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Electrochemistry

Electrochemistry. Important Class of Chemical Reactions:. Explains rusting of metals Photography Biological reactions (operation of mitochondria; transfer of O 2 by hemoglobin) Operation of a car battery Industrial production of chemicals such as Cl 2 , NaOH, F 2 and Al

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Electrochemistry

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  1. Electrochemistry

  2. Important Class of Chemical Reactions: • Explains rusting of metals • Photography • Biological reactions (operation of mitochondria; transfer of O2 by hemoglobin) • Operation of a car battery • Industrial production of chemicals such as Cl2, NaOH, F2 and Al • Horsepower of a race car (oxidation of fuels)

  3. Characterizing Oxidation and Reduction • Mg(s) + O2(g)  2MgO(s) • Net IONIC Equation: • Mg(s) + O2(g)2Mg+2(s) + 2O-2(s) • Notice that: • Mg(s)  Mg+2 = Lost 2 electrons • O2(g)  O-2 = gained 2 electrons

  4. Electron Transfer Reactions • Electron transfer reactions are oxidation-reduction or redox reactions. • Results in the generation of an electric current (electricity) or be caused by imposing an electric current. • Therefore, this field of chemistry is often called ELECTROCHEMISTRY.

  5. Demo: Zn(s) + CuSO4(aq)  _______________ ACTIVITY SERIES OF METALS • Balanced Chemical Equation • Total Net Ionic Equation • Net Ionic Equation

  6. Let’s Examine Reaction… • Zn (s) + Cu+2 + SO4-2 Zn+2 + SO4-2 + Cu(s) Spectator Ions Lost 2 e Gained 2 e Zn – lost 2 e  Zn got OXIDAZED Cu – gained 2 e  Cu got REDUCED

  7. You can’t have one… without the other! • Reduction (gaining electrons) can’t happen without an oxidation to provide the electrons. • You can’t have 2 oxidations or 2 reductions in the same equation. Reduction has to occur at the cost of oxidation LEO the lion says GER! ose lectrons xidation ain lectrons eduction GER!

  8. To understand electrochemistry, we need to be able to keep track of electrons, we do so by assigning oxidation numbers.

  9. Terminology for Redox Reactions • OXIDATION—loss of electron(s) by a species; increase in oxidation number; increase in oxygen. • REDUCTION—gain of electron(s); decrease in oxidation number; decrease in oxygen; increase in hydrogen. • OXIDIZING AGENT—electron acceptor; species is reduced. (an agent facilitates something; ex. Travel agents don’t travel, they facilitate travel) • REDUCING AGENT—electron donor; species is oxidized.

  10. Reducing vs. Oxidizing AGENTSZn (s) + Cu+2 Zn+2 + Cu(s) Oxidizing agent Reducing agent Donates electrons Undergoes oxidation • Accepts electrons • Undergoes reduction

  11. LEARNING CHECK #1 • What is the difference between oxidation and reduction? • Explain why an oxidation reaction must be accomplished by a reduction reaction.

  12. Assigning Oxidation Numbers At the end of this section, please complete WORKSHEET #1

  13. Review of Oxidation numbers The charge the atom would have in a molecule (or an ionic compound) if electrons were completely transferred. • Free elements (uncombined state) have an oxidation number of zero. Na, Be, K, Pb, H2, O2, P4 = 0 • In monatomic ions, the oxidation number is equal to the charge on the ion. Li+, Li = +1; Fe3+, Fe = +3; O2-, O = -2 • The oxidation number of oxygen is usually–2. In H2O2 and O22- it is –1. 4.4

  14. Oxidation numbers of all the atoms in HCO3- ? • The oxidation number of hydrogen is +1except when it is bonded to metals in binary compounds. In these cases, its oxidation number is –1. (Example: NaH) • Group IA metals are +1, IIA metals are +2 and fluorine is always –1. 6. The sum of the oxidation numbers of all the atoms in a molecule or ion is equal to the charge on the molecule or ion. HCO3- O = -2 H = +1 3x(-2) + 1 + ? = -1 C = +4 4.4

  15. Rules for assigning Oxidation Numbers  pg.658 Table 1 • Examples: • Cl2 • Na2O • Fe2O3 • PbSO4 • KNO2 • Fe(NO3)3 • Supply the oxidation number of the underlined element in the following formulas; a) Zn3(PO4)2, b) NaNO2, c) SnBr2, d) HSbO2, e) Mg(MnO4)2, f) NH4NO3

  16. STOP: Turn to Worksheet #1 • K/U = 16 marks + 10 marks • Find the oxidation number of each atom • Identify the reactant oxidized and the reactant reduced in each of the following equations. Which of the following equations represent redox reactions? http://www2.stetson.edu/mahjongchem/

  17. REDOX reactionsMcGrawhill (tutorial)

  18. http://www2.stetson.edu/mahjongchem/ - GAME

  19. Redox Reactions Involving Ionic Compounds Writing Balanced Half-Reactions Balancing (Half method; Acidic solution; Basic Solution)

