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Chapter 19 Redox Equilibria

Chapter 19 Redox Equilibria. 19.1 Redox Reactions (reduction-oxidation). is an equilibrium of the competition for e - between the 2 species reducing agent - undergo oxidation oxidizing agent - undergo reduction. 19.2 Oxidation States. Ex. 19.1.

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Chapter 19 Redox Equilibria

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  1. Chapter 19Redox Equilibria

  2. 19.1 Redox Reactions (reduction-oxidation) • is an equilibrium of the competition for e- between the 2 species • reducing agent - undergo oxidation oxidizing agent - undergo reduction

  3. 19.2 Oxidation States

  4. Ex. 19.1 • a) KMnO4 0 =(+1) + x + 4(-2) x=+7 oxidation number of Mn is +7 • b) POCl3 0 = x + (-2) + 3(-1) x=+5 oxidation number of P is +5 • c) K2Cr2O7 0 = 2(+1) + 2x + 7(-2) x=+6 oxidation number of Cr is +6 • d) CuSO4 0 = +2 + x + 4(-2) x=+6 oxidation number of S is +6

  5. Ex. 19.1 • e) CaH2 0 = +2 + 2x x=-1 oxidation number of H is -1 • f) Na2S2O3 0 = 2(+1)+2x +3(-2) x=+2 oxidation number of S is +2 • g) Na2S2O8 0 = 2(+1)+2x +8(-2) x=+7 oxidation number of S is +7 • h) Na2S4O6 0 = 2(+1)+4x +6(-2 ) x=+2.5 oxidation number of S is +2.5 ??

  6. -2 -2 -2 -2 Thiosulphate ion S2O32- Average oxidation number of S is +2

  7. +6 +6 Peroxodisulphate ion S2O82- oxidation number of S is +6

  8. Tetrathionate ion S4O62- Average oxidation number of S is +4

  9. Ex. 19.2 • a) Mg(s) + H2O(l) → MgO(s) + H2(g)0 +1 +2 0reducing agent oxidizing agent • b) Cr2O72-(aq)+2OH-(aq)→2CrO42-(aq)+H2O(l)+6 +1 +6 +1Not a redox reaction - only conversion under diff. pH • c) 2CuCl(aq)→ Cu(s) + CuCl2(aq)+1 0 +2 CuCl acts as the reducing agent and the oxidizing agent at the same time.

  10. 2CuCl(aq)→ Cu(s) + CuCl2(aq)+1 0 +2 • Cu+(aq) + e- → Cu(s) reductionCu+ (aq)→ Cu2+(aq) + e- oxidation___________________________________ • Cu+(aq)+ Cu+ (aq)→ Cu(s)+ Cu2+(aq) • Disproportionation is a chemical change in which a particular chemical species is simultaneously oxidized and reduced.

  11. 19.3 Balancing Redox Equations • a) SO32- +H2O→ SO42-+2H++2e- • b) VO2++ H2O→ VO3-+ 2H+ (not redox) • c) MnO4- +4H+ + 3e-→ MnO2 + 2H2OMnO4- +4H2O + 3e-→ MnO2+2H2O+4OH- MnO4- + 2H2O + 3e-→ MnO2+ 4OH-

  12. d) MnO4- + 8H+ + 5e-→ Mn2+ + 4H2O • e) BrO3-+6H++ 6e-→ Br- +3H2O • f) N2H4→ N2+ 4H+ + 4e- • g) Cl2+ 6H2O→2ClO3- + 12H+ + 4e- • h) NO3- + 4H+ +3e-→ NO + 2H2O

  13. Ex. 19.4 • a) Br2 +2I-→2Br- + I2 • b) 3S2O82- + 2Cr3++7H2O →6SO42- + Cr2O72- + 14H+ • c) IO3- + 6Fe2++ 6H+→ I- + 6Fe3+ +3H2O

  14. ne- M M ne- Mn+ Mn+ 19.4 Electrochemical Cells • When a metal is dipped into a solution containing ions of the same metal, a metal/metal ion system is set up. • Mn+(aq) + ne- → M(s)M(s) →Mn+(aq) + ne-

  15. Mn+(aq) + ne- <======> M(s) • The equilibrium position depends on - nature of the M / Mn+ system,- concentration of ions- temperature • Equilibrium lies on the right oxidation ? Reduction? predominates • Charge on electrode? positive ? Negative?

