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Chapter 4

Chapter 4. Red uction- Ox idation Reactions Redox Reactions. Sodium chloride. Sodium chloride. Na  Na + + e . Cl 2 + 2e   2Cl . Involves 2 processes: Oxidation = Loss of Electrons Na  Na + + e  Oxidation Half-Reaction Reduction = Gain of electrons

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Chapter 4

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  1. Chapter 4 Reduction-Oxidation Reactions Redox Reactions

  2. Sodium chloride

  3. Sodium chloride Na  Na+ + e Cl2 + 2e 2Cl

  4. Involves 2 processes: Oxidation = Loss of Electrons Na  Na+ + eOxidation Half-Reaction Reduction = Gain of electrons Cl2 + 2e 2ClReduction Half-Reaction Net reaction: 2Na + Cl2 2Na+ + 2Cl Oxidation & reduction always occur together Can't have one without the other Oxidation–Reduction Reactions

  5. Oxidation Reduction Reaction Oxidizing Agent - Substance that accepts e's • Accepts e's from another substance • Substance that is reduced Cl2 + 2e 2Cl– Reducing Agent - Substance that donates e's • Releases e's to another substance • Substance that is oxidized Na  Na+ + e–

  6. Your Turn! Which species functions as the oxidizing agent in the following oxidation-reduction reaction? Zn(s) + Pt2+(aq) Pt(s) + Zn2+(aq) Pt(s) Zn2+(aq) Pt2+(aq) Zn(s) None of these, as this is not a redox reaction.

  7. Redox Reactions • Very common • Batteries—car, flashlight, cell phone, computer • Metabolism of food • Combustion • Chlorine Bleach • Dilute NaOCl solution • Cleans through redox reaction • Oxidizing agent • Destroys stains by oxidizing them

  8. Redox Reaction Ex. Fireworks displays Net: 2Mg + O2 2MgO Oxidation: Mg  Mg2+ + 2e • Loses electrons = Oxidized • Reducing agent Reduction: O2 + 4e 2O2 • Gains electrons = Reduced • Oxidizing agent

  9. Redox Reaction? S + O2 SO2 • Combustion: Oxidation in old sense, reaction with oxygen • But n no ions in SO2 • How can we decide which loses and which gains!! • Oxidation number (state)! • If compound were ionic, what would the charges have been.

  10. Rules for Oxidation States (Numbers) • The sum of Oxidation numbers equals to the charge on molecule, formula unit or ion. • The oxidation state of elements is zero. • Oxidation state for monoatomic ions are the same as their charge. • In its compounds fluorine is always –1. • Hydrogen is assigned the oxidation state +1. • Oxygen is assigned an oxidation state of -2 in its covalent compounds (except as a peroxide). • If two rules conflict, apply higher rule.

  11. Oxidation States • Assign the oxidation states to each element in the following. • CO2 • NO3- • H2SO4 • Fe2O3 • Fe3O4 • Cr2O72- • O2F2 • H2O2 • LiH • BaO2

  12. Oxidation-Reduction Transfer electrons, so the oxidation states change. Ox Ox Oxidation Increase in Oxid. number Reduction decrease in Oxid. number Red Red

  13. Ox Ox Red Red Red C (CH4) oxidized → CH4 reducing agent O2 reduced → O2 oxidizing agent

  14. Red Ox PbS has been oxidized, PbS is the reducing agent. O2 has been reduced, O2 is the oxidizing agent.

  15. Red Ox PbO has been reduced, PbO is the oxidizing agent. CO has been oxidized, CO is the reducing agent.

  16. Identify the 1) Oxidizing agent 2) Reducing agent 3) Substance oxidized 4) Substance reduced in the following reactions • Fe(s) + O2(g) ® Fe2O3(s) • Fe2O3(s)+ 3 CO(g) ® 2 Fe(l) + 3 CO2(g) • SO3- + H+ + MnO4- ® SO4- + H2O + Mn+2

  17. Balancing Redox ReactionsIon-Electron Method – Acidic Solution 1. Divide equation into 2 half-reactions 2. Balance atoms other than H & O 3. Balance O by adding H2O to side that needs O 4. Balance H by adding H+ to side that needs H 5. Balance net charge by adding e– 6. Make e– gain equal e– loss; then add half-reactions 7. Cancel anything that is the same on both sides

  18. Balance in Acidic Solution Cr2O72– + Fe2+ Cr3+ + Fe3+ 1.Break into half-reactions Cr2O72 Cr3+ Fe2+ Fe3+ 2.Balance atoms other than H & O Cr2O722Cr3+ • Put in 2 coefficient to balance Cr Fe2+ Fe3+ • Fe already balanced

  19. 3. Balance O by adding H2O to the side that needs O. Cr2O72 2Cr3+ • Right side has 7 O atoms • Left side has none • Add 7 H2O to left side Fe2+ Fe3+ • No O to balance + 7 H2O

  20. 4.Balance H by adding H+ to side that needs H Cr2O72  2Cr3+ + 7H2O • Left side has 14 H atoms • Right side has none • Add 14 H+ to right side Fe2+ Fe3+ • No H to balance 14H++

  21. 5.Balance net charge by adding electrons. 14H+ + Cr2O72 2Cr3+ + 7H2O • 6 electrons must be added to reactant side Fe2+  Fe3+ • 1 electron must be added to product side • Now both half-reactions balanced for mass & charge 6e + Net Charge = 14(+1) (–2) = 12 Net Charge = 2(+3)+7(0) = 6 + e

  22. 6. Make e– gain equal e– loss; then add half-reactions 6e + 14H++ Cr2O72– 2Cr3+ + 7H2O Fe2+ Fe3++ e 7. Cancel anything that's the same on both sides 6[ ] 6Fe2+ + 14H+ + Cr2O72 6Fe3++ 2Cr3+ + 7H2O 6e + 6Fe2++ 14H+ + Cr2O72 6Fe3++ 2Cr3++ 7H2O+ 6e  

  23. Practice • The following reactions occur in acidic aqueous solution. Balance them: • MnO4-+ Fe2+® Mn2++ Fe3+ • Cu + NO3- ®Cu2++ NO(g) • Pb + PbO2 + SO42-® PbSO4 • Mn2++ NaBiO3®Bi3++ MnO4- • Cr2O72- + C2H5OH ® Cr3+ + CO2

  24. Ion-Electron method in Basic Solution • The simplest way to balance an equation in basicsolution Use steps 1-7 above, then 8. Add the same number of OH– to both sides of the equation as there are H+. 9. CombineH+ & OH– to form H2O 10. Cancel any H2O that you can from both sides

  25. Basic Solution • Ag + CN- +O2® Ag(CN)2- • Cr(OH)3 + OCl- + OH-® CrO42- + Cl- + H2O • CrI3+ Cl2® CrO4-+ IO4- + Cl-

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