  20. Writing Balanced Half-Reactions • Zn (s) + Cu+2 Zn+2 + Cu(s) • Oxidation Half-Reaction: Zn(s)  Zn+2 • Reduction Half-Reaction:Cu+2 Cu(s) Zn0(s)  Zn+2 Cu+2 Cu0(s) + 2e- + 2e- Zn(s) + Cu+2 Zn+2 + Cu(s)

  21. H+ H2 Step 1: H+ H2 Step 2: 2H+ H20 Step 3: +2 0 (not equal) Step 4: +2 add 2e-  0 Step 5: count again 0 = 0

  22. Learning Check #3 Write balanced half-reactions based on each of the following ionic equations: • Al(s) + Fe+3 Al+3 + Fe(s) • Fe(s) + Cu+2  Fe+2 + Cu(s) • Sn(s) + Pb+2  Sn+2 + Pb(s)

  23. Balancing EQ using Half-Reactions with Differing Number of Electrons • K(s) + Cl2(g)  KCl(s) • RULES: • Identify the key element that undergoes an oxidation state change. • Balance the number of atoms of the key element on both sides. • Add the appropriate number of electrons to compensate for the change of oxidation state. • Add two equations

  24. K(s) + Cl2(g)  KCl(s) • Oxidation: K(s)  K+1 • Reduction: Cl2(g)  Cl-1 Balance the atoms: • K(s)  K+1 - K is balanced on both sides • Cl2  Cl-1 • There are TWO Cl on the left and only ONE of the right ...multiply by 2 • Cl2 2Cl-1 Balance charges Eq1: K(s) K+1 + 1e- Eq2: Cl2+ 2e-  2Cl-1 Make sure electrons cancel out....therefore multiply eq.1 by 2 2K(s) 2K+1 + 2e- Cl2+ 2e-  2Cl-1 Add two equations 2K(s) + Cl2 2K+1 + 2Cl-1

  25. Al +Cu2+ --> Cu + Al 3+ Start by writting half reactions (Oxidation and reduction) Oxidation: Al --> Al 3+ + 3e- Reduction: 2e- +Cu2+ --> Cu • Balance the electrons by finding the common multiple and multiply the half reactions accordingly. The common multiple of the electrons is 6 so • Oxidation: 2 x (Al --> Al 3+ + 3e-) • Reduction: 3 x ( 2e- +Cu2+ --> Cu) • Recombine the reactions 6e- + 2 Al +3 Cu2+--> 2 Al 3++3Cu + 6e- The electrons must cancel. 2 Al +3 Cu2+--> 2 Al 3+ + 3Cu Atoms and charges must be conserved.

  26. Example 1 (Acidic Solution)MnO4-1 + I-1 --> I2 + Mn2+ • Step 1 Half Reactions MnO4-1  Mn2+ I-1  I2 • Lets balance the reduction one first • for every Oxygen add a water on the other side • For every hydrogenadd a H+ to the other side • Balance the imbalance of charge with electrons (+7 vs. +2) • MnO4- --> Mn2+ + 4H2O • 8H++ MnO4- --> Mn2+ + 4H2O +8-1=+7  +2+0=+2 • 5e-+ 8H+ + MnO4- --> Mn2+ + 4H2O -5+8-1=+2  +2+0=+2

  27. Continued... • Now for the oxidation • Balance the atoms • Balance the imbalance of charge with electrons (-2 vs. 0) • Step 2 Balance the electrons by finding the common multiple and multiply the half reactions accordingly. Common Multiple here is 10. • I- 1--> I2 • 2I-1 --> I2 • 2I- --> I2 + 2e- -2 0 -2 • 2( 5e- + 8H+ + MnO4- --> Mn2+ + 4H2O ) • 5( 2I- --> I2 + 2e- )

  28. Step 3 Check electrons, atoms and charge. Clean it up 10e- + 16H+ + 2MnO4- + 10I--->5I2 + 2Mn2+ + 8H2O + 10e- 16H+ + 2MnO4- + 10I--->5I2 + 2Mn2+ + 8H2O

  29. Fe2+ + Cr2O72- Fe3+ + Cr3+ +2 +3 Fe2+ Fe3+ +6 +3 Cr2O72- Cr3+ Cr2O72- 2Cr3+ Example 2 (Acidic Solution) The oxidation of Fe2+ to Fe3+ by Cr2O72- in acid solution? • Write the unbalanced equation for the reaction ion ionic form. • Separate the equation into two half-reactions. Oxidation: Reduction: • Balance the atoms other than O and H in each half-reaction. 19.1