  16. M ne- ne- M M M ne- ne- Mn+ Mn+ Mn+ Mn+ • a separation of charge  a potential difference between the electrode and the ions in solution • metal/metal ion system is called half cell

  17. V e- X e- Y Y+ X+ Metal Y Metal X • connect 2 different half cells externally  e-s can flow from one electrode to another • Current flow through the external conducting wire • after the charge is built up between the two system, the current stop???? Solution??? Y+ solution X+ solution

  18. Solutions: • Using porous partition or salt bridge • Functions:- complete the circuit by allowing ions flow- without extensive mixing of solutions of the 2 half-cell

  19. combination of the two half cell systems is called electrochemical cell • Also called galvanic cell • is a device which converts chemical energy into electrical energy • Each half-cell has a tendency to accept electrons from the other • the half-cell with a stronger ability to gain electrons will win the competition

  20. Each half-cell system has its own electrode potential that cannot be measured. • Only potential difference between 2 half-cells can be measured • The maximum potential difference which the cell can produce, called the electromotive force (e.m.f.)

  21. e- e- Daniel cell: • Zn has a higher tendency to lose e- than Cu • e-s are then pumped out from the Zn electrode through the external circuit to the cell at Cu electrode

  22. Daniel cell: • Oxidation (Anode / negative terminal) : • Zn(s)  Zn2+(aq) + 2e- • Reduction (Cathode / positive terminal): • Cu2+(aq) +2e-  Cu(s) • Overall Redox equation: • Zn(s) + Cu2+(aq)  Zn2+(aq) + Cu(s)

  23. Ex 19.5 • An electrochemical cell consists of Ag electrode with AgNO3 as electrolyte and Cu electrode with CuSO4 as electrolyte. The salt bridge is made of agar gel with dissolved NaCl. However, no current flow through the external wire. Draw a diagram to show this electrochemical cell and explain why there is no current? Ag+ ion from the silver nitrate solution may react with Cl- ion in the salt bridge to form silver chloride which is insoluble in water. Thus, ions cannot flow between the two system and thus the current stops.

  24. Explain why salt bridge of NaCl or Na2SO3 should not be used when the electrolyte is acidified KMnO4. It is because chloride ion and sulphite ion (sulphate(IV)) ion may be oxidized by acidified KMnO4.

  25. Cell Diagrams • A representation of an electrochemical cell using the IUPAC conventions • For example, cell diagram of Daniel cell:Zn(s)∣Zn2+(aq) Cu2+(aq)∣Cu(s) E =+1.1V • solid vertical line (∣)=phase boundary • a pair of vertical broken lines ( )= salt bridge • single vertical broken line ( ) =porous partition • If the salt bridge is made of KCl,Zn(s)∣Zn2+(aq) KCl Cu2+(aq)∣Cu(s) E=+1.1V • E represents the e.m.f. of the cell in volts

  26. Zn(s)∣Zn2+(aq) Cu2+(aq)∣Cu(s) E=+1.1VMeaning??? L.H.S. : Anode (negative electrode): Zn(s)  Zn2+(aq) + 2e- R.H.S.: Cathode (positive electrode): Cu2+(aq) +2e-  Cu(s) • By convention, the electrode on LHS is considered as the anode while the one on RHS is considered to be the cathode • sign of the e.m.f. indicates the polarity of the right-hand electrode • For a +ve e.m.f., the reaction proceeds from left to right

  27. Ex. 19.6Cu(s)∣Cu2+(aq) Zn2+(aq)∣Zn(s) E = ? • E = -1.1V • that means the reaction proceeds from RHS to LHS • or Cu + Zn2+ Cu2+ + Zn is not spontaneous. But Cu2+ + Zn  Cu + Zn2+is spontaneous.

