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Chapter 18 Electrochemistry Part 2

Chapter 18 Electrochemistry Part 2. Current = # e-’s that flow thru a system per sec unit = Ampere 1 A of current = 1 C/sec 1 A = 6.242 x 10 18 e-’s per second Electrode surface area dictates # e-’s that can flow larger batteries produce larger currents.

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Chapter 18 Electrochemistry Part 2

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  1. Chapter 18 ElectrochemistryPart 2

  2. Current=# e-’sthat flow thru a system per sec unit = Ampere • 1 A of current = 1 C/sec • 1 A = 6.242 x 1018e-’s per second • Electrode surface area dictates # e-’s that can flow • larger batteries produce larger currents Tro: Chemistry: A Molecular Approach, 2/e

  3. Voltage Diff in PE btwn reactants & products = 1 V = 1 J of energy per Coulomb of charge the voltage needed to drive electrons through the external circuit Amount of force pushing e-’s thru the wire = Tro: Chemistry: A Molecular Approach, 2/e

  4. Cell Potential • Diff in PE btwnthe anode &the cathode in a voltaic cell = the cell potential • Cell potential depends on relative ease that Tro: Chemistry: A Molecular Approach, 2/e

  5. Cell Potential • Cell potential under standard conditions = the standard emf, E°cell Tro: Chemistry: A Molecular Approach, 2/e

  6. Cell Notation • Shorthand description of a voltaic cell • Electrode| electrolyte || electrolyte | electrode • Ox half-cell on left (anode) • red half-cell on right (cathode) • Single | = • if multiple electrolytes in same phase, comma is used • often use an inert electrode • Double line || = Tro: Chemistry: A Molecular Approach, 2/e

  7. Voltaic Cell Tro: Chemistry: A Molecular Approach, 2/e

  8. Electrodes • Often anode is made of metal that is oxidized and cathode is made of same metal as is produced by reduction Tro: Chemistry: A Molecular Approach, 2/e

  9. Electrodes • However, if the redox rxtn involves oxidation or reduction of an ion to a or oxidation or reduction of a we may use an • an inert electrode does not participate in rxtn, just provides surface for transfer of e-’s Tro: Chemistry: A Molecular Approach, 2/e

  10. Because half-rxtninvolves reducing Mnoxidation state from +7 to +2, we use an electrode that provides a surface for e- transfer without reacting with MnO4−. Fe(s) | Fe2+(aq) || MnO4(aq), Mn2+(aq), H+(aq) | Pt(s) Tro: Chemistry: A Molecular Approach, 2/e

  11. Cell Potential • Half-rxtnwith strong tendency to occur has a • When two half-cells are connected, e-’s will flow so that half-rxtnwith stronger tendency will occur Tro: Chemistry: A Molecular Approach, 2/e

  12. Which Way Will Electrons flow? Under standard conditions, Zn has a stronger tendency to oxidize than Cu Zn → Zn2+ + 2 e− Electron Flow Potential Energy Cu2+ + 2 e− → Cu Zn → Zn2+ + 2 e− E°= +0.76 Cu → Cu2+ + 2 e− E°= −0.34 Tro: Chemistry: A Molecular Approach, 2/e

  13. Which Way Will Electrons flow? Tro: Chemistry: A Molecular Approach, 2/e

  14. Standard Reduction Potential • We cannot measure absolute tendency of a half-rxtn • we can only measure it relative to another half-reaction • Select as a standard half-rxtn reduction of H+ to H2 under standard conditions, assigned a potential difference = Tro: Chemistry: A Molecular Approach, 2/e

  15. Zn(s) | Zn2+(1 M) || H+(1 M) | H2(g) | Pt(s) Tro: Chemistry: A Molecular Approach, 2/e

  16. Half-Cell Potentials • SHE reduction potential defined to be • Standard Reduction Potentials compare tendency for a particular reduction half-rxtn to occur relative to Tro: Chemistry: A Molecular Approach, 2/e