  30. Fe2+ Fe3+ + 1e- 6Fe2+ 6Fe3+ + 6e- 6e- + 14H+ + Cr2O72- 2Cr3+ + 7H2O 6e- + 14H+ + Cr2O72- 2Cr3+ + 7H2O 14H+ + Cr2O72- 2Cr3+ + 7H2O Cr2O72- 2Cr3+ + 7H2O Balancing Redox Equations • For reactions in acid, add H2O to balance O atoms and H+ to balance H atoms. • Add electrons to one side of each half-reaction to balance the charges on the half-reaction. • If necessary, equalize the number of electrons in the two half-reactions by multiplying the half-reactions by appropriate coefficients. 19.1

  31. 14H+ + Cr2O72- + 6Fe2+ 6Fe3+ + 2Cr3+ + 7H2O 6Fe2+ 6Fe3+ + 6e- 6e- + 14H+ + Cr2O72- 2Cr3+ + 7H2O Balancing Redox Equations • Add the two half-reactions together and balance the final equation by inspection. The number of electrons on both sides must cancel. You should also cancel like species. Oxidation: Reduction: • Verify that the number of atoms and the charges are balanced. 14x1 – 2 + 6x2 = 24 = 6x3 + 2x3 • For reactions in basic solutions, add OH- to both sides of the equation for every H+ that appears in the final equation. You should combine H+ and OH- to make H2O. 19.1

  32. Balancing redox reactions under Basic Conditions Cr(OH)3 + ClO3-1 --> CrO42- + Cl-1 (basic) • Lets balance the reduction one first • for every Oxygen add a water on the other side • For every hydrogen add a H+to the other side • Each H+ will react with an OH-on both sides (whatever you do to one side...you do the same to the other) • H+ and OH- make water • cancel the waters • Balance the imbalance of charge with electrons (-1 vs. -7) • ClO3- --> Cl- • ClO3- --> Cl- + 3H2O • 6H+ + ClO3- --> Cl- + 3H2O • 6 OH-+6H+ + ClO3- --> Cl- + 3H2O + 6 OH- • 6H2O + ClO3- --> Cl- + 3H2O + 6 OH- • 3H2O + ClO3- --> Cl- + 6 OH- • 6e- + 3H2O + ClO3- --> Cl- + 6 OH-

  33. Continued... • Now for the oxidation • for every Oxygen add a water on the other side • For every hydrogenadd a H+to the other side • Each H+ will react with an OH- on both sides • H+ and OH- make water • cancel the waters • Balance the imbalance of chagre with electrons (-2 vs. 0) • Cr(OH)3 --> CrO42- • H2O + Cr(OH)3--> CrO42- • H2O + Cr(OH)3--> CrO42- +5H+ • 5 OH- + H2O + Cr(OH)3 --> CrO42- +5H++ 5OH- • 5 OH- + H2O + Cr(OH)3 --> CrO42- + 5H2O • 5 OH- + Cr(OH)3 --> CrO42- + 4H2O • 5 OH- + Cr(OH)3 --> CrO42- + 4H2O + 3e-

  34. Step 2 Balance the electrons by finding the common multiple and multiply the half reactions accordingly. Common Multiple here is 6 • 1(6e-+ 3H2O + ClO3- --> Cl- + 6OH- ) • 2(5 OH- + Cr(OH)3 --> CrO42- + 4H2O + 3e-) Step 3 Check electrons, atoms and charge then clean it up. 6e- + 3H2O + ClO3- + 10 OH- + 2Cr(OH)3 -->Cl- + 6OH-+ 2CrO42- + 8H2O + 6e- ClO3- + 4 OH- + 2Cr(OH)3 -->Cl- + 2CrO42- + 5H2O

  35. Redox ReactionsOxidizing/Reducing Agents

  36. The Spontaneity of REDOX reactions • Zn (s) + Cu+2 Zn+2 + Cu(s) OBSERVATIONS: • Spontaneous (it proceeded with no addition of energy or any other stimulus) • Reaction did not proceed in the reverse direction • Zn(s) is stronger reducing agent than Cu(s) • Activity SERIES???

  37. Relative Strength of Oxidizing and Reducing agents According to this experiment Mg(s) is the most reactive metal and thus the strongest reducing agent ... And Ag(s) is the weakest reducing agent

  38. Activity Series of METALS • Li – strongest reducing agent (easily loses its e-) • Au – weakest reducing agent

  39. The process for Predicting the SPONTANEITY of REDOX RXN • Write Net Ionic equation • Identify “who” lost/gained electrons • Identify reducing/oxidizing agents • Look at the ACTIVITY Series and predict whether reaction will proceed (SD)

  40. Reactions occur spontaneously only when the oxidizing agent is combined with a reducing agent that is below it on the redox table.

  41. LEARNING CHECK #2 • Which of the following reaction will proceed spontaneously? • Al(s) with copper (II) sulfate ANS: Al is above copper on the Activity Series and thus reaction between Al and CuSO4 will proceed spontaneously • Aqueous calcium nitrate and solid nickel

  42. Answers to the Practice Questions on the previous slide

  43. Turn to Worksheet #2 and #3a, #3b

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