  28. (a) Ex. 19.7 • Sn(s)∣Sn2+(aq) Cu2+(aq)∣Cu(s) E=+0.50V • Oxidation (anode):Sn(s) Sn2+(aq) + 2e- • Reduction (cathode)Cu2+(aq) + 2e-  Cu(s) • Overall equation:Sn(s) + Cu2+(aq)  Sn2+(aq) + Cu(s)

  29. (b) • Cu(s)∣Cu2+(aq) Mg2+(aq)∣Mg(s) E=-2.70V • Oxidation (anode):Mg(s) Mg2+(aq) + 2e- • Reduction (cathode)Cu2+(aq) + 2e-  Cu(s) • Overall equation:Mg(s) + Cu2+(aq)  Mg2+(aq) + Cu(s)

  30. (c) • Ni(s)∣Ni2+(aq) Zn2+(aq)∣Zn(s) E=-0.70V • Oxidation (anode):Zn(s) Zn2+(aq) + 2e- • Reduction (cathode)Ni2+(aq) + 2e-  Ni(s) • Overall equation:Zn(s) + Ni2+(aq)  Zn2+(aq) + Ni(s)

  31. (d) • Ni(s)∣Ni2+(aq) Ag+(aq)∣Ag(s) E=+0.95V • Oxidation (anode):Ni(s) Ni2+(aq) + 2e- • Reduction (cathode)Ag+(aq) + e- Ag (s) • Overall equation:Ni(s) + 2Ag+(aq)  Ni2+(aq) + 2Ag(s)

  32. Types of Half-cell • 1. Metal in contact with an aqueous solution of its metal ion • Zn2+(aq)+2e- <====> Zn(s) • Cell diagram : Zn(s)∣Zn2+(aq) Zn2+(aq)∣Zn(s)

  33. 2. Metal in contact with its insoluble salt and an aqueous solution of the anion • AgCl(s) + e- <====> Ag(s) +Cl-(aq) • Cell diagram : Pt(s)∣[Ag(s) + Cl-(aq)], AgCl(s) AgCl(s), [Ag(s) + Cl-(aq)]∣Pt(s) • PbSO4(s) + 2e- <====> Pb(s) +SO42-(aq) • Cell diagram : Pt(s)∣[Pb(s) + SO42-(aq,1M)], PbSO4(s) PbSO4(s), [Pb(s) + SO42-(aq,1M)]∣Pt(s)

  34. 3. An inert electrode (e.g. Pt) and a gas in contact with an aqueous solution from which the gas can be generated • 2H+(aq) + 2e- <=====> H2(g) • Cell diagram :Pt(s)∣H2(g)∣2H+(aq) or Pt(s) [H2(g)]∣2H+(aq) 2H+(aq)∣H2(g)∣Pt(s) or 2H+(aq)∣[H2(g)] Pt(s)

  35. O2(g) +4e- +2H2O(l) <=====> 4OH-(aq) • Cell diagram : Pt(s)∣4OH-(aq), [O2(g) + 2H2O(l)] [O2(g) + 2H2O(l)], 4OH-(aq)∣Pt(s)

  36. 4. An inert electrode and a solution containing both the oxidized and reduced formsof the species • Fe3+(aq) + e- ===== Fe2+(aq) • Cell diagram : Pt(s)∣Fe2+(aq), Fe3+(aq) Fe3+(aq), Fe2+(aq)∣Pt(s)

  37. I2(aq) + 2e- <====> 2I-(aq) • Cell diagram : Pt(s)∣2I-(aq), I2(aq) I2(aq), 2I-(aq)∣Pt(s) • MnO4-(aq)+8H+(aq)+5e- <==>Mn2+(aq)+4H2O(l) • Cell diagram :Pt(s)∣[Mn2+(aq)+4H2O(l)],[MnO4-(aq) +8H+(aq)] [MnO4-(aq) +8H+(aq)], [Mn2+(aq)+4H2O(l)] ∣ Pt(s) • The most reduced form is written next to the inert electrode (i.e.Pt)