  17. Half-rxtnswith a stronger tendency toward reduction than the SHE have a (+) value for E°red • Half-rxtnswith a stronger tendency toward oxidation than • For an oxidation half-reaction, Tro: Chemistry: A Molecular Approach, 2/e

  18. Tro: Chemistry: A Molecular Approach, 2/e

  19. Tro: Chemistry: A Molecular Approach, 2/e

  20. Calculating Cell Potentials under Standard Conditions • E°cell = E°oxidation + E°reduction • When adding E°values for the half-cells, Tro: Chemistry: A Molecular Approach, 2/e

  21. Calculate Ecell for the reaction at 25 CAl(s) + NO3−(aq) + 4 H+(aq)  Al3+(aq) + NO(g) + 2 H2O(l) Tro: Chemistry: A Molecular Approach, 2/e

  22. Try one: Calculate Ecell for the reaction at 25 CIO3–(aq)+ 6 H+(aq)+ 5 I−(aq) → 3 I2(s)+ 3 H2O(l) Tro: Chemistry: A Molecular Approach, 2/e

  23. Calculate Ecell for the reaction at 25 C 2 IO3– + 12 H+ + 10 I−→ 6 I2 + 6 H2O Tro: Chemistry: A Molecular Approach, 2/e

  24. Tendencies from Table of Standard Reduction Potentials A redox rxtnwill be spontaneous when there is a strong tendency for Tro: Chemistry: A Molecular Approach, 2/e

  25. Tendencies from Table of Standard Reduction Potentials ___________on table = stronger tendency for the reactant to be reduced ___________on table = stronger tendency for the product to be oxidized Tro: Chemistry: A Molecular Approach, 2/e

  26. Predicting Spontaneity of Redox Reactions • A spontaneous reaction will take place when a reduction half-reaction is paired with an oxidation half-reaction Tro: Chemistry: A Molecular Approach, 2/e

  27. Predicting Spontaneity of Redox Reactions • If paired the other way, the reverse reaction is spontaneous Cu2+(aq) + 2 e− Cu(s)Ered = +0.34 V Zn2+(aq) + 2 e− Zn(s)Ered = −0.76 V Tro: Chemistry: A Molecular Approach, 2/e

  28. Predict if the following reaction is spontaneous under standard conditionsFe(s)+ Mg2+(aq)  Fe2+(aq) + Mg(s) Tro: Chemistry: A Molecular Approach, 2/e

  29. Tro: Chemistry: A Molecular Approach, 2/e

  30. Practice – Decide whether each of the following will be spontaneous as written or in the reverse direction F2(g) + 2 I−(aq)I2(s) + 2 F−(aq) Mg(s) + 2 Ag+(aq)Mg2+(aq) + 2 Ag(s) Cu2+(aq) + 2 I−(aq)I2(s) + Cu(s) Cu2+(aq) + 2 Cr2+(aq) Cu(s) + 2 Cr3+(aq) Tro: Chemistry: A Molecular Approach, 2/e

  31. Clicker: Which of the following materials can be used to oxidize Cu withoutoxidizing Ag? F− I− I2 Cr3+ F− I− I2 Cr3+ Tro: Chemistry: A Molecular Approach, 2/e

  32. Try this: Sketch and label a voltaic cellin which one half-cell has Ag(s) immersed in 1 M AgNO3, and other half-cell has a Pt electrode immersed in 1 M Cr(NO3)2 and 1 M Cr(NO3)3 Write half-reactions and overall reaction, and determine cell potential under standard conditions. Tro: Chemistry: A Molecular Approach, 2/e

  33. e− → e− → e− → e− → salt bridge anode = Pt cathode = Ag Cr2+ Ag+ Cr3+ Tro: Chemistry: A Molecular Approach, 2/e

  34. Predicting Whether a Metal Will Dissolve in an Acid • Metals dissolve in acids if reduction of metal ion is easier than reduction of H+(aq) • Metals whose ion reduction rxtnlies below H+ reduction on the table will dissolve in acid Tro: Chemistry: A Molecular Approach, 2/e

