • Fe  Fe3+ + 3e- (oxidation) • O2 + 4e-  2 O2- (reduction)

  19. Half reactions • Example. • Half-reactions can be of the ‘net ionic’ form. Balance the follow • Fe3+ + Zn (s) Fe2+ + Zn2+ • 2 ( Fe3+ + e- Fe2+) (reduction) • Zn(s) Zn2+ + 2e- (oxidation) • 2Fe3+ + Zn(s) 2Fe2+ + Zn2+

  20. Example 2Fe(NO3)3 (aq) + Zn(s) 2Fe(NO3)2 (aq) Zn(NO3)2 (aq) +3 0 +2 +2 Fe3+ is reduced to Fe2+ Zn is oxidized to Zn2+ NO3- is a spectator ion.

  21. Zinc - Copper Voltaic Cell

  22. A Zinc-Copper Voltaic Cell Cu electrode Cathode (+) Reduction Half Reaction: Cu2+ + 2 e- Cu Zn electrode Anode (-) Oxidation Half Reaction: Zn  Zn2+ + 2 e-

  23. Voltaic Cell cont… • In a voltaic cell the redox reaction is spontaneous and produces an electric current. This current can be measured using a voltmeter.

  24. Standard Electrode Potentials • A standard electrode potential, Eo, is based on the tendency for reduction to occur at the electrode. • The cell voltage, called the standard cellpotential (Eocell), is the difference between the standard potential of the cathode and that of the anode. • Eocell = Eo (cathode) – Eo (anode)

  25. Standard Electrode Potentials

  26. Balancing REDOX equations • Many REDOX equations can be balanced by inspection. • H2S (g) + H2O2(aq) S (s) + 2 H2O (l) • However, others are more difficult. • 2KMnO4 (aq) + H2O2 (l) + 3H2SO4 (aq) • 2MnSO4 (aq) + K2SO4 (aq) + 3O2 (g) + 4H2O(l)

  27. Balancing REDOX equations • Half-Reaction method. • With this approach, the reaction is broken into two parts. • Oxidation half-reaction. The portion of the reaction where electrons are lost. • A An+ + ne- • Reduction half-reaction. The portion of the reaction where electrons are gained. • me- + B Bm-

  28. Balancing REDOX equations • The goal is then to make sure that the same number of electrons are being produced and consumed. • (m) ( A An+ + ne- ) • (n) (me- + B Bm- ) • nB + mA mAn+ + nBm+ • When properly balanced, the electrons will cancel out.

  29. Half reactions • Another Example • Determine the balanced equation for the reaction of Fe2+ with Cr2O72- in an acidic solution. • Fe2+ + Cr2O72- Fe3+ + Cr3+ • The two half-reactions would be: • Fe2+ Fe3+ • Cr2O72- Cr3+ H+

  30. Half reactions • First, balance each half-reaction for all elements except hydrogen and oxygen. • Fe2+ Fe3+ • Cr2O72- 2Cr3+ • Next, balance each half-reaction with respect to oxygen by adding an appropriate number of H2O. • Fe2+ Fe3+ • Cr2O72- 2Cr3+ + 7H2O

  31. Half reactions • Remember that this reaction occurs in an acid solution so we can add H+ as needed. • Fe2+ Fe3+ • 14H+ + Cr2O72- 2Cr3+ + 7H2O • Now we need to know how many electrons are produced or consumed and place them in our half-reactions. • For iron, one e- is produced. • For dichromate, six e- are consumed.

  32. Half-reactions Fe2+ Fe3+ + e- • 6e- + 14 H+ + Cr2O72- 2Cr3+ + 7H2O • We need the same number of electrons produced and consumed so: • 6Fe2+ 6Fe3+ + 6e- • 6e- + 14H+ + Cr2O72- 2Cr3+ + 7H2O • As our final step, we need to combine the half-reactions and cancel out the electrons.

  33. Half-reactions • 6Fe2+ + 14H+ + Cr2O72- • 6Fe3+ + 2Cr3+ + 7H2O • In this reaction, Fe2+ is oxidized and the dichromate ion is reduced. • This reaction is used for the determination of iron by titration.

  34. Disproportionation reactions • In some reactions, the same species is both oxidized and reduced. • Examples • 2H2O2 (l) 2H2O (l) + O2 (g) • 3Br2 (aq) +6OH- (aq) BrO3-(aq) +5Br-(aq) +3H2O(l) • For this to occur, the species must be in an intermediate oxidation state. Both a higher and lower oxidation state must exit.

  35. Oxidation-reduction titrations • REDOX reactions can also serve as the basis for titrations. • For example, we can determine the amount of iron in an ore by titration. • Initially, we must dissolve the sample. This results in both iron(II) and iron(III) being produced in solution. • The first step is to get all of the iron into one oxidation state.

  36. Sample preparation • One option is to use a reductor. You slowly wash your sample through the column with water. • Jones Reductor • Zn(Hg) Zn2+ + Hg(l) + 2e • An amalgam is used to prevent • Zn + 2H+ Zn2+ + H2(g) • For iron, we get • Fe2+ + Fe3+ Fe2+ reductor

  37. Titrants • Now that all of our iron is in a single oxidation state, we’re ready to do a titration. • We need an oxidizing agent to convert all the iron from Fe2+ to Fe3+. • Primary standard • A material that is available in pure form. • Ideally, it should be something that we can directly weigh out, dissolve and then titrate with.

  38. Common titrants • Oxidizing titrants • Dichromate -Cr2O72- • The potassium salt is a primary standard material. • Very stable solutions. If air is kept out, it can last for years. • It is a very strong oxidizing agent. • Need an indicator such as diphenylamine sulfonic acid.

  39. Common titrants • Oxidizing titrants • Permanganate - MnO4- • The potassium salt is the most commonly used. It is not a primary standard. • Solutions must be standardized - typically use Na2C2O4 ( a primary standard material.) • Reagent slowly degrades and MnO2 must be removed • No indicator is needed - excess reagent produces a pink solution.

  40. Common titrants • The standardization of MnO42- with oxalate involves the following reaction. • 2MnO42-(aq) + 5C2O42-(aq) + 8H+(aq) • 2Mn2+(aq) + 10CO2 (g) + 8H2O (l) • Since MnO42- in an intense magenta color and Mn2+ is a faint green, detecting the endpoint for the titration is easy.

  41. Oxidation by oxygen • Oxygen is not the strongest of oxidizing agents but it is about 19% of our atmosphere. • It is able to react with all other elements except: noble gases • halogens • noble metals like gold • One very common reaction for oxygen is: • Combustion - rapid oxidation accompanied by heat and usually light.

  42. Combustion • Examples • CH4(g) + 2O2(g) CO2(g) + 2H2O(g) • S(s) + O2(g) SO2(g) • N2(g) + O2(g) 2 NO(g) • 2NO(g) + O2 (g) 2 NO2(g) • Note. All of these oxides contribute to air pollution. CO2 contributes to the “greenhouse effect.”

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