  38. 19.5 Standard Electrode and Relative Electrode Potentials from E.m.f Measurements • Standard Electrode • Electrode potential of a half cell cannot be measured. • standard hydrogen electrode (s.h.e.) - selected as the reference electrode- its electrode potential is assigned as 0V at 25oC, 1atm and 1.0M H+. • With the reference, the relative scale of electrode potential is established for different systems in order to compare their tendency to release e-s

  39. Standard hydrogen electrode: A. Inlet for 1atm H2 at 25oC B. Platinumcoated with platinum black D. outlet for H2 C. 1M H+

  40. has an electrode made of a piece of platinum coated with finely divided platinum black • function of this special electrode: • catalyses the half-cell reaction : H2(g) <=> 2H+(aq) + 2e- • provides a surface on which the hydrogen can be adsorbed • provides an electrical connection to the voltmeter • with the concentration of HCl(aq) of 1 mol dm-3 • a slow stream of pure hydrogen gas at 1atm is bubbled over the platinized surface

  41. cell diagram • Pt(s)∣H2(g)∣2H+(aq) or Pt(s) [H2(g)]∣2H+(aq) • Half -cell reaction: • Reduction: 2H+(aq) + 2e- ===> H2(g) • Oxidation: H2(g) ===> 2H+(aq) + 2e-

  42. As the potential of an electrode system depends on ,and. The electrode potential of the reference electrode can only be assigned as 0V under a specified conditions: • H2 gas at 1atm • [H+] = 1 mol dm-3 • under temperature of 298K • When the hydrogen electrode is used under the above conditions, the hydrogen electrode is known as a standard hydrogen electrode (s.h.e.)

  43. s.h.e. is always assumed to be the anode of the system. (the left hand side in the cell diagram) • Apart from hydrogen electrode, a calomel electrode can also be used: • Cell diagram :Pt(s)∣[2Hg(l) + 2Cl-(aq,1M)], Hg2Cl2(s) • Equation : Hg2Cl2(s) + 2e- 2Hg(l) + 2Cl-(aq)

  44. Relative Electrode Potential from e.m.f. Measurements • e.m.f. for the cell is a measure of the relative tendencies of half cells to release or gain e-s • e.m.f. of a cell depends on temperature, concentration and also pressure • all the measurement should be measured under the standard conditions:- temp = 298K (25oC) - concentration of all solution = 1.0M - pressure of all gases = 1.0 atm

  45. standard electrode potential (or standard reduction potential / Eo)is the electrode potential (e.m.f.) of a half cell measured when connected to a standard hydrogen electrode under the standard conditions

  46. Cell diagram : Pt(s)∣H2(g, 1atm)∣2H+(aq,1M) Mn+(aq,1M)∣M(s) Eo=? • Assumed Anode? Assumed Cathode? • If s.h.e. undergoes oxidation, sign of Eo?

  47. For the standard reduction potential of the half equation:Cu2+(aq) + 2e- Cu(s) Eo = +0.34VMeaning?? • Pt(s)∣H2(g, 1atm)∣2H+(aq,1M) Cu2+(aq,1M)∣Cu(s) Eo = +0.34V • Anode: H2(g)  2H+(aq) + 2e- • Cathode: Cu2+(aq) + 2e- Cu(s) • Overall reaction: H2(g) + Cu2+(aq) 2H+(aq) + Cu(s)

  48. for the following half equation:Ni2+(aq) + 2e- Ni(s) Eo = -0.25V • cell diagram:Pt(s)∣H2(g, 1atm)∣2H+(aq,1M) Ni2+(aq,1M)∣Ni(s) Eo = -0.25V • The value of Eo is , the polarity of the nickel electrode is . That means nickel is the and undergoes . • Anode: Ni(s)  Ni2+(aq) + 2e- • Cathode 2H+(aq) + 2e- H2(g) • Overall Reaction: Ni(s) + 2H+(aq)  Ni2+(aq) + H2(g)

  49. e.m.f. of this cell, Eocell = ER.H.S. - EL.H.S. • The electrode system with the greatest negative electrode potential is the strongest reducing agent. • By comparing the value of the standard reduction potentials of different electodes, the electrochemical series can be obtained.

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