  35. Predicting Whether a Metal Will Dissolve in an Acid • Almost all metals will dissolve in HNO3 • having Nreduced rather than H • NO3−(aq) + 4H+(aq) + 3e− →NO(g) + 2H2O(l) Tro: Chemistry: A Molecular Approach, 2/e

  36. Practice – Which of the following metals will dissolve in HC2H3O2(aq)? Write the reaction. Ag Cu Fe Cr Tro: Chemistry: A Molecular Approach, 2/e

  37. Practice – Which of the following metals will dissolve in HC2H3O2(aq)? Write the reaction. Ag Cu Fe Cr Tro: Chemistry: A Molecular Approach, 2/e

  38. E°cell,DG° and K • For a spontaneous reaction one that proceeds in the forward direction with the chemicals in their standard states DG° E° K Tro: Chemistry: A Molecular Approach, 2/e

  39. DG° = −RTlnK = −nFE°cell n is the number of electrons F = Faraday’s Constant = Tro: Chemistry: A Molecular Approach, 2/e

  40. E°ox, E°red E°cell DG° Calculate DG° for the reactionI2(s) + 2 Br−(aq) →Br2(l) + 2 I−(aq) Given: Find: I2(s) + 2 Br−(aq) →Br2(l) + 2 I−(aq) DG°, (J) Conceptual Plan: Relationships: Solve: Answer: Tro: Chemistry: A Molecular Approach, 2/e

  41. Try one: Calculate DG for the reaction at 25C2IO3–(aq)+ 12H+(aq)+ 10 I−(aq) → 6I2(s)+ 6H2O(l) Tro: Chemistry: A Molecular Approach, 2/e

  42. Calculate DG for the reaction at 25 C2IO3–(aq)+ 12H+(aq)+ 10 I−(aq) 6 I2(s)+ 6 H2O(l) E°ox, E°red E°cell G° Given: Find: 2IO3−(aq)+12H+(aq)+10I−(aq) →6 I2(s) + 6 H2O(l) G°, (J) Conceptual Plan: Relationships: Solve: Answer: because DG° is −, the reaction is spontaneous in the forward direction under standard conditions Tro: Chemistry: A Molecular Approach, 2/e

  43. E°ox, E°red E°cell K Calculate Kat 25 °C for the reactionCu(s) + 2 H+(aq) →H2(g) + Cu2+(aq) Given: Find: Cu(s) + 2 H+(aq) →H2(g) + Cu2+(aq) K Conceptual Plan: Relationships: Solve: Answer: Tro: Chemistry: A Molecular Approach, 2/e

  44. Practice – Calculate K for the reaction at 25C2IO3–(aq)+ 12H+(aq)+ 10 I−(aq) → 6I2(s)+ 6H2O(l) Tro: Chemistry: A Molecular Approach, 2/e

  45. Practice – Calculate K for the reaction at 25 C2IO3–(aq)+ 12H+(aq)+ 10 I−(aq) → 6I2(s)+ 6H2O(l) E°ox, E°red E°cell K Given: Find: 2 IO3−(aq)+12 H+(aq)+10 I−(aq)→ 6 I2(s) + 6 H2O(l) K Conceptual Plan: Relationships: Solve: ox : 2 I−(s) → I2(aq) + 2e−Eº = −0.54 V red: IO3−(aq) + 6 H+(aq) + 5e−→ ½ I2(s) + 3H2O(l) Eº = 1.20 V tot: 2IO3−(aq) + 12H+(aq) + 10I−(aq)→ 6I2(s) + 6H2O(l)Eº = 0.66 V Answer: Tro: Chemistry: A Molecular Approach, 2/e

  46. Practice – Calculate K for the reaction at 25 C2IO3–(aq)+ 12H+(aq)+ 10 I−(aq) → 6I2(s)+ 6H2O(l) E°ox, E°red E°cell K Given: Find: 2 IO3−(aq)+12 H+(aq)+10 I−(aq)→ 6 I2(s) + 6 H2O(l) K Conceptual Plan: Relationships: Solve: Answer: Tro: Chemistry: A Molecular Approach, 2